Elementary Linear Algebra 5th Edition Larson Solutions Manual
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About This Presentation
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Elementary Linear Algebra 5th Edition Larson Solutions Manual
Slide Content
Instructor’s Manual
with Sample Tests for
Elementary
Linear Algebra
Fifth
Edition
Stephen Andrilli
David H ecker
Elementary Linear Algebra 5th Edition Larson Solutions Manual
Full Download: http://alibabadownload.com/product/elementary-linear-algebra-5th-edition-larson-solutions-manual/
This sample only, Download all chapters at: alibabadownload.com
with Sample Tests for
Elementary
Linear Algebra
Fifth
Edition
Stephen Andrilli
David H ecker
Elementary Linear Algebra 5th Edition Larson Solutions Manual
Full Download: http://alibabadownload.com/product/elementary-linear-algebra-5th-edition-larson-solutions-manual/
This sample only, Download all chapters at: alibabadownload.com
Dedication
To all the instructors who have used the various
editions of our book over the years
To all the instructors who have used the various
editions of our book over the years
ii
Table of Contents
Preface ....................................................................... .iii
Answers to Exercises………………………………….1
Chapter 1 Answer Key ……………………………….1
Chapter 2 Answer Key ……………………………...19
Chapter 3 Answer Key ……………………………...3 3
Chapter 4 Answer Key ……………………………...5 2
Chapter 5 Answer Key ……………………………... 87
Chapter 6 Answer Key …………………………….1 18
Chapter 7 Answer Key …………………………….1 28
Chapter 8 Answer Key …………………………….1 45
Chapter 9 Answer Key …………………………….1 70
Appendix B Answer Key …………………………. 185
Appendix C Answer Key …………………………. 186
Appendix D Answer Key …………………………. 187
Chapter Tests ……..………………………………. 190
Chapter 1 – Chapter Tests ………………………… 191
Chapter 2 – Chapter Tests ………………………… 194
Chapter 3 – Chapter Tests ………………………… 197
Chapter 4 – Chapter Tests ………………………… 201
Chapter 5 – Chapter Tests ………………………… 204
Chapter 6 – Chapter Tests ………………………… 210
Chapter 7 – Chapter Tests ………………………… 213
Answers to Chapter Tests ………………………….2 19
Chapter 1 – Answers for Chapter Tests …………...219
Chapter 2 – Answers for Chapter Tests …………...222
Chapter 3 – Answers for Chapter Tests …………...225
Chapter 4 – Answers for Chapter Tests …………...228
Chapter 5 – Answers for Chapter Tests …………...234
Chapter 6 – Answers for Chapter Tests …………...240
Chapter 7 – Answers for Chapter Tests …………...245
Table of Contents
Preface ....................................................................... .iii
Answers to Exercises………………………………….1
Chapter 1 Answer Key ……………………………….1
Chapter 2 Answer Key ……………………………...19
Chapter 3 Answer Key ……………………………...3 3
Chapter 4 Answer Key ……………………………...5 2
Chapter 5 Answer Key ……………………………... 87
Chapter 6 Answer Key …………………………….1 18
Chapter 7 Answer Key …………………………….1 28
Chapter 8 Answer Key …………………………….1 45
Chapter 9 Answer Key …………………………….1 70
Appendix B Answer Key …………………………. 185
Appendix C Answer Key …………………………. 186
Appendix D Answer Key …………………………. 187
Chapter Tests ……..………………………………. 190
Chapter 1 – Chapter Tests ………………………… 191
Chapter 2 – Chapter Tests ………………………… 194
Chapter 3 – Chapter Tests ………………………… 197
Chapter 4 – Chapter Tests ………………………… 201
Chapter 5 – Chapter Tests ………………………… 204
Chapter 6 – Chapter Tests ………………………… 210
Chapter 7 – Chapter Tests ………………………… 213
Answers to Chapter Tests ………………………….2 19
Chapter 1 – Answers for Chapter Tests …………...219
Chapter 2 – Answers for Chapter Tests …………...222
Chapter 3 – Answers for Chapter Tests …………...225
Chapter 4 – Answers for Chapter Tests …………...228
Chapter 5 – Answers for Chapter Tests …………...234
Chapter 6 – Answers for Chapter Tests …………...240
Chapter 7 – Answers for Chapter Tests …………...245
iii
Preface
This Instructor’s Manual with Sample Tests is designed
to accompany Elementary Linear Algebra, 5
th
edition, by
Stephen Andrilli and David Hecker.
This manual contains answers for all the computational
exercises in the textbook and detailed solutions for
virtually all of the problems that ask students for proofs.
The exceptions are typically those exercises that ask
students to verify a particular computation, or that ask for
a proof for which detailed hints have already been
supplied in the textbook. A few proofs that are extremely
trivial in nature have also been omitted.
This manual also contains sample Chapter Tests for the
material in Chapters 1 through 7, as well as answer keys
for these tests.
Additional information regarding the textbook, this
manual, the Student Solutions Manual, and linear algebra
in general can be found at the web site for the textbook,
where you found this manual.
Thank you for using our textbook.
Stephen Andrilli
David Hecker
August 2015
Preface
This Instructor’s Manual with Sample Tests is designed
to accompany Elementary Linear Algebra, 5
th
edition, by
Stephen Andrilli and David Hecker.
This manual contains answers for all the computational
exercises in the textbook and detailed solutions for
virtually all of the problems that ask students for proofs.
The exceptions are typically those exercises that ask
students to verify a particular computation, or that ask for
a proof for which detailed hints have already been
supplied in the textbook. A few proofs that are extremely
trivial in nature have also been omitted.
This manual also contains sample Chapter Tests for the
material in Chapters 1 through 7, as well as answer keys
for these tests.
Additional information regarding the textbook, this
manual, the Student Solutions Manual, and linear algebra
in general can be found at the web site for the textbook,
where you found this manual.
Thank you for using our textbook.
Stephen Andrilli
David Hecker
August 2015
Answers to Exercises Section 1.1
Answers to Exercises
Chapter 1
Section 1.1
(1) (a)[9−4],distance=
√
97
(b)[−611],distance=
√
38
(c)[−1−12−3−4],distance=
√
31
(2) (a)(342)(see Figure 1)
(b)(053)(see Figure 2)
(c)(1−20)(see Figure 3)
(d)(300)(see Figure 4)
(3) (a)(7−13) (b)(64−8) (c)(−13−146)
(4) (a)
¡
16
3
−
13
3
8
¢
(b)(−
20
3
−1−6−1)
(5) (a)
h
3
√
70
−
5
√
70
6
√
70
i
; shorter, since length of original vector is1
(b)
£
−
6
7
2
7
0−
3
7
¤
; shorter, since length of original vector is1
(c)[06−08]; neither, since given vector is a unit vector
(d)
h
1
√
11
−
2
√
11
−
1
√
11
1
√
11
2
√
11
i
; longer, since length of original vector =√
11
5
1
(6) (a) Parallel (b) Parallel (c) Not parallel (d) Not parallel
(7) (a)[−61215]
(b)[106−12]
(c)[−7111]
(d)[−9−24]
(e)[−10−32−1]
(f)[−35320]
(8) (a)x+y=[11],x−y=[−39],y−x=[3−9](see Figure 5)
(b)x+y=[3−5],x−y=[171],y−x=[−17
−1](see Figure 6)
(c)x+y=[18−5],x−y=[32−1],y−x=[−3−21](see Figure 7)
(d)x+y=[−2−44],x−y=[406],y−x=[−40−6](see Figure 8)
(9) With=(7−36),=(11−53),and=(10−78), the length of side=lengthofside
=
√
29. The triangle is isosceles, but not equilateral, since the length of sideis
√
30.
(10) (a)[10−10] (b)[−5
√
3−15] (c)[00] =0
(11) See Figures 1.7 and 1.8 in Section 1.1 of the textbook.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 1
Answers to Exercises
Chapter 1
Section 1.1
(1) (a)[9−4],distance=
√
97
(b)[−611],distance=
√
38
(c)[−1−12−3−4],distance=
√
31
(2) (a)(342)(see Figure 1)
(b)(053)(see Figure 2)
(c)(1−20)(see Figure 3)
(d)(300)(see Figure 4)
(3) (a)(7−13) (b)(64−8) (c)(−13−146)
(4) (a)
¡
16
3
−
13
3
8
¢
(b)(−
20
3
−1−6−1)
(5) (a)
h
3
√
70
−
5
√
70
6
√
70
i
; shorter, since length of original vector is1
(b)
£
−
6
7
2
7
0−
3
7
¤
; shorter, since length of original vector is1
(c)[06−08]; neither, since given vector is a unit vector
(d)
h
1
√
11
−
2
√
11
−
1
√
11
1
√
11
2
√
11
i
; longer, since length of original vector =√
11
5
1
(6) (a) Parallel (b) Parallel (c) Not parallel (d) Not parallel
(7) (a)[−61215]
(b)[106−12]
(c)[−7111]
(d)[−9−24]
(e)[−10−32−1]
(f)[−35320]
(8) (a)x+y=[11],x−y=[−39],y−x=[3−9](see Figure 5)
(b)x+y=[3−5],x−y=[171],y−x=[−17
−1](see Figure 6)
(c)x+y=[18−5],x−y=[32−1],y−x=[−3−21](see Figure 7)
(d)x+y=[−2−44],x−y=[406],y−x=[−40−6](see Figure 8)
(9) With=(7−36),=(11−53),and=(10−78), the length of side=lengthofside
=
√
29. The triangle is isosceles, but not equilateral, since the length of sideis
√
30.
(10) (a)[10−10] (b)[−5
√
3−15] (c)[00] =0
(11) See Figures 1.7 and 1.8 in Section 1.1 of the textbook.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 1
Answers to Exercises Section 1.1
Copyrightc°2016 Elsevier Ltd. All rights reserved. 2
Copyrightc°2016 Elsevier Ltd. All rights reserved. 2
Answers to Exercises Section 1.1
Copyrightc°2016 Elsevier Ltd. All rights reserved. 3
Copyrightc°2016 Elsevier Ltd. All rights reserved. 3
Answers to Exercises Section 1.1
(12) See Figure 9. Both represent the same diagonal vector by the associative law of addition for vectors.
Figure 9:x+(y+z)
(13)[05−06
√
2−04
√
2]≈[−03485−05657]
(14) Net velocity =[−2
√
2−3+2
√
2], resultant speed≈283km/hr
(15) Net velocity =
h
−
3
√
2
2
8−3
√
2
2
i
; speed≈283km/hr
(16)[−8−
√
2−
√
2]
(17) Acceleration =
1
20
[
12
13
−
344
65
392
65
]≈[00462−0264603015]
(18) Acceleration =
£
13
2
0
4
3
¤
(19)
180
√
14
[−231]≈[−9622144324811]
(20)a=[
−
1+
√
3
1+
√
3
];b=[
1+
√
3
√
3
1+
√
3
]
(21) Leta=[
1].
(a)kak
2
=
2
1
+···+
2
is a sum of squares, which must be nonnegative. But then||a||≥0because
the square root of a nonnegative real number is a nonnegative real number.
(b) If||a||=0,thenkak
2
=
2
1
+···+
2
=0, which is only possible if every =0.Thus,a=0.
(22) In each part, suppose thatx=[
1],y=[ 1],andz=[ 1].
(a)x+(y+z)=[
1]+[( 1+1)( +)] = [( 1+(1+1))( +(+))]
=[((
1+1)+ 1)(( +)+ )] = [( 1+1)( +)] + [ 1]=(x+y)+z
(b)x+(−x)=[
1]+[− 1− ]=[( 1+(− 1))( +(− ))] = [00].Also,
(−x)+x=x+(−x)(by part (1) of Theorem 1.3)=0,bytheabove.
(c)(x+y)=([(
1+1)( +)]) = [( 1+1)( +)] = [( 1+1)( +)] =
[
1]+[ 1]=x+y
Copyrightc°2016 Elsevier Ltd. All rights reserved. 4
(12) See Figure 9. Both represent the same diagonal vector by the associative law of addition for vectors.
Figure 9:x+(y+z)
(13)[05−06
√
2−04
√
2]≈[−03485−05657]
(14) Net velocity =[−2
√
2−3+2
√
2], resultant speed≈283km/hr
(15) Net velocity =
h
−
3
√
2
2
8−3
√
2
2
i
; speed≈283km/hr
(16)[−8−
√
2−
√
2]
(17) Acceleration =
1
20
[
12
13
−
344
65
392
65
]≈[00462−0264603015]
(18) Acceleration =
£
13
2
0
4
3
¤
(19)
180
√
14
[−231]≈[−9622144324811]
(20)a=[
−
1+
√
3
1+
√
3
];b=[
1+
√
3
√
3
1+
√
3
]
(21) Leta=[
1].
(a)kak
2
=
2
1
+···+
2
is a sum of squares, which must be nonnegative. But then||a||≥0because
the square root of a nonnegative real number is a nonnegative real number.
(b) If||a||=0,thenkak
2
=
2
1
+···+
2
=0, which is only possible if every =0.Thus,a=0.
(22) In each part, suppose thatx=[
1],y=[ 1],andz=[ 1].
(a)x+(y+z)=[
1]+[( 1+1)( +)] = [( 1+(1+1))( +(+))]
=[((
1+1)+ 1)(( +)+ )] = [( 1+1)( +)] + [ 1]=(x+y)+z
(b)x+(−x)=[
1]+[− 1− ]=[( 1+(− 1))( +(− ))] = [00].Also,
(−x)+x=x+(−x)(by part (1) of Theorem 1.3)=0,bytheabove.
(c)(x+y)=([(
1+1)( +)]) = [( 1+1)( +)] = [( 1+1)( +)] =
[
1]+[ 1]=x+y
Copyrightc°2016 Elsevier Ltd. All rights reserved. 4
Answers to Exercises Section 1.2
(d)()x=[(() 1)(() )] = [(( 1))(( ))] =[( 1)( )]
=([
1]) =(x)
(23) If=0, done. Otherwise,(
1
)(x)=
1
(0)=⇒(
1
·)x=0(by part (7) of Theorem 1.3)=⇒x=0
Thus either=0orx=0.
(24)
1x= 2x=⇒ 1x− 2x=0=⇒( 1−2)x=0=⇒( 1−2)=0orx=0by Theorem 1.4. But
since
16 =2(1−2)6 =0Hence,x=0
(25) (a) F (b) T (c) T (d) F (e) T (f) F (g) F (h) F
Section 1.2
(1) (a)arccos(−
27
5
√
37
),orabout1526
◦
,or266radians
(b)arccos(
46
√
74
√
29
),orabout68
◦
,or012radians
(c)arccos(0),whichis90
◦
,or
2
radians
(d)arccos(−
435
√
2175
√
87
) = arccos(−1),whichis180
◦
,orradians (sincex=−5y)
(2) The vector from
1to 2is[2−7−3], and the vector from 1to 3is[54−6]. These vectors are
orthogonal.
(3) (a)[ ]·[− ]=(−)+=0Similarly,[−]·[ ]=0
(b) A vector in the direction of the line++=0is[−], while a vector in the direction of
−+=0is[ ].
(4) (a)15joules (b)
1040
√
5
9
≈2584joules (c) −
189
√
15
5
≈−1464joules
(5) Note thaty·zis a scalar, sox·(y·z)is not defined.
(6) In all parts, letx=[
12]y=[ 12]andz=[ 12]
(a)x·y=[
12]·[12]= 11+···+ =11+···+
=[12]·[12]=y·x
(b)x·x=[
12]·[12]= 11+···+ =
2
1
+···+
2
.Now
2
1
+···+
2
is a sum of squares, each of which must be nonnegative. Hence, the sum is also nonnegative, and
soitssquarerootisdefined. Thus,0≤x·x=
2
1
+···+
2
=
³p
2
1
+···+
2
´
2
=kxk
2
.
(c) Supposex·x=0.Frompart(b),0=x·x=
2
1
+···+
2
≥
2
,foreach,sincealltermsinthe
sum are nonnegative. Hence,0≥
2
for each.But
2
≥0, because it is a square. Hence each
=0. Therefore,x=0.
(d)(x·y)=([
12]·[12]) =( 11+···+ )
=
11+···+ =[12]·[12]=(x)·y.
Next,(x·y)=(y·x)(by part (a))=(y)·x(from above)=x·(y), by part (a)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 5
(d)()x=[(() 1)(() )] = [(( 1))(( ))] =[( 1)( )]
=([
1]) =(x)
(23) If=0, done. Otherwise,(
1
)(x)=
1
(0)=⇒(
1
·)x=0(by part (7) of Theorem 1.3)=⇒x=0
Thus either=0orx=0.
(24)
1x= 2x=⇒ 1x− 2x=0=⇒( 1−2)x=0=⇒( 1−2)=0orx=0by Theorem 1.4. But
since
16 =2(1−2)6 =0Hence,x=0
(25) (a) F (b) T (c) T (d) F (e) T (f) F (g) F (h) F
Section 1.2
(1) (a)arccos(−
27
5
√
37
),orabout1526
◦
,or266radians
(b)arccos(
46
√
74
√
29
),orabout68
◦
,or012radians
(c)arccos(0),whichis90
◦
,or
2
radians
(d)arccos(−
435
√
2175
√
87
) = arccos(−1),whichis180
◦
,orradians (sincex=−5y)
(2) The vector from
1to 2is[2−7−3], and the vector from 1to 3is[54−6]. These vectors are
orthogonal.
(3) (a)[ ]·[− ]=(−)+=0Similarly,[−]·[ ]=0
(b) A vector in the direction of the line++=0is[−], while a vector in the direction of
−+=0is[ ].
(4) (a)15joules (b)
1040
√
5
9
≈2584joules (c) −
189
√
15
5
≈−1464joules
(5) Note thaty·zis a scalar, sox·(y·z)is not defined.
(6) In all parts, letx=[
12]y=[ 12]andz=[ 12]
(a)x·y=[
12]·[12]= 11+···+ =11+···+
=[12]·[12]=y·x
(b)x·x=[
12]·[12]= 11+···+ =
2
1
+···+
2
.Now
2
1
+···+
2
is a sum of squares, each of which must be nonnegative. Hence, the sum is also nonnegative, and
soitssquarerootisdefined. Thus,0≤x·x=
2
1
+···+
2
=
³p
2
1
+···+
2
´
2
=kxk
2
.
(c) Supposex·x=0.Frompart(b),0=x·x=
2
1
+···+
2
≥
2
,foreach,sincealltermsinthe
sum are nonnegative. Hence,0≥
2
for each.But
2
≥0, because it is a square. Hence each
=0. Therefore,x=0.
(d)(x·y)=([
12]·[12]) =( 11+···+ )
=
11+···+ =[12]·[12]=(x)·y.
Next,(x·y)=(y·x)(by part (a))=(y)·x(from above)=x·(y), by part (a)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 5
Answers to Exercises Section 1.2
(e)(x+y)·z=([ 12]+[ 12])·[ 12]
=[
1+12+2+]·[12]
=(
1+1)1+( 2+2)2+···+( +)
=( 11+22+···+ )+( 11+22+···+ )
Also,(x·z)+(y·z)=([
12]·[12]) + ([ 12]·[12])
=(
11+22+···+ )+( 11+22+···+ ).
Hence,(x+y)·z=(x·z)+(y·z).
(7) No; considerx=[10],y=[01],andz=[11].
(8) A method similar to thefirst part of the proof of Lemma 1.6 in the textbook yields:
ka−bk
2
≥0=⇒(a·a)−(b·a)−(a·b)+(b·b)≥0=⇒1−2(a·b)+1≥0=⇒a·b≤1.
(9) Note that(x+y)·(x−y)=(x·x)+(y·x)−(x·y)−(y·y)=kxk
2
−kyk
2
. Hence,(x+y)·(x−y)=0
implieskxk
2
=kyk
2
,whichmeanskxk=kyk(since both are nonnegative)
(10) Note thatkx+yk
2
=kxk
2
+2(x·y)+kyk
2
, whilekx−yk
2
=kxk
2
−2(x·y)+kyk
2
.
Hence,
1
2
(kx+yk
2
+kx−yk
2
)=
1
2
(2kxk
2
+2kyk
2
)=kxk
2
+kyk
2
(11) (a) From thefirst equation in the solution to Exercise 10 above,kx+yk
2
=kxk
2
+kyk
2
implies
2(x·y)=0which meansx·y=0.
(b) From thefirst equation in the solution to Exercise 10 above,x·y=0implieskx+yk
2
=kxk
2
+kyk
2
.
(12) Note thatkx+y+zk
2
=k(x+y)+zk
2
=kx+yk
2
+2((x+y)·z)+kzk
2
=kxk
2
+2(x·y)+kyk
2
+2(x·z)+2(y·z)+kzk
2
=kxk
2
+kyk
2
+kzk
2
sincex,y,zare mutually orthogonal.
(13) From thefirst two equations in the solution for Exercise 10 above,
1
4
(kx+yk
2
−kx−yk
2
)=
1
4
(4(x·y)) =x·y.
(14) Sincexis orthogonal to bothyandz,wehavex·(
1y+ 2z)= 1(x·y)+ 2(x·z)= 1(0) + 2(0) = 0
(15) Supposey=x,forsome6 =0Then,x·y=x·(x)=(x·x)=kxk
2
=kxk(kxk)=kxk(±||kxk)
=±kxkkxk=±kxkkyk
(16) (a) Length =
√
3 (b) angle =arccos(
√
3
3
)≈547
◦
,or0955radians
(17) (a)proj
ab=[−
3
5
−
3
10
−
3
2
];b−proj
ab=[
8
5
43
10
−
3
2
]; (b−proj
ab)·a=0
(b)proj
ab=[−
6
5
1
2
5
];b−proj
ab=[−
14
5
−4
8
5
]; (b−proj
ab)·a=0
(c)proj
ab=[
1
6
0−
1
6
1
3
];b−proj
ab=[
17
6
−1
1
6
−
4
3
]; (b−proj
ab)·a=0
(d)proj
ab=[−1
3
2
−2−
3
2
];b−proj
ab=[6−
1
2
−6
7
2
]; (b−proj
ab)·a=0
(18) (a)0(zero vector). The dropped perpendicular travels alongbto the common initial point ofaand
b.
(b) The vectorb. The terminal point ofblies on the line througha, so the dropped perpendicular
has length zero.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 6
(e)(x+y)·z=([ 12]+[ 12])·[ 12]
=[
1+12+2+]·[12]
=(
1+1)1+( 2+2)2+···+( +)
=( 11+22+···+ )+( 11+22+···+ )
Also,(x·z)+(y·z)=([
12]·[12]) + ([ 12]·[12])
=(
11+22+···+ )+( 11+22+···+ ).
Hence,(x+y)·z=(x·z)+(y·z).
(7) No; considerx=[10],y=[01],andz=[11].
(8) A method similar to thefirst part of the proof of Lemma 1.6 in the textbook yields:
ka−bk
2
≥0=⇒(a·a)−(b·a)−(a·b)+(b·b)≥0=⇒1−2(a·b)+1≥0=⇒a·b≤1.
(9) Note that(x+y)·(x−y)=(x·x)+(y·x)−(x·y)−(y·y)=kxk
2
−kyk
2
. Hence,(x+y)·(x−y)=0
implieskxk
2
=kyk
2
,whichmeanskxk=kyk(since both are nonnegative)
(10) Note thatkx+yk
2
=kxk
2
+2(x·y)+kyk
2
, whilekx−yk
2
=kxk
2
−2(x·y)+kyk
2
.
Hence,
1
2
(kx+yk
2
+kx−yk
2
)=
1
2
(2kxk
2
+2kyk
2
)=kxk
2
+kyk
2
(11) (a) From thefirst equation in the solution to Exercise 10 above,kx+yk
2
=kxk
2
+kyk
2
implies
2(x·y)=0which meansx·y=0.
(b) From thefirst equation in the solution to Exercise 10 above,x·y=0implieskx+yk
2
=kxk
2
+kyk
2
.
(12) Note thatkx+y+zk
2
=k(x+y)+zk
2
=kx+yk
2
+2((x+y)·z)+kzk
2
=kxk
2
+2(x·y)+kyk
2
+2(x·z)+2(y·z)+kzk
2
=kxk
2
+kyk
2
+kzk
2
sincex,y,zare mutually orthogonal.
(13) From thefirst two equations in the solution for Exercise 10 above,
1
4
(kx+yk
2
−kx−yk
2
)=
1
4
(4(x·y)) =x·y.
(14) Sincexis orthogonal to bothyandz,wehavex·(
1y+ 2z)= 1(x·y)+ 2(x·z)= 1(0) + 2(0) = 0
(15) Supposey=x,forsome6 =0Then,x·y=x·(x)=(x·x)=kxk
2
=kxk(kxk)=kxk(±||kxk)
=±kxkkxk=±kxkkyk
(16) (a) Length =
√
3 (b) angle =arccos(
√
3
3
)≈547
◦
,or0955radians
(17) (a)proj
ab=[−
3
5
−
3
10
−
3
2
];b−proj
ab=[
8
5
43
10
−
3
2
]; (b−proj
ab)·a=0
(b)proj
ab=[−
6
5
1
2
5
];b−proj
ab=[−
14
5
−4
8
5
]; (b−proj
ab)·a=0
(c)proj
ab=[
1
6
0−
1
6
1
3
];b−proj
ab=[
17
6
−1
1
6
−
4
3
]; (b−proj
ab)·a=0
(d)proj
ab=[−1
3
2
−2−
3
2
];b−proj
ab=[6−
1
2
−6
7
2
]; (b−proj
ab)·a=0
(18) (a)0(zero vector). The dropped perpendicular travels alongbto the common initial point ofaand
b.
(b) The vectorb. The terminal point ofblies on the line througha, so the dropped perpendicular
has length zero.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 6
Answers to Exercises Section 1.2
(19)i,j,k
(20) (a) Parallel:[
20
29
−
30
29
40
29
], orthogonal:[−
194
29
88
29
163
29
]
(b) Parallel:[−
1
2
1−
1
2
], orthogonal:[−
11
2
1
15
2
]
(c) Parallel:[
60
49
−
40
49
120
49
], orthogonal:[−
354
49
138
49
223
49
]
(21) From the lower triangle in thefigure, we have(proj
rx)+(proj
rx−x)=reflection ofx(see Figure
10).
Figure 10: Reflection of a vectorxthrough a line.
(22) For the casekxk≤kyk:|kxk−kyk|=kyk−kxk=kx+y+(−x)k−kxk≤kx+yk+k−xk−kxk
(by the Triangle Inequality)=kx+yk+kxk−kxk=kx+yk.
The casekxk≥kykis done similarly, with the roles ofxandyreversed.
(23) (a) Note thatproj
xy=[
8
5
−
6
5
2] =
2
5
xandy−proj
xy=[
7
5
−
24
5
−4]is orthogonal tox.
Letw=y−proj
xyThen sincey=proj
xy+(y−proj
xy)we havey=
2
5
x+wwherewis
orthogonal tox.
(b) Let=(x·y)(kxk
2
)(so thatx=proj
xy), and letw=y−proj
xywhich is orthogonal tox
by the argument before Theorem 1.11. Theny=x+w,wherewis orthogonal tox.
(c) Supposex+w=x+v.Then(−)x=w−v,and(−)x·(w−v)=(w−v)·(w−v)=kw−vk
2
.
But(−)x·(w−v)=(−)(x·w)−(−)(x·v)=0,sincevandware orthogonal tox. Hence
kw−vk
2
=0=⇒w−v=0=⇒w=v. Then,x=x=⇒=, from Theorem 1.4, sincexis
nonzero.
(24) Ifis the angle betweenxandy,andis the angle betweenproj
xyandproj
yx,then
cos=
proj
xy·proj
yx
kproj
xyk
°
°
proj
yx
°
°
=
³
x·y
kxk
2
´
x·
³
y·x
kyk
2
´
y
°
°
°
³
x·y
kxk
2
´
x
°
°
°
°
°
°
³
y·x
kyk
2
´
y
°
°
°
=
³
x·y
kxk
2
´³
y·x
kyk
2
´
(x·y)
³
|x·y|
kxk
2
´
kxk
³
|y·x|
kyk
2
´
kyk
=
³
(x·y)
3
kxk
2
kyk
2
´
³
(x·y)
2
kxkkyk
´=
(x·y)
kxkkyk
=cos
Hence=.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 7
(19)i,j,k
(20) (a) Parallel:[
20
29
−
30
29
40
29
], orthogonal:[−
194
29
88
29
163
29
]
(b) Parallel:[−
1
2
1−
1
2
], orthogonal:[−
11
2
1
15
2
]
(c) Parallel:[
60
49
−
40
49
120
49
], orthogonal:[−
354
49
138
49
223
49
]
(21) From the lower triangle in thefigure, we have(proj
rx)+(proj
rx−x)=reflection ofx(see Figure
10).
Figure 10: Reflection of a vectorxthrough a line.
(22) For the casekxk≤kyk:|kxk−kyk|=kyk−kxk=kx+y+(−x)k−kxk≤kx+yk+k−xk−kxk
(by the Triangle Inequality)=kx+yk+kxk−kxk=kx+yk.
The casekxk≥kykis done similarly, with the roles ofxandyreversed.
(23) (a) Note thatproj
xy=[
8
5
−
6
5
2] =
2
5
xandy−proj
xy=[
7
5
−
24
5
−4]is orthogonal tox.
Letw=y−proj
xyThen sincey=proj
xy+(y−proj
xy)we havey=
2
5
x+wwherewis
orthogonal tox.
(b) Let=(x·y)(kxk
2
)(so thatx=proj
xy), and letw=y−proj
xywhich is orthogonal tox
by the argument before Theorem 1.11. Theny=x+w,wherewis orthogonal tox.
(c) Supposex+w=x+v.Then(−)x=w−v,and(−)x·(w−v)=(w−v)·(w−v)=kw−vk
2
.
But(−)x·(w−v)=(−)(x·w)−(−)(x·v)=0,sincevandware orthogonal tox. Hence
kw−vk
2
=0=⇒w−v=0=⇒w=v. Then,x=x=⇒=, from Theorem 1.4, sincexis
nonzero.
(24) Ifis the angle betweenxandy,andis the angle betweenproj
xyandproj
yx,then
cos=
proj
xy·proj
yx
kproj
xyk
°
°
proj
yx
°
°
=
³
x·y
kxk
2
´
x·
³
y·x
kyk
2
´
y
°
°
°
³
x·y
kxk
2
´
x
°
°
°
°
°
°
³
y·x
kyk
2
´
y
°
°
°
=
³
x·y
kxk
2
´³
y·x
kyk
2
´
(x·y)
³
|x·y|
kxk
2
´
kxk
³
|y·x|
kyk
2
´
kyk
=
³
(x·y)
3
kxk
2
kyk
2
´
³
(x·y)
2
kxkkyk
´=
(x·y)
kxkkyk
=cos
Hence=.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 7
Answers to Exercises Section 1.3
(25) (a) T (b) T (c) F (d) F (e) T (f) F
Section 1.3
(1) (a) We havek4x+7yk≤k4xk+k7yk=4kxk+7kyk≤7kxk+7kyk=7(kxk+kyk).
(b) Let=max{||||}.Thenkx±yk≤(kxk+kyk).
(2) (a) Note that6−5 = 3(2(−1)) + 1.Let=2(−1).
(b) Consider the number4.
(3) Note that sincex6 =0andy6 =0,
proj
xy=0iff(x·y)(kxk
2
)=0iffx·y=0iffy·x=0iff
(y·x)(kyk
2
)=0iffproj
yx=0.
(4) Ify=x(for0), thenkx+yk=kx+xk=(1+)kxk=kxk+kxk=kxk+kxk=kxk+kyk.
On the other hand, ifkx+yk=kxk+kyk,thenkx+yk
2
=(kxk+kyk)
2
.Nowkx+yk
2
=
(x+y)·(x+y)=kxk
2
+2(x·y)+kyk
2
, while(kxk+kyk)
2
=kxk
2
+2kxkkyk+kyk
2
. Hence
x·y=kxkkyk.ByResult4,y=x,forsome0.
(5) (a) Supposey6 =0. We must show thatxis not orthogonal toy.Nowkx+yk
2
=kxk
2
,so
kxk
2
+2(x·y)+kyk
2
=kxk
2
. Hencekyk
2
=−2(x·y).Sincey6 =0,wehavekyk
2
6 =0,andso
x·y6 =0.
(b) Supposexis not a unit vector. We must show thatx·y6 =1.
Nowproj
xy=x=⇒((x·y)(kxk
2
))x=1x=⇒(x·y)(kxk
2
)=1=⇒x·y=kxk
2
.
But thenkxk6 =1=⇒kxk
2
6 =1=⇒x·y6 =1.
(6) (a) Considerx=[100]andy=[110].
(b) Ifx6 =y,thenx·y6 =kxk
2
.
(c) Yes
(7) See the answer for Exercise 11(a) in Section 1.2.
(8) Ifkxkkyk,thenkxk
2
kyk
2
,andsokxk
2
−kyk
2
0But then(x+y)·(x−y)0and so the
cosine of the angle between(x+y)and(x−y)is positive. Thus the angle between(x+y)and(x−y)
is acute.
(9) (a) Contrapositive: Ifx=0,thenxis not a unit vector.
Converse: Ifxis nonzero, thenxis a unit vector.
Inverse: Ifxis not a unit vector, thenx=0.
(b) (Letxandybe nonzero vectors.)
Contrapositive: Ify6 =proj
xy,thenxis not parallel toy.
Converse: Ify=proj
xy,thenxky.
Inverse: Ifxis not parallel toy,theny6 =proj
xy.
(c) (Letx,ybe nonzero vectors.)
Contrapositive: Ifproj
yx6 =0,thenproj xy6 =0.
Converse: Ifproj
yx=0,thenproj xy=0.
Inverse: Ifproj
xy6 =0,thenproj yx6 =0.
(10) (a) Converse: Letxandybe nonzero vectors inR
.Ifkx+ykkyk,thenx·y≥0.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 8
(25) (a) T (b) T (c) F (d) F (e) T (f) F
Section 1.3
(1) (a) We havek4x+7yk≤k4xk+k7yk=4kxk+7kyk≤7kxk+7kyk=7(kxk+kyk).
(b) Let=max{||||}.Thenkx±yk≤(kxk+kyk).
(2) (a) Note that6−5 = 3(2(−1)) + 1.Let=2(−1).
(b) Consider the number4.
(3) Note that sincex6 =0andy6 =0,
proj
xy=0iff(x·y)(kxk
2
)=0iffx·y=0iffy·x=0iff
(y·x)(kyk
2
)=0iffproj
yx=0.
(4) Ify=x(for0), thenkx+yk=kx+xk=(1+)kxk=kxk+kxk=kxk+kxk=kxk+kyk.
On the other hand, ifkx+yk=kxk+kyk,thenkx+yk
2
=(kxk+kyk)
2
.Nowkx+yk
2
=
(x+y)·(x+y)=kxk
2
+2(x·y)+kyk
2
, while(kxk+kyk)
2
=kxk
2
+2kxkkyk+kyk
2
. Hence
x·y=kxkkyk.ByResult4,y=x,forsome0.
(5) (a) Supposey6 =0. We must show thatxis not orthogonal toy.Nowkx+yk
2
=kxk
2
,so
kxk
2
+2(x·y)+kyk
2
=kxk
2
. Hencekyk
2
=−2(x·y).Sincey6 =0,wehavekyk
2
6 =0,andso
x·y6 =0.
(b) Supposexis not a unit vector. We must show thatx·y6 =1.
Nowproj
xy=x=⇒((x·y)(kxk
2
))x=1x=⇒(x·y)(kxk
2
)=1=⇒x·y=kxk
2
.
But thenkxk6 =1=⇒kxk
2
6 =1=⇒x·y6 =1.
(6) (a) Considerx=[100]andy=[110].
(b) Ifx6 =y,thenx·y6 =kxk
2
.
(c) Yes
(7) See the answer for Exercise 11(a) in Section 1.2.
(8) Ifkxkkyk,thenkxk
2
kyk
2
,andsokxk
2
−kyk
2
0But then(x+y)·(x−y)0and so the
cosine of the angle between(x+y)and(x−y)is positive. Thus the angle between(x+y)and(x−y)
is acute.
(9) (a) Contrapositive: Ifx=0,thenxis not a unit vector.
Converse: Ifxis nonzero, thenxis a unit vector.
Inverse: Ifxis not a unit vector, thenx=0.
(b) (Letxandybe nonzero vectors.)
Contrapositive: Ify6 =proj
xy,thenxis not parallel toy.
Converse: Ify=proj
xy,thenxky.
Inverse: Ifxis not parallel toy,theny6 =proj
xy.
(c) (Letx,ybe nonzero vectors.)
Contrapositive: Ifproj
yx6 =0,thenproj xy6 =0.
Converse: Ifproj
yx=0,thenproj xy=0.
Inverse: Ifproj
xy6 =0,thenproj yx6 =0.
(10) (a) Converse: Letxandybe nonzero vectors inR
.Ifkx+ykkyk,thenx·y≥0.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 8
Answers to Exercises Section 1.3
(b) Letx=[2−1]andy=[02].
(11) (a) Converse: Letx,y,andzbe vectors inR
.Ify=z,thenx·y=x·z. The converse is obviously
true, but the original statement is false in general, with counterexamplex=[11],y=[1−1],
andz=[−11].
(b) Converse: Letxandybe vectors inR
.Ifkx+yk≥kyk,thenx·y=0. The original
statement is true, but the converse is false in general. Proof of the original statement follows from
kx+yk
2
=(x+y)·(x+y)=kxk
2
+2(x·y)+kyk
2
=kxk
2
+kyk
2
≥kyk
2
.
Counterexample to converse: letx=[10],y=[11].
(c) Converse: Letx,ybe vectors inR
,with1.Ifx=0ory=0,thenx·y=0.Theconverse
is obviously true, but the original statement is false in general, with counterexamplex=[1−1]
andy=[11].
(12) Supposex⊥yandis odd. Thenx·y=0.Nowx·y=
P
=1
. But each product equals
either1or−1.Ifexactlyof these products equal1,thenx·y=−(−)=−+2. Hence
−+2=0,andso=2, contradictingodd.
(13) Suppose that[65],[−23],and[
12]are mutually orthogonal, with[ 12]6 =[00].Then6 1+52=
0and−2
1+3 2=0. Multiplying the latter equation by3,weobtain−6 1+9 2=0Adding this
to thefirst equation gives14
2=0,whichmeans 2=0, and hence 1=0.Thus,[ 12]=[00],a
contradiction.
(14) Assume that||x||=1.Weknowthat ||proj
xy||6 =x·yHowever,proj
xy=
¡
(x·y)||x||
2
¢
xso
||proj
xy||=|x·y|||x||
2
=|x·y|Butx·y0because the angle betweenxandyis acute, so
||proj
xy||=x·y, a contradiction. Thus,||x||6 =1
(15) Base Step (=1):x
1=x1.
Inductive Step: Assumex
1+x2+···+x −1+x=x+x−1+···+x 2+x1,forsome≥1
Prove:x
1+x2+···+x −1+x+x+1=x+1+x+x−1+···+x 2+x1
But,
x
1+x2+···+x −1+x+x+1=(x 1+x2+···+x −1+x)+x +1
=x +1+(x 1+x2+···+x −1+x)
=x
+1+(x +x−1+···+x 2+x1)
(by the inductive hypothesis)
=x
+1+x+x−1+···+x 2+x1
(16) Base Step (=1):kx
1k≤kx 1k.
Inductive Step: Assumekx
1+···+x k≤kx 1k+···+kx k,forsome≥1.
Prove:kx
1+···+x +x+1k≤kx 1k+···+kx k+kx +1k.
But, by the Triangle Inequality,
k(x
1+···+x )+x +1k≤kx 1+···+x k+kx +1k
≤kx
1k+···+kx k+kx +1k
by the inductive hypothesis.
(17) Base Step (=1):kx
1k
2
=kx 1k
2
.
Inductive Step: Assumekx
1+···+x k
2
=kx 1k
2
+···+kx k
2
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 9
(b) Letx=[2−1]andy=[02].
(11) (a) Converse: Letx,y,andzbe vectors inR
.Ify=z,thenx·y=x·z. The converse is obviously
true, but the original statement is false in general, with counterexamplex=[11],y=[1−1],
andz=[−11].
(b) Converse: Letxandybe vectors inR
.Ifkx+yk≥kyk,thenx·y=0. The original
statement is true, but the converse is false in general. Proof of the original statement follows from
kx+yk
2
=(x+y)·(x+y)=kxk
2
+2(x·y)+kyk
2
=kxk
2
+kyk
2
≥kyk
2
.
Counterexample to converse: letx=[10],y=[11].
(c) Converse: Letx,ybe vectors inR
,with1.Ifx=0ory=0,thenx·y=0.Theconverse
is obviously true, but the original statement is false in general, with counterexamplex=[1−1]
andy=[11].
(12) Supposex⊥yandis odd. Thenx·y=0.Nowx·y=
P
=1
. But each product equals
either1or−1.Ifexactlyof these products equal1,thenx·y=−(−)=−+2. Hence
−+2=0,andso=2, contradictingodd.
(13) Suppose that[65],[−23],and[
12]are mutually orthogonal, with[ 12]6 =[00].Then6 1+52=
0and−2
1+3 2=0. Multiplying the latter equation by3,weobtain−6 1+9 2=0Adding this
to thefirst equation gives14
2=0,whichmeans 2=0, and hence 1=0.Thus,[ 12]=[00],a
contradiction.
(14) Assume that||x||=1.Weknowthat ||proj
xy||6 =x·yHowever,proj
xy=
¡
(x·y)||x||
2
¢
xso
||proj
xy||=|x·y|||x||
2
=|x·y|Butx·y0because the angle betweenxandyis acute, so
||proj
xy||=x·y, a contradiction. Thus,||x||6 =1
(15) Base Step (=1):x
1=x1.
Inductive Step: Assumex
1+x2+···+x −1+x=x+x−1+···+x 2+x1,forsome≥1
Prove:x
1+x2+···+x −1+x+x+1=x+1+x+x−1+···+x 2+x1
But,
x
1+x2+···+x −1+x+x+1=(x 1+x2+···+x −1+x)+x +1
=x +1+(x 1+x2+···+x −1+x)
=x
+1+(x +x−1+···+x 2+x1)
(by the inductive hypothesis)
=x
+1+x+x−1+···+x 2+x1
(16) Base Step (=1):kx
1k≤kx 1k.
Inductive Step: Assumekx
1+···+x k≤kx 1k+···+kx k,forsome≥1.
Prove:kx
1+···+x +x+1k≤kx 1k+···+kx k+kx +1k.
But, by the Triangle Inequality,
k(x
1+···+x )+x +1k≤kx 1+···+x k+kx +1k
≤kx
1k+···+kx k+kx +1k
by the inductive hypothesis.
(17) Base Step (=1):kx
1k
2
=kx 1k
2
.
Inductive Step: Assumekx
1+···+x k
2
=kx 1k
2
+···+kx k
2
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 9
Answers to Exercises Section 1.3
Prove:kx 1+···+x +x+1k
2
=kx 1k
2
+···+kx k
2
+kx +1k
2
.
We have
k(x
1+···+x )+x +1k
2
=kx 1+···+x k
2
+2((x 1+···+x )·x+1)+kx +1k
2
=kx 1+···+x k
2
+kx +1k
2
(sincex +1is orthogonal to all ofx 1x )
=kx
1k
2
+···+kx k
2
+kx +1k
2
by the inductive hypothesis.
(18) Base Step (=1): We must show(
1x1)·y≤| 1|kyk.
But,
(
1x1)·y≤|( 1x1)·y|≤k 1x1kkyk
(by the Cauchy-Schwarz Inequality)
=|
1|kx1kkyk=| 1|kyk
sincex
1is a unit vector.
Inductive Step: Assume(
1x1+···+ x)·y≤(| 1|+···+| |)kyk,forsome≥1.
Prove:(
1x1+···+ x++1x+1)·y≤(| 1|+···+| |+| +1|)kyk.
We have
((
1x1+···+ x)+ +1x+1)·y=( 1x1+···+ x)·y+( +1x+1)·y
≤(|
1|+···+| |)kyk+( +1x+1)·y
(by the inductive hypothesis)
≤(|
1|+···+| |)kyk+| +1|||y||
(by an argument similar to the Base Step)
=(|
1|+···+| |+| +1|)kyk
(19) Step 1 cannot be reversed, becausecould equal±(
2
+2).
Step 2 cannot be reversed, because
2
could equal
4
+4
2
+.
Step 4 cannot be reversed, because in generaldoes not have to equal
2
+2.
Step 6 cannot be reversed, since
could equal2+.
All other steps remain true when reversed.
(20) (a) For every unit vectorxinR
3
,x·[1−23]6 =0.
(b)x6 =0andx·y≤0, for some vectorsxandyinR
.
(c)x=0orkx+yk6 =kyk, for all vectorsxandyinR
.
(d) There is some vectorx∈R
for whichx·x≤0.
(e) There is anx∈R
3
such that for every nonzeroy∈R
3
,x·y6 =0.
(f) For everyx∈R
4
,thereissomey∈R
4
such thatx·y6 =0.
(21) (a) Contrapositive: Ifx6 =0andkx−yk≤kyk,thenx·y6 =0.
Converse: Ifx=0orkx−ykkyk,thenx·y=0.
Inverse: Ifx·y6 =0,thenx6 =0andkx−yk≤kyk.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 10
Prove:kx 1+···+x +x+1k
2
=kx 1k
2
+···+kx k
2
+kx +1k
2
.
We have
k(x
1+···+x )+x +1k
2
=kx 1+···+x k
2
+2((x 1+···+x )·x+1)+kx +1k
2
=kx 1+···+x k
2
+kx +1k
2
(sincex +1is orthogonal to all ofx 1x )
=kx
1k
2
+···+kx k
2
+kx +1k
2
by the inductive hypothesis.
(18) Base Step (=1): We must show(
1x1)·y≤| 1|kyk.
But,
(
1x1)·y≤|( 1x1)·y|≤k 1x1kkyk
(by the Cauchy-Schwarz Inequality)
=|
1|kx1kkyk=| 1|kyk
sincex
1is a unit vector.
Inductive Step: Assume(
1x1+···+ x)·y≤(| 1|+···+| |)kyk,forsome≥1.
Prove:(
1x1+···+ x++1x+1)·y≤(| 1|+···+| |+| +1|)kyk.
We have
((
1x1+···+ x)+ +1x+1)·y=( 1x1+···+ x)·y+( +1x+1)·y
≤(|
1|+···+| |)kyk+( +1x+1)·y
(by the inductive hypothesis)
≤(|
1|+···+| |)kyk+| +1|||y||
(by an argument similar to the Base Step)
=(|
1|+···+| |+| +1|)kyk
(19) Step 1 cannot be reversed, becausecould equal±(
2
+2).
Step 2 cannot be reversed, because
2
could equal
4
+4
2
+.
Step 4 cannot be reversed, because in generaldoes not have to equal
2
+2.
Step 6 cannot be reversed, since
could equal2+.
All other steps remain true when reversed.
(20) (a) For every unit vectorxinR
3
,x·[1−23]6 =0.
(b)x6 =0andx·y≤0, for some vectorsxandyinR
.
(c)x=0orkx+yk6 =kyk, for all vectorsxandyinR
.
(d) There is some vectorx∈R
for whichx·x≤0.
(e) There is anx∈R
3
such that for every nonzeroy∈R
3
,x·y6 =0.
(f) For everyx∈R
4
,thereissomey∈R
4
such thatx·y6 =0.
(21) (a) Contrapositive: Ifx6 =0andkx−yk≤kyk,thenx·y6 =0.
Converse: Ifx=0orkx−ykkyk,thenx·y=0.
Inverse: Ifx·y6 =0,thenx6 =0andkx−yk≤kyk.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 10
Answers to Exercises Section 1.4
(b) Contrapositive: Ifkx−yk≤kyk, then eitherx=0orx·y6 =0.
Converse: Ifkx−ykkyk,thenx6 =0andx·y=0.
Inverse: Ifx=0orx·y6 =0,thenkx−yk≤kyk.
(22) Supposex6 =0.Wemustprovex·y6 =0for some vectory∈R
.Lety=x.
(23) Letx=[11],y=[1−1].
(24) Lety=[1−22].Thensincex·y≥0, Result 3 implies thatkx+ykkyk=3.
(25) (a)F (b)T (c)T (d)F (e)F (f)F (g)F (h)T (i)F
Section 1.4
(1) (a)
⎡
⎣
21 3
27−5
90−1
⎤
⎦
(b) Impossible
(c)
⎡
⎣
−16 8 12
020−4
24 4−8
⎤
⎦
(d)
⎡
⎣
−32 8 6
−82 14
06−8
⎤
⎦
(e) Impossible
(f)
⎡
⎣
−70 8
−13 1
99−5
⎤
⎦
(g)
⎡
⎣
−23 14 −9
−588
−9−18 1
⎤
⎦
(h) Impossible
(i)
⎡
⎣
−1112
−158
8−3−4
⎤
⎦
(j)
⎡
⎣
−1112
−158
8−3−4
⎤
⎦
(k)
∙
−12 8 −6
10−826
¸
(l) Impossible
(m) Impossible
(n)
⎡
⎣
13−62
3−3−5
351
⎤
⎦
(2) Square:BCEFGHJKLMNPQ
Diagonal:BGN
Upper triangular:BGLN
Lower triangular:BGMNQ
Symmetric:BFGJNP
Skew-symmetric:H(but notECK)
Transposes:A
=
∙
−106
410
¸
,B
=B,C
=
∙
−1−1
11
¸
,andsoon
(3) (a)
⎡
⎢
⎢
⎣
3−
1
2
5
2
−
1
2
21
5
2
12
⎤
⎥
⎥
⎦
+
⎡
⎢
⎢
⎣
0−
1
2
3
2
1
2
04
−
3
2
−40
⎤
⎥
⎥
⎦
(b)
⎡
⎢
⎢
⎣
1
3
2
0
3
2
3−1
0−10
⎤
⎥
⎥
⎦
+
⎡
⎢
⎢
⎣
0−
3
2
−4
3
2
00
400
⎤
⎥
⎥
⎦
(c)
⎡
⎢
⎢
⎣
20 00
05 00
00−20
00 05
⎤
⎥
⎥
⎦
+
⎡
⎢
⎢
⎣
034 −1
−30 −12
−4100
1−200
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 11
(b) Contrapositive: Ifkx−yk≤kyk, then eitherx=0orx·y6 =0.
Converse: Ifkx−ykkyk,thenx6 =0andx·y=0.
Inverse: Ifx=0orx·y6 =0,thenkx−yk≤kyk.
(22) Supposex6 =0.Wemustprovex·y6 =0for some vectory∈R
.Lety=x.
(23) Letx=[11],y=[1−1].
(24) Lety=[1−22].Thensincex·y≥0, Result 3 implies thatkx+ykkyk=3.
(25) (a)F (b)T (c)T (d)F (e)F (f)F (g)F (h)T (i)F
Section 1.4
(1) (a)
⎡
⎣
21 3
27−5
90−1
⎤
⎦
(b) Impossible
(c)
⎡
⎣
−16 8 12
020−4
24 4−8
⎤
⎦
(d)
⎡
⎣
−32 8 6
−82 14
06−8
⎤
⎦
(e) Impossible
(f)
⎡
⎣
−70 8
−13 1
99−5
⎤
⎦
(g)
⎡
⎣
−23 14 −9
−588
−9−18 1
⎤
⎦
(h) Impossible
(i)
⎡
⎣
−1112
−158
8−3−4
⎤
⎦
(j)
⎡
⎣
−1112
−158
8−3−4
⎤
⎦
(k)
∙
−12 8 −6
10−826
¸
(l) Impossible
(m) Impossible
(n)
⎡
⎣
13−62
3−3−5
351
⎤
⎦
(2) Square:BCEFGHJKLMNPQ
Diagonal:BGN
Upper triangular:BGLN
Lower triangular:BGMNQ
Symmetric:BFGJNP
Skew-symmetric:H(but notECK)
Transposes:A
=
∙
−106
410
¸
,B
=B,C
=
∙
−1−1
11
¸
,andsoon
(3) (a)
⎡
⎢
⎢
⎣
3−
1
2
5
2
−
1
2
21
5
2
12
⎤
⎥
⎥
⎦
+
⎡
⎢
⎢
⎣
0−
1
2
3
2
1
2
04
−
3
2
−40
⎤
⎥
⎥
⎦
(b)
⎡
⎢
⎢
⎣
1
3
2
0
3
2
3−1
0−10
⎤
⎥
⎥
⎦
+
⎡
⎢
⎢
⎣
0−
3
2
−4
3
2
00
400
⎤
⎥
⎥
⎦
(c)
⎡
⎢
⎢
⎣
20 00
05 00
00−20
00 05
⎤
⎥
⎥
⎦
+
⎡
⎢
⎢
⎣
034 −1
−30 −12
−4100
1−200
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 11
Answers to Exercises Section 1.4
(d)
⎡
⎢
⎢
⎣
−37 −2−1
743 −6
−235 −8
−1−6−8−5
⎤
⎥
⎥
⎦
+
⎡
⎢
⎢
⎣
0−47−3
4025
−7−20−6
3−56 0
⎤
⎥
⎥
⎦
(4) IfA
=B
,then
¡
A
¢
=
¡
B
¢
, implyingA=B, by part (1) of Theorem 1.13.
(5) (a) IfAis an×matrix, thenA
is×.IfA=A
,then=.
(b) IfAis a diagonal matrix and if6 =,then
=0= .
(c) Follows directly from part (b), sinceI
is diagonal.
(d) The matrix must be a square zero matrix; that is,O
,forsome.
(6) (a) If6 =,then
+=0+0=0 .
(b) Use the fact that
+=+.
(7) Use induction on.BaseStep(=1): Obvious. Inductive Step: AssumeA
1A +1andB=
P
=1
Aare upper triangular. Prove thatD=
P
+1
=1
Ais upper triangular. LetC=A +1.Then
D=B+C. Hence,
=+=0+0=0 if(by the inductive hypothesis). HenceDis
upper triangular.
(8) (a) LetB=A
.Then ===.LetD=A.Then ===.
(b) LetB=A
.Then ==− =− .
LetD=A.Then
==(− )=− =− .
(9) (a) Part (4): LetB=A+(−A)(=(−A)+A, by part (1)). Then
=+(− )=0.
(b) Part (5): LetD=(A+B)and letE=A+B. Then,
=( +)= +=.
(c) Part (7): LetB=()Aand letE=(A).Then
=() =( )= .
(10) (a) Part (1):(A
)
andAare both×matrices. The( )entry of(A
)
=( )entry ofA
=( )entry ofAThus,(A
)
=A
(b) Part (2), Subtraction Case:(A−B)
andA
−B
are both×matrices. The( )entry
of(A−B)
=( )entry ofA−B=( )entry ofA−()entry ofB=( )entry ofA
−
()entry ofB
=( )entry ofA
−B
Thus,(A−B)
=A
−B
(c) Part (3):(A)
and(A
)are both×matrices. The( )entry of(A)
=( )entry of
A=(( )entry ofA)=(( )entry ofA
)=( )entry of(A
)Thus,(A)
=(A
)
(11) Assume6 =0. We must showA=O
.Butforall ,1≤≤,1≤≤, =0with6 =0.
Hence, all
=0.
(12) (a)S+V=
1
2
(A+A
)+
1
2
(A−A
)=
1
2
(2A)=A;
S
=(
1
2
(A+A
))
=
1
2
(A+A
)
=
1
2
(A
+(A
)
)=
1
2
(A
+A)=
1
2
(A+A
)=S;
V
=(
1
2
(A−A
))
=
1
2
(A−A
)
=
1
2
(A
−(A
)
)=
1
2
(A
−A)=−
1
2
(A−A
)=−V
(b)S
1−V1=S2−V2
(c) Follows immediately from part (b).
(d) Part (a) showsAcan be decomposed as the sum of a symmetric matrix and a skew-symmetric
matrix, while parts (b) and (c) show that the decomposition forAis unique.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 12
(d)
⎡
⎢
⎢
⎣
−37 −2−1
743 −6
−235 −8
−1−6−8−5
⎤
⎥
⎥
⎦
+
⎡
⎢
⎢
⎣
0−47−3
4025
−7−20−6
3−56 0
⎤
⎥
⎥
⎦
(4) IfA
=B
,then
¡
A
¢
=
¡
B
¢
, implyingA=B, by part (1) of Theorem 1.13.
(5) (a) IfAis an×matrix, thenA
is×.IfA=A
,then=.
(b) IfAis a diagonal matrix and if6 =,then
=0= .
(c) Follows directly from part (b), sinceI
is diagonal.
(d) The matrix must be a square zero matrix; that is,O
,forsome.
(6) (a) If6 =,then
+=0+0=0 .
(b) Use the fact that
+=+.
(7) Use induction on.BaseStep(=1): Obvious. Inductive Step: AssumeA
1A +1andB=
P
=1
Aare upper triangular. Prove thatD=
P
+1
=1
Ais upper triangular. LetC=A +1.Then
D=B+C. Hence,
=+=0+0=0 if(by the inductive hypothesis). HenceDis
upper triangular.
(8) (a) LetB=A
.Then ===.LetD=A.Then ===.
(b) LetB=A
.Then ==− =− .
LetD=A.Then
==(− )=− =− .
(9) (a) Part (4): LetB=A+(−A)(=(−A)+A, by part (1)). Then
=+(− )=0.
(b) Part (5): LetD=(A+B)and letE=A+B. Then,
=( +)= +=.
(c) Part (7): LetB=()Aand letE=(A).Then
=() =( )= .
(10) (a) Part (1):(A
)
andAare both×matrices. The( )entry of(A
)
=( )entry ofA
=( )entry ofAThus,(A
)
=A
(b) Part (2), Subtraction Case:(A−B)
andA
−B
are both×matrices. The( )entry
of(A−B)
=( )entry ofA−B=( )entry ofA−()entry ofB=( )entry ofA
−
()entry ofB
=( )entry ofA
−B
Thus,(A−B)
=A
−B
(c) Part (3):(A)
and(A
)are both×matrices. The( )entry of(A)
=( )entry of
A=(( )entry ofA)=(( )entry ofA
)=( )entry of(A
)Thus,(A)
=(A
)
(11) Assume6 =0. We must showA=O
.Butforall ,1≤≤,1≤≤, =0with6 =0.
Hence, all
=0.
(12) (a)S+V=
1
2
(A+A
)+
1
2
(A−A
)=
1
2
(2A)=A;
S
=(
1
2
(A+A
))
=
1
2
(A+A
)
=
1
2
(A
+(A
)
)=
1
2
(A
+A)=
1
2
(A+A
)=S;
V
=(
1
2
(A−A
))
=
1
2
(A−A
)
=
1
2
(A
−(A
)
)=
1
2
(A
−A)=−
1
2
(A−A
)=−V
(b)S
1−V1=S2−V2
(c) Follows immediately from part (b).
(d) Part (a) showsAcan be decomposed as the sum of a symmetric matrix and a skew-symmetric
matrix, while parts (b) and (c) show that the decomposition forAis unique.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 12
Answers to Exercises Section 1.5
(13) (a) Trace (B) = 1, trace (C) = 0, trace (E)=−6,trace(F) = 2, trace (G) = 18, trace (H)=0,
trace (J) = 1, trace (K) = 4, trace (L) = 3, trace (M) = 0, trace (N) = 3, trace (P)=0,
trace (Q)=1
(b) Part (i): LetD=A+B.Thentrace(D)=
P
=1
=
P
=1
+
P
=1
=trace(A)+trace(B).
Part (ii): LetB=A.Thentrace(B)=
P
=1
=
P
=1
=
P
=1
=(trace(A)).
Part (iii): LetB=A
.Thentrace(B)=
P
=1
=
P
=1
(since =for all)=trace(A).
(c) Not necessarily: consider the matricesLandNin Exercise 2. (Note: If=1, the statement is
true.)
(14) (a) F (b) T (c) F (d) T (e) T
Section 1.5
(1) (a) Impossible
(b)
⎡
⎣
34−24
42 49
8−22
⎤
⎦
(c) Impossible
(d)
⎡
⎣
73−34
77−25
19−14
⎤
⎦
(e)[−38]
(f)
⎡
⎣
−24 48−16
3−62
−12 24 −8
⎤
⎦
(g) Impossible
(h)[56−8]
(i) Impossible
(j) Impossible
(k)
⎡
⎣
22 9 −6
97318
2−64
⎤
⎦
(l)
⎡
⎢
⎢
⎣
5325
4131
1102
4131
⎤
⎥
⎥
⎦
(m)[226981]
(n)
⎡
⎣
146 5 −603
154 27−560
38−9−193
⎤
⎦
(o) Impossible
(2) (a) No (b) Yes (c) No (d) Yes (e) No
(3) (a)
[15−13−8]
(b)
⎡
⎣
11
6
3
⎤
⎦
(c)[4]
(d)[28−212]
(4) (a) Valid, by Theorem 1.16, part (1)
(b) Invalid
(c) Valid, by Theorem 1.16, part (1)
(d) Valid, by Theorem 1.16, part (2)
(e) Valid, by Theorem 1.18
(f) Invalid
(g) Valid, by Theorem 1.16, part (3)
(h) Valid, by Theorem 1.16, part (2)
(i) Invalid
(j) Valid, by Theorem 1.16, part (3),
and Theorem 1.18
(5)
Outlet 1
Outlet 2
Outlet 3
Outlet 4
Salary Fringe Benefits
⎡
⎢
⎢
⎣
$367500 $78000
$225000 $48000
$765000 $162000
$360000 $76500
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 13
(13) (a) Trace (B) = 1, trace (C) = 0, trace (E)=−6,trace(F) = 2, trace (G) = 18, trace (H)=0,
trace (J) = 1, trace (K) = 4, trace (L) = 3, trace (M) = 0, trace (N) = 3, trace (P)=0,
trace (Q)=1
(b) Part (i): LetD=A+B.Thentrace(D)=
P
=1
=
P
=1
+
P
=1
=trace(A)+trace(B).
Part (ii): LetB=A.Thentrace(B)=
P
=1
=
P
=1
=
P
=1
=(trace(A)).
Part (iii): LetB=A
.Thentrace(B)=
P
=1
=
P
=1
(since =for all)=trace(A).
(c) Not necessarily: consider the matricesLandNin Exercise 2. (Note: If=1, the statement is
true.)
(14) (a) F (b) T (c) F (d) T (e) T
Section 1.5
(1) (a) Impossible
(b)
⎡
⎣
34−24
42 49
8−22
⎤
⎦
(c) Impossible
(d)
⎡
⎣
73−34
77−25
19−14
⎤
⎦
(e)[−38]
(f)
⎡
⎣
−24 48−16
3−62
−12 24 −8
⎤
⎦
(g) Impossible
(h)[56−8]
(i) Impossible
(j) Impossible
(k)
⎡
⎣
22 9 −6
97318
2−64
⎤
⎦
(l)
⎡
⎢
⎢
⎣
5325
4131
1102
4131
⎤
⎥
⎥
⎦
(m)[226981]
(n)
⎡
⎣
146 5 −603
154 27−560
38−9−193
⎤
⎦
(o) Impossible
(2) (a) No (b) Yes (c) No (d) Yes (e) No
(3) (a)
[15−13−8]
(b)
⎡
⎣
11
6
3
⎤
⎦
(c)[4]
(d)[28−212]
(4) (a) Valid, by Theorem 1.16, part (1)
(b) Invalid
(c) Valid, by Theorem 1.16, part (1)
(d) Valid, by Theorem 1.16, part (2)
(e) Valid, by Theorem 1.18
(f) Invalid
(g) Valid, by Theorem 1.16, part (3)
(h) Valid, by Theorem 1.16, part (2)
(i) Invalid
(j) Valid, by Theorem 1.16, part (3),
and Theorem 1.18
(5)
Outlet 1
Outlet 2
Outlet 3
Outlet 4
Salary Fringe Benefits
⎡
⎢
⎢
⎣
$367500 $78000
$225000 $48000
$765000 $162000
$360000 $76500
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 13
Answers to Exercises Section 1.5
(6)
June
July
August
Tickets Food Souvenirs
⎡
⎣
$1151300 $3056900 $2194400
$1300700 $3456700 $2482400
$981100 $2615900 $1905100
⎤
⎦
(7)
Nitrogen
Phosphate
Potash
Field 1 Field 2 Field 3
⎡
⎣
100 045 065
090 035 075
095 035 085
⎤
⎦(in tons)
(8)
Rocket 1
Rocket 2
Rocket 3
Rocket 4
Chip 1 Chip 2 Chip 3 Chip 4
⎡
⎢
⎢
⎣
2131 1569 1839 2750
2122 1559 1811 2694
2842 2102 2428 3618
2456 1821 2097 3124
⎤
⎥
⎥
⎦
(9) (a) One example:
∙
11
0−1
¸
(b) One example:
⎡
⎣
110
0−10
001
⎤
⎦
(c) Consider
⎡
⎣
001
100
010
⎤
⎦
(10) (a) Third row, fourth column entry ofAB
(b) Fourth row,first column entry ofAB
(c) Third row, second column entry ofBA
(d) Second row,fifthcolumnentryofBA
(11) (a)
P
=1
32 (b)
P
=1
41
(12) (a)[−2743−56]
(b)
⎡
⎣
56
−57
18
⎤
⎦
(13) (a)[−6115209]
(b)
⎡
⎣
43
−41
−12
⎤
⎦
(14) (a) LetB=Ai.ThenBis×1and
1=
P
=1
1=( 1)(1) + ( 2)(0) + ( 3)(0) = 1.
(b)Ae
=th column ofA.
(c) By part (b), each column ofAis easily seen to be the zero vector by lettingxequal each of
e
1e in turn.
(15) (a) Proof of Part (2): The( )entry ofA(B+C)
=(th row ofA)·(th column of(B+C))
=(th row ofA)·(th column ofB+th column ofC)
=(th row ofA)·(th column ofB)
+(th row ofA)·(th column ofC)
=(( )entry ofAB)+(( )entry ofAC)
=( )entry of(AB+AC).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 14
(6)
June
July
August
Tickets Food Souvenirs
⎡
⎣
$1151300 $3056900 $2194400
$1300700 $3456700 $2482400
$981100 $2615900 $1905100
⎤
⎦
(7)
Nitrogen
Phosphate
Potash
Field 1 Field 2 Field 3
⎡
⎣
100 045 065
090 035 075
095 035 085
⎤
⎦(in tons)
(8)
Rocket 1
Rocket 2
Rocket 3
Rocket 4
Chip 1 Chip 2 Chip 3 Chip 4
⎡
⎢
⎢
⎣
2131 1569 1839 2750
2122 1559 1811 2694
2842 2102 2428 3618
2456 1821 2097 3124
⎤
⎥
⎥
⎦
(9) (a) One example:
∙
11
0−1
¸
(b) One example:
⎡
⎣
110
0−10
001
⎤
⎦
(c) Consider
⎡
⎣
001
100
010
⎤
⎦
(10) (a) Third row, fourth column entry ofAB
(b) Fourth row,first column entry ofAB
(c) Third row, second column entry ofBA
(d) Second row,fifthcolumnentryofBA
(11) (a)
P
=1
32 (b)
P
=1
41
(12) (a)[−2743−56]
(b)
⎡
⎣
56
−57
18
⎤
⎦
(13) (a)[−6115209]
(b)
⎡
⎣
43
−41
−12
⎤
⎦
(14) (a) LetB=Ai.ThenBis×1and
1=
P
=1
1=( 1)(1) + ( 2)(0) + ( 3)(0) = 1.
(b)Ae
=th column ofA.
(c) By part (b), each column ofAis easily seen to be the zero vector by lettingxequal each of
e
1e in turn.
(15) (a) Proof of Part (2): The( )entry ofA(B+C)
=(th row ofA)·(th column of(B+C))
=(th row ofA)·(th column ofB+th column ofC)
=(th row ofA)·(th column ofB)
+(th row ofA)·(th column ofC)
=(( )entry ofAB)+(( )entry ofAC)
=( )entry of(AB+AC).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 14
Answers to Exercises Section 1.5
(b) Proof of Part (3): The( )entry of(A+B)C
=(th row of(A+B))·(th column ofC)
=(th row ofA+th row ofB)·(th column ofC)
=(th row ofA)·(th column ofC)
+(th row ofB)·(th column ofC)
=(( )entry ofAC)+(( )entry ofBC)
=( )entry of(AC+BC).
(c) For thefirst equation in Part (4), the( )entry of(AB
)
=((th row ofA)·(th column ofB))
=((th row ofA))·(th column ofB)
=(th row ofA)·(th column ofB)
=( )entry of(A)B.
For the second equation in Part (4), the( )entry of(A)B
=(th row ofA)·(th column ofB)
=((th row ofA))·(th column ofB)
=(th row ofA)·((th column of
B))
=(th row ofA)·(th column ofB)
=( )entry ofA(B).
.
(16) LetB=AO
. ClearlyBis an×matrix and =
P
=1
0=0.
(17) Proof thatAI
=A:LetB=AI .ThenBis clearly an×matrix and =
P
=1
=,
since
=0unless=,inwhichcase =1. The proof thatI A=Ais similar.
(18) (a) We need to show
=0if6 =.Now,if6 =, both factors in each term of the formula for
in the formal definition of matrix multiplication are zero, except possibly for the terms and
. But since the factors and also equal zero, all terms in the formula for equal zero.
(b) Assume. Consider the term
in the formula for .If,then =0.If≤,
then,so
=0. Hence all terms in the formula for equal zero.
(c) LetL
1andL 2be lower triangular matrices.U 1=L
1
andU 2=L
2
are then upper triangular,
andL
1L2=U
1
U
2
=(U 2U1)
(by Theorem 1.18). But by part (b),U 2U1is upper triangular.
SoL
1L2is lower triangular.
(19) Base Step: Clearly,(A)
1
=
1
A
1
.
Inductive Step: Assume(A)
=
A
,andprove(A)
+1
=
+1
A
+1
.Now,(A)
+1
=(A)
(A)
=(
A
)(A)(by the inductive hypothesis) =
A
A(by part (4) of Theorem 1.16) =
+1
A
+1
.
(20) (a) Proof of Part (1): Base Step:A
+0
=A
=A
I=A
A
0
.
Inductive Step: AssumeA
+
=A
A
for some≥0.WemustproveA
+(+1)
=A
A
+1
.But
A
+(+1)
=A
(+)+1
=A
+
A=(A
A
)A(by the inductive hypothesis)=A
(A
A)=A
A
+1
.
(b) Proof of Part (2): (We only need prove(A
)
=A
.) Base Step:(A
)
0
=I=A
0
=A
0
.
Inductive Step: Assume(A
)
=A
for some integer≥0.Wemustprove(A
)
+1
=A
(+1)
.
But(A
)
+1
=(A
)
A
=A
A
(by the inductive hypothesis)=A
+
(by part (1))=A
(+1)
.
(21)A
A
=A
+
(by part (1) of Theorem 1.17)=A
+
=A
A
(by part (1) of Theorem 1.17).
(22) (a)A(
1B1+2B2+···+ B)= 1AB1+2AB2+···+ AB(by parts (2) and (4) of Theorem
1.16)=
1B1A+ 2B2A+···+ BA(sinceAcommutes withB 1,B2,,B )=( 1B1+
2B2+···+ B)A(by part (3) of Theorem 1.16).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 15
(b) Proof of Part (3): The( )entry of(A+B)C
=(th row of(A+B))·(th column ofC)
=(th row ofA+th row ofB)·(th column ofC)
=(th row ofA)·(th column ofC)
+(th row ofB)·(th column ofC)
=(( )entry ofAC)+(( )entry ofBC)
=( )entry of(AC+BC).
(c) For thefirst equation in Part (4), the( )entry of(AB
)
=((th row ofA)·(th column ofB))
=((th row ofA))·(th column ofB)
=(th row ofA)·(th column ofB)
=( )entry of(A)B.
For the second equation in Part (4), the( )entry of(A)B
=(th row ofA)·(th column ofB)
=((th row ofA))·(th column ofB)
=(th row ofA)·((th column of
B))
=(th row ofA)·(th column ofB)
=( )entry ofA(B).
.
(16) LetB=AO
. ClearlyBis an×matrix and =
P
=1
0=0.
(17) Proof thatAI
=A:LetB=AI .ThenBis clearly an×matrix and =
P
=1
=,
since
=0unless=,inwhichcase =1. The proof thatI A=Ais similar.
(18) (a) We need to show
=0if6 =.Now,if6 =, both factors in each term of the formula for
in the formal definition of matrix multiplication are zero, except possibly for the terms and
. But since the factors and also equal zero, all terms in the formula for equal zero.
(b) Assume. Consider the term
in the formula for .If,then =0.If≤,
then,so
=0. Hence all terms in the formula for equal zero.
(c) LetL
1andL 2be lower triangular matrices.U 1=L
1
andU 2=L
2
are then upper triangular,
andL
1L2=U
1
U
2
=(U 2U1)
(by Theorem 1.18). But by part (b),U 2U1is upper triangular.
SoL
1L2is lower triangular.
(19) Base Step: Clearly,(A)
1
=
1
A
1
.
Inductive Step: Assume(A)
=
A
,andprove(A)
+1
=
+1
A
+1
.Now,(A)
+1
=(A)
(A)
=(
A
)(A)(by the inductive hypothesis) =
A
A(by part (4) of Theorem 1.16) =
+1
A
+1
.
(20) (a) Proof of Part (1): Base Step:A
+0
=A
=A
I=A
A
0
.
Inductive Step: AssumeA
+
=A
A
for some≥0.WemustproveA
+(+1)
=A
A
+1
.But
A
+(+1)
=A
(+)+1
=A
+
A=(A
A
)A(by the inductive hypothesis)=A
(A
A)=A
A
+1
.
(b) Proof of Part (2): (We only need prove(A
)
=A
.) Base Step:(A
)
0
=I=A
0
=A
0
.
Inductive Step: Assume(A
)
=A
for some integer≥0.Wemustprove(A
)
+1
=A
(+1)
.
But(A
)
+1
=(A
)
A
=A
A
(by the inductive hypothesis)=A
+
(by part (1))=A
(+1)
.
(21)A
A
=A
+
(by part (1) of Theorem 1.17)=A
+
=A
A
(by part (1) of Theorem 1.17).
(22) (a)A(
1B1+2B2+···+ B)= 1AB1+2AB2+···+ AB(by parts (2) and (4) of Theorem
1.16)=
1B1A+ 2B2A+···+ BA(sinceAcommutes withB 1,B2,,B )=( 1B1+
2B2+···+ B)A(by part (3) of Theorem 1.16).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 15
Answers to Exercises Section 1.5
(b) By Exercise 21,A
commutes withA
,for=0to. Therefore, by part (a),A
commutes with
I+1A+ 2A
2
+···+ A
.
(c) By part (b), (
I+1A+ 2A
2
+···+ A
)commuteswithA
for every nonnegative integer.
Therefore, by part (a), (
I+1A+ 2A
2
+···+ A
)commuteswith( I+1A+ 2A
2
+
···+
A
).
(23) (a) IfAis an×matrix, andBis a×matrix, then the fact thatABexists means=,
and the fact thatBAexists means=.ThenABis an×matrix, whileBAis an×
matrix. IfAB=BA,then=.
(b) Note that by the Distributive Law,(A+B)
2
=A
2
+AB+BA+B
2
.
(24)(AB)C=A(BC)=A(CB)=(AC)B=(CA)B=C(AB).
(25) Use the fact thatA
B
=(BA)
,whileB
A
=(AB)
.
(26)(AA
)
=(A
)
A
(by Theorem 1.18)=AA
. (Similarly forA
A.)
(27) (a) IfABare both skew-symmetric, then(AB)
=B
A
=(−B)(−A)=(−1)(−1)BA=BA.
The symmetric case is similar.
(b) Use the fact that(AB)
=B
A
=BA,sinceABare symmetric.
(28) (a) The( )entry ofAA
=(th row ofA)·(th column ofA
)=(th row ofA)·(th row ofA)
= sum of the squares of the entries in theth row ofA. Hence, trace(AA
) is the sum of the
squares of the entries from all rows ofA.
(c) Trace(AB)=
P
=1
(
P
=1
)=
P
=1P
=1
=
P
=1P
=1
.
Reversing the roles of the dummy variablesandgives
P
=1P
=1
,whichisequalto
trace(BA).
(29) (a) Consider any matrix of the form
∙
10
0
¸
.
(c)(I
−A)
2
=I
2
−IA−AI +A
2
=I−A−A+A=I −A.
(d)
⎡
⎣
2−1−1
10 −1
1−10
⎤
⎦
(e)A
2
=(A)A=(AB)A=A(BA)=AB=A.
(30) (a) (th row ofA)·(th column ofB)=( )th entry ofO
=0.
(b) ConsiderA=
∙
12−1
24−2
¸
andB=
⎡
⎣
1−2
01
10
⎤
⎦.
(c) LetC=
⎡
⎣
2−4
02
20
⎤
⎦.
(31) A2×2matrixA=
∙
1112
2122
¸
that commutes with every other2×2matrix must have the form
A=I
2.For,ifB=
∙
01
00
¸
,thenAB=
∙
0
11
0 21
¸
, which must equalBA=
∙
2122
00
¸
. Hence,
Copyrightc°2016 Elsevier Ltd. All rights reserved. 16
(b) By Exercise 21,A
commutes withA
,for=0to. Therefore, by part (a),A
commutes with
I+1A+ 2A
2
+···+ A
.
(c) By part (b), (
I+1A+ 2A
2
+···+ A
)commuteswithA
for every nonnegative integer.
Therefore, by part (a), (
I+1A+ 2A
2
+···+ A
)commuteswith( I+1A+ 2A
2
+
···+
A
).
(23) (a) IfAis an×matrix, andBis a×matrix, then the fact thatABexists means=,
and the fact thatBAexists means=.ThenABis an×matrix, whileBAis an×
matrix. IfAB=BA,then=.
(b) Note that by the Distributive Law,(A+B)
2
=A
2
+AB+BA+B
2
.
(24)(AB)C=A(BC)=A(CB)=(AC)B=(CA)B=C(AB).
(25) Use the fact thatA
B
=(BA)
,whileB
A
=(AB)
.
(26)(AA
)
=(A
)
A
(by Theorem 1.18)=AA
. (Similarly forA
A.)
(27) (a) IfABare both skew-symmetric, then(AB)
=B
A
=(−B)(−A)=(−1)(−1)BA=BA.
The symmetric case is similar.
(b) Use the fact that(AB)
=B
A
=BA,sinceABare symmetric.
(28) (a) The( )entry ofAA
=(th row ofA)·(th column ofA
)=(th row ofA)·(th row ofA)
= sum of the squares of the entries in theth row ofA. Hence, trace(AA
) is the sum of the
squares of the entries from all rows ofA.
(c) Trace(AB)=
P
=1
(
P
=1
)=
P
=1P
=1
=
P
=1P
=1
.
Reversing the roles of the dummy variablesandgives
P
=1P
=1
,whichisequalto
trace(BA).
(29) (a) Consider any matrix of the form
∙
10
0
¸
.
(c)(I
−A)
2
=I
2
−IA−AI +A
2
=I−A−A+A=I −A.
(d)
⎡
⎣
2−1−1
10 −1
1−10
⎤
⎦
(e)A
2
=(A)A=(AB)A=A(BA)=AB=A.
(30) (a) (th row ofA)·(th column ofB)=( )th entry ofO
=0.
(b) ConsiderA=
∙
12−1
24−2
¸
andB=
⎡
⎣
1−2
01
10
⎤
⎦.
(c) LetC=
⎡
⎣
2−4
02
20
⎤
⎦.
(31) A2×2matrixA=
∙
1112
2122
¸
that commutes with every other2×2matrix must have the form
A=I
2.For,ifB=
∙
01
00
¸
,thenAB=
∙
0
11
0 21
¸
, which must equalBA=
∙
2122
00
¸
. Hence,
Copyrightc°2016 Elsevier Ltd. All rights reserved. 16
Answers to Exercises Chapter 1 Review
21=0and 11=22.Let= 11=22.ThenA=
∙
12
0
¸
.Also,ifD=
∙
00
10
¸
,then
AD=
∙
120
0
¸
must equalDA=
∙
00
12
¸
,whichgives
12=0,andsoA=I 2.
Finally, note thatI
2actually does commute with every2×2matrixM,since(I 2)M=(I 2M)=
M=(MI
2)=M(I 2).
(32) (a) T (b) T (c) T (d) F (e) F (f) F (g) F
Chapter 1 Review Exercises
(1) Yes. Vectors corresponding to adjacent sides are orthogonal. Vectors corresponding to opposite sides
are parallel, with one pair having slope
3
5
and the other pair having slope−
5
3
.
(2)u=
h
5
√
394
−
12
√
394
15
√
394
i
≈[02481−0595507444]; slightly longer.
(3) Net velocity=
£
4
√
2−5−4
√
2
¤
≈[06569−56569]; speed≈56947mi/hr.
(4)a=[−10910]m/sec
2
(5)|x·y|=74≤kxkkyk≈909
(6)≈136
◦
(7)proj
ab=
£
114
25
−
38
25
19
25
57
25
¤
=[456−152076228];
b−proj
ab=
£
−
14
25
−
62
25
56
25
−
32
25
¤
=[−056−248224−128];
a·(b−proj
ab)=0.
(8)−1782joules
(9) We must prove that(x+y)·(x−y)=0=⇒kxk=kykBut,
(x+y)·(x−y)=0=⇒x·x−x·y+y·x−y·y=0
=⇒kxk
2
−kyk
2
=0=⇒kxk
2
=kyk
2
=⇒kxk=kyk
(10) First,x6 =0,orelseproj
xyis not defined. Also,y6 =0,sincethatwouldimplyproj
xy=y.Now,
assumexky. Then, there is a scalar6 =0such thaty=x. Hence,
proj
xy=
Ã
x·y
kxk
2
!
x=
Ã
x·x
kxk
2
!
x=
Ã
kxk
2
kxk
2
!
x=x=y
contradicting the assumption thaty6 =proj
xy.
(11) (a)3A−4C
=
∙
3213
−11−19 0
¸
;AB=
∙
15−21−4
22−30 11
¸
;BAis not defined;
AC=
∙
23 14
−523
¸
;CA=
⎡
⎣
30−11 17
2018
−11 5 16
⎤
⎦;A
3
is not defined;
B
3
=
⎡
⎣
97−128 24
−284 375 −92
268−354 93
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 17
21=0and 11=22.Let= 11=22.ThenA=
∙
12
0
¸
.Also,ifD=
∙
00
10
¸
,then
AD=
∙
120
0
¸
must equalDA=
∙
00
12
¸
,whichgives
12=0,andsoA=I 2.
Finally, note thatI
2actually does commute with every2×2matrixM,since(I 2)M=(I 2M)=
M=(MI
2)=M(I 2).
(32) (a) T (b) T (c) T (d) F (e) F (f) F (g) F
Chapter 1 Review Exercises
(1) Yes. Vectors corresponding to adjacent sides are orthogonal. Vectors corresponding to opposite sides
are parallel, with one pair having slope
3
5
and the other pair having slope−
5
3
.
(2)u=
h
5
√
394
−
12
√
394
15
√
394
i
≈[02481−0595507444]; slightly longer.
(3) Net velocity=
£
4
√
2−5−4
√
2
¤
≈[06569−56569]; speed≈56947mi/hr.
(4)a=[−10910]m/sec
2
(5)|x·y|=74≤kxkkyk≈909
(6)≈136
◦
(7)proj
ab=
£
114
25
−
38
25
19
25
57
25
¤
=[456−152076228];
b−proj
ab=
£
−
14
25
−
62
25
56
25
−
32
25
¤
=[−056−248224−128];
a·(b−proj
ab)=0.
(8)−1782joules
(9) We must prove that(x+y)·(x−y)=0=⇒kxk=kykBut,
(x+y)·(x−y)=0=⇒x·x−x·y+y·x−y·y=0
=⇒kxk
2
−kyk
2
=0=⇒kxk
2
=kyk
2
=⇒kxk=kyk
(10) First,x6 =0,orelseproj
xyis not defined. Also,y6 =0,sincethatwouldimplyproj
xy=y.Now,
assumexky. Then, there is a scalar6 =0such thaty=x. Hence,
proj
xy=
Ã
x·y
kxk
2
!
x=
Ã
x·x
kxk
2
!
x=
Ã
kxk
2
kxk
2
!
x=x=y
contradicting the assumption thaty6 =proj
xy.
(11) (a)3A−4C
=
∙
3213
−11−19 0
¸
;AB=
∙
15−21−4
22−30 11
¸
;BAis not defined;
AC=
∙
23 14
−523
¸
;CA=
⎡
⎣
30−11 17
2018
−11 5 16
⎤
⎦;A
3
is not defined;
B
3
=
⎡
⎣
97−128 24
−284 375 −92
268−354 93
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 17
Answers to Exercises Chapter 1 Review
(b) Third row ofBC=[58].
(12)S=
⎡
⎢
⎢
⎣
4−
1
2
11
2
−
1
2
7−1
11
2
−1−2
⎤
⎥
⎥
⎦
;V=
⎡
⎢
⎢
⎣
0−
5
2
−
1
2
5
2
0−2
1
2
20
⎤
⎥
⎥
⎦
(13) Now,(3(A−B)
)
=3
¡
(A−B)
¢
(by part (3) of Theorem 1.13)=3(A−B)(by part (1) of
Theorem 1.13). Also,−(3(A−B)
)=3(−1)(A
−B
)(parts (2) and (3) of Theorem 1.13)=
3(−1)((−A)−(−B))(definition of skew-symmetric)=3(A−B). Hence,(3(A−B)
)
=−(3(A−B)
),
and so3(A−B)
is skew-symmetric.
(14) LetC=A+B.Now,
=+.Butfor, ==0. Hence, for, =0.Thus,C
is lower triangular.
(15)
Company I
Company II
Company III
Price Shipping Cost
⎡
⎣
$168500 $24200
$202500 $29100
$155000 $22200
⎤
⎦.
(16) Take the transpose of both sides ofA
B
=B
A
to getBA=AB.
Then,(AB)
2
=(AB)(AB)=A(BA)B=A(AB)B=A
2
B
2
.
(17) Negation: For every square matrixA,A
2
=A. Counterexample:A=[−1].
(18) IfA6 =O
22, then some row ofA,saytheth row, is nonzero. Apply Result 5 in Section 1.3 with
x=(th row ofA).
(19) Base Step(=2): SupposeAandBare upper triangular×matrices, and letC=AB.Then
==0,for. Hence, for,
=
X
=1
=
−1
X
=1
0·++
X
=+1
·0= (0) = 0
Thus,Cis upper triangular.
Inductive Step: LetA
1A +1be upper triangular matrices. Then, the productC=A 1···A
is upper triangular by the Inductive Hypothesis, and so the productA 1···A +1=CA +1is upper
triangular by the Base Step.
(20) (a) F
(b) T
(c) F
(d) F
(e) F
(f) T
(g) F
(h) F
(i) F
(j) T
(k) T
(l) T
(m) T
(n) F
(o) F
(p) F
(q) F
(r) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 18
(b) Third row ofBC=[58].
(12)S=
⎡
⎢
⎢
⎣
4−
1
2
11
2
−
1
2
7−1
11
2
−1−2
⎤
⎥
⎥
⎦
;V=
⎡
⎢
⎢
⎣
0−
5
2
−
1
2
5
2
0−2
1
2
20
⎤
⎥
⎥
⎦
(13) Now,(3(A−B)
)
=3
¡
(A−B)
¢
(by part (3) of Theorem 1.13)=3(A−B)(by part (1) of
Theorem 1.13). Also,−(3(A−B)
)=3(−1)(A
−B
)(parts (2) and (3) of Theorem 1.13)=
3(−1)((−A)−(−B))(definition of skew-symmetric)=3(A−B). Hence,(3(A−B)
)
=−(3(A−B)
),
and so3(A−B)
is skew-symmetric.
(14) LetC=A+B.Now,
=+.Butfor, ==0. Hence, for, =0.Thus,C
is lower triangular.
(15)
Company I
Company II
Company III
Price Shipping Cost
⎡
⎣
$168500 $24200
$202500 $29100
$155000 $22200
⎤
⎦.
(16) Take the transpose of both sides ofA
B
=B
A
to getBA=AB.
Then,(AB)
2
=(AB)(AB)=A(BA)B=A(AB)B=A
2
B
2
.
(17) Negation: For every square matrixA,A
2
=A. Counterexample:A=[−1].
(18) IfA6 =O
22, then some row ofA,saytheth row, is nonzero. Apply Result 5 in Section 1.3 with
x=(th row ofA).
(19) Base Step(=2): SupposeAandBare upper triangular×matrices, and letC=AB.Then
==0,for. Hence, for,
=
X
=1
=
−1
X
=1
0·++
X
=+1
·0= (0) = 0
Thus,Cis upper triangular.
Inductive Step: LetA
1A +1be upper triangular matrices. Then, the productC=A 1···A
is upper triangular by the Inductive Hypothesis, and so the productA 1···A +1=CA +1is upper
triangular by the Base Step.
(20) (a) F
(b) T
(c) F
(d) F
(e) F
(f) T
(g) F
(h) F
(i) F
(j) T
(k) T
(l) T
(m) T
(n) F
(o) F
(p) F
(q) F
(r) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 18
Answers to Exercises Section 2.1
Chapter 2
Section 2.1
(1) (a) Consistent;{(−236)}
(b) Consistent;{(5−42)}
(c) Inconsistent;{}
(d) Consistent;{(+72−36)|∈R};(7−306)(8−116)(9126)
(e) Consistent;{(2−−42+52)| ∈R};(−40502)(−215
02)(−50712)
(f) Consistent;{(+3−28)| ∈R};(−2008)(−1108)(1018)
(g) Consistent;{(6−13)}
(h) Inconsistent;{}
(2) (a){(3+13+46++13−2+5)| ∈R}
(b){(3−12+50
−132−8+3−2)| ∈R}
(c){(−20+9−153−687−2+37+154+2)| ∈R}
(d){}
(3) 52 nickels, 64 dimes, 32 quarters
(4)=2
2
−+5
(5)=−3
3
+4
2
−5+6
(6)
2
+
2
−6−8=0,or(−3)
2
+(−4)
2
=25
(7) In each part,(AB)=((A))B, which equals the given matrix.
(a)
⎡
⎢
⎢
⎣
26 15 −6
64 1
0−612
10 4 −14
⎤
⎥
⎥
⎦
(b)
⎡
⎢
⎢
⎣
26 15−6
10 4−14
18 6 15
64 1
⎤
⎥
⎥
⎦
(8) (a) To save space, we writehCi
for theth row of a matrixC.
For the Type (I) operation:hi←hi:Now,h(AB)i
=hABi =hAi B(by the hint
in the text)=h(A)i
B=h(A)Bi .But,if6 =,h(AB)i =hABi =hAi B(by the hint
in the text)=h(A)i
B=h(A)Bi .
For the Type (II) operation:hi←hi+hi:Now,h(AB)i
=hABi +hABi =
hAi
B+hAi B(by the hint in the text)=(hAi +hAi )B=h(A)i B=h(A)Bi .But,if
6 =,h(AB)i
=hABi =hAi B(by the hint in the text)=h(A)i B=h(A)Bi .
For the Type (III) operation:hi←→hi:Now,h(AB)i
=hABi =hAi B(by the
hint in the text)=h(A)i
B=h(A)Bi . Similarly,h(AB)i =hABi =hAi B(by the hint
in the text)=h(A)i
B=h(A)Bi .And,if6 =and6 =,h(AB)i =hABi =hAi B
(by the hint in the text)=h(A)i
B=h(A)Bi .
Copyrightc°2016 Elsevier Ltd. All rights reserved. 19
Chapter 2
Section 2.1
(1) (a) Consistent;{(−236)}
(b) Consistent;{(5−42)}
(c) Inconsistent;{}
(d) Consistent;{(+72−36)|∈R};(7−306)(8−116)(9126)
(e) Consistent;{(2−−42+52)| ∈R};(−40502)(−215
02)(−50712)
(f) Consistent;{(+3−28)| ∈R};(−2008)(−1108)(1018)
(g) Consistent;{(6−13)}
(h) Inconsistent;{}
(2) (a){(3+13+46++13−2+5)| ∈R}
(b){(3−12+50
−132−8+3−2)| ∈R}
(c){(−20+9−153−687−2+37+154+2)| ∈R}
(d){}
(3) 52 nickels, 64 dimes, 32 quarters
(4)=2
2
−+5
(5)=−3
3
+4
2
−5+6
(6)
2
+
2
−6−8=0,or(−3)
2
+(−4)
2
=25
(7) In each part,(AB)=((A))B, which equals the given matrix.
(a)
⎡
⎢
⎢
⎣
26 15 −6
64 1
0−612
10 4 −14
⎤
⎥
⎥
⎦
(b)
⎡
⎢
⎢
⎣
26 15−6
10 4−14
18 6 15
64 1
⎤
⎥
⎥
⎦
(8) (a) To save space, we writehCi
for theth row of a matrixC.
For the Type (I) operation:hi←hi:Now,h(AB)i
=hABi =hAi B(by the hint
in the text)=h(A)i
B=h(A)Bi .But,if6 =,h(AB)i =hABi =hAi B(by the hint
in the text)=h(A)i
B=h(A)Bi .
For the Type (II) operation:hi←hi+hi:Now,h(AB)i
=hABi +hABi =
hAi
B+hAi B(by the hint in the text)=(hAi +hAi )B=h(A)i B=h(A)Bi .But,if
6 =,h(AB)i
=hABi =hAi B(by the hint in the text)=h(A)i B=h(A)Bi .
For the Type (III) operation:hi←→hi:Now,h(AB)i
=hABi =hAi B(by the
hint in the text)=h(A)i
B=h(A)Bi . Similarly,h(AB)i =hABi =hAi B(by the hint
in the text)=h(A)i
B=h(A)Bi .And,if6 =and6 =,h(AB)i =hABi =hAi B
(by the hint in the text)=h(A)i
B=h(A)Bi .
Copyrightc°2016 Elsevier Ltd. All rights reserved. 19
Answers to Exercises Section 2.2
(b) Use induction on, the number of row operations used.
Base Step: The case=1is part (1) of Theorem 2.1, and is proven in part (a) of this exercise.
Inductive Step: Assume that
(···( 2(1(AB)))···)= (···( 2(1(A)))···)B
and prove that
+1((···( 2(1(AB)))···)) = +1((···( 2(1(A)))···))B
Now,
+1((···( 2(1(AB)))···)) = +1((···( 2(1(A)))···)B)
(by the inductive hypothesis) =
+1((···( 2(1(A)))···))B(by part (a)).
(9) Multiplying a row by zero changes all of its entries to zero, essentially erasing all of the information in
the row.
(10) Suppose thatA,B,X
1,andX 2are as given in the problem.
(a) Letbe a scalar. Then,
A(X
1+(X 2−X1)) =AX 1+A(X 2−X1)=B+AX 2−AX 1=B+B−B=B
Hence,X
1+(X 2−X1)is a solution toAX=B.
(b) SupposeX
1+(X 2−X1)=X 1+(X 2−X1). Then,
X
1+(X 2−X1)=X 1+(X 2−X1)
=⇒(X
2−X1)=(X 2−X1)
=⇒(−)(X
2−X1)=0
=⇒(−)=0or(X
2−X1)=0
However,X
26 =X 1,andso=.
(c) Parts (a) and (b) show that, for each different real number,X
1+(X 2−X 1)is a different
solution toAX=B. Therefore, since there are an infinite number of real numbers,AX=Bhas
an infinite number of solutions.
(11) (a) T (b) F (c) F (d) F (e) T (f) T (g) F
Section 2.2
(1) Matrices in (a), (b), (c), (d), and (f) are not in reduced row echelon form.
Matrix in (a) fails condition 2 of the definition.
Matrix in (b) fails condition 4 of the definition.
Matrix in (c) fails condition 1 of the definition.
Matrix in (d) fails conditions 1, 2, and 3 of the definition.
Matrix in (f) fails condition 3 of the definition.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 20
(b) Use induction on, the number of row operations used.
Base Step: The case=1is part (1) of Theorem 2.1, and is proven in part (a) of this exercise.
Inductive Step: Assume that
(···( 2(1(AB)))···)= (···( 2(1(A)))···)B
and prove that
+1((···( 2(1(AB)))···)) = +1((···( 2(1(A)))···))B
Now,
+1((···( 2(1(AB)))···)) = +1((···( 2(1(A)))···)B)
(by the inductive hypothesis) =
+1((···( 2(1(A)))···))B(by part (a)).
(9) Multiplying a row by zero changes all of its entries to zero, essentially erasing all of the information in
the row.
(10) Suppose thatA,B,X
1,andX 2are as given in the problem.
(a) Letbe a scalar. Then,
A(X
1+(X 2−X1)) =AX 1+A(X 2−X1)=B+AX 2−AX 1=B+B−B=B
Hence,X
1+(X 2−X1)is a solution toAX=B.
(b) SupposeX
1+(X 2−X1)=X 1+(X 2−X1). Then,
X
1+(X 2−X1)=X 1+(X 2−X1)
=⇒(X
2−X1)=(X 2−X1)
=⇒(−)(X
2−X1)=0
=⇒(−)=0or(X
2−X1)=0
However,X
26 =X 1,andso=.
(c) Parts (a) and (b) show that, for each different real number,X
1+(X 2−X 1)is a different
solution toAX=B. Therefore, since there are an infinite number of real numbers,AX=Bhas
an infinite number of solutions.
(11) (a) T (b) F (c) F (d) F (e) T (f) T (g) F
Section 2.2
(1) Matrices in (a), (b), (c), (d), and (f) are not in reduced row echelon form.
Matrix in (a) fails condition 2 of the definition.
Matrix in (b) fails condition 4 of the definition.
Matrix in (c) fails condition 1 of the definition.
Matrix in (d) fails conditions 1, 2, and 3 of the definition.
Matrix in (f) fails condition 3 of the definition.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 20
Answers to Exercises Section 2.2
(2) (a)
⎡
⎣
140
00 1
000
¯
¯
¯
¯
¯
¯
−13
−3
0
⎤
⎦
(b)I
4
(c)
⎡
⎢
⎢
⎣
1−20 11 −23
00 1−25
0000 0
0000 0
⎤
⎥
⎥
⎦
(d)
⎡
⎢
⎢
⎢
⎢
⎣
10
01
00
00
00
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
0
0
1
0 0
⎤
⎥
⎥
⎥
⎥
⎦
(e)
∙
1−20 2 −1
00 1−13
¯
¯
¯
¯
1
2
¸
(f)I
5
(3) (a)
⎡
⎣
100
010
001
¯
¯
¯
¯
¯
¯
−2
3
6
⎤
⎦;(−236)
(e)
⎡
⎣
1−20 10
001 −20
00001
¯
¯
¯
¯
¯
¯
−4
5
2
⎤
⎦;{(2−−42+52)| ∈R}
(g)
⎡
⎢
⎢
⎣
100
010
001
000
¯
¯
¯
¯
¯
¯
¯
¯
6
−1
3
0
⎤
⎥
⎥
⎦
;{(6−13)}
(4) (a) Solution set ={(−2−
3 )| ∈R}; one particular solution =(−3−612)
(b) Solution set ={(−2−−3 )| ∈R}; one particular solution =(113−11)
(c) Solution set ={(−4+2−−3+2−2)| ∈R}; one particular solution =
(−3102−63)
(5) (a){(2−4 )|∈R}=
{(2−41)|∈R}
(b){(000)}
(c){(0000)}
(d){(3−2 )|∈R}
={(3−211)|∈R}
(6) (a)=2,=15,=12,=6
(b)=2,=25,=16,=18
(c)=4,=2,=4,=1,=4
(d)
=3,=11,=2,=3,=2,=6
(7) (a)=3,=4,=−2 (b)=4,=−3,=1,=0
(8) Solution for systemAX=B
1:(6−5121); solution for systemAX=B 2:(
35
3
−98
79
2
).
(9) Solution for systemAX=B
1:(−1−6149); solution for systemAX=B 2:(−10568
10
3
).
(10) (a)
1:(III):h1i↔h2i
2:(I):h1i←
1
2
h1i
3:(II):h2i←
1
3
h2i
4:(II):h1i←−2h2i+h1i;
(b)AB=
∙
57−21
56−20
¸
;
4(3(2(1(AB)))) = 4(3(2(1(A))))B=
∙
−10 4
19−7
¸
(11) (a)A(X
1+X2)=AX 1+AX 2=0+0=0;A(X 1)=(AX 1)=0=0.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 21
(2) (a)
⎡
⎣
140
00 1
000
¯
¯
¯
¯
¯
¯
−13
−3
0
⎤
⎦
(b)I
4
(c)
⎡
⎢
⎢
⎣
1−20 11 −23
00 1−25
0000 0
0000 0
⎤
⎥
⎥
⎦
(d)
⎡
⎢
⎢
⎢
⎢
⎣
10
01
00
00
00
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
0
0
1
0 0
⎤
⎥
⎥
⎥
⎥
⎦
(e)
∙
1−20 2 −1
00 1−13
¯
¯
¯
¯
1
2
¸
(f)I
5
(3) (a)
⎡
⎣
100
010
001
¯
¯
¯
¯
¯
¯
−2
3
6
⎤
⎦;(−236)
(e)
⎡
⎣
1−20 10
001 −20
00001
¯
¯
¯
¯
¯
¯
−4
5
2
⎤
⎦;{(2−−42+52)| ∈R}
(g)
⎡
⎢
⎢
⎣
100
010
001
000
¯
¯
¯
¯
¯
¯
¯
¯
6
−1
3
0
⎤
⎥
⎥
⎦
;{(6−13)}
(4) (a) Solution set ={(−2−
3 )| ∈R}; one particular solution =(−3−612)
(b) Solution set ={(−2−−3 )| ∈R}; one particular solution =(113−11)
(c) Solution set ={(−4+2−−3+2−2)| ∈R}; one particular solution =
(−3102−63)
(5) (a){(2−4 )|∈R}=
{(2−41)|∈R}
(b){(000)}
(c){(0000)}
(d){(3−2 )|∈R}
={(3−211)|∈R}
(6) (a)=2,=15,=12,=6
(b)=2,=25,=16,=18
(c)=4,=2,=4,=1,=4
(d)
=3,=11,=2,=3,=2,=6
(7) (a)=3,=4,=−2 (b)=4,=−3,=1,=0
(8) Solution for systemAX=B
1:(6−5121); solution for systemAX=B 2:(
35
3
−98
79
2
).
(9) Solution for systemAX=B
1:(−1−6149); solution for systemAX=B 2:(−10568
10
3
).
(10) (a)
1:(III):h1i↔h2i
2:(I):h1i←
1
2
h1i
3:(II):h2i←
1
3
h2i
4:(II):h1i←−2h2i+h1i;
(b)AB=
∙
57−21
56−20
¸
;
4(3(2(1(AB)))) = 4(3(2(1(A))))B=
∙
−10 4
19−7
¸
(11) (a)A(X
1+X2)=AX 1+AX 2=0+0=0;A(X 1)=(AX 1)=0=0.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 21
Answers to Exercises Section 2.3
(b) Any nonhomogeneous system with two equations and two unknowns that has a unique solution
will serve as a counterexample. For instance, consider
½
+=1
−=1
This system has a unique solution:(10).Let(
12)and( 12)both equal(10).Thenthesum
of solutions is not a solution in this case. Also, if6 =1, then the scalar multiple of a solution by
is not a solution.
(c)A(X
1+X2)=AX 1+AX 2=B+0=B.
(d) LetX
1betheuniquesolutiontoAX=B. SupposeAX=0has a nontrivial solutionX 2. Then,
by (c),X
1+X2is a solution toAX=B,andX 16 =X 1+X2sinceX 26 =0. This contradicts the
uniqueness ofX
1.
(12) If6 =0, then pivoting in thefirst column of
∙
¯
¯
¯
¯
0
0
¸
yields
"
1
0−
¡
¢
¯
¯
¯
¯
¯
0
0
#
.ByTheorem
2.2, there is a nontrivial solution if and only if the(22)entry of this matrix is zero, which occurs if
and only if−=0.
If=0and6 =0,then
∙
¯
¯
¯
¯
0
0
¸
=
∙
0
¯
¯
¯
¯
0
0
¸
Swapping thefirst and second rows
andthenpivotinginthefirst column yields
"
1
0
¯
¯
¯
¯
¯
0
0
#
. By Theorem 2.2, there is a nontrivial
solution if and only if the(22)entry of this matrix (that is,) equals zero. Butis zero if and only
if−=0since=0and6 =0.
Finally, if bothandequal0,then−=0and(10)is a nontrivial solution.
(13) (a) The contrapositive is: IfAX=0has only the trivial solution, thenA
2
X=0has only the trivial
solution.
LetX
1be a solution toA
2
X=0.Then0=A
2
X1=A(AX 1).ThusAX 1is a solution to
AX=0. HenceAX
1=0by the premise. Thus,X 1=0, using the premise again.
(b) The contrapositive is: Letbe a positive integer. IfAX=0has only the trivial solution, then
A
X=0has only the trivial solution.
Proceed by induction. The statement is clearly true when=1, completing the Base Step.
Inductive Step: Assume that ifAX=0has only the trivial solution, thenA
X=0has only the
trivial solution. We must prove that ifAX=0has only the trivial solution, thenA
+1
X=0
has only the trivial solution.
LetX
1be a solution toA
+1
X=0.ThenA(A
X1)=0.ThusA
X1=0,sinceA
X1is a
solution toAX=0.ButthenX
1=0by the inductive hypothesis.
(14) (a) T (b) T (c) F (d) T (e) F (f) F
Section 2.3
(1) (a) A row operation of Type (I) convertsAtoB:h2i←−5h2i.
(b) A row operation of Type (III) convertsAtoB:h1i↔h3i.
(c) A row operation of Type (II) convertsAtoB:h2i←h3i+h2i.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 22
(b) Any nonhomogeneous system with two equations and two unknowns that has a unique solution
will serve as a counterexample. For instance, consider
½
+=1
−=1
This system has a unique solution:(10).Let(
12)and( 12)both equal(10).Thenthesum
of solutions is not a solution in this case. Also, if6 =1, then the scalar multiple of a solution by
is not a solution.
(c)A(X
1+X2)=AX 1+AX 2=B+0=B.
(d) LetX
1betheuniquesolutiontoAX=B. SupposeAX=0has a nontrivial solutionX 2. Then,
by (c),X
1+X2is a solution toAX=B,andX 16 =X 1+X2sinceX 26 =0. This contradicts the
uniqueness ofX
1.
(12) If6 =0, then pivoting in thefirst column of
∙
¯
¯
¯
¯
0
0
¸
yields
"
1
0−
¡
¢
¯
¯
¯
¯
¯
0
0
#
.ByTheorem
2.2, there is a nontrivial solution if and only if the(22)entry of this matrix is zero, which occurs if
and only if−=0.
If=0and6 =0,then
∙
¯
¯
¯
¯
0
0
¸
=
∙
0
¯
¯
¯
¯
0
0
¸
Swapping thefirst and second rows
andthenpivotinginthefirst column yields
"
1
0
¯
¯
¯
¯
¯
0
0
#
. By Theorem 2.2, there is a nontrivial
solution if and only if the(22)entry of this matrix (that is,) equals zero. Butis zero if and only
if−=0since=0and6 =0.
Finally, if bothandequal0,then−=0and(10)is a nontrivial solution.
(13) (a) The contrapositive is: IfAX=0has only the trivial solution, thenA
2
X=0has only the trivial
solution.
LetX
1be a solution toA
2
X=0.Then0=A
2
X1=A(AX 1).ThusAX 1is a solution to
AX=0. HenceAX
1=0by the premise. Thus,X 1=0, using the premise again.
(b) The contrapositive is: Letbe a positive integer. IfAX=0has only the trivial solution, then
A
X=0has only the trivial solution.
Proceed by induction. The statement is clearly true when=1, completing the Base Step.
Inductive Step: Assume that ifAX=0has only the trivial solution, thenA
X=0has only the
trivial solution. We must prove that ifAX=0has only the trivial solution, thenA
+1
X=0
has only the trivial solution.
LetX
1be a solution toA
+1
X=0.ThenA(A
X1)=0.ThusA
X1=0,sinceA
X1is a
solution toAX=0.ButthenX
1=0by the inductive hypothesis.
(14) (a) T (b) T (c) F (d) T (e) F (f) F
Section 2.3
(1) (a) A row operation of Type (I) convertsAtoB:h2i←−5h2i.
(b) A row operation of Type (III) convertsAtoB:h1i↔h3i.
(c) A row operation of Type (II) convertsAtoB:h2i←h3i+h2i.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 22
Answers to Exercises Section 2.3
(2) (a)B=I 3. The sequence of row operations convertingAtoBis:
(I):h1i←
1
4
h1i
(II):h2i←2h1i+h2i
(II):h3i←−3h1i+h3i
(III):h2i↔h3i
(II):h1i←5h3i+h1i
(b) The sequence of row operations convertingBtoAis:
(II):h1i←−5h3i+h1i
(III):h2i↔h3i
(II):h3i←3h1i+h3i
(II):h2i←−2h1i+h2
i
(I):h1i←4h1i
(3) (a) LetCbe the common reduced row echelon form matrix forAandB. Then,AandBare both
row equivalent toC. Also, by part (1) of Theorem 2.4,Cis row equivalent toB.But,sinceAis
row equivalent toC,andCis row equivalent toB, part (2) of Theorem 2.4 asserts thatAis row
equivalent toB.
(b) The common reduced row echelon form isI
3.
(c) The sequence of row operations is:
(II):h3i←2h2i+h3i
(I):h3i←−1h3i
(II):h1i←−9h3i+h1i
(II):h2i←3h3i+h2i
(II):h3i←−
9
5
h2i+h3i
(II):h1i←−
3
5
h2i+h1i
(I):h2i←−
1
5
h2i
(II):h3i←−3h1i+h3i
(II):h2i←−2h1i+h2i
(I):h1i←−5h1i
(4) (a) AssumeBis row equivalent toA.LetCbe the reduced row echelon form matrix forA.ThenA
is row equivalent toC.But,sinceBis row equivalent toA,andAis row equivalent toC,part
(2) of Theorem 2.4 asserts thatAis row equivalent toC. Now, by Theorem 2.6, the reduced row
echelon form matrix forBis unique, so it must beC.ThusAandChavethesamereducedrow
echelon form matrix.
(b) The reduced row echelon form matrices forAandBare, respectively,
⎡
⎢
⎢
⎣
100 −23
010 −10
001 10
000 00
⎤
⎥
⎥
⎦
and
⎡
⎢
⎢
⎣
100
−20
010 −10
001 10
000 01
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 23
(2) (a)B=I 3. The sequence of row operations convertingAtoBis:
(I):h1i←
1
4
h1i
(II):h2i←2h1i+h2i
(II):h3i←−3h1i+h3i
(III):h2i↔h3i
(II):h1i←5h3i+h1i
(b) The sequence of row operations convertingBtoAis:
(II):h1i←−5h3i+h1i
(III):h2i↔h3i
(II):h3i←3h1i+h3i
(II):h2i←−2h1i+h2
i
(I):h1i←4h1i
(3) (a) LetCbe the common reduced row echelon form matrix forAandB. Then,AandBare both
row equivalent toC. Also, by part (1) of Theorem 2.4,Cis row equivalent toB.But,sinceAis
row equivalent toC,andCis row equivalent toB, part (2) of Theorem 2.4 asserts thatAis row
equivalent toB.
(b) The common reduced row echelon form isI
3.
(c) The sequence of row operations is:
(II):h3i←2h2i+h3i
(I):h3i←−1h3i
(II):h1i←−9h3i+h1i
(II):h2i←3h3i+h2i
(II):h3i←−
9
5
h2i+h3i
(II):h1i←−
3
5
h2i+h1i
(I):h2i←−
1
5
h2i
(II):h3i←−3h1i+h3i
(II):h2i←−2h1i+h2i
(I):h1i←−5h1i
(4) (a) AssumeBis row equivalent toA.LetCbe the reduced row echelon form matrix forA.ThenA
is row equivalent toC.But,sinceBis row equivalent toA,andAis row equivalent toC,part
(2) of Theorem 2.4 asserts thatAis row equivalent toC. Now, by Theorem 2.6, the reduced row
echelon form matrix forBis unique, so it must beC.ThusAandChavethesamereducedrow
echelon form matrix.
(b) The reduced row echelon form matrices forAandBare, respectively,
⎡
⎢
⎢
⎣
100 −23
010 −10
001 10
000 00
⎤
⎥
⎥
⎦
and
⎡
⎢
⎢
⎣
100
−20
010 −10
001 10
000 01
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 23
Answers to Exercises Section 2.3
(5) (a)2 (b)1 (c)2 (d)3 (e)3 (f)2
(6) (a) Rank=3. Theorem 2.7 predicts that there is onlythe trivial solution. Solution set ={(000)}
(b) Rank=2. Theorem 2.7 predicts that nontrivial solutions exist.
Solution set={(3−4 )|∈R}
(7) In the following answers, the asterisk represents any real entry:
(a) Smallest rank=1 Largest rank=4
⎡
⎢
⎢
⎣
1∗∗
000
000
000
¯
¯
¯
¯
¯
¯
¯
¯
∗
0
0
0
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
100
010
001
000
¯
¯
¯
¯
¯
¯
¯
¯
0
0
0
1
⎤
⎥
⎥
⎦
(b) Smallest rank=1 Largest rank=3
⎡
⎣
1∗∗∗
0000
0000
¯
¯
¯
¯
¯
¯
∗
0
0
⎤
⎦
⎡ ⎣
100 ∗
010 ∗
001 ∗
¯
¯
¯
¯
¯
¯
∗
∗
∗
⎤
⎦
(c) Smallest rank=2 Largest rank=3
⎡
⎣
1∗∗∗
0000
0000
¯
¯
¯
¯
¯
¯
0
1
0
⎤
⎦
⎡ ⎣
10∗∗
01∗∗
0000
¯
¯
¯
¯
¯
¯
0
0
1
⎤
⎦
(d) Smallest rank=1 Largest rank=3
⎡
⎢
⎢
⎢
⎢
⎣
1∗∗
000
000
000
000
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
∗
0
0
0
0
⎤
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎣
100
010
001
000
000
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
∗
∗
∗
0
0
⎤
⎥
⎥
⎥
⎥
⎦
(8) (a)x=−
21
11
a1+
6
11
a2
(b)x=3a 1−4a 2+a3
(c) Not possible
(d)x=
1
2
a1−
1
2
a2+
1
2
a3
(e) The answer is not unique; one possible answer isx=−3a 1+2a 2+0a 3.
(f) Not possible (g) x=2a
1−a2−a3 (h) Not possible
(9) (a) Yes:5(row 1)−3(row 2)−1(row 3)
(b) Not in row space
(c) Not in row space
(d) Yes:−3(row 1)+1(row 2)
(e) Yes, but the linear combination of the rows
is not unique; one possible expression for the
given vector is
−3(row 1) +1(row 2) +0(row 3).
(10) (a)[13−2360] =−2q
1+q2+3q 3
(b)q 1=3r 1−r2−2r 3;q2=2r 1+2r 2−5r 3;q3=r1−6r 2+4r 3
Copyrightc°2016 Elsevier Ltd. All rights reserved. 24
(5) (a)2 (b)1 (c)2 (d)3 (e)3 (f)2
(6) (a) Rank=3. Theorem 2.7 predicts that there is onlythe trivial solution. Solution set ={(000)}
(b) Rank=2. Theorem 2.7 predicts that nontrivial solutions exist.
Solution set={(3−4 )|∈R}
(7) In the following answers, the asterisk represents any real entry:
(a) Smallest rank=1 Largest rank=4
⎡
⎢
⎢
⎣
1∗∗
000
000
000
¯
¯
¯
¯
¯
¯
¯
¯
∗
0
0
0
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
100
010
001
000
¯
¯
¯
¯
¯
¯
¯
¯
0
0
0
1
⎤
⎥
⎥
⎦
(b) Smallest rank=1 Largest rank=3
⎡
⎣
1∗∗∗
0000
0000
¯
¯
¯
¯
¯
¯
∗
0
0
⎤
⎦
⎡ ⎣
100 ∗
010 ∗
001 ∗
¯
¯
¯
¯
¯
¯
∗
∗
∗
⎤
⎦
(c) Smallest rank=2 Largest rank=3
⎡
⎣
1∗∗∗
0000
0000
¯
¯
¯
¯
¯
¯
0
1
0
⎤
⎦
⎡ ⎣
10∗∗
01∗∗
0000
¯
¯
¯
¯
¯
¯
0
0
1
⎤
⎦
(d) Smallest rank=1 Largest rank=3
⎡
⎢
⎢
⎢
⎢
⎣
1∗∗
000
000
000
000
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
∗
0
0
0
0
⎤
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎣
100
010
001
000
000
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
∗
∗
∗
0
0
⎤
⎥
⎥
⎥
⎥
⎦
(8) (a)x=−
21
11
a1+
6
11
a2
(b)x=3a 1−4a 2+a3
(c) Not possible
(d)x=
1
2
a1−
1
2
a2+
1
2
a3
(e) The answer is not unique; one possible answer isx=−3a 1+2a 2+0a 3.
(f) Not possible (g) x=2a
1−a2−a3 (h) Not possible
(9) (a) Yes:5(row 1)−3(row 2)−1(row 3)
(b) Not in row space
(c) Not in row space
(d) Yes:−3(row 1)+1(row 2)
(e) Yes, but the linear combination of the rows
is not unique; one possible expression for the
given vector is
−3(row 1) +1(row 2) +0(row 3).
(10) (a)[13−2360] =−2q
1+q2+3q 3
(b)q 1=3r 1−r2−2r 3;q2=2r 1+2r 2−5r 3;q3=r1−6r 2+4r 3
Copyrightc°2016 Elsevier Ltd. All rights reserved. 24
Answers to Exercises Section 2.3
(c)[13−2360] =−r 1−14r 2+11r 3
(11) (a) (i)B=
⎡
⎣
10−12
01 3 2
00 0 0
⎤
⎦;
(ii)[10−12] =−
7
8
[04128] +
1
2
[271918] + 0[1256];
[0132] =
1
4
[04128] + 0[271918] + 0[1256]
(iii)[04128] = 0[10−12] + 4[0132];
[2171918] = 2[10−12] + 7[0132];
[1256] = 1[10−12]+ 2[0132]
(b) (i)B=
⎡
⎢
⎢
⎣
1230 −1
0001 5
0000 0
0000 0
⎤
⎥
⎥
⎦
;
(ii)[12
30−1] =−
5
3
[123−4−21]−
4
3
[−2−4−6527]+0[132639512]+0[246−1−7];
[00015] =−
2
3
[123−4−21]−
1
3
[−2−4−6527] + 0[132639512] + 0[246−1−7]
(iii)[123−4−21] = 1[1230−1]−4[00015];
[−2−4−6527] =−2[1230−1] + 5[00015];
[1326395
12] = 13[1230−1] + 5[00015];
[246−1−7] = 2[1230−1]−1[00015]
(12) Suppose that all main diagonal entries ofAare nonzero. Then, for each, perform the row operation
hi←(1
)hion the matrixA. This will convertAintoI . We prove the converse by contrapositive.
Suppose some diagonal entry
equals0.Thentheth column ofAhas all zero entries. No step in
the row reduction process will alter this column of zeroes, and so the unique reduced row echelon form
for the matrix must contain at least one column of zeroes, and so cannot equalI
.
(13) (a) Suppose we are performing row operations on an×matrixA. Throughout this part, we will
writehBi
for theth row of a matrixB.
For the Type (I) operation:hi←hi:Now
−1
ishi←
1
hi. Clearly,and
−1
change
only theth row ofA. We want to show that
−1
leaveshAi unchanged. Buth
−1
((A))i
=
1
h(A)i =
1
(hAi )=hAi .
For the Type (II) operation:hi←hi+hi:Now
−1
ishi←−hi+hi. Again,and
−1
change only theth row ofA, and we need to show that
−1
leaveshAi unchanged. But
h
−1
((A))i =−h(A)i +h(A)i =−hAi +h(A)i =−hAi +hAi +hAi =hAi .
For the Type (III) operation:hi↔hi:Now,
−1
=.Also,changes only theth and
th rows ofA, and these get swapped. Obviously, a second application ofswaps them back to
where they were, proving thatis indeed its own inverse.
(b) SupposeCis row equivalent toD,andDis row equivalent toE. Then by the definition of
row equivalence, there is a sequence
12 of row operations convertingCtoD,anda
sequence
12 of row operations convertingDtoE. But then the combined sequence
12 12 of row operations convertsCtoE,andsoCis row equivalent toE.
(c) An approach similar to that used for Type (II) operations in the abridged proof of Theorem 2.5 in
the text works just as easily for Type (I) and Type (III) operations. However, here is a different
approach: Supposeis a row operation, and letXsatisfyAX=B. Multiplying both sides of
this matrix equation by the matrix(I)yields(I)AX=(I)B, implying(IA)X=(IB),
Copyrightc°2016 Elsevier Ltd. All rights reserved. 25
(c)[13−2360] =−r 1−14r 2+11r 3
(11) (a) (i)B=
⎡
⎣
10−12
01 3 2
00 0 0
⎤
⎦;
(ii)[10−12] =−
7
8
[04128] +
1
2
[271918] + 0[1256];
[0132] =
1
4
[04128] + 0[271918] + 0[1256]
(iii)[04128] = 0[10−12] + 4[0132];
[2171918] = 2[10−12] + 7[0132];
[1256] = 1[10−12]+ 2[0132]
(b) (i)B=
⎡
⎢
⎢
⎣
1230 −1
0001 5
0000 0
0000 0
⎤
⎥
⎥
⎦
;
(ii)[12
30−1] =−
5
3
[123−4−21]−
4
3
[−2−4−6527]+0[132639512]+0[246−1−7];
[00015] =−
2
3
[123−4−21]−
1
3
[−2−4−6527] + 0[132639512] + 0[246−1−7]
(iii)[123−4−21] = 1[1230−1]−4[00015];
[−2−4−6527] =−2[1230−1] + 5[00015];
[1326395
12] = 13[1230−1] + 5[00015];
[246−1−7] = 2[1230−1]−1[00015]
(12) Suppose that all main diagonal entries ofAare nonzero. Then, for each, perform the row operation
hi←(1
)hion the matrixA. This will convertAintoI . We prove the converse by contrapositive.
Suppose some diagonal entry
equals0.Thentheth column ofAhas all zero entries. No step in
the row reduction process will alter this column of zeroes, and so the unique reduced row echelon form
for the matrix must contain at least one column of zeroes, and so cannot equalI
.
(13) (a) Suppose we are performing row operations on an×matrixA. Throughout this part, we will
writehBi
for theth row of a matrixB.
For the Type (I) operation:hi←hi:Now
−1
ishi←
1
hi. Clearly,and
−1
change
only theth row ofA. We want to show that
−1
leaveshAi unchanged. Buth
−1
((A))i
=
1
h(A)i =
1
(hAi )=hAi .
For the Type (II) operation:hi←hi+hi:Now
−1
ishi←−hi+hi. Again,and
−1
change only theth row ofA, and we need to show that
−1
leaveshAi unchanged. But
h
−1
((A))i =−h(A)i +h(A)i =−hAi +h(A)i =−hAi +hAi +hAi =hAi .
For the Type (III) operation:hi↔hi:Now,
−1
=.Also,changes only theth and
th rows ofA, and these get swapped. Obviously, a second application ofswaps them back to
where they were, proving thatis indeed its own inverse.
(b) SupposeCis row equivalent toD,andDis row equivalent toE. Then by the definition of
row equivalence, there is a sequence
12 of row operations convertingCtoD,anda
sequence
12 of row operations convertingDtoE. But then the combined sequence
12 12 of row operations convertsCtoE,andsoCis row equivalent toE.
(c) An approach similar to that used for Type (II) operations in the abridged proof of Theorem 2.5 in
the text works just as easily for Type (I) and Type (III) operations. However, here is a different
approach: Supposeis a row operation, and letXsatisfyAX=B. Multiplying both sides of
this matrix equation by the matrix(I)yields(I)AX=(I)B, implying(IA)X=(IB),
Copyrightc°2016 Elsevier Ltd. All rights reserved. 25
Answers to Exercises Section 2.3
by Theorem 2.1. Thus,(A)X=(B),showingthatXis a solution to the new linear system
obtained fromAX=Bafter the row operationis performed.
(14) The zero vector is a solution toAX=0, but it is not a solution forAX=B.
(15) Consider the systems
½
+=1
+=0
and
½
−=1
−=2
The reduced row echelon matrices for theseinconsistent systems are, respectively,
∙
11
00
¯
¯
¯
¯
0
1
¸
and
∙
1−1
00
¯
¯
¯
¯
0
1
¸
Thus, the original augmented matrices are not row equivalent, since their reduced row echelon forms
are different.
(16) (a) Plug each of the 5 points in turn into the equation for the conic. This will give a homogeneous
system of 5 equations in the 6 variables,,,,,and. This system has a nontrivial solution,
by Corollary 2.2.
(b) Yes. In this case, there will be even fewer equations, so Corollary 2.3 again applies.
(17) BecauseAandBare row equivalent,
A= (···( 2(1(B)))···)for some row operations 1.
Now, ifDis the unique reduced row echelon form matrix to whichAis row equivalent, then for some
additional row operations
+1+,
D=
+(···( +2(+1(A)))···)
=
+(···( +2(+1((···( 2(1(B)))···))))···)
showing thatBalso hasDas its reduced echelon form matrix. Therefore, by the definition of rank,
rank(B)=the number of nonzero rows inD=rank(A).
(18) (a)
(···( 1(A))···)andAare clearly row equivalent. Use Exercise 17.
(b) Since the row reduction process has no effect on rows having all zeroes, at leastof therows
in the reduced row echelon form ofAare rows of all zeroes. Thus,rank(A)≤−
(c) SinceAis in reduced row echelon form and hasrows of zeroes,rank(A)=−.ButABhas
at leastrows of zeroes, sorank(AB)≤−by part (b).
(d) LetA=
(···( 2(1(D)))···),whereDis in reduced row echelon form. Thenrank(AB)=
rank(
(···( 2(1(D)))···)B)=rank( (···( 2(1(DB)))···)) (by part (2) of Theorem 2.1)
=rank(DB)(bypart(a))≤rank(D)(bypart(c))=rank(A)(bydefinition of rank).
(19) As in the abridged proof of Theorem 2.9 in the text, leta
1a represent the rows ofA,andlet
b
1b represent the rows ofB.
For the Type (I) operation:hi←hi:Nowb
=0a 1+0a 2+···+a +0a +1+···+0a ,
and, for6 =,b
=0a 1+0a 2+···+1a +0a +1+···+0a . Hence, each row ofBis a linear
combination of the rows ofA, implying it is in the row space ofA.
For the Type (II) operation:hi←hi+hi:Nowb
=0a 1+0a 2+···+a +0a +1+···+
a
+0a +1++0a , where our notation assumes. (An analogous argument works for.)
And, for6 =,b
=0a 1+0a 2+···+1a +0a +1+···+0a . Hence, each row ofBis a linear
Copyrightc°2016 Elsevier Ltd. All rights reserved. 26
by Theorem 2.1. Thus,(A)X=(B),showingthatXis a solution to the new linear system
obtained fromAX=Bafter the row operationis performed.
(14) The zero vector is a solution toAX=0, but it is not a solution forAX=B.
(15) Consider the systems
½
+=1
+=0
and
½
−=1
−=2
The reduced row echelon matrices for theseinconsistent systems are, respectively,
∙
11
00
¯
¯
¯
¯
0
1
¸
and
∙
1−1
00
¯
¯
¯
¯
0
1
¸
Thus, the original augmented matrices are not row equivalent, since their reduced row echelon forms
are different.
(16) (a) Plug each of the 5 points in turn into the equation for the conic. This will give a homogeneous
system of 5 equations in the 6 variables,,,,,and. This system has a nontrivial solution,
by Corollary 2.2.
(b) Yes. In this case, there will be even fewer equations, so Corollary 2.3 again applies.
(17) BecauseAandBare row equivalent,
A= (···( 2(1(B)))···)for some row operations 1.
Now, ifDis the unique reduced row echelon form matrix to whichAis row equivalent, then for some
additional row operations
+1+,
D=
+(···( +2(+1(A)))···)
=
+(···( +2(+1((···( 2(1(B)))···))))···)
showing thatBalso hasDas its reduced echelon form matrix. Therefore, by the definition of rank,
rank(B)=the number of nonzero rows inD=rank(A).
(18) (a)
(···( 1(A))···)andAare clearly row equivalent. Use Exercise 17.
(b) Since the row reduction process has no effect on rows having all zeroes, at leastof therows
in the reduced row echelon form ofAare rows of all zeroes. Thus,rank(A)≤−
(c) SinceAis in reduced row echelon form and hasrows of zeroes,rank(A)=−.ButABhas
at leastrows of zeroes, sorank(AB)≤−by part (b).
(d) LetA=
(···( 2(1(D)))···),whereDis in reduced row echelon form. Thenrank(AB)=
rank(
(···( 2(1(D)))···)B)=rank( (···( 2(1(DB)))···)) (by part (2) of Theorem 2.1)
=rank(DB)(bypart(a))≤rank(D)(bypart(c))=rank(A)(bydefinition of rank).
(19) As in the abridged proof of Theorem 2.9 in the text, leta
1a represent the rows ofA,andlet
b
1b represent the rows ofB.
For the Type (I) operation:hi←hi:Nowb
=0a 1+0a 2+···+a +0a +1+···+0a ,
and, for6 =,b
=0a 1+0a 2+···+1a +0a +1+···+0a . Hence, each row ofBis a linear
combination of the rows ofA, implying it is in the row space ofA.
For the Type (II) operation:hi←hi+hi:Nowb
=0a 1+0a 2+···+a +0a +1+···+
a
+0a +1++0a , where our notation assumes. (An analogous argument works for.)
And, for6 =,b
=0a 1+0a 2+···+1a +0a +1+···+0a . Hence, each row ofBis a linear
Copyrightc°2016 Elsevier Ltd. All rights reserved. 26
Answers to Exercises Section 2.4
combination of the rows ofA, implying it is in the row space ofA.
For the Type (III) operation:hi↔hi:Now,b
=0a 1+0a2+···+1a +0a+1+···+0a ,b=
0a
1+0a2+···+1a +0a+1+···+0a , and, for6 =,6 =,b =0a 1+0a2+···+1a +0a+1+···+0a .
Hence, each row ofBis a linear combination of the rows ofA, implying it is in the row space ofA.
(20) Letbe the number of matricesbetweenAandBwhen performing row operations to get fromAto
B. Use a proof by induction on.
Base Step: If=0, then there are no intermediary matrices, and Exercise 19 shows that the row
space ofBis contained in the row space ofA.
Inductive Step: Given the chain
A→D
1→D 2→···→D →D +1→B
we must show that the row space ofBiscontainedintherowspaceofA. The inductive hypothesis
shows that the row space ofD
+1is in the row space ofA, since there are onlymatrices between
AandD
+1in the chain. Thus, each row ofD +1can be expressed as a linear combination of the
rows ofA. But by Exercise 19, each row ofBcan be expressed as a linear combination of the rows of
D
+1. Hence, by Lemma 2.8, each row ofBcan be expressed as a linear combination of the rows of
A, and therefore is in the row space ofA. By Lemma 2.8 again, the row space ofBis contained in
therowspaceofA.
(21) Let
represent theth coordinate ofx . The corresponding homogeneous system in variables
1+1is
⎧
⎪
⎨
⎪
⎩
111+ 221+···+ +1+11 =0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
11+ 22+···+ +1+1 =0
which has a nontrivial solution for
1+1, by Corollary 2.3.
(22) (a) T (b) T (c) F (d) F (e) F (f) T
Section 2.4
(1) The product of each given pair of matrices equalsI.
(2) (a) Rank=2; nonsingular
(b) Rank=2; singular
(c) Rank=3; nonsingular
(d) Rank=4; nonsingular
(e) Rank=3; singular
(3) No inverse exists for (b), (e) and (f).
(a)
"
1
10
1
15
3
10
−
2
15
#
(c)
"
−
221
−
5
84
1
7
−
1
28
#
(d)
"
3
11
2
11
4
11
−
1
11
#
Copyrightc°2016 Elsevier Ltd. All rights reserved. 27
combination of the rows ofA, implying it is in the row space ofA.
For the Type (III) operation:hi↔hi:Now,b
=0a 1+0a2+···+1a +0a+1+···+0a ,b=
0a
1+0a2+···+1a +0a+1+···+0a , and, for6 =,6 =,b =0a 1+0a2+···+1a +0a+1+···+0a .
Hence, each row ofBis a linear combination of the rows ofA, implying it is in the row space ofA.
(20) Letbe the number of matricesbetweenAandBwhen performing row operations to get fromAto
B. Use a proof by induction on.
Base Step: If=0, then there are no intermediary matrices, and Exercise 19 shows that the row
space ofBis contained in the row space ofA.
Inductive Step: Given the chain
A→D
1→D 2→···→D →D +1→B
we must show that the row space ofBiscontainedintherowspaceofA. The inductive hypothesis
shows that the row space ofD
+1is in the row space ofA, since there are onlymatrices between
AandD
+1in the chain. Thus, each row ofD +1can be expressed as a linear combination of the
rows ofA. But by Exercise 19, each row ofBcan be expressed as a linear combination of the rows of
D
+1. Hence, by Lemma 2.8, each row ofBcan be expressed as a linear combination of the rows of
A, and therefore is in the row space ofA. By Lemma 2.8 again, the row space ofBis contained in
therowspaceofA.
(21) Let
represent theth coordinate ofx . The corresponding homogeneous system in variables
1+1is
⎧
⎪
⎨
⎪
⎩
111+ 221+···+ +1+11 =0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
11+ 22+···+ +1+1 =0
which has a nontrivial solution for
1+1, by Corollary 2.3.
(22) (a) T (b) T (c) F (d) F (e) F (f) T
Section 2.4
(1) The product of each given pair of matrices equalsI.
(2) (a) Rank=2; nonsingular
(b) Rank=2; singular
(c) Rank=3; nonsingular
(d) Rank=4; nonsingular
(e) Rank=3; singular
(3) No inverse exists for (b), (e) and (f).
(a)
"
1
10
1
15
3
10
−
2
15
#
(c)
"
−
221
−
5
84
1
7
−
1
28
#
(d)
"
3
11
2
11
4
11
−
1
11
#
Copyrightc°2016 Elsevier Ltd. All rights reserved. 27
Answers to Exercises Section 2.4
(4) No inverse exists for (b) and (e).
(a)
⎡
⎣
13 2
−10 2
22−1
⎤
⎦
(c)
⎡
⎢
⎢
⎣
3
2
0
1
2
−3
1
2
−
1
2
−
8
3
1
3
−
2
3
⎤
⎥
⎥
⎦
(d)
⎡
⎢
⎢
⎣
1−112
−75 −10−19
−21 −2−4
3−248
⎤
⎥
⎥
⎦
(f)
⎡
⎢
⎢
⎣
4−13 −6
−31 −510
10−201
10 −36
⎤
⎥
⎥
⎦
(5) (a)
∙
1
11
0
0
1
22
¸
(b)
⎡
⎣
1
11
00
0
1
22
0
00
1
33
⎤
⎦
(c)
⎡
⎢
⎢
⎢
⎣
1
11
0···0
0
1
22
···0
.
.
.
.
.
.
.
.
.
.
.
.
00 ···
1
⎤
⎥
⎥
⎥
⎦
(6) (a) The general inverse is
∙
cossin
−sincos
¸
.When=
6
,theinverseis
"
√
3
2
1
2
−
1
2
√
3
2
#
.
When=
4
,theinverseis
"
√
2
2
√
2
2
−
√
2
2
√
2
2
#
.When=
2
,theinverseis
∙
01
−10
¸
.
(b) The general inverse is
⎡
⎣
cossin0
−sincos0
001
⎤
⎦.When=
6
,theinverseis
⎡
⎢
⎢
⎣
√
3
2
1
2
0
−
1
2
√
3
2
0
001
⎤
⎥
⎥
⎦
.
When=
4
,theinverseis
⎡
⎢
⎢
⎣
√
2
2
√
2
2
0
−
√
2
2
√
2
2
0
001
⎤
⎥
⎥
⎦
.When=
2
,theinverseis
⎡
⎣
010
−100
001
⎤
⎦.
(7) (a) Inverse =
"
2
3
1
3
7
3
5
3
#
;solutionset={(3−5)}
(b) Inverse =
⎡
⎢
⎢
⎣
10 −3
8
5
2
5
−
17
5
1
5
−
1
5
−
4
5
⎤
⎥
⎥
⎦
;solutionset={(−24−3)}
(c) Inverse =
⎡
⎢
⎢
⎣
1−13−15 5
−330 −7
−121 −3
0−4−51
⎤
⎥
⎥
⎦
;solutionset={(5−82−1)}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 28
(4) No inverse exists for (b) and (e).
(a)
⎡
⎣
13 2
−10 2
22−1
⎤
⎦
(c)
⎡
⎢
⎢
⎣
3
2
0
1
2
−3
1
2
−
1
2
−
8
3
1
3
−
2
3
⎤
⎥
⎥
⎦
(d)
⎡
⎢
⎢
⎣
1−112
−75 −10−19
−21 −2−4
3−248
⎤
⎥
⎥
⎦
(f)
⎡
⎢
⎢
⎣
4−13 −6
−31 −510
10−201
10 −36
⎤
⎥
⎥
⎦
(5) (a)
∙
1
11
0
0
1
22
¸
(b)
⎡
⎣
1
11
00
0
1
22
0
00
1
33
⎤
⎦
(c)
⎡
⎢
⎢
⎢
⎣
1
11
0···0
0
1
22
···0
.
.
.
.
.
.
.
.
.
.
.
.
00 ···
1
⎤
⎥
⎥
⎥
⎦
(6) (a) The general inverse is
∙
cossin
−sincos
¸
.When=
6
,theinverseis
"
√
3
2
1
2
−
1
2
√
3
2
#
.
When=
4
,theinverseis
"
√
2
2
√
2
2
−
√
2
2
√
2
2
#
.When=
2
,theinverseis
∙
01
−10
¸
.
(b) The general inverse is
⎡
⎣
cossin0
−sincos0
001
⎤
⎦.When=
6
,theinverseis
⎡
⎢
⎢
⎣
√
3
2
1
2
0
−
1
2
√
3
2
0
001
⎤
⎥
⎥
⎦
.
When=
4
,theinverseis
⎡
⎢
⎢
⎣
√
2
2
√
2
2
0
−
√
2
2
√
2
2
0
001
⎤
⎥
⎥
⎦
.When=
2
,theinverseis
⎡
⎣
010
−100
001
⎤
⎦.
(7) (a) Inverse =
"
2
3
1
3
7
3
5
3
#
;solutionset={(3−5)}
(b) Inverse =
⎡
⎢
⎢
⎣
10 −3
8
5
2
5
−
17
5
1
5
−
1
5
−
4
5
⎤
⎥
⎥
⎦
;solutionset={(−24−3)}
(c) Inverse =
⎡
⎢
⎢
⎣
1−13−15 5
−330 −7
−121 −3
0−4−51
⎤
⎥
⎥
⎦
;solutionset={(5−82−1)}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 28
Answers to Exercises Section 2.4
(8) (a) Consider
∙
01
10
¸
.
(b) Consider
⎡
⎣
010
100
001
⎤
⎦.
(c)A=A
−1
ifAis involutory.
(9) (a)A=I
2,B=−I 2,A+B=O 2
(b)A=
∙
10
00
¸
,B=
∙
00
01
¸
,A+B=
∙
10
01
¸
(c) IfA=B=I
2,thenA+B=2I 2,A
−1
=B
−1
=I2,A
−1
+B
−1
=2I 2,and(A+B)
−1
=
1
2
I2,
soA
−1
+B
−1
6 =(A+B)
−1
.
(10) (a)Bmust be the zero matrix.
(b) No.AB=I
impliesA
−1
exists (and equalsB). Multiply both sides ofAC=O on the left
byA
−1
.
(11)A
−9
A
−5
A
−1
A
3
A
7
A
11
(12)B
−1
Ais the inverse ofA
−1
B.
(13)(A
−1
)
=(A
)
−1
(by Theorem 2.12, part (4))=A
−1
.
(14) (a) Supposefirst that the matrix contains a column of zeroes. No row operations can alter a column
of zeroes, so the unique reduced row echelon form for the matrix must also contain a column of
zeroes, and thus cannot equalI
. Hence the matrix has rank less thanBut by Theorem 2.15,
an×matrix is nonsingular if and only if its rank equals. Thus,thematrixmustbesingular.
Suppose a matrix contains a row of all zeroes. We assume that it is nonsingular and derive a
contradiction. If the matrix is nonsingular, then part (4) of Theorem 2.12 says that its transpose
is also nonsingular. But the transpose contains a column of all zeroes. This contradicts the result
obtained in thefirst paragraph.
(b) Such a matrix will contain a column of all zeroes. Use part (a).
(c) When pivoting in theth column of the matrix during row reduction, the( )entry will be
nonzero, allowing a pivot in that position. Then, since all entries in that column below the
main diagonal are already zero, none of the rows below theth row are changed in that column.
Hence, none of the entries below theth row are changed when that row is used as the pivot row.
Thus, none of the nonzero entries on the main diagonal are affected by the row reduction steps
on previous columns (and the matrix stays in upper triangular form throughout the process).
Therefore the matrix row reduces toI
.
(d) IfAis lower triangular with no zeroes on the main diagonal, thenA
is upper triangular with no
zeroes on the main diagonal. By part (c),A
is nonsingular. HenceA=(A
)
is nonsingular
by part (4) of Theorem 2.12.
(e) Note that when row reducing theth column ofA, all of the following occur:
1. The pivot
is nonzero, so we use a Type (I) operation to change it to a1. This changes
only row.
2. No nonzero targets appear below row, so all Type (II) row operations only change rows
above row.
3. Because of (2), no entries are changed below the main diagonal, and no main diagonal entries
are changed by Type (II) operations.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 29
(8) (a) Consider
∙
01
10
¸
.
(b) Consider
⎡
⎣
010
100
001
⎤
⎦.
(c)A=A
−1
ifAis involutory.
(9) (a)A=I
2,B=−I 2,A+B=O 2
(b)A=
∙
10
00
¸
,B=
∙
00
01
¸
,A+B=
∙
10
01
¸
(c) IfA=B=I
2,thenA+B=2I 2,A
−1
=B
−1
=I2,A
−1
+B
−1
=2I 2,and(A+B)
−1
=
1
2
I2,
soA
−1
+B
−1
6 =(A+B)
−1
.
(10) (a)Bmust be the zero matrix.
(b) No.AB=I
impliesA
−1
exists (and equalsB). Multiply both sides ofAC=O on the left
byA
−1
.
(11)A
−9
A
−5
A
−1
A
3
A
7
A
11
(12)B
−1
Ais the inverse ofA
−1
B.
(13)(A
−1
)
=(A
)
−1
(by Theorem 2.12, part (4))=A
−1
.
(14) (a) Supposefirst that the matrix contains a column of zeroes. No row operations can alter a column
of zeroes, so the unique reduced row echelon form for the matrix must also contain a column of
zeroes, and thus cannot equalI
. Hence the matrix has rank less thanBut by Theorem 2.15,
an×matrix is nonsingular if and only if its rank equals. Thus,thematrixmustbesingular.
Suppose a matrix contains a row of all zeroes. We assume that it is nonsingular and derive a
contradiction. If the matrix is nonsingular, then part (4) of Theorem 2.12 says that its transpose
is also nonsingular. But the transpose contains a column of all zeroes. This contradicts the result
obtained in thefirst paragraph.
(b) Such a matrix will contain a column of all zeroes. Use part (a).
(c) When pivoting in theth column of the matrix during row reduction, the( )entry will be
nonzero, allowing a pivot in that position. Then, since all entries in that column below the
main diagonal are already zero, none of the rows below theth row are changed in that column.
Hence, none of the entries below theth row are changed when that row is used as the pivot row.
Thus, none of the nonzero entries on the main diagonal are affected by the row reduction steps
on previous columns (and the matrix stays in upper triangular form throughout the process).
Therefore the matrix row reduces toI
.
(d) IfAis lower triangular with no zeroes on the main diagonal, thenA
is upper triangular with no
zeroes on the main diagonal. By part (c),A
is nonsingular. HenceA=(A
)
is nonsingular
by part (4) of Theorem 2.12.
(e) Note that when row reducing theth column ofA, all of the following occur:
1. The pivot
is nonzero, so we use a Type (I) operation to change it to a1. This changes
only row.
2. No nonzero targets appear below row, so all Type (II) row operations only change rows
above row.
3. Because of (2), no entries are changed below the main diagonal, and no main diagonal entries
are changed by Type (II) operations.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 29
Answers to Exercises Section 2.4
When row reducing[A|I ], we use the exact same row operations we use to reduceA.SinceI
is also upper triangular, fact (3) above shows that all the zeroes below the main diagonal ofI
remain zero when the row operations are applied. Thus, the result of the row operations, namely
A
−1
, is upper triangular.
(15) (a) Part (1): SinceAA
−1
=I,wemusthave(A
−1
)
−1
=A.
Part (2): For0,toshow(A
)
−1
=(A
−1
)
, we must show thatA
(A
−1
)
=I. We proceed
by induction on.
Base Step: For=1,clearlyAA
−1
=I.
Inductive Step: AssumeA
(A
−1
)
=I.ProveA
+1
(A
−1
)
+1
=I.
Now,A
+1
(A
−1
)
+1
=AA
(A
−1
)
A
−1
=AI nA
−1
=AA
−1
=I. This concludes the proof
for0.
We now showA
(A
−1
)
=Ifor≤0.
For=0, clearlyA
0
(A
−1
)
0
=II=I.Thecase=−1is covered by part (1) of the theorem.
For≤−2,(A
)
−1
=((A
−1
)
−
)
−1
(by definition)=((A
−
)
−1
)
−1
(by the0case)=A
−
(by part (1)).
(b) To show(A
1···A )
−1
=A
−1
···A
−1
1
,wemustprovethat
(A
1···A )(A
−1
···A
−1
1
)=I . Use induction on.
Base Step: For=1,clearlyA
1A
−1
1
=I.
Inductive Step: Assume that(A
1···A )(A
−1
···A
−1
1
)=I .
Prove that(A
1···A +1)(A
−1
+1
···A
−1
1
)=I .
Now,(A
1···A +1)(A
−1
+1
···A
−1
1
)=(A 1···A )A+1A
−1
+1
(A
−1
···A
−1
1
)
=(A
1···A )I(A
−1
···A
−1
1
)=(A 1···A )(A
−1
···A
−1
1
)=I .
(16) We must prove that(A)(
1
A
−1
)=I.But,(A)(
1
A
−1
)=
1
AA
−1
=1I=I.
(17) (a) Let=−,=−.Then 0.Now,A
+
=A
−(+)
=(A
−1
)
+
=(A
−1
)
(A
−1
)
(by
Theorem 1.17)=A
−
A
−
=A
A
.
(b) Let=−.Then(A
)
=(A
)
−
=((A
)
−1
)
=((A
−1
)
)
(by Theorem 2.12, part (2))=
(A
−1
)
(by Theorem 1.17)=A
−
(by Theorem 2.12, part (2))=A
(−)
=A
. Similarly,
(A
)
=((A
−1
)
)
(as before)=((A
−1
)
)
(by Theorem 1.17)=(A
−
)
=(A
)
.
(18) First assumeAB=BA.Then(AB)
2
=ABAB=A(BA)B=A(AB)B=A
2
B
2
.
Conversely, if(AB)
2
=A
2
B
2
,thenABAB=AABB=⇒A
−1
ABABB
−1
=A
−1
AABBB
−1
=⇒
BA=AB.
(19) If(AB)
=A
B
for all≥2,use=2and the proof in Exercise 18 to showBA=AB.
Conversely, we need to show thatBA=AB=⇒(AB)
=A
B
for all≥2.
First, we prove thatBA=AB=⇒AB
=B
Afor all≥2. We use induction on.
Base Step (=2):AB
2
=A(BB)=(AB)B=(BA)B=B(AB)=B(BA)=(BB)A=B
2
A.
Inductive Step:AB
+1
=A(B
B)=(AB
)B=(B
A)B(by the inductive hypothesis)=B
(AB)
=B
(BA)=(B
B)A=B
+1
A.
Now we use this “lemma” (BA=AB=⇒AB
=B
Afor all≥2)toproveBA=AB=⇒
(AB)
=A
B
for all≥2. Again, we proceed by induction on.
Base Step(=2):(AB)
2
=(AB)(AB)=A(BA)B=A(AB)B=A
2
B
2
.
Inductive Step:(AB)
+1
=(AB)
(AB)=(A
B
)(AB)(by the inductive hypothesis)=A
(B
A)B
=A
(AB
)B(by the lemma)=(A
A)(B
B)=A
+1
B
+1
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 30
When row reducing[A|I ], we use the exact same row operations we use to reduceA.SinceI
is also upper triangular, fact (3) above shows that all the zeroes below the main diagonal ofI
remain zero when the row operations are applied. Thus, the result of the row operations, namely
A
−1
, is upper triangular.
(15) (a) Part (1): SinceAA
−1
=I,wemusthave(A
−1
)
−1
=A.
Part (2): For0,toshow(A
)
−1
=(A
−1
)
, we must show thatA
(A
−1
)
=I. We proceed
by induction on.
Base Step: For=1,clearlyAA
−1
=I.
Inductive Step: AssumeA
(A
−1
)
=I.ProveA
+1
(A
−1
)
+1
=I.
Now,A
+1
(A
−1
)
+1
=AA
(A
−1
)
A
−1
=AI nA
−1
=AA
−1
=I. This concludes the proof
for0.
We now showA
(A
−1
)
=Ifor≤0.
For=0, clearlyA
0
(A
−1
)
0
=II=I.Thecase=−1is covered by part (1) of the theorem.
For≤−2,(A
)
−1
=((A
−1
)
−
)
−1
(by definition)=((A
−
)
−1
)
−1
(by the0case)=A
−
(by part (1)).
(b) To show(A
1···A )
−1
=A
−1
···A
−1
1
,wemustprovethat
(A
1···A )(A
−1
···A
−1
1
)=I . Use induction on.
Base Step: For=1,clearlyA
1A
−1
1
=I.
Inductive Step: Assume that(A
1···A )(A
−1
···A
−1
1
)=I .
Prove that(A
1···A +1)(A
−1
+1
···A
−1
1
)=I .
Now,(A
1···A +1)(A
−1
+1
···A
−1
1
)=(A 1···A )A+1A
−1
+1
(A
−1
···A
−1
1
)
=(A
1···A )I(A
−1
···A
−1
1
)=(A 1···A )(A
−1
···A
−1
1
)=I .
(16) We must prove that(A)(
1
A
−1
)=I.But,(A)(
1
A
−1
)=
1
AA
−1
=1I=I.
(17) (a) Let=−,=−.Then 0.Now,A
+
=A
−(+)
=(A
−1
)
+
=(A
−1
)
(A
−1
)
(by
Theorem 1.17)=A
−
A
−
=A
A
.
(b) Let=−.Then(A
)
=(A
)
−
=((A
)
−1
)
=((A
−1
)
)
(by Theorem 2.12, part (2))=
(A
−1
)
(by Theorem 1.17)=A
−
(by Theorem 2.12, part (2))=A
(−)
=A
. Similarly,
(A
)
=((A
−1
)
)
(as before)=((A
−1
)
)
(by Theorem 1.17)=(A
−
)
=(A
)
.
(18) First assumeAB=BA.Then(AB)
2
=ABAB=A(BA)B=A(AB)B=A
2
B
2
.
Conversely, if(AB)
2
=A
2
B
2
,thenABAB=AABB=⇒A
−1
ABABB
−1
=A
−1
AABBB
−1
=⇒
BA=AB.
(19) If(AB)
=A
B
for all≥2,use=2and the proof in Exercise 18 to showBA=AB.
Conversely, we need to show thatBA=AB=⇒(AB)
=A
B
for all≥2.
First, we prove thatBA=AB=⇒AB
=B
Afor all≥2. We use induction on.
Base Step (=2):AB
2
=A(BB)=(AB)B=(BA)B=B(AB)=B(BA)=(BB)A=B
2
A.
Inductive Step:AB
+1
=A(B
B)=(AB
)B=(B
A)B(by the inductive hypothesis)=B
(AB)
=B
(BA)=(B
B)A=B
+1
A.
Now we use this “lemma” (BA=AB=⇒AB
=B
Afor all≥2)toproveBA=AB=⇒
(AB)
=A
B
for all≥2. Again, we proceed by induction on.
Base Step(=2):(AB)
2
=(AB)(AB)=A(BA)B=A(AB)B=A
2
B
2
.
Inductive Step:(AB)
+1
=(AB)
(AB)=(A
B
)(AB)(by the inductive hypothesis)=A
(B
A)B
=A
(AB
)B(by the lemma)=(A
A)(B
B)=A
+1
B
+1
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 30
Answers to Exercises Chapter 2 Review
(20) Base Step (=0):I =(A
1
−I)(A−I )
−1
.
Inductive Step: AssumeI
+A+A
2
+···+A
=(A
+1
−I)(A−I )
−1
for some
ProveI
+A+A
2
+···+A
+A
+1
=(A
+2
−I)(A−I )
−1
Now,I
+A+A
2
+···+A
+A
+1
=(I +A+A
2
+···+A
)+A
+1
=(A
+1
−I)(A−I )
−1
+A
+1
(A−I )(A−I )
−1
(where thefirst term is obtained from the
inductive hypothesis)=((A
+1
−I)+A
+1
(A−I ))(A−I )
−1
=(A
+2
−I)(A−I )
−1
(21) SupposeAis an×matrix andBis a×matrix. Suppose, further, thatAB=I
and.By
Corollary 2.3, the homogeneous system havingBas its matrix of coefficients has a nontrivial solution
X. That is, there is a nonzero vectorXsuch thatBX=0.ButthenX=I
X=(AB)X=A(BX)
=A0=0,acontradiction.
(22) (a) F (b) T (c) T (d) F (e) F (f) T
Chapter 2 Review Exercises
(1) (a) staircase pattern of pivots after Gauss-Jordan Method:
⎡
⎢
⎢
⎣
100
010
00 1
000
¯
¯
¯
¯
¯
¯
¯
¯
−6
8
−5
0
⎤
⎥
⎥
⎦
;
complete solution set={(−6,8,−5)}
(b) staircase pattern of pivots after Gauss-Jordan Method:
⎡
⎢
⎢
⎢
⎢
⎣
100
19
6
010
1
6
00 1
7
6
000 0
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
0
0
0
1
⎤
⎥
⎥
⎥
⎥
⎥
⎦
; no solutions
(c) staircase pattern of pivots after Gauss-Jordan Method:
⎡
⎣
1010 −1
0120 1
000 1−2
¯
¯
¯
¯
¯
¯
−5
1
1
⎤
⎦;
complete solution set={[−5−+1−2− 1+2 ]| ∈R}
(2)=−2
3
+5
2
−6+3
(3) (a) No. Entries above pivots need to be zero. (b) No. Rows 3 and 4 should be switched.
(4)=4,=7,=4,=6
(5) (i)
1=−308, 2=−18, 3=−641, 4=108 (ii) 1=−29, 2=−19, 3=−88, 4=36
(6) Corollary 2.3 applies since there are more variables than equations.
(7) (a) (I):h3i←−6h3i (b) (II):h2i←3h4i+h2i (c) (III):h2i↔h3i
(8) (a)rank(A)=2,rank(B)=4,rank(C)=3
(b)AX=0andCX=0:infinite number of solutions;BX=0:onesolution
Copyrightc°2016 Elsevier Ltd. All rights reserved. 31
(20) Base Step (=0):I =(A
1
−I)(A−I )
−1
.
Inductive Step: AssumeI
+A+A
2
+···+A
=(A
+1
−I)(A−I )
−1
for some
ProveI
+A+A
2
+···+A
+A
+1
=(A
+2
−I)(A−I )
−1
Now,I
+A+A
2
+···+A
+A
+1
=(I +A+A
2
+···+A
)+A
+1
=(A
+1
−I)(A−I )
−1
+A
+1
(A−I )(A−I )
−1
(where thefirst term is obtained from the
inductive hypothesis)=((A
+1
−I)+A
+1
(A−I ))(A−I )
−1
=(A
+2
−I)(A−I )
−1
(21) SupposeAis an×matrix andBis a×matrix. Suppose, further, thatAB=I
and.By
Corollary 2.3, the homogeneous system havingBas its matrix of coefficients has a nontrivial solution
X. That is, there is a nonzero vectorXsuch thatBX=0.ButthenX=I
X=(AB)X=A(BX)
=A0=0,acontradiction.
(22) (a) F (b) T (c) T (d) F (e) F (f) T
Chapter 2 Review Exercises
(1) (a) staircase pattern of pivots after Gauss-Jordan Method:
⎡
⎢
⎢
⎣
100
010
00 1
000
¯
¯
¯
¯
¯
¯
¯
¯
−6
8
−5
0
⎤
⎥
⎥
⎦
;
complete solution set={(−6,8,−5)}
(b) staircase pattern of pivots after Gauss-Jordan Method:
⎡
⎢
⎢
⎢
⎢
⎣
100
19
6
010
1
6
00 1
7
6
000 0
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
0
0
0
1
⎤
⎥
⎥
⎥
⎥
⎥
⎦
; no solutions
(c) staircase pattern of pivots after Gauss-Jordan Method:
⎡
⎣
1010 −1
0120 1
000 1−2
¯
¯
¯
¯
¯
¯
−5
1
1
⎤
⎦;
complete solution set={[−5−+1−2− 1+2 ]| ∈R}
(2)=−2
3
+5
2
−6+3
(3) (a) No. Entries above pivots need to be zero. (b) No. Rows 3 and 4 should be switched.
(4)=4,=7,=4,=6
(5) (i)
1=−308, 2=−18, 3=−641, 4=108 (ii) 1=−29, 2=−19, 3=−88, 4=36
(6) Corollary 2.3 applies since there are more variables than equations.
(7) (a) (I):h3i←−6h3i (b) (II):h2i←3h4i+h2i (c) (III):h2i↔h3i
(8) (a)rank(A)=2,rank(B)=4,rank(C)=3
(b)AX=0andCX=0:infinite number of solutions;BX=0:onesolution
Copyrightc°2016 Elsevier Ltd. All rights reserved. 31
Answers to Exercises Chapter 2 Review
(9) The reduced row echelon form matrices forAandBare both
⎡
⎣
10−30 2
01 20 −4
00 01 3
⎤
⎦. Therefore,A
andBare row equivalent by an argument similar to that in Exercise 3(b) of Section 2.3.
(10) (a) Yes.[−3429−21] = 5[2−3−1] + 2[5−21]−6[9−83]
(b) Yes.[−3429−21]is a linear combination of the rows of the matrix.
(11)A
−1
=
"
−
2
9
1
9
−
1
6
1
3
#
(12) (a) Nonsingular.A
−1
=
⎡
⎢
⎣
−
1
2
5
2
2
−164
1−5−3
⎤
⎥
⎦
(b) Singular
(13) This is true by the Inverse Method, which is justified in Section 2.4.
(14) No, by part (1) of Theorem 2.16.
(15) The inverse of the coefficient matrix is
⎡
⎣
3−1−4
2−1−3
1−2−2
⎤
⎦
The solution set is
1=−27, 2=−21, 3=−1
(16) (a) BecauseBis nonsingular,Bis row equivalent toI
(see Exercise 13). Thus, there is a sequence
of row operations,
1such that 1(···( (B))···)=I . Hence, by part (2) of Theorem
2.1,
1(···( (BA))···)= 1(···( (B))···)A=I A=A
Therefore,BAis row equivalent toA. Thus, by Exercise 17 in Section 2.3,BAandAhave the
same rank.
(b) By part (d) of Exercise 18 in Section 2.3,rank(AC)≤rank(A).
Similarly,rank(A)=rank((AC)C
−1
)≤rank(AC). Hence,rank(AC)=rank(A).
(17) (a) F
(b) F
(c) F
(d) T
(e) F
(f) T
(g) T
(h) T
(i) F
(j) T
(k) F
(l) F
(m) T
(n) T
(o) F
(p) T
(q) T
(r) T
(s) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 32
(9) The reduced row echelon form matrices forAandBare both
⎡
⎣
10−30 2
01 20 −4
00 01 3
⎤
⎦. Therefore,A
andBare row equivalent by an argument similar to that in Exercise 3(b) of Section 2.3.
(10) (a) Yes.[−3429−21] = 5[2−3−1] + 2[5−21]−6[9−83]
(b) Yes.[−3429−21]is a linear combination of the rows of the matrix.
(11)A
−1
=
"
−
2
9
1
9
−
1
6
1
3
#
(12) (a) Nonsingular.A
−1
=
⎡
⎢
⎣
−
1
2
5
2
2
−164
1−5−3
⎤
⎥
⎦
(b) Singular
(13) This is true by the Inverse Method, which is justified in Section 2.4.
(14) No, by part (1) of Theorem 2.16.
(15) The inverse of the coefficient matrix is
⎡
⎣
3−1−4
2−1−3
1−2−2
⎤
⎦
The solution set is
1=−27, 2=−21, 3=−1
(16) (a) BecauseBis nonsingular,Bis row equivalent toI
(see Exercise 13). Thus, there is a sequence
of row operations,
1such that 1(···( (B))···)=I . Hence, by part (2) of Theorem
2.1,
1(···( (BA))···)= 1(···( (B))···)A=I A=A
Therefore,BAis row equivalent toA. Thus, by Exercise 17 in Section 2.3,BAandAhave the
same rank.
(b) By part (d) of Exercise 18 in Section 2.3,rank(AC)≤rank(A).
Similarly,rank(A)=rank((AC)C
−1
)≤rank(AC). Hence,rank(AC)=rank(A).
(17) (a) F
(b) F
(c) F
(d) T
(e) F
(f) T
(g) T
(h) T
(i) F
(j) T
(k) F
(l) F
(m) T
(n) T
(o) F
(p) T
(q) T
(r) T
(s) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 32
Answers to Exercises Section 3.1
Chapter 3
Section 3.1
(1) (a)−17
(b)6
(c)0
(d)1
(e)−108
(f)156
(g)−40
(h)−60
(i)0
(j)−3
(2) (a)
¯
¯
¯
¯
43
−24
¯
¯
¯
¯
=22
(b)
¯
¯
¯
¯
¯
¯
02 −3
142
4−11
¯
¯
¯
¯
¯
¯
=65 (c)
¯
¯
¯
¯
¯
¯
−305
2−14
640
¯
¯
¯
¯
¯
¯
= 118
(3) (a)(−1)
2+2
¯
¯
¯
¯
4−3
9−7
¯
¯
¯
¯
=−1
(b)(−1)
2+3
¯
¯
¯
¯
−96
43
¯
¯
¯
¯
=51
(c)(−1)
4+3
¯
¯
¯
¯
¯
¯
−5213
−8222
−6−3−16
¯
¯
¯
¯
¯
¯
= 222
(d)(−1)
1+2
¯
¯
¯
¯
−4−3
−1+2
¯
¯
¯
¯
=−2+11
(4) Same answers as Exercise 1.
(5) (a)0 (b)−251 (c)−60 (d)352
(6)
11223344+11233442+11243243+12213443+12233144+12243341
+13213244+13223441+13243142+14213342+14223143+14233241
−11223443−11233244−11243342−11213344−12233441−12243143
−13213442−13223144−13243241−14213243−14223341−14233142
(7) LetA=
∙
11
11
¸
,andletB=
∙
10
01
¸
.
(8) (a) Perform the basketweaving method.
(b)(a×b)·a=(
23−32)1+( 31−13)2+( 12−21)3
=123−132+231−123+132−231=0.
Similarly,(a×b)·b=0.
(9) (a)7 (b)18 (c)12 (d)0
(10) Letx=[
12]andy=[ 12].Then
proj
xy=
µ
x·y
kxk
2
¶
x=
1
kxk
2
[(11+22)1(11+22)2]
Hence,
y−proj
xy=
1
kxk
2
(kxk
2
y)−proj
xy
=
1
kxk
2
[(
2
1
+
2
2
)1(
2
1
+
2
2
)2]−
1
kxk
2
[(11+22)1(11+22)2]
=
1
kxk
2
[
2
1
1+
2
2
1−
2
1
1−122
2
1
2+
2
2
2−121−
2
2
2]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 33
Chapter 3
Section 3.1
(1) (a)−17
(b)6
(c)0
(d)1
(e)−108
(f)156
(g)−40
(h)−60
(i)0
(j)−3
(2) (a)
¯
¯
¯
¯
43
−24
¯
¯
¯
¯
=22
(b)
¯
¯
¯
¯
¯
¯
02 −3
142
4−11
¯
¯
¯
¯
¯
¯
=65 (c)
¯
¯
¯
¯
¯
¯
−305
2−14
640
¯
¯
¯
¯
¯
¯
= 118
(3) (a)(−1)
2+2
¯
¯
¯
¯
4−3
9−7
¯
¯
¯
¯
=−1
(b)(−1)
2+3
¯
¯
¯
¯
−96
43
¯
¯
¯
¯
=51
(c)(−1)
4+3
¯
¯
¯
¯
¯
¯
−5213
−8222
−6−3−16
¯
¯
¯
¯
¯
¯
= 222
(d)(−1)
1+2
¯
¯
¯
¯
−4−3
−1+2
¯
¯
¯
¯
=−2+11
(4) Same answers as Exercise 1.
(5) (a)0 (b)−251 (c)−60 (d)352
(6)
11223344+11233442+11243243+12213443+12233144+12243341
+13213244+13223441+13243142+14213342+14223143+14233241
−11223443−11233244−11243342−11213344−12233441−12243143
−13213442−13223144−13243241−14213243−14223341−14233142
(7) LetA=
∙
11
11
¸
,andletB=
∙
10
01
¸
.
(8) (a) Perform the basketweaving method.
(b)(a×b)·a=(
23−32)1+( 31−13)2+( 12−21)3
=123−132+231−123+132−231=0.
Similarly,(a×b)·b=0.
(9) (a)7 (b)18 (c)12 (d)0
(10) Letx=[
12]andy=[ 12].Then
proj
xy=
µ
x·y
kxk
2
¶
x=
1
kxk
2
[(11+22)1(11+22)2]
Hence,
y−proj
xy=
1
kxk
2
(kxk
2
y)−proj
xy
=
1
kxk
2
[(
2
1
+
2
2
)1(
2
1
+
2
2
)2]−
1
kxk
2
[(11+22)1(11+22)2]
=
1
kxk
2
[
2
1
1+
2
2
1−
2
1
1−122
2
1
2+
2
2
2−121−
2
2
2]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 33
Answers to Exercises Section 3.1
=
1
kxk
2
[
2
2
1−122
2
1
2−121]
=
1kxk
2
[2(21−12)1(12−21)]
=
12−21
kxk
2
[−21]
Thus,
kxkky−proj
xyk=kxk
|
12−21|
kxk
2
q
2
2
+
2
1
=|12−21|
=absolute value of
¯
¯
¯
¯
12
12
¯
¯
¯
¯
(11) (a)18 (b)7 (c)63 (d)8
(12) First, notice from the definition ofx×yin Exercise 8, that
kx×yk=
p(23−32)
2
+( 13−31)
2
+( 12−21)
2
We will verify thatkx×ykis equal to the area of the parallelogram determined byxandyNow, from
the solution to Exercise 10 above, the area of this parallelogram is equal tokxkky−proj
xyk.One
can verifykx×yk=kxkky−proj
xykby a tedious, brute force, argument. (Algebraically expand
and simplifykxk
2
ky−proj
xyk
2
to get( 23−32)
2
+(13−31)
2
+(12−21)
2
)Analternate
approach is the following: Note that
kx×yk
2
=( 23−32)
2
+( 13−31)
2
+( 12−21)
2
=
2
2
2
3
−2 2323+
2
3
2
2
+
2
1
2
3
−2 1313+
2
3
2
1
+
2
1
2
2
−2 1212+
2
2
2
1
Using some algebraic manipulation, this can be expressed as
kx×yk
2
=(
2
1
+
2
2
+
2
3
)(
2
1
+
2
2
+
2
3
)−( 11+22+33)
2
Therefore,
kx×yk
2
=kxk
2
kyk
2
−(x·y)
2
=
kxk
2
kxk
2
³
kxk
2
kyk
2
−(x·y)
2
´
=kxk
2
Ã
kxk
2
kyk
2
kxk
2
−
(x·y)
2
kxk
2
!
=kxk
2
Ã
kxk
2
(y·y)
kxk
2
−2
Ã
(x·y)
2
kxk
2
!
+
(x·y)
2
kxk
2
!
=kxk
2
⎛
⎝(y·y)−2
Ã
x·y
kxk
2
!
(x·y)+
Ã
x·y
kxk
2
!
2
(x·x)
⎞
⎠
Copyrightc°2016 Elsevier Ltd. All rights reserved. 34
=
1
kxk
2
[
2
2
1−122
2
1
2−121]
=
1kxk
2
[2(21−12)1(12−21)]
=
12−21
kxk
2
[−21]
Thus,
kxkky−proj
xyk=kxk
|
12−21|
kxk
2
q
2
2
+
2
1
=|12−21|
=absolute value of
¯
¯
¯
¯
12
12
¯
¯
¯
¯
(11) (a)18 (b)7 (c)63 (d)8
(12) First, notice from the definition ofx×yin Exercise 8, that
kx×yk=
p(23−32)
2
+( 13−31)
2
+( 12−21)
2
We will verify thatkx×ykis equal to the area of the parallelogram determined byxandyNow, from
the solution to Exercise 10 above, the area of this parallelogram is equal tokxkky−proj
xyk.One
can verifykx×yk=kxkky−proj
xykby a tedious, brute force, argument. (Algebraically expand
and simplifykxk
2
ky−proj
xyk
2
to get( 23−32)
2
+(13−31)
2
+(12−21)
2
)Analternate
approach is the following: Note that
kx×yk
2
=( 23−32)
2
+( 13−31)
2
+( 12−21)
2
=
2
2
2
3
−2 2323+
2
3
2
2
+
2
1
2
3
−2 1313+
2
3
2
1
+
2
1
2
2
−2 1212+
2
2
2
1
Using some algebraic manipulation, this can be expressed as
kx×yk
2
=(
2
1
+
2
2
+
2
3
)(
2
1
+
2
2
+
2
3
)−( 11+22+33)
2
Therefore,
kx×yk
2
=kxk
2
kyk
2
−(x·y)
2
=
kxk
2
kxk
2
³
kxk
2
kyk
2
−(x·y)
2
´
=kxk
2
Ã
kxk
2
kyk
2
kxk
2
−
(x·y)
2
kxk
2
!
=kxk
2
Ã
kxk
2
(y·y)
kxk
2
−2
Ã
(x·y)
2
kxk
2
!
+
(x·y)
2
kxk
2
!
=kxk
2
⎛
⎝(y·y)−2
Ã
x·y
kxk
2
!
(x·y)+
Ã
x·y
kxk
2
!
2
(x·x)
⎞
⎠
Copyrightc°2016 Elsevier Ltd. All rights reserved. 34
Answers to Exercises Section 3.1
=kxk
2
Ã
y−
(x·y)x
kxk
2
!
·
Ã
y−
(x·y)x
kxk
2
!
=kxk
2
°
°
°
°
°
y−
Ã
x·y
kxk
2
!
x
°
°
°
°
°
2
=kxk
2
ky−proj
xyk
2
Hence,kx×yk=kxkky−proj
xyk, the area of the parallelogram determined byxandy.
Now, we determine the volume of the parallelepiped. As in the hint, leth=proj
(x×y)z,the
perpendicular fromzto the parallelogram determined byxandy. Since the area of this parallelogram
equalskx×yk, the volume of the parallelepiped equals
khkkx×yk=kproj
(x×y)zkkx×yk=
¯
¯
¯
¯
z·(x×y)
kx×yk
¯
¯
¯
¯
kx×yk=|z·(x×y)|=|(x×y)·z|
But from the definition in Exercise 8,
|(x×y)·z|=|(
23−32)1+( 31−13)2+( 12−21)3|
=|
123+231+312−321−132−213|
=absolute value of
¯
¯
¯
¯
¯
¯
123
123
123
¯
¯
¯
¯
¯
¯
(13) (a) Base Step: If=1,thenA=[
11],soA=[ 11],and|A|= 11=|A|.
Inductive Step: Assume that ifAis an×matrix andis a scalar, then|A|=
|A|.Prove
that ifAis an(+1)×(+1)matrix, andis a scalar, then|A|=
+1
|A|.LetB=A.
Then
|B|=
+11B+11 ++12B+12 +···+ +1B+1++1+1 B+1+1
EachB
+1=(−1)
+1+
|B+1|=
(−1)
+1+
|A+1|(by the inductive hypothesis, since
B
+1is an×matrix)=
A+1Thus,
|A|=|B|=
+11(
A+11)+ +12(
A+12)+···+ +1(
A+1)
+
+1+1 (
A+1+1 )
=
+1
(+11A+11 ++12A+12 +···+ +1A+1++1+1 A+1+1 )
=
+1
|A|
(b)|2A|=2
3
|A|(sinceAis a3×3matrix)=8|A|
(14)
¯
¯
¯
¯
¯
¯
¯
¯
−10 0
0−10
00 −1
0123+
¯
¯
¯
¯
¯
¯
¯
¯
=
0(−1)
4+1
¯
¯
¯
¯
¯
¯
−100
−10
0−1
¯
¯
¯
¯
¯
¯
+
1(−1)
4+2
¯
¯
¯
¯
¯
¯
00
0−10
0−1
¯
¯
¯
¯
¯
¯
+
2(−1)
4+3
¯
¯
¯
¯
¯
¯
−10
00
00 −1
¯
¯
¯
¯
¯
¯
+
3(−1)
4+4
¯
¯
¯
¯
¯
¯
−10
0−1
00
¯
¯
¯
¯
¯
¯
Copyrightc°2016 Elsevier Ltd. All rights reserved. 35
=kxk
2
Ã
y−
(x·y)x
kxk
2
!
·
Ã
y−
(x·y)x
kxk
2
!
=kxk
2
°
°
°
°
°
y−
Ã
x·y
kxk
2
!
x
°
°
°
°
°
2
=kxk
2
ky−proj
xyk
2
Hence,kx×yk=kxkky−proj
xyk, the area of the parallelogram determined byxandy.
Now, we determine the volume of the parallelepiped. As in the hint, leth=proj
(x×y)z,the
perpendicular fromzto the parallelogram determined byxandy. Since the area of this parallelogram
equalskx×yk, the volume of the parallelepiped equals
khkkx×yk=kproj
(x×y)zkkx×yk=
¯
¯
¯
¯
z·(x×y)
kx×yk
¯
¯
¯
¯
kx×yk=|z·(x×y)|=|(x×y)·z|
But from the definition in Exercise 8,
|(x×y)·z|=|(
23−32)1+( 31−13)2+( 12−21)3|
=|
123+231+312−321−132−213|
=absolute value of
¯
¯
¯
¯
¯
¯
123
123
123
¯
¯
¯
¯
¯
¯
(13) (a) Base Step: If=1,thenA=[
11],soA=[ 11],and|A|= 11=|A|.
Inductive Step: Assume that ifAis an×matrix andis a scalar, then|A|=
|A|.Prove
that ifAis an(+1)×(+1)matrix, andis a scalar, then|A|=
+1
|A|.LetB=A.
Then
|B|=
+11B+11 ++12B+12 +···+ +1B+1++1+1 B+1+1
EachB
+1=(−1)
+1+
|B+1|=
(−1)
+1+
|A+1|(by the inductive hypothesis, since
B
+1is an×matrix)=
A+1Thus,
|A|=|B|=
+11(
A+11)+ +12(
A+12)+···+ +1(
A+1)
+
+1+1 (
A+1+1 )
=
+1
(+11A+11 ++12A+12 +···+ +1A+1++1+1 A+1+1 )
=
+1
|A|
(b)|2A|=2
3
|A|(sinceAis a3×3matrix)=8|A|
(14)
¯
¯
¯
¯
¯
¯
¯
¯
−10 0
0−10
00 −1
0123+
¯
¯
¯
¯
¯
¯
¯
¯
=
0(−1)
4+1
¯
¯
¯
¯
¯
¯
−100
−10
0−1
¯
¯
¯
¯
¯
¯
+
1(−1)
4+2
¯
¯
¯
¯
¯
¯
00
0−10
0−1
¯
¯
¯
¯
¯
¯
+
2(−1)
4+3
¯
¯
¯
¯
¯
¯
−10
00
00 −1
¯
¯
¯
¯
¯
¯
+
3(−1)
4+4
¯
¯
¯
¯
¯
¯
−10
0−1
00
¯
¯
¯
¯
¯
¯
Copyrightc°2016 Elsevier Ltd. All rights reserved. 35
Answers to Exercises Section 3.2
The four3×3determinants in the previous equation can be calculated using “basketweaving” as−1,
,−
2
,and
3
, respectively. Therefore, the original4×4determinant equals
(−
0)(−1) + ( 1)()+(− 2)(−
2
)+( 3)(
3
)= 0+1+ 2
2
+3
3
(15) (a)=−5or=2 (b)=−2or=−1 (c)=3,=1,or=2
(16) (a) Use “basketweaving” and factor. (b) 20
(17) (a) Area of=
√
3
2
4
(b) Suppose one side ofextends from( )to( ),where( )and( )are lattice points. Then
2
=(−)
2
+(−)
2
an integer. Hence,
2
4
is rational, and the area of=
√
3
2
4
is irrational.
(c) Suppose the vertices ofare lattice points=( ),=( ),=( ).Thenside
is expressed using vector[− −], and sideis expressed using vector[− −].
Hence, the area of=
1
2
(area of the parallelogram formed by[− −]and[− −])
=
1
2
¯
¯
¯
¯
−−
−−
¯
¯
¯
¯
(d)
1
2
¯
¯
¯
¯
−−
−−
¯
¯
¯
¯
=
1
2
((−)(−)−(−)(−))which is
1
2
times the difference of two
products of integers, and hence, is rational.
(18) (a) F (b) T (c) F (d) F (e) T
Section 3.2
(1) (a) (II):h1i←−3h2i+h1i; determinant=1
(b) (III):h2i↔h3i; determinant=−1
(c) (I):h3i←−4h3i; determinant=−4
(d) (II):h2i←2h1i+h2i; determinant=1
(e) (I):h1i←
1
2
h1i; determinant=
1
2
(f) (III):h1i↔h2i; determinant=−1
(2) (a)30 (b)−3 (c)−4 (d)3 (e)35 (f)20
(3) (a) Determinant=−2; nonsingular
(b) Determinant=1; nonsingular
(c) Determinant=−79; nonsingular
(d) Determinant=0;singular
(4) (a) Determinant=−1; the system has only the trivial solution.
(b) Determinant=0; the system has nontrivial solutions. (One nontrivial solution is(173)).
(c) Determinant=−42; the system has only the trivial solution.
(5) Use Theorem 3.2.
(6) Use row operations to reverse the order of the rows ofA. This can be done using 3 Type (III)
operations, an odd number. (These operations areh1i↔h6i,h2i↔h5iandh3i↔h4i.) Hence, by
part (3) of Theorem 3.3,|A|=−
162534435261.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 36
The four3×3determinants in the previous equation can be calculated using “basketweaving” as−1,
,−
2
,and
3
, respectively. Therefore, the original4×4determinant equals
(−
0)(−1) + ( 1)()+(− 2)(−
2
)+( 3)(
3
)= 0+1+ 2
2
+3
3
(15) (a)=−5or=2 (b)=−2or=−1 (c)=3,=1,or=2
(16) (a) Use “basketweaving” and factor. (b) 20
(17) (a) Area of=
√
3
2
4
(b) Suppose one side ofextends from( )to( ),where( )and( )are lattice points. Then
2
=(−)
2
+(−)
2
an integer. Hence,
2
4
is rational, and the area of=
√
3
2
4
is irrational.
(c) Suppose the vertices ofare lattice points=( ),=( ),=( ).Thenside
is expressed using vector[− −], and sideis expressed using vector[− −].
Hence, the area of=
1
2
(area of the parallelogram formed by[− −]and[− −])
=
1
2
¯
¯
¯
¯
−−
−−
¯
¯
¯
¯
(d)
1
2
¯
¯
¯
¯
−−
−−
¯
¯
¯
¯
=
1
2
((−)(−)−(−)(−))which is
1
2
times the difference of two
products of integers, and hence, is rational.
(18) (a) F (b) T (c) F (d) F (e) T
Section 3.2
(1) (a) (II):h1i←−3h2i+h1i; determinant=1
(b) (III):h2i↔h3i; determinant=−1
(c) (I):h3i←−4h3i; determinant=−4
(d) (II):h2i←2h1i+h2i; determinant=1
(e) (I):h1i←
1
2
h1i; determinant=
1
2
(f) (III):h1i↔h2i; determinant=−1
(2) (a)30 (b)−3 (c)−4 (d)3 (e)35 (f)20
(3) (a) Determinant=−2; nonsingular
(b) Determinant=1; nonsingular
(c) Determinant=−79; nonsingular
(d) Determinant=0;singular
(4) (a) Determinant=−1; the system has only the trivial solution.
(b) Determinant=0; the system has nontrivial solutions. (One nontrivial solution is(173)).
(c) Determinant=−42; the system has only the trivial solution.
(5) Use Theorem 3.2.
(6) Use row operations to reverse the order of the rows ofA. This can be done using 3 Type (III)
operations, an odd number. (These operations areh1i↔h6i,h2i↔h5iandh3i↔h4i.) Hence, by
part (3) of Theorem 3.3,|A|=−
162534435261.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 36
Answers to Exercises Section 3.2
(7) If|A|6 =0,A
−1
exists. HenceAB=AC=⇒A
−1
AB=A
−1
AC=⇒B=C.
(8) (a) Base Step: When=1,A=[
11].ThenB=(A)(with=1)=[ 11]. Hence,|B|= 11=
|A|
(b) Inductive Hypothesis: Assume that ifAis an×matrix, andis the row operation
(I):hi←hiandB=(A),then|B|=|A|.
Note: In parts (c) and (d), we must prove that ifAis an(+1)×(+1)matrix andis the row
operation (I):hi←hiandB=(A),then|B|=|A|.
(c) Inductive Step when6 =:
|B|=
+11B+11 +···+ +1B+1++1+1 B+1+1
= +11(−1)
+1+1
|B+11|+···+ +1(−1)
+1+
|B+1|
+
+1+1 (−1)
+1++1
|B+1+1 |
But since|B
+1|is the determinant of the×matrixA +1after the row operation (I):
hi←hihas been performed (since6 =), the inductive hypothesis shows that each|B
+1|=
|A
+1|Since each +1=+1,wehave
|B|=
+11(−1)
+1+1
|A+11|+···+ +1(−1)
+1+
|A+1|
+
+1+1 (−1)
+1++1
|A+1+1 |
=|A|
(d) Inductive Step when=: Again,
|B|=
+11B+11 +···+ +1B+1++1+1 B+1+1
Since=,each
+1=+1, while eachB +1=A+1(since thefirstrows ofAare left
unchanged by the row operation in this case). Hence,
|B|=(
+11A+11)+···+( +1A+1)+( +1+1 A+1+1 )=|A|
(9) (a) In order to add a multiple of one row to another, we need at least two rows. Hence=2here.
(b) LetA=
∙
1112
2122
¸
Then the row operation (II):h1i←h2i+h1iproduces
B=
∙
21+1122+12
21 22
¸
and|B|=(
21+11)22−( 22+12)21which reduces to 1122−1221=|A|
(c) LetA=
∙
1112
2122
¸
Then the row operation (II):h2i←h1i+h2iproduces
B=
∙
11 12
11+2112+22
¸
and|B|=
11(12+22)− 12(11+21)which reduces to 1122−1221=|A|
Copyrightc°2016 Elsevier Ltd. All rights reserved. 37
(7) If|A|6 =0,A
−1
exists. HenceAB=AC=⇒A
−1
AB=A
−1
AC=⇒B=C.
(8) (a) Base Step: When=1,A=[
11].ThenB=(A)(with=1)=[ 11]. Hence,|B|= 11=
|A|
(b) Inductive Hypothesis: Assume that ifAis an×matrix, andis the row operation
(I):hi←hiandB=(A),then|B|=|A|.
Note: In parts (c) and (d), we must prove that ifAis an(+1)×(+1)matrix andis the row
operation (I):hi←hiandB=(A),then|B|=|A|.
(c) Inductive Step when6 =:
|B|=
+11B+11 +···+ +1B+1++1+1 B+1+1
= +11(−1)
+1+1
|B+11|+···+ +1(−1)
+1+
|B+1|
+
+1+1 (−1)
+1++1
|B+1+1 |
But since|B
+1|is the determinant of the×matrixA +1after the row operation (I):
hi←hihas been performed (since6 =), the inductive hypothesis shows that each|B
+1|=
|A
+1|Since each +1=+1,wehave
|B|=
+11(−1)
+1+1
|A+11|+···+ +1(−1)
+1+
|A+1|
+
+1+1 (−1)
+1++1
|A+1+1 |
=|A|
(d) Inductive Step when=: Again,
|B|=
+11B+11 +···+ +1B+1++1+1 B+1+1
Since=,each
+1=+1, while eachB +1=A+1(since thefirstrows ofAare left
unchanged by the row operation in this case). Hence,
|B|=(
+11A+11)+···+( +1A+1)+( +1+1 A+1+1 )=|A|
(9) (a) In order to add a multiple of one row to another, we need at least two rows. Hence=2here.
(b) LetA=
∙
1112
2122
¸
Then the row operation (II):h1i←h2i+h1iproduces
B=
∙
21+1122+12
21 22
¸
and|B|=(
21+11)22−( 22+12)21which reduces to 1122−1221=|A|
(c) LetA=
∙
1112
2122
¸
Then the row operation (II):h2i←h1i+h2iproduces
B=
∙
11 12
11+2112+22
¸
and|B|=
11(12+22)− 12(11+21)which reduces to 1122−1221=|A|
Copyrightc°2016 Elsevier Ltd. All rights reserved. 37
Answers to Exercises Section 3.2
(10) (a) Inductive Hypothesis: Assume that ifAis an(−1)×(−1)matrix andis a Type (II) row
operation, then|(A)|=|A|.
To complete the Inductive Step, we must prove that ifAis an×matrix, andis a Type (II)
row operation, then|(A)|=|A|
In what follows, letbetheType(II)rowoperationhi←hi+hi(so,6 =).
(b) Inductive Step when6 = 6 =:LetB=(A).Then|B|=
1B1+···+ BBut since
theth row ofAis not affected by,each
=Also, eachB =(−1)
+
|B|and|B |
is the determinant of the(−1)×(−1)matrix(A
). But by the inductive hypothesis, each
|B
|=|A |. Hence,|B|=|A|.
(c) Since6 =, we can assume6 =.Let=,andlet6 =
Theth row of
1(2(1(A)))
=th row of
2(1(A))
=th row of
1(A)+(th row of 1(A))
=th row ofA+(th row ofA)
=th row of(A).
Theth row of
1(2(1(A)))
=th row of
2(1(A))
=th row of
1(A)(since6 =)
=th row ofA=th row of(A).
Theth row (where6 = )of
1(2(1(A)))
=th row of
2(1(A))(since6 = )
=th row of
1(A)(since6 =)
=th row ofA(since6 = )
=th row of(A)(since6 ==).
(d) Inductive Step when=: From part (c), we have:
|(A)|=|
1(2(1(A)))|=−| 2(1(A))|(by part (3) of Theorem 3.3)
=−|
1(A)|(by part (b), since 6 ==)
=−(−|A|)(by part (3) of Theorem 3.3)=|A|
(e) Since6 =, we can assume6 =Now,
theth row of
1(3(1(A)))
=th row of
3(1(A))
=th row of
1(A)+(th row of 1(A))
=th row ofA+(th row ofA)(since6 =)
=th row of(A)(since=).
Theth row of
1(3(1(A)))
=th row of
3(1(A))
=th row of
1(A)(since6 =)
=th row ofA=th row of(A)(since6 =).
Theth row of
1(3(1(A)))
=th row of
3(1(A))
=th row of
1(A)(since6 =)=th row ofA
=th row of(A)(since6 =).
Theth row (where6 = )of
1(3(1(A)))
=th row of
3(1(A))
=th row of
1(A)
=th row ofA
=th row of(A).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 38
(10) (a) Inductive Hypothesis: Assume that ifAis an(−1)×(−1)matrix andis a Type (II) row
operation, then|(A)|=|A|.
To complete the Inductive Step, we must prove that ifAis an×matrix, andis a Type (II)
row operation, then|(A)|=|A|
In what follows, letbetheType(II)rowoperationhi←hi+hi(so,6 =).
(b) Inductive Step when6 = 6 =:LetB=(A).Then|B|=
1B1+···+ BBut since
theth row ofAis not affected by,each
=Also, eachB =(−1)
+
|B|and|B |
is the determinant of the(−1)×(−1)matrix(A
). But by the inductive hypothesis, each
|B
|=|A |. Hence,|B|=|A|.
(c) Since6 =, we can assume6 =.Let=,andlet6 =
Theth row of
1(2(1(A)))
=th row of
2(1(A))
=th row of
1(A)+(th row of 1(A))
=th row ofA+(th row ofA)
=th row of(A).
Theth row of
1(2(1(A)))
=th row of
2(1(A))
=th row of
1(A)(since6 =)
=th row ofA=th row of(A).
Theth row (where6 = )of
1(2(1(A)))
=th row of
2(1(A))(since6 = )
=th row of
1(A)(since6 =)
=th row ofA(since6 = )
=th row of(A)(since6 ==).
(d) Inductive Step when=: From part (c), we have:
|(A)|=|
1(2(1(A)))|=−| 2(1(A))|(by part (3) of Theorem 3.3)
=−|
1(A)|(by part (b), since 6 ==)
=−(−|A|)(by part (3) of Theorem 3.3)=|A|
(e) Since6 =, we can assume6 =Now,
theth row of
1(3(1(A)))
=th row of
3(1(A))
=th row of
1(A)+(th row of 1(A))
=th row ofA+(th row ofA)(since6 =)
=th row of(A)(since=).
Theth row of
1(3(1(A)))
=th row of
3(1(A))
=th row of
1(A)(since6 =)
=th row ofA=th row of(A)(since6 =).
Theth row of
1(3(1(A)))
=th row of
3(1(A))
=th row of
1(A)(since6 =)=th row ofA
=th row of(A)(since6 =).
Theth row (where6 = )of
1(3(1(A)))
=th row of
3(1(A))
=th row of
1(A)
=th row ofA
=th row of(A).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 38
Answers to Exercises Section 3.2
(f) Inductive Step when=:
|(A)|=|
1(3(1(A)))|=−| 3(1(A))|(by part (3) of Theorem 3.3)
=−|
1(A)|(by part (b))
=−(−|A|)(by part (3) of Theorem 3.3)=|A|
(11) (a) Suppose theth row ofAis a row of all zeroes. Letbe the row operation (I):hi←2hi.Then
|(A)|=2|A|by part (1) of Theorem 3.3. But also|(A)|=|A|sincehas no effect onA.
Hence,2|A|=|A|,so|A|=0.
(b) IfAcontains a row of zeroes, rank(A)(since the reduced row echelon form ofAalso contains
at least one row of zeroes). Hence by Corollary 3.6,|A|=0.
(12) (a) Let rows ofAbe identical. Letbe the row operation (III):h
i↔hiThen,(A)=A,
and so|(A)|=|A|But by part (3) of Theorem 3.3,|(A)|=−|A|Hence,|A|=−|A|and
so|A|=0
(b) If two rows ofAare identical, then subtracting one of these rows from the other shows thatAis
row equivalent to a matrix with a row of zeroes. Use part (b) of Exercise 11.
(13) (a) Suppose theth row ofA=(th row ofA). Using the row operation (II):hi←−hi+hi
shows thatAis row equivalent to a matrix with a row of zeroes. Use part (a) of Exercise 11.
(b) Use the given hint, Theorem 2.7 and Corollary 3.6.
(14) (a) We present two differentproofsofthisresult.
First proof: LetA=
∙
BC
OD
¸
.If
|B|=0, then, when row reducingAinto upper triangular
form, one of thefirstcolumns will not be a pivot column. Similarly, if|B|6 =0and|D|=0,
one of the last(−)columns will not be a pivot column. In either case, some column will not
contain a pivot, and so rank(A). Hence, in both cases,|A|=0=|B||D|.
Now suppose that both|B|and|D|are nonzero. Then the row operations for thefirst
pivots used to putAinto upper triangular form will be the same as the row operations for the
firstpivots used to putBinto upper triangular form (putting1’s along the entire main diagonal
ofB). These operations convertAinto the matrix
∙
UG
OD
¸
,whereUis an×upper
triangular matrix with1’s on the main diagonal, andGis some×(−)matrix. Hence,
at this point in the computation of|B|, the multiplicative factor
that we use to calculate the
determinant (as in Example 5 in the text) equals1|B|.
The remaining(−)pivots to putAinto upper triangular form will involve the same
row operations needed to putDinto upper triangular form (with1’s all along the main diagonal).
Clearly, this means that the multiplicative factormust be multiplied by1|D|,andsothenew
value ofequals1(|B||D|).Sincethefinal matrix is in upper triangular form with1’s all along
the main diagonal, its determinant is1. Multiplying this by the reciprocal of thefinal value of
(as in Example 5) yields|A|=|B||D|, completing thefirst proof.
Second proof: LetA=
∙
BC
OD
¸
,whereBis an×submatrix,Cis an×(−)
submatrix,Dis an(−)×(
−)submatrix, andOis an(−)×zero submatrix. Let
=−,sothatDis a×matrix. We will prove|A|=|B||D|by induction on.
Base Step(=1):
|A|=
1A1+···+ A=0A 1+···+0A
(−1) +11|B|=|D||B|
Copyrightc°2016 Elsevier Ltd. All rights reserved. 39
(f) Inductive Step when=:
|(A)|=|
1(3(1(A)))|=−| 3(1(A))|(by part (3) of Theorem 3.3)
=−|
1(A)|(by part (b))
=−(−|A|)(by part (3) of Theorem 3.3)=|A|
(11) (a) Suppose theth row ofAis a row of all zeroes. Letbe the row operation (I):hi←2hi.Then
|(A)|=2|A|by part (1) of Theorem 3.3. But also|(A)|=|A|sincehas no effect onA.
Hence,2|A|=|A|,so|A|=0.
(b) IfAcontains a row of zeroes, rank(A)(since the reduced row echelon form ofAalso contains
at least one row of zeroes). Hence by Corollary 3.6,|A|=0.
(12) (a) Let rows ofAbe identical. Letbe the row operation (III):h
i↔hiThen,(A)=A,
and so|(A)|=|A|But by part (3) of Theorem 3.3,|(A)|=−|A|Hence,|A|=−|A|and
so|A|=0
(b) If two rows ofAare identical, then subtracting one of these rows from the other shows thatAis
row equivalent to a matrix with a row of zeroes. Use part (b) of Exercise 11.
(13) (a) Suppose theth row ofA=(th row ofA). Using the row operation (II):hi←−hi+hi
shows thatAis row equivalent to a matrix with a row of zeroes. Use part (a) of Exercise 11.
(b) Use the given hint, Theorem 2.7 and Corollary 3.6.
(14) (a) We present two differentproofsofthisresult.
First proof: LetA=
∙
BC
OD
¸
.If
|B|=0, then, when row reducingAinto upper triangular
form, one of thefirstcolumns will not be a pivot column. Similarly, if|B|6 =0and|D|=0,
one of the last(−)columns will not be a pivot column. In either case, some column will not
contain a pivot, and so rank(A). Hence, in both cases,|A|=0=|B||D|.
Now suppose that both|B|and|D|are nonzero. Then the row operations for thefirst
pivots used to putAinto upper triangular form will be the same as the row operations for the
firstpivots used to putBinto upper triangular form (putting1’s along the entire main diagonal
ofB). These operations convertAinto the matrix
∙
UG
OD
¸
,whereUis an×upper
triangular matrix with1’s on the main diagonal, andGis some×(−)matrix. Hence,
at this point in the computation of|B|, the multiplicative factor
that we use to calculate the
determinant (as in Example 5 in the text) equals1|B|.
The remaining(−)pivots to putAinto upper triangular form will involve the same
row operations needed to putDinto upper triangular form (with1’s all along the main diagonal).
Clearly, this means that the multiplicative factormust be multiplied by1|D|,andsothenew
value ofequals1(|B||D|).Sincethefinal matrix is in upper triangular form with1’s all along
the main diagonal, its determinant is1. Multiplying this by the reciprocal of thefinal value of
(as in Example 5) yields|A|=|B||D|, completing thefirst proof.
Second proof: LetA=
∙
BC
OD
¸
,whereBis an×submatrix,Cis an×(−)
submatrix,Dis an(−)×(
−)submatrix, andOis an(−)×zero submatrix. Let
=−,sothatDis a×matrix. We will prove|A|=|B||D|by induction on.
Base Step(=1):
|A|=
1A1+···+ A=0A 1+···+0A
(−1) +11|B|=|D||B|
Copyrightc°2016 Elsevier Ltd. All rights reserved. 39
Answers to Exercises Section 3.3
Inductive Step: Assume|A|=|B||D|wheneverDis a(−1)×(−1)matrix. We must
prove|A|=|B||D|whenDis a×matrix. Now,|A|=
1A1+···+ A=0A 1+···+0A
(−) +1(−1)
2−+1
¯
¯A
(−+1)
¯
¯+···+
(−1)
2
|A|
ButA
(−+) =
∙
BG
OD
¸
whereG
is the matrixCwith itsth column removed. Hence,
by the inductive hypothesis|A|=
1(−1)
2−+1
|B||D 1|+···+ (−1)
2
|B||D |.Now,+
can be expressed as(2−+)+2(−), and so has the same parity (even or odd) as2−+,and
hence(−1)
2−+
=(−1)
+
. Therefore,|A|= 1(−1)
+1
|B||D 1|+···+ (−1)
2
|B||D |
=|B|
³
1(−1)
+1
|D1|+···+ (−1)
2
|D|
´
=|B||D|, completing the second proof.
(b)(18−18)(20−3) = 0(17) = 0
(15) Follow the hint in the text. IfAis row equivalent toBin reduced row echelon form, follow the method
in Example 5 in the text tofind determinants using row reductionby maintaining a multiplicative
factorthroughout the process. But, since the same rules work forwith regard to row operations
as for the determinant, the same factorwill be produced for.IfB=I
,then(B)=1=|B|.
Since the factoristhesameforand the determinant,(A)=(1)(B)=(1)|B|=|A|.If,
instead,B6 =I
, then theth row ofBwill be all zeroes. Perform the operation (I):hi←2hion
B,whichyieldsB. Then,(B)=2(B), implying(B)=0. Hence,(A)=(1)0 = 0.
(16) By Corollary 3.6 and part (1) of Theorem 2.7, the homogeneous systemAX=0has nontrivial
solutions. LetBbe any×matrix such that every column ofBis a nontrivial solution forAX=0.
Then theth column ofAB=A(th column ofB)=0for every. Hence,AB=O
.
(17) (a) F (b) T (c) F (d) F (e) F (f) T
Section 3.3
(1) (a) 31(−1)
3+1
|A31|+ 32(−1)
3+2
|A32|+ 33(−1)
3+3
|A33|+ 34(−1)
3+4
|A34|
(b)
11(−1)
1+1
|A11|+ 12(−1)
1+2
|A12|+ 13(−1)
1+3
|A13|+ 14(−1)
1+4
|A14|
(c)
14(−1)
1+4
|A14|+ 24(−1)
2+4
|A24|+ 34(−1)
3+4
|A34|+ 44(−1)
4+4
|A44|
(d)
11(−1)
1+1
|A11|+ 21(−1)
2+1
|A21|+ 31(−1)
3+1
|A31|+ 41(−1)
4+1
|A41|
(2) (a)−76 (b)465 (c)102 (d)410
(3) (a){(−43−7)} (b){(5−1−3)} (c){(64−5)} (d){(4−1−36)}
(4) (a) Supposeis the Type (I) operationhi←hi.Then(I
)is the diagonal matrix having1’s
in every entry on the main diagonal, except for the( )entry, which has the value.But,since
(I
)is a diagonal matrix, it is symmetric.
(b) SupposeistheType(III)operationhi↔hi. Then the only nonzero entries offthe main
diagonal for(I
)are the( )entry and the( )entry, which are both1’s. Hence, all entries
across the main diagonal from each other in(I
)are equal, and so(I )is symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 40
Inductive Step: Assume|A|=|B||D|wheneverDis a(−1)×(−1)matrix. We must
prove|A|=|B||D|whenDis a×matrix. Now,|A|=
1A1+···+ A=0A 1+···+0A
(−) +1(−1)
2−+1
¯
¯A
(−+1)
¯
¯+···+
(−1)
2
|A|
ButA
(−+) =
∙
BG
OD
¸
whereG
is the matrixCwith itsth column removed. Hence,
by the inductive hypothesis|A|=
1(−1)
2−+1
|B||D 1|+···+ (−1)
2
|B||D |.Now,+
can be expressed as(2−+)+2(−), and so has the same parity (even or odd) as2−+,and
hence(−1)
2−+
=(−1)
+
. Therefore,|A|= 1(−1)
+1
|B||D 1|+···+ (−1)
2
|B||D |
=|B|
³
1(−1)
+1
|D1|+···+ (−1)
2
|D|
´
=|B||D|, completing the second proof.
(b)(18−18)(20−3) = 0(17) = 0
(15) Follow the hint in the text. IfAis row equivalent toBin reduced row echelon form, follow the method
in Example 5 in the text tofind determinants using row reductionby maintaining a multiplicative
factorthroughout the process. But, since the same rules work forwith regard to row operations
as for the determinant, the same factorwill be produced for.IfB=I
,then(B)=1=|B|.
Since the factoristhesameforand the determinant,(A)=(1)(B)=(1)|B|=|A|.If,
instead,B6 =I
, then theth row ofBwill be all zeroes. Perform the operation (I):hi←2hion
B,whichyieldsB. Then,(B)=2(B), implying(B)=0. Hence,(A)=(1)0 = 0.
(16) By Corollary 3.6 and part (1) of Theorem 2.7, the homogeneous systemAX=0has nontrivial
solutions. LetBbe any×matrix such that every column ofBis a nontrivial solution forAX=0.
Then theth column ofAB=A(th column ofB)=0for every. Hence,AB=O
.
(17) (a) F (b) T (c) F (d) F (e) F (f) T
Section 3.3
(1) (a) 31(−1)
3+1
|A31|+ 32(−1)
3+2
|A32|+ 33(−1)
3+3
|A33|+ 34(−1)
3+4
|A34|
(b)
11(−1)
1+1
|A11|+ 12(−1)
1+2
|A12|+ 13(−1)
1+3
|A13|+ 14(−1)
1+4
|A14|
(c)
14(−1)
1+4
|A14|+ 24(−1)
2+4
|A24|+ 34(−1)
3+4
|A34|+ 44(−1)
4+4
|A44|
(d)
11(−1)
1+1
|A11|+ 21(−1)
2+1
|A21|+ 31(−1)
3+1
|A31|+ 41(−1)
4+1
|A41|
(2) (a)−76 (b)465 (c)102 (d)410
(3) (a){(−43−7)} (b){(5−1−3)} (c){(64−5)} (d){(4−1−36)}
(4) (a) Supposeis the Type (I) operationhi←hi.Then(I
)is the diagonal matrix having1’s
in every entry on the main diagonal, except for the( )entry, which has the value.But,since
(I
)is a diagonal matrix, it is symmetric.
(b) SupposeistheType(III)operationhi↔hi. Then the only nonzero entries offthe main
diagonal for(I
)are the( )entry and the( )entry, which are both1’s. Hence, all entries
across the main diagonal from each other in(I
)are equal, and so(I )is symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 40
Answers to Exercises Section 3.3
(c) SupposeistheType(II)operationhi←hi+hiandistheType(II)operation
hi←hi+hi.Then(I
)is the×matrix having1’s on the main diagonal and0’s
everywhere else, except for the( )entry, which has the value. Similarly,(I
)has1’s all
along the main diagonal and0’s everywhere else, except for the( )entry, which has the value
. However, this exactly describes the matrix((I
))
.
(5) (a) IfAis nonsingular,(A
)
−1
=(A
−1
)
by part (4) of Theorem 2.12.
For the converse,A
−1
=((A
)
−1
)
.
(b)|AB|=|A||B|=|B||A|=|BA|.
(6) (a) Use the fact that|AB|=|A||B|.
(b)|AB|=|−BA|=⇒|AB|=(−1)
|BA|(by Corollary 3.4)=⇒|A||B|=(−1)|B||A|(sinceis
odd). Hence, either|A|=0or|B|=0.
(7) (a)|AA
|=|A||A
|=|A||A|=|A|
2
≥0.(b) |AB
|=|A||B
|=|A
||B|
(8) (a)A
=−A,so|A|=|A
|=|−A|=(−1)
|A|(by Corollary 3.4)=(−1)|A|(sinceis odd).
Hence|A|=0.
(b) ConsiderA=
∙
0−1
10
¸
.
(9) (a)I
=I=I
−1
.
(b) Consider
⎡
⎣
010
100
001
⎤
⎦.
(c)|A
|=|A
−1
|=⇒|A|=1|A|
=⇒|A|
2
=1 =⇒|A|=±1.
(10) Note that
B=
⎡
⎣
90−3
32−1
−60 1
⎤
⎦
has a determinant equal to−18. Hence, ifB=A
2
,then|A|
2
is negative, a contradiction.
(11) (a) Base Step:=2:|A
1A2|=|A 1||A2|by Theorem 3.7.
Inductive Step: Assume|A
1A2···A |=|A 1||A2|···|A |, for any×matricesA 1A2A
Prove|A
1A2···A A+1|=|A 1||A2|···|A ||A+1|, for any×matricesA 1A2A A+1
But,|A
1A2···A A+1|=|(A 1A2···A )A+1|=|A 1A2···A ||A+1|(by Theorem 3.7)=
|A
1||A2|···|A ||A+1|by the inductive hypothesis.
(b) Use part (a) withA
=A,for1≤≤Or, for an induction proof:
Base Step (=1): Obvious.
Inductive Step:|A
+1
|=|A
A|=|A
||A|=|A|
|A|=|A|
+1
(c) Use part (b):|A|
=|A
|=|O |=0,andso|A|=0. Or, for an induction proof:
Base Step (=1): Obvious.
Inductive Step: Assume.A
+1
=O .IfAis singular, then|A|=0, and we are done. IfAis
nonsingular,A
+1
=O =⇒A
+1
A
−1
=O A
−1
=⇒A
=O =⇒|A|=0by the inductive
hypothesis, which impliesAis singular, a contradiction.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 41
(c) SupposeistheType(II)operationhi←hi+hiandistheType(II)operation
hi←hi+hi.Then(I
)is the×matrix having1’s on the main diagonal and0’s
everywhere else, except for the( )entry, which has the value. Similarly,(I
)has1’s all
along the main diagonal and0’s everywhere else, except for the( )entry, which has the value
. However, this exactly describes the matrix((I
))
.
(5) (a) IfAis nonsingular,(A
)
−1
=(A
−1
)
by part (4) of Theorem 2.12.
For the converse,A
−1
=((A
)
−1
)
.
(b)|AB|=|A||B|=|B||A|=|BA|.
(6) (a) Use the fact that|AB|=|A||B|.
(b)|AB|=|−BA|=⇒|AB|=(−1)
|BA|(by Corollary 3.4)=⇒|A||B|=(−1)|B||A|(sinceis
odd). Hence, either|A|=0or|B|=0.
(7) (a)|AA
|=|A||A
|=|A||A|=|A|
2
≥0.(b) |AB
|=|A||B
|=|A
||B|
(8) (a)A
=−A,so|A|=|A
|=|−A|=(−1)
|A|(by Corollary 3.4)=(−1)|A|(sinceis odd).
Hence|A|=0.
(b) ConsiderA=
∙
0−1
10
¸
.
(9) (a)I
=I=I
−1
.
(b) Consider
⎡
⎣
010
100
001
⎤
⎦.
(c)|A
|=|A
−1
|=⇒|A|=1|A|
=⇒|A|
2
=1 =⇒|A|=±1.
(10) Note that
B=
⎡
⎣
90−3
32−1
−60 1
⎤
⎦
has a determinant equal to−18. Hence, ifB=A
2
,then|A|
2
is negative, a contradiction.
(11) (a) Base Step:=2:|A
1A2|=|A 1||A2|by Theorem 3.7.
Inductive Step: Assume|A
1A2···A |=|A 1||A2|···|A |, for any×matricesA 1A2A
Prove|A
1A2···A A+1|=|A 1||A2|···|A ||A+1|, for any×matricesA 1A2A A+1
But,|A
1A2···A A+1|=|(A 1A2···A )A+1|=|A 1A2···A ||A+1|(by Theorem 3.7)=
|A
1||A2|···|A ||A+1|by the inductive hypothesis.
(b) Use part (a) withA
=A,for1≤≤Or, for an induction proof:
Base Step (=1): Obvious.
Inductive Step:|A
+1
|=|A
A|=|A
||A|=|A|
|A|=|A|
+1
(c) Use part (b):|A|
=|A
|=|O |=0,andso|A|=0. Or, for an induction proof:
Base Step (=1): Obvious.
Inductive Step: Assume.A
+1
=O .IfAis singular, then|A|=0, and we are done. IfAis
nonsingular,A
+1
=O =⇒A
+1
A
−1
=O A
−1
=⇒A
=O =⇒|A|=0by the inductive
hypothesis, which impliesAis singular, a contradiction.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 41
Answers to Exercises Section 3.3
(12) (a)|A
|=|A|
.If|A|=0,then|A|
=0is not prime. Similarly, if|A|=±1,then|A|
=±1and
is not prime. Otherwise, letting=|A|,wehave||1,and|A|
has||as a positive proper
divisor, for≥2.
(b)|A|
=|A
|=|I|=1.Since|A|is an integer,|A|=±1.Ifis odd, then|A|6 =−1,so|A|=1.
(13) (a) LetP
−1
AP=B. SupposePis an×matrix. ThenP
−1
is also an×matrix. Hence
P
−1
APis not defined unlessAis also an×matrix, and thus, the productB=P
−1
APis
also×.
(b) For example, consider
B=
∙
21
11
¸
−1
A
∙
21
11
¸
=
∙
−6−4
16 11
¸
and,
B=
∙
2−5
−13
¸
−1
A
∙
2−5
−13
¸
=
∙
10−12
4−5
¸
(c)A=I
AI=I
−1
AI
(d) IfP
−1
AP=B,thenPP
−1
APP
−1
=PB P
−1
=⇒A=PBP
−1
. Hence(P
−1
)
−1
BP
−1
=A.
(e)Asimilar toBimplies there is a matrixPsuch thatA=P
−1
BP.Bsimilar toCimplies
there is a matrixQsuch thatB=Q
−1
CQ. Hence,A=P
−1
BP=P
−1
(Q
−1
CQ)P=
(P
−1
Q
−1
)C(QP)=(QP)
−1
C(QP),andsoAis similar toC.
(f)Asimilar toI
implies there is a matrixPsuch thatA=P
−1
IP. Hence,A=P
−1
P=I .
(g)|B|=|P
−1
AP|=|P
−1
||A||P|=|P
−1
||P||A|=(1|P|)|P||A|=|A|.
(14) (a) LetbetheType(III)rowoperationhi↔h−1i,letAbe an×matrix and letB=(A).
Then, by part (3) of Theorem 3.3,|B|=(−1)|A|. Next, notice that the submatrixA
(−1) =B
because the(−1)st row ofAbecomes theth row ofB, implying the same row is being eliminated
from both matrices, and since all other rows maintain their original relative positions (notice the
same column is being eliminated in both cases). Hence,
(−1)1A
(−1)1+
(−1)2A
(−1)2+···+
(−1) A
(−1)
=1(−1)
(−1)+1
|A
(−1)1|+ 2(−1)
(−1)+2
|A
(−1)2|+···+ (−1)
(−1)+
|A
(−1) |
(because
(−1) =for1≤≤)
=
1(−1)
|B1|+ 2(−1)
+1
|B2|+···+ (−1)
+−1
|B|
=(−1)(
1(−1)
+1
|B1|+ 2(−1)
+2
|B2|+···+ (−1)
+
|B|)
=(−1)|B|(by applying part (1) of Theorem 3.11 along theth row ofB,
since we have assumed this result is valid for=)
=(−1)|(A)|
=(−1)(−1)|A|
=|A|,finishing the proof.
(b) The definition of the determinant is the Base Step for an induction proof on the row number,
counting down fromto1. Part (a) is the Inductive Step.
(15) (a) Case 1: Assume1≤and1≤. Then the( )entry of(A
)
=( )entry ofA =( )entry ofA=( )entry ofA
=( )entry of(A
).
Case 2: Assume≤and1≤. Then the( )entry of(A
)
=( )entry ofA =(+1)entry ofA=( +1)entry ofA
=( )entry of(A
).
Case 3: Assume1≤and≤. Then the( )entry of(A
)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 42
(12) (a)|A
|=|A|
.If|A|=0,then|A|
=0is not prime. Similarly, if|A|=±1,then|A|
=±1and
is not prime. Otherwise, letting=|A|,wehave||1,and|A|
has||as a positive proper
divisor, for≥2.
(b)|A|
=|A
|=|I|=1.Since|A|is an integer,|A|=±1.Ifis odd, then|A|6 =−1,so|A|=1.
(13) (a) LetP
−1
AP=B. SupposePis an×matrix. ThenP
−1
is also an×matrix. Hence
P
−1
APis not defined unlessAis also an×matrix, and thus, the productB=P
−1
APis
also×.
(b) For example, consider
B=
∙
21
11
¸
−1
A
∙
21
11
¸
=
∙
−6−4
16 11
¸
and,
B=
∙
2−5
−13
¸
−1
A
∙
2−5
−13
¸
=
∙
10−12
4−5
¸
(c)A=I
AI=I
−1
AI
(d) IfP
−1
AP=B,thenPP
−1
APP
−1
=PB P
−1
=⇒A=PBP
−1
. Hence(P
−1
)
−1
BP
−1
=A.
(e)Asimilar toBimplies there is a matrixPsuch thatA=P
−1
BP.Bsimilar toCimplies
there is a matrixQsuch thatB=Q
−1
CQ. Hence,A=P
−1
BP=P
−1
(Q
−1
CQ)P=
(P
−1
Q
−1
)C(QP)=(QP)
−1
C(QP),andsoAis similar toC.
(f)Asimilar toI
implies there is a matrixPsuch thatA=P
−1
IP. Hence,A=P
−1
P=I .
(g)|B|=|P
−1
AP|=|P
−1
||A||P|=|P
−1
||P||A|=(1|P|)|P||A|=|A|.
(14) (a) LetbetheType(III)rowoperationhi↔h−1i,letAbe an×matrix and letB=(A).
Then, by part (3) of Theorem 3.3,|B|=(−1)|A|. Next, notice that the submatrixA
(−1) =B
because the(−1)st row ofAbecomes theth row ofB, implying the same row is being eliminated
from both matrices, and since all other rows maintain their original relative positions (notice the
same column is being eliminated in both cases). Hence,
(−1)1A
(−1)1+
(−1)2A
(−1)2+···+
(−1) A
(−1)
=1(−1)
(−1)+1
|A
(−1)1|+ 2(−1)
(−1)+2
|A
(−1)2|+···+ (−1)
(−1)+
|A
(−1) |
(because
(−1) =for1≤≤)
=
1(−1)
|B1|+ 2(−1)
+1
|B2|+···+ (−1)
+−1
|B|
=(−1)(
1(−1)
+1
|B1|+ 2(−1)
+2
|B2|+···+ (−1)
+
|B|)
=(−1)|B|(by applying part (1) of Theorem 3.11 along theth row ofB,
since we have assumed this result is valid for=)
=(−1)|(A)|
=(−1)(−1)|A|
=|A|,finishing the proof.
(b) The definition of the determinant is the Base Step for an induction proof on the row number,
counting down fromto1. Part (a) is the Inductive Step.
(15) (a) Case 1: Assume1≤and1≤. Then the( )entry of(A
)
=( )entry ofA =( )entry ofA=( )entry ofA
=( )entry of(A
).
Case 2: Assume≤and1≤. Then the( )entry of(A
)
=( )entry ofA =(+1)entry ofA=( +1)entry ofA
=( )entry of(A
).
Case 3: Assume1≤and≤. Then the( )entry of(A
)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 42
Answers to Exercises Section 3.3
=( )entry ofA =( +1)entry ofA=(+1)entry ofA
=( )entry of(A
).
Case 4: Assume≤and≤.Thenthe( )entry of(A
)
=( )entry ofA =(+1+1)entry ofA=(+1+1)entry ofA
=( )entry of(A
).
Hence, the corresponding entries of(A
)
and(A
)are all equal, proving that the matrices
themselves are equal.
(b)
1A1+2A2+···+ A=1(−1)
1+
|A1|+ 2(−1)
2+
|A2|+···+ (−1)
+
|A|
=
1(−1)
1+
¯
¯
¯(A
1)
¯
¯
¯+
2(−1)
2+
¯
¯
¯(A
2)
¯
¯
¯+···+
(−1)
+
¯
¯
¯(A
)
¯
¯
¯(by Theorem 3.10)
=
1(−1)
+1
¯
¯
¯
¡
A
¢
1
¯
¯
¯+
2(−1)
+2
¯
¯
¯
¡
A
¢
2
¯
¯
¯+···+
(−1)
+
¯
¯
¯
¡
A
¢
¯
¯
¯(by part (a))
=
¯
¯
A
¯
¯
(using part (1) of Theorem 3.11 along theth row ofA
)=|A|, (by Theorem 3.10).
(16) LetA, andBbe as given in the exercise and its hint. Then by Exercise 12 in Section 3.2,|B|=0,
since itsth andth rows are the same. Also, since every row ofAequals the corresponding row ofB,
with the exception of theth row, the submatricesA
andB are equal for1≤≤. Hence,A =
B
for1≤≤. Now, computing the determinant ofBusing a cofactor expansion along theth row
(part (1) of Theorem 3.11) yields0=|B|=
1B1+2B2+···+ B=1B1+2B2+···+ B
(because theth row ofBequals theth row ofA)= 1A1+2A2+···+ A, completing the
proof.
(17) For6 =,the( )entry ofAB
is1A1+2A2+···+ A.Thisequals0by Exercise 16. The
( )entry ofAB
is1A1+2A2+···+ A,whichequals|A|, by part (1) of Theorem 3.11.
Hence,AB
=(|A|)I .
(18) Assume in what follows thatAis an×matrix. LetA
be theth matrix (as defined in Theorem
3.12) for the matrixA.
(a) LetC=([A|B])We must show that for each,1≤≤the matrixC
(as defined in Theorem
3.12) forCis identical to(A
)
First we consider the columns ofC
other than theth column. If1≤≤with6 =,
then theth column ofC
is the same as theth column ofC=([A|B]). But since theth
column of[A|B]is identical to theth column ofA
, it follows that theth column of([A|B])
is identical to theth column of(A
). Therefore, for1≤≤with6 =,theth column of
C
isthesameastheth column of(A ).
Finally, we consider theth column ofC
.Now,theth column ofC is identical to the last
column of([A|B]) = [(A)|(B)],whichis(B)But since theth column ofA
equalsB,it
follows that theth column of(A
)=(B).Thus,theth column ofC is identical to theth
column of(A
).
Therefore, sinceC
and(A )agree in every column, we haveC =(A )
(b) Consider each type of operation in turn. First, ifistheType(I)operationhi←hifor some
6 =0, then by part (1) of Theorem 3.3,
|(A
)|
|(A)|
=
|A
|
|A|
=
|A
|
|A|
Ifis any Type (II) operation, then by part (2) of Theorem 3.3,
|(A
)|
|(A)|
=
|A
|
|A|
Copyrightc°2016 Elsevier Ltd. All rights reserved. 43
=( )entry ofA =( +1)entry ofA=(+1)entry ofA
=( )entry of(A
).
Case 4: Assume≤and≤.Thenthe( )entry of(A
)
=( )entry ofA =(+1+1)entry ofA=(+1+1)entry ofA
=( )entry of(A
).
Hence, the corresponding entries of(A
)
and(A
)are all equal, proving that the matrices
themselves are equal.
(b)
1A1+2A2+···+ A=1(−1)
1+
|A1|+ 2(−1)
2+
|A2|+···+ (−1)
+
|A|
=
1(−1)
1+
¯
¯
¯(A
1)
¯
¯
¯+
2(−1)
2+
¯
¯
¯(A
2)
¯
¯
¯+···+
(−1)
+
¯
¯
¯(A
)
¯
¯
¯(by Theorem 3.10)
=
1(−1)
+1
¯
¯
¯
¡
A
¢
1
¯
¯
¯+
2(−1)
+2
¯
¯
¯
¡
A
¢
2
¯
¯
¯+···+
(−1)
+
¯
¯
¯
¡
A
¢
¯
¯
¯(by part (a))
=
¯
¯
A
¯
¯
(using part (1) of Theorem 3.11 along theth row ofA
)=|A|, (by Theorem 3.10).
(16) LetA, andBbe as given in the exercise and its hint. Then by Exercise 12 in Section 3.2,|B|=0,
since itsth andth rows are the same. Also, since every row ofAequals the corresponding row ofB,
with the exception of theth row, the submatricesA
andB are equal for1≤≤. Hence,A =
B
for1≤≤. Now, computing the determinant ofBusing a cofactor expansion along theth row
(part (1) of Theorem 3.11) yields0=|B|=
1B1+2B2+···+ B=1B1+2B2+···+ B
(because theth row ofBequals theth row ofA)= 1A1+2A2+···+ A, completing the
proof.
(17) For6 =,the( )entry ofAB
is1A1+2A2+···+ A.Thisequals0by Exercise 16. The
( )entry ofAB
is1A1+2A2+···+ A,whichequals|A|, by part (1) of Theorem 3.11.
Hence,AB
=(|A|)I .
(18) Assume in what follows thatAis an×matrix. LetA
be theth matrix (as defined in Theorem
3.12) for the matrixA.
(a) LetC=([A|B])We must show that for each,1≤≤the matrixC
(as defined in Theorem
3.12) forCis identical to(A
)
First we consider the columns ofC
other than theth column. If1≤≤with6 =,
then theth column ofC
is the same as theth column ofC=([A|B]). But since theth
column of[A|B]is identical to theth column ofA
, it follows that theth column of([A|B])
is identical to theth column of(A
). Therefore, for1≤≤with6 =,theth column of
C
isthesameastheth column of(A ).
Finally, we consider theth column ofC
.Now,theth column ofC is identical to the last
column of([A|B]) = [(A)|(B)],whichis(B)But since theth column ofA
equalsB,it
follows that theth column of(A
)=(B).Thus,theth column ofC is identical to theth
column of(A
).
Therefore, sinceC
and(A )agree in every column, we haveC =(A )
(b) Consider each type of operation in turn. First, ifistheType(I)operationhi←hifor some
6 =0, then by part (1) of Theorem 3.3,
|(A
)|
|(A)|
=
|A
|
|A|
=
|A
|
|A|
Ifis any Type (II) operation, then by part (2) of Theorem 3.3,
|(A
)|
|(A)|
=
|A
|
|A|
Copyrightc°2016 Elsevier Ltd. All rights reserved. 43
Answers to Exercises Section 3.3
Ifis any Type (III) operation, then
|(A
)|
|(A)|
=
(−1)|A
|
(−1)|A|
=
|A
|
|A|
(c) In the special case when=1,wehaveA=[1].LetB=[]. Hence,A
1(as definedinTheorem
3.12)=B=[]Then the formula in Theorem 3.12 gives
1=
|A
1|
|A|
=
1
=
which is the correct solution to the equationAX=Bin this case. For the remainder of the proof,
we assume1.
IfA=I
, then the solution to the system isX=B. Therefore we must show that, for each
,1≤≤, the formula given in Theorem 3.12 yields
=(theth entry ofB). First, note
that|A|=|I
|=1.Now,theth matrixA (as defined in Theorem 3.12) forAis identical to
I
except in itsth column, which equalsB. Therefore, theth row ofA has as itsth entry,
and zeroes elsewhere. Thus, a cofactor expansion along theth row ofA
yields
|A
|= (−1)
+
|I−1|=
Hence, for each, the formula in Theorem 3.12 produces
=
|A
|
|A|
=
1
=
completing the proof.
(d) BecauseAis nonsingular,[A|B]row reduces to[I
|X],whereXis the unique solution to the sys-
tem. Let[C|D]represent any intermediate augmented matrix during this row reduction process.
Now, by part (a) and repeated use of part (b), the ratio
|C
|
|C|
is identical to the ratio
|A
|
|A|
obtained from the original augmented matrix, for each,1≤≤. But part (c) proves that for
thefinal augmented matrix,[I
|X], this common ratio gives the correct solution for ,foreach
,1≤≤. Since all of the systems corresponding to these intermediate matrices have the same
unique solution, the formula in Theorem 3.12 gives the correct solution for the original system,
[A|B],aswell.Thus,Cramer’sRuleisvalidated.
(19) SupposeAis an×matrix with|A|=0. Then, by Theorem 3.10,
¯
¯
A
¯
¯
=0. Thus, by Exercise 16
in Section 3.2, there is an×matrixCsuch thatA
C=O . Taking the transpose of both sides
of this equation yieldsC
A=O
=O .LettingB=C
completes the proof.
(20) (a) T (b) T (c) F (d) T (e) T (f) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 44
Ifis any Type (III) operation, then
|(A
)|
|(A)|
=
(−1)|A
|
(−1)|A|
=
|A
|
|A|
(c) In the special case when=1,wehaveA=[1].LetB=[]. Hence,A
1(as definedinTheorem
3.12)=B=[]Then the formula in Theorem 3.12 gives
1=
|A
1|
|A|
=
1
=
which is the correct solution to the equationAX=Bin this case. For the remainder of the proof,
we assume1.
IfA=I
, then the solution to the system isX=B. Therefore we must show that, for each
,1≤≤, the formula given in Theorem 3.12 yields
=(theth entry ofB). First, note
that|A|=|I
|=1.Now,theth matrixA (as defined in Theorem 3.12) forAis identical to
I
except in itsth column, which equalsB. Therefore, theth row ofA has as itsth entry,
and zeroes elsewhere. Thus, a cofactor expansion along theth row ofA
yields
|A
|= (−1)
+
|I−1|=
Hence, for each, the formula in Theorem 3.12 produces
=
|A
|
|A|
=
1
=
completing the proof.
(d) BecauseAis nonsingular,[A|B]row reduces to[I
|X],whereXis the unique solution to the sys-
tem. Let[C|D]represent any intermediate augmented matrix during this row reduction process.
Now, by part (a) and repeated use of part (b), the ratio
|C
|
|C|
is identical to the ratio
|A
|
|A|
obtained from the original augmented matrix, for each,1≤≤. But part (c) proves that for
thefinal augmented matrix,[I
|X], this common ratio gives the correct solution for ,foreach
,1≤≤. Since all of the systems corresponding to these intermediate matrices have the same
unique solution, the formula in Theorem 3.12 gives the correct solution for the original system,
[A|B],aswell.Thus,Cramer’sRuleisvalidated.
(19) SupposeAis an×matrix with|A|=0. Then, by Theorem 3.10,
¯
¯
A
¯
¯
=0. Thus, by Exercise 16
in Section 3.2, there is an×matrixCsuch thatA
C=O . Taking the transpose of both sides
of this equation yieldsC
A=O
=O .LettingB=C
completes the proof.
(20) (a) T (b) T (c) F (d) T (e) T (f) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 44
Answers to Exercises Section 3.4
Section 3.4
(1) (a)
2
−7+14
(b)
3
−6
2
+3+10
(c)
3
−8
2
+21−18
(d)
3
−8
2
+7−5
(e)
4
−3
3
−4
2
+12
(2) (a)
2={[11]} (b) 2={[−110]} (c) −1={[120] +[001]}
(3) In the answers for this exercise,,,andrepresent arbitrary scalars.
(a)=1,
1={[10]}, algebraic multiplicity of=2
(b)
1=2, 2={[10]}, algebraic multiplicity of 1=1;
2=3, 3={[1−1]}, algebraic multiplicity of 2=1
(c)
1=1, 1={[100]}, algebraic multiplicity of 1=1;
2=2, 2={[010]}, algebraic multiplicity of 2=1;
3=−5, −5={[−
1
6
3
7
1]}={[−71842]}, algebraic multiplicity of 3=1
(d)
1=1, 1={[31]}, algebraic multiplicity of 1=1;
2=−1, −1={[73]}, algebraic multiplicity of 2=1
(e)
1=0, 0={[132]}, algebraic multiplicity of 1=1;
2=2, 2={[010] +[101]}, algebraic multiplicity of 2=2
(f)
1=13, 13={[413]}, algebraic multiplicity of 1=1;
2=−13, −13={[1−40] +[30−4]}, algebraic multiplicity of 2=2
(g)
1=1, 1={[2010]}, algebraic multiplicity of 1=1;
2=−1, −1={[02−11]}, algebraic multiplicity of 2=1
(h)
1=0, 0={[−1110] +[0−101]}, algebraic multiplicity of 1=2;
2=−3, −3={[−1022]}, algebraic multiplicity of 2=2
(4) (a)P=
∙
32
11
¸
D=
∙
30
0−5
¸
(b)P=
∙
25
12
¸
,D=
∙
20
0−2
¸
(c) Not diagonalizable
(d)P=
⎡
⎣
611
221
511
⎤
⎦,D=
⎡
⎣
100
0−10
002
⎤
⎦
(e) Not diagonalizable
(f) Not diagonalizable
(g)P=
⎡
⎣
21 0
30−1
03 1
⎤
⎦,D=
⎡
⎣
200
020
003
⎤
⎦
(h)P=
⎡
⎣
631
210
−101
⎤
⎦,D=
⎡
⎣
000
010
001
⎤
⎦
(i)
P=
⎡
⎢
⎢
⎣
211 1
202 −1
101 0
010 1
⎤
⎥
⎥
⎦
,
D=
⎡
⎢
⎢
⎣
10 00
01 00
00−10
00 00
⎤
⎥
⎥
⎦
(5) (a)
∙
32770−65538
32769−65537
¸
(b)
⎡
⎣
−17624
−15620
−9313
⎤
⎦
(c)A
49
=A
Copyrightc°2016 Elsevier Ltd. All rights reserved. 45
Section 3.4
(1) (a)
2
−7+14
(b)
3
−6
2
+3+10
(c)
3
−8
2
+21−18
(d)
3
−8
2
+7−5
(e)
4
−3
3
−4
2
+12
(2) (a)
2={[11]} (b) 2={[−110]} (c) −1={[120] +[001]}
(3) In the answers for this exercise,,,andrepresent arbitrary scalars.
(a)=1,
1={[10]}, algebraic multiplicity of=2
(b)
1=2, 2={[10]}, algebraic multiplicity of 1=1;
2=3, 3={[1−1]}, algebraic multiplicity of 2=1
(c)
1=1, 1={[100]}, algebraic multiplicity of 1=1;
2=2, 2={[010]}, algebraic multiplicity of 2=1;
3=−5, −5={[−
1
6
3
7
1]}={[−71842]}, algebraic multiplicity of 3=1
(d)
1=1, 1={[31]}, algebraic multiplicity of 1=1;
2=−1, −1={[73]}, algebraic multiplicity of 2=1
(e)
1=0, 0={[132]}, algebraic multiplicity of 1=1;
2=2, 2={[010] +[101]}, algebraic multiplicity of 2=2
(f)
1=13, 13={[413]}, algebraic multiplicity of 1=1;
2=−13, −13={[1−40] +[30−4]}, algebraic multiplicity of 2=2
(g)
1=1, 1={[2010]}, algebraic multiplicity of 1=1;
2=−1, −1={[02−11]}, algebraic multiplicity of 2=1
(h)
1=0, 0={[−1110] +[0−101]}, algebraic multiplicity of 1=2;
2=−3, −3={[−1022]}, algebraic multiplicity of 2=2
(4) (a)P=
∙
32
11
¸
D=
∙
30
0−5
¸
(b)P=
∙
25
12
¸
,D=
∙
20
0−2
¸
(c) Not diagonalizable
(d)P=
⎡
⎣
611
221
511
⎤
⎦,D=
⎡
⎣
100
0−10
002
⎤
⎦
(e) Not diagonalizable
(f) Not diagonalizable
(g)P=
⎡
⎣
21 0
30−1
03 1
⎤
⎦,D=
⎡
⎣
200
020
003
⎤
⎦
(h)P=
⎡
⎣
631
210
−101
⎤
⎦,D=
⎡
⎣
000
010
001
⎤
⎦
(i)
P=
⎡
⎢
⎢
⎣
211 1
202 −1
101 0
010 1
⎤
⎥
⎥
⎦
,
D=
⎡
⎢
⎢
⎣
10 00
01 00
00−10
00 00
⎤
⎥
⎥
⎦
(5) (a)
∙
32770−65538
32769−65537
¸
(b)
⎡
⎣
−17624
−15620
−9313
⎤
⎦
(c)A
49
=A
Copyrightc°2016 Elsevier Ltd. All rights reserved. 45
Answers to Exercises Section 3.4
(d)
⎡
⎣
352246−4096 354294
−175099 2048 −175099
−175099 4096 −177147
⎤
⎦ (e)
⎡
⎣
4188163 6282243−9421830
4192254 6288382−9432060
4190208 6285312−9426944
⎤
⎦
(6)Asimilar toBimplies there is a nonsingular matrixPsuch thatA=P
−1
BP. Hence,
A()= |I −A|=|I −P
−1
BP|=|P
−1
IP−P
−1
BP|
=|P
−1
(I−B)P|=|P
−1
||(I −B)||P|=
1
|P|
|(I
−B)||P|
=
1
|P|
|P||(I
−B)|=|(I −B)|= B()
(7) (a) SupposeA=PDP
−1
for some diagonal matrixD.LetCbe the diagonal matrix with =( )
1
3.
ThenC
3
=D. Clearly then,(PCP
−1
)
3
=A.
(b) IfAhas all eigenvalues nonnegative, thenAhas a square root. Proceed as in part (a), only taking
square roots instead of cube roots.
(8) LetB=
⎡
⎣
15−14−14
−13 16 17
20−22−23
⎤
⎦.Wefind a matrixPand a diagonal matrixDsuch thatB=PDP
−1
.
IfCis the diagonal matrix with
=
3
√
,thenC
3
=D.So,ifA=PCP
−1
,thenA
3
=(PCP
−1
)
3
=
PC
3
P
−1
=PDP
−1
=B. Hence, to solve this problem, wefirst need tofindPandD. Todothis,we
diagonalizeB.
Step 1:
B()=|I 3−B|=
3
−8
2
−+8=(−1)(+1)(−8).
Step 2: The three eigenvalues ofBare
1=1, 2=−1and 3=8.
Step 3: Now we solve for fundamental eigenvectors.
Eigenvalue
1=1:[(1I 3−B)|0]reduces to
⎡
⎣
101
012
000
¯
¯
¯
¯
¯
¯
0
0
0
⎤
⎦. This yields the fundamental eigenvector
[−1−21].
Eigenvalue
2=−1:[((−1)I 3−B)|0]reduces to
⎡
⎣
100
011
000
¯
¯
¯
¯
¯
¯
0
0
0
⎤
⎦. This produces the fundamental
eigenvector[0−11].
Eigenvalue
3=8:[(8I 3−B)|0]reduces to
⎡
⎢
⎣
10−1
01
1
2
00 0
¯
¯
¯
¯
¯
¯
¯
0
0
0
⎤
⎥
⎦. This yields the fundamental eigen-
vector[2−12].
Step 4: Since=3, and we have found3fundamental eigenvectors,Bis diagonalizable.
Step 5:P=
⎡
⎣
−102
−2−1−1
112
⎤
⎦.
Step 6:D=
⎡
⎣
100
0−10
008
⎤
⎦.Also,P
−1
=
⎡
⎣
1−2−2
−345
1−1−1
⎤
⎦.
The diagonal matrixCwhose main diagonal entries are the cube roots of the eigenvalues ofB.Thatis,
Copyrightc°2016 Elsevier Ltd. All rights reserved. 46
(d)
⎡
⎣
352246−4096 354294
−175099 2048 −175099
−175099 4096 −177147
⎤
⎦ (e)
⎡
⎣
4188163 6282243−9421830
4192254 6288382−9432060
4190208 6285312−9426944
⎤
⎦
(6)Asimilar toBimplies there is a nonsingular matrixPsuch thatA=P
−1
BP. Hence,
A()= |I −A|=|I −P
−1
BP|=|P
−1
IP−P
−1
BP|
=|P
−1
(I−B)P|=|P
−1
||(I −B)||P|=
1
|P|
|(I
−B)||P|
=
1
|P|
|P||(I
−B)|=|(I −B)|= B()
(7) (a) SupposeA=PDP
−1
for some diagonal matrixD.LetCbe the diagonal matrix with =( )
1
3.
ThenC
3
=D. Clearly then,(PCP
−1
)
3
=A.
(b) IfAhas all eigenvalues nonnegative, thenAhas a square root. Proceed as in part (a), only taking
square roots instead of cube roots.
(8) LetB=
⎡
⎣
15−14−14
−13 16 17
20−22−23
⎤
⎦.Wefind a matrixPand a diagonal matrixDsuch thatB=PDP
−1
.
IfCis the diagonal matrix with
=
3
√
,thenC
3
=D.So,ifA=PCP
−1
,thenA
3
=(PCP
−1
)
3
=
PC
3
P
−1
=PDP
−1
=B. Hence, to solve this problem, wefirst need tofindPandD. Todothis,we
diagonalizeB.
Step 1:
B()=|I 3−B|=
3
−8
2
−+8=(−1)(+1)(−8).
Step 2: The three eigenvalues ofBare
1=1, 2=−1and 3=8.
Step 3: Now we solve for fundamental eigenvectors.
Eigenvalue
1=1:[(1I 3−B)|0]reduces to
⎡
⎣
101
012
000
¯
¯
¯
¯
¯
¯
0
0
0
⎤
⎦. This yields the fundamental eigenvector
[−1−21].
Eigenvalue
2=−1:[((−1)I 3−B)|0]reduces to
⎡
⎣
100
011
000
¯
¯
¯
¯
¯
¯
0
0
0
⎤
⎦. This produces the fundamental
eigenvector[0−11].
Eigenvalue
3=8:[(8I 3−B)|0]reduces to
⎡
⎢
⎣
10−1
01
1
2
00 0
¯
¯
¯
¯
¯
¯
¯
0
0
0
⎤
⎥
⎦. This yields the fundamental eigen-
vector[2−12].
Step 4: Since=3, and we have found3fundamental eigenvectors,Bis diagonalizable.
Step 5:P=
⎡
⎣
−102
−2−1−1
112
⎤
⎦.
Step 6:D=
⎡
⎣
100
0−10
008
⎤
⎦.Also,P
−1
=
⎡
⎣
1−2−2
−345
1−1−1
⎤
⎦.
The diagonal matrixCwhose main diagonal entries are the cube roots of the eigenvalues ofB.Thatis,
Copyrightc°2016 Elsevier Ltd. All rights reserved. 46
Answers to Exercises Section 3.4
C=
⎡
⎣
100
0−10
002
⎤
⎦. Finally,A=PCP
−1
=
⎡
⎣
3−2−2
−71011
8−10−11
⎤
⎦Direct calculation ofA
3
verifies
thatA
3
=B.
(9) The characteristic polynomial of the matrix is
2
−(+)+(−), whose discriminant simplifies
to(−)
2
+4.
(10) (a) Letvbe an eigenvector forAcorresponding to.ThenAv=v
A
2
v=A(Av)=A(v)=(Av)=(v)=
2
v
and an analogous induction argument showsA
v=
v, for any integer≥1.
(b) Consider the matrixA=
∙
0−1
10
¸
. AlthoughAhas no eigenvalues,A
4
=I2has1as an
eigenvalue.
(11)(−)
|A
−1
|A(
1
)=(−)
|A
−1
||
1
I−A|=(−)
|A
−1
(
1
I−A)|
=(−)
|
1
A
−1
−I|=|(−)(
1
A
−1
−I)|=|I −A
−1
|=
A
−1().
(12) Both parts are true, becauseI
−Ais also upper triangular, and so
A()=|I −A|=(− 11)(− 22)···(− ), by Theorem 3.2.
(13)
A
()=|I −A
|=|I
−A
|=|(I −A)
|=|I −A|= A()
(14)A
X=Xsince the entries of each row ofA
sum to1. Hence=1is an eigenvalue forA
,and
hence forA, by Exercise 13.
(15) Base Step:=1. Use the argument in the text directly after the definition of similarity.
Inductive Step: Assume that ifP
−1
AP=D,thenforsome0,A
=PD
P
−1
.Wemustprove
that ifP
−1
AP=DthenA
+1
=PD
+1
P
−1
ButA
+1
=AA
=(PDP
−1
)(PD
P
−1
)(by the
inductive hypothesis)=PD
+1
P
−1
(16) SinceAis upper triangular, so isI
−A.Thus,byTheorem3.2,|I −A|equals the product of the
main diagonal entries ofI
−A,whichis(− 11)(− 22)···(− ). Hence the main diagonal
entries ofAare the eigenvalues ofA.Thus,Ahasdistinct eigenvalues. Then, by the Diagonalization
Method given in Section 3.4, the necessary matrixPcan be constructed (since only one fundamental
eigenvector for each eigenvalue is needed in this case), and soAis diagonalizable.
(17) AssumeAis an×matrix. ThenAis singular⇐⇒|A|=0⇐⇒(−1)
|A|=0⇐⇒|−A|=0⇐⇒
|0I
−A|=0⇐⇒ A(0) = 0⇐⇒=0is an eigenvalue forA.
(18) LetD=P
−1
AP.ThenD=D
=(P
−1
AP)
=P
A
(P
−1
)
=((P
)
−1
)
−1
A
(P
)
−1
,soA
is
diagonalizable.
(19)D=P
−1
AP=⇒D
−1
=(P
−1
AP)
−1
=P
−1
A
−1
P.ButD
−1
is a diagonal matrix whose main
diagonal entries are the reciprocals of the main diagonal entries ofD(Since the main diagonal entries
ofDare the nonzero eigenvalues ofA, their reciprocals exist.) Thus,A
−1
is diagonalizable.
(20) (a) Theth column ofAP=A(th column ofP)=AP
.SinceP is an eigenvector for we have
AP
=P,foreach.
(b)P
−1
P=(P
−1
(th column ofP)) = (th column ofI )= e
Copyrightc°2016 Elsevier Ltd. All rights reserved. 47
C=
⎡
⎣
100
0−10
002
⎤
⎦. Finally,A=PCP
−1
=
⎡
⎣
3−2−2
−71011
8−10−11
⎤
⎦Direct calculation ofA
3
verifies
thatA
3
=B.
(9) The characteristic polynomial of the matrix is
2
−(+)+(−), whose discriminant simplifies
to(−)
2
+4.
(10) (a) Letvbe an eigenvector forAcorresponding to.ThenAv=v
A
2
v=A(Av)=A(v)=(Av)=(v)=
2
v
and an analogous induction argument showsA
v=
v, for any integer≥1.
(b) Consider the matrixA=
∙
0−1
10
¸
. AlthoughAhas no eigenvalues,A
4
=I2has1as an
eigenvalue.
(11)(−)
|A
−1
|A(
1
)=(−)
|A
−1
||
1
I−A|=(−)
|A
−1
(
1
I−A)|
=(−)
|
1
A
−1
−I|=|(−)(
1
A
−1
−I)|=|I −A
−1
|=
A
−1().
(12) Both parts are true, becauseI
−Ais also upper triangular, and so
A()=|I −A|=(− 11)(− 22)···(− ), by Theorem 3.2.
(13)
A
()=|I −A
|=|I
−A
|=|(I −A)
|=|I −A|= A()
(14)A
X=Xsince the entries of each row ofA
sum to1. Hence=1is an eigenvalue forA
,and
hence forA, by Exercise 13.
(15) Base Step:=1. Use the argument in the text directly after the definition of similarity.
Inductive Step: Assume that ifP
−1
AP=D,thenforsome0,A
=PD
P
−1
.Wemustprove
that ifP
−1
AP=DthenA
+1
=PD
+1
P
−1
ButA
+1
=AA
=(PDP
−1
)(PD
P
−1
)(by the
inductive hypothesis)=PD
+1
P
−1
(16) SinceAis upper triangular, so isI
−A.Thus,byTheorem3.2,|I −A|equals the product of the
main diagonal entries ofI
−A,whichis(− 11)(− 22)···(− ). Hence the main diagonal
entries ofAare the eigenvalues ofA.Thus,Ahasdistinct eigenvalues. Then, by the Diagonalization
Method given in Section 3.4, the necessary matrixPcan be constructed (since only one fundamental
eigenvector for each eigenvalue is needed in this case), and soAis diagonalizable.
(17) AssumeAis an×matrix. ThenAis singular⇐⇒|A|=0⇐⇒(−1)
|A|=0⇐⇒|−A|=0⇐⇒
|0I
−A|=0⇐⇒ A(0) = 0⇐⇒=0is an eigenvalue forA.
(18) LetD=P
−1
AP.ThenD=D
=(P
−1
AP)
=P
A
(P
−1
)
=((P
)
−1
)
−1
A
(P
)
−1
,soA
is
diagonalizable.
(19)D=P
−1
AP=⇒D
−1
=(P
−1
AP)
−1
=P
−1
A
−1
P.ButD
−1
is a diagonal matrix whose main
diagonal entries are the reciprocals of the main diagonal entries ofD(Since the main diagonal entries
ofDare the nonzero eigenvalues ofA, their reciprocals exist.) Thus,A
−1
is diagonalizable.
(20) (a) Theth column ofAP=A(th column ofP)=AP
.SinceP is an eigenvector for we have
AP
=P,foreach.
(b)P
−1
P=(P
−1
(th column ofP)) = (th column ofI )= e
Copyrightc°2016 Elsevier Ltd. All rights reserved. 47
Answers to Exercises Section 3.4
(c) Theth column ofP
−1
AP=P
−1
(th column ofAP)=P
−1
(P)(by part (a))=( e)(by
part (b)). Hence,P
−1
APis a diagonal matrix with main diagonal entries 12
(21) SincePD=AP,theth column ofPDequals theth column ofAP.But,
th column ofPD
=P(th column ofD)
=P(
e)= Pe=(P(th column ofI ))
=
(th column ofPI )
=
(th column ofP)= P
Also, theth column ofAP=A(th column ofP)=AP
Thus, is an eigenvalue for theth column
ofP.
(22) Use induction on.
Base Step(=1):|C|=
11+ 11, which has degree1if=1(implying 116 =0), and degree0if
=0.
Inductive Step: Assume true for(−1)×(−1)matrices and prove for×matrices. Using a
cofactor expansion along thefirst row yields
|C|=(
11+ 11)(−1)
2
|C11|+( 12+ 12)(−1)
3
|C12|+···+( 1+ 1)(−1)
+1
|C1|
Case 1: Suppose thefirst row ofAis all zeroes (so each
1=0). Then, each of the submatricesA 1has
at mostnonzero rows. Hence, by the inductive hypothesis, each|C
1|=|A 1+B 1|is a polynomial
of degree at most. Therefore,
|C|=
11(−1)
2
|C11|+ 12(−1)
3
|C12|+···+ 1(−1)
+1
|C1|
is a sum of constants multiplied by such polynomials, and hence is a polynomial of degree at most.
Case 2: Suppose some
16 =0. Then, at most−1of the rows from2throughofAhave a
nonzero entry. Hence, each submatrixA
1has at most−1nonzero rows, and, by the inductive
hypothesis, each of|C
1|=|A 1+B 1|is a polynomial of degree at most−1. Therefore, each term
(
1+ 1)(−1)
+1
|C1|is a polynomial of degree at most. The desired result easily follows.
(23) (a)|I
2−A|=
¯
¯
¯
¯
−
11−12
−21− 22
¯
¯
¯
¯
=(−
11)(− 22)− 1221
=
2
−( 11+22)+( 1122−1221)=
2
−(trace(A))+|A|
(b) Use induction on.
Base Step(=1):Now
A()=− 11,whichisafirst degree polynomial with the coefficient
of itsterm equal to1.
Inductive Step: Assume that the statement is true for all(−1)×(−1)matrices, and prove it
is true for an×matrixA.Consider
C=I
−A=
⎡
⎢
⎢
⎢
⎣
(−
11)− 12 ···− 1
−21 (− 22)···− 2
.
.
.
.
.
.
.
.
.
.
.
.
−
1 −2 ···(− )
⎤
⎥
⎥
⎥
⎦
.
Using a cofactor expansion along thefirst row,
A()=|C|=
(−
11)(−1)
1+1
|C11|+(− 12)(−1)
1+2
|C12|+(− 13)(−1)
1+3
|C13|+···+(− 1)(−1)
1+
|C1|.
Now|C
11|= A11(),andso,bytheinductivehypothesis,isan−1degree polynomial with
Copyrightc°2016 Elsevier Ltd. All rights reserved. 48
(c) Theth column ofP
−1
AP=P
−1
(th column ofAP)=P
−1
(P)(by part (a))=( e)(by
part (b)). Hence,P
−1
APis a diagonal matrix with main diagonal entries 12
(21) SincePD=AP,theth column ofPDequals theth column ofAP.But,
th column ofPD
=P(th column ofD)
=P(
e)= Pe=(P(th column ofI ))
=
(th column ofPI )
=
(th column ofP)= P
Also, theth column ofAP=A(th column ofP)=AP
Thus, is an eigenvalue for theth column
ofP.
(22) Use induction on.
Base Step(=1):|C|=
11+ 11, which has degree1if=1(implying 116 =0), and degree0if
=0.
Inductive Step: Assume true for(−1)×(−1)matrices and prove for×matrices. Using a
cofactor expansion along thefirst row yields
|C|=(
11+ 11)(−1)
2
|C11|+( 12+ 12)(−1)
3
|C12|+···+( 1+ 1)(−1)
+1
|C1|
Case 1: Suppose thefirst row ofAis all zeroes (so each
1=0). Then, each of the submatricesA 1has
at mostnonzero rows. Hence, by the inductive hypothesis, each|C
1|=|A 1+B 1|is a polynomial
of degree at most. Therefore,
|C|=
11(−1)
2
|C11|+ 12(−1)
3
|C12|+···+ 1(−1)
+1
|C1|
is a sum of constants multiplied by such polynomials, and hence is a polynomial of degree at most.
Case 2: Suppose some
16 =0. Then, at most−1of the rows from2throughofAhave a
nonzero entry. Hence, each submatrixA
1has at most−1nonzero rows, and, by the inductive
hypothesis, each of|C
1|=|A 1+B 1|is a polynomial of degree at most−1. Therefore, each term
(
1+ 1)(−1)
+1
|C1|is a polynomial of degree at most. The desired result easily follows.
(23) (a)|I
2−A|=
¯
¯
¯
¯
−
11−12
−21− 22
¯
¯
¯
¯
=(−
11)(− 22)− 1221
=
2
−( 11+22)+( 1122−1221)=
2
−(trace(A))+|A|
(b) Use induction on.
Base Step(=1):Now
A()=− 11,whichisafirst degree polynomial with the coefficient
of itsterm equal to1.
Inductive Step: Assume that the statement is true for all(−1)×(−1)matrices, and prove it
is true for an×matrixA.Consider
C=I
−A=
⎡
⎢
⎢
⎢
⎣
(−
11)− 12 ···− 1
−21 (− 22)···− 2
.
.
.
.
.
.
.
.
.
.
.
.
−
1 −2 ···(− )
⎤
⎥
⎥
⎥
⎦
.
Using a cofactor expansion along thefirst row,
A()=|C|=
(−
11)(−1)
1+1
|C11|+(− 12)(−1)
1+2
|C12|+(− 13)(−1)
1+3
|C13|+···+(− 1)(−1)
1+
|C1|.
Now|C
11|= A11(),andso,bytheinductivehypothesis,isan−1degree polynomial with
Copyrightc°2016 Elsevier Ltd. All rights reserved. 48
Answers to Exercises Chapter 3 Review
coefficient of
−1
equal to1. Hence,(− 11)|C11|is a degreepolynomial having coefficient
1for its
term. But, for≥2,C 1is an(−1)×(−1)matrix having exactly−2rows
containing entries which are linear polynomials, since the linear polynomials at the(11)and( )
entries ofChave been eliminated. Therefore, by Exercise 22,|C
1|is a polynomial of degree at
most−2. Summing to get|C|then gives andegree polynomial with coefficient of the
term
equal to1, since only the term(−
11)|C11|in|C|contributes to the
term in A().
(c) Note that the constant term of
A()is A(0) =|0I −A|=|−A|=(−1)
|A|.
(d) Use induction on.
Base Step(=1):Now
A()=− 11.Thedegree0term is− 11=−trace(A).
Inductive Step: Assume that the statement is true for all(−1)×(−1)matrices, and prove
it is true for an×matrixA. We know that
A()is a polynomial of degreefrom part (b).
Then, letC=I
−Aas in part (b). Then, arguing as in part (b), only the term(− 11)|C11|in
|C|contributes to the
and
−1
terms in A(), because all of the terms of the form± 1|C1|
are polynomials of degree≤(−2),for≥2(using Exercise 22). Also, since|C
11|= A11(),
the inductive hypothesis and part (b) imply that
|C
11|=
−1
−trace(A 11)
−2
+(terms of degree≤(−3))
Hence,
(−
11)|C11|=(− 11)(
−1
−trace(A 11)
−2
+(terms of degree≤(−3)))
=
−11
−1
−trace(A 11)
−1
+(terms of degree≤(−2))
=
−trace(A)
−1
+(terms of degree≤(−2))
since
11+ trace(A 11) = trace(A). This completes the proof.
(24) (a) T (b) F (c) T (d) T (e) F (f) T (g) T (h) F
Chapter 3 Review Exercises
(1) (a)(34)minor=|A 34|=−30
(b)(34)cofactor=A
34=−|A 34|=30
(c)|A|=−830
(d)|A|=−830
(2)|A|=−262
(3)|A|=−42
(4) Volume=45
(5) (a)|B|=60 (b)|B|=−15 (c)|B|=15
(6) (a) Yes (b) 4 (c) Yes (d) Yes
(7)378
Copyrightc°2016 Elsevier Ltd. All rights reserved. 49
coefficient of
−1
equal to1. Hence,(− 11)|C11|is a degreepolynomial having coefficient
1for its
term. But, for≥2,C 1is an(−1)×(−1)matrix having exactly−2rows
containing entries which are linear polynomials, since the linear polynomials at the(11)and( )
entries ofChave been eliminated. Therefore, by Exercise 22,|C
1|is a polynomial of degree at
most−2. Summing to get|C|then gives andegree polynomial with coefficient of the
term
equal to1, since only the term(−
11)|C11|in|C|contributes to the
term in A().
(c) Note that the constant term of
A()is A(0) =|0I −A|=|−A|=(−1)
|A|.
(d) Use induction on.
Base Step(=1):Now
A()=− 11.Thedegree0term is− 11=−trace(A).
Inductive Step: Assume that the statement is true for all(−1)×(−1)matrices, and prove
it is true for an×matrixA. We know that
A()is a polynomial of degreefrom part (b).
Then, letC=I
−Aas in part (b). Then, arguing as in part (b), only the term(− 11)|C11|in
|C|contributes to the
and
−1
terms in A(), because all of the terms of the form± 1|C1|
are polynomials of degree≤(−2),for≥2(using Exercise 22). Also, since|C
11|= A11(),
the inductive hypothesis and part (b) imply that
|C
11|=
−1
−trace(A 11)
−2
+(terms of degree≤(−3))
Hence,
(−
11)|C11|=(− 11)(
−1
−trace(A 11)
−2
+(terms of degree≤(−3)))
=
−11
−1
−trace(A 11)
−1
+(terms of degree≤(−2))
=
−trace(A)
−1
+(terms of degree≤(−2))
since
11+ trace(A 11) = trace(A). This completes the proof.
(24) (a) T (b) F (c) T (d) T (e) F (f) T (g) T (h) F
Chapter 3 Review Exercises
(1) (a)(34)minor=|A 34|=−30
(b)(34)cofactor=A
34=−|A 34|=30
(c)|A|=−830
(d)|A|=−830
(2)|A|=−262
(3)|A|=−42
(4) Volume=45
(5) (a)|B|=60 (b)|B|=−15 (c)|B|=15
(6) (a) Yes (b) 4 (c) Yes (d) Yes
(7)378
Copyrightc°2016 Elsevier Ltd. All rights reserved. 49
Answers to Exercises Chapter 3 Review
(8) (a)|A|=0
(b) No. A nontrivial solution is
∙
1
−8
¸
(9) From Exercise 17 in Section 3.3,AB
=|A|I . Hence,A
³
1
|A|
B
´
=I
. Therefore, by Theorem
2.10,
³
1
|A|
B
´
A=I
,andsoB
A=|A|I =AB
.
(10)
1=−4, 2=−3, 3=5
(11) (a) The determinant of the given matrix is−289.Thus,wewouldneed |A|
4
=−289. But no real
number raised to the fourth power is negative.
(b) The determinant of the given matrix is zero, making the given matrix singular. Hence it can not
be the inverse of any matrix.
(12)Bsimilar toAimplies there is a matrixPsuch thatB=P
−1
AP.
(a) We will prove thatB
=P
−1
A
Pby induction on.
Base Step:=1.B=P
−1
APis given.
Inductive Step: Assume thatB
=P
−1
A
Pfor some.Then
B
+1
=BB
=(P
−1
AP)(P
−1
A
P)=P
−1
A(PP
−1
)A
P
=P
−1
A(I)A
P=P
−1
AA
P=P
−1
A
+1
P
soB
+1
is similar toA
+1
.
(b)|B
|=|B|=|P
−1
AP|=|P
−1
||A||P|=
1
|P|
|A||P|=|A|=|A
|
(c) IfAis nonsingular,
B
−1
=(P
−1
AP)
−1
=P
−1
A
−1
(P
−1
)
−1
=P
−1
A
−1
P
soBis nonsingular.
Note thatA=PBP
−1
.So,ifBis nonsingular,
A
−1
=(PBP
−1
)
−1
=(P
−1
)
−1
B
−1
P
−1
=PB
−1
P
−1
soAis nonsingular.
(d) By part (c),A
−1
=(P
−1
)
−1
B
−1
(P
−1
),provingthatA
−1
is similar toB
−1
.
(e)B+I
=P
−1
AP+I =P
−1
AP+P
−1
IP=P
−1
(A+I )P.
(f) Exercise 28(c) in Section 1.5 states thattrace(AB) = trace(BA)for any two×matricesA
andB. In this particular case,
trace(B) = trace(P
−1
(AP)) = trace((AP)P
−1
)=trace(A(PP
−1
))
= trace(AI
) = trace(A)
(g) IfBis diagonalizable, then there is a matrixQsuch thatD=Q
−1
BQis diagonal. Thus,
D=Q
−1
(P
−1
AP)Q=(Q
−1
P
−1
)A(PQ)=(PQ)
−1
A(PQ),andsoAis diagonalizable. Since
Ais similar toBif and only ifBis similar toA(see Exercise 13(d) in Section 3.3), an analogous
argument shows thatAis diagonalizable=⇒Bis diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 50
(8) (a)|A|=0
(b) No. A nontrivial solution is
∙
1
−8
¸
(9) From Exercise 17 in Section 3.3,AB
=|A|I . Hence,A
³
1
|A|
B
´
=I
. Therefore, by Theorem
2.10,
³
1
|A|
B
´
A=I
,andsoB
A=|A|I =AB
.
(10)
1=−4, 2=−3, 3=5
(11) (a) The determinant of the given matrix is−289.Thus,wewouldneed |A|
4
=−289. But no real
number raised to the fourth power is negative.
(b) The determinant of the given matrix is zero, making the given matrix singular. Hence it can not
be the inverse of any matrix.
(12)Bsimilar toAimplies there is a matrixPsuch thatB=P
−1
AP.
(a) We will prove thatB
=P
−1
A
Pby induction on.
Base Step:=1.B=P
−1
APis given.
Inductive Step: Assume thatB
=P
−1
A
Pfor some.Then
B
+1
=BB
=(P
−1
AP)(P
−1
A
P)=P
−1
A(PP
−1
)A
P
=P
−1
A(I)A
P=P
−1
AA
P=P
−1
A
+1
P
soB
+1
is similar toA
+1
.
(b)|B
|=|B|=|P
−1
AP|=|P
−1
||A||P|=
1
|P|
|A||P|=|A|=|A
|
(c) IfAis nonsingular,
B
−1
=(P
−1
AP)
−1
=P
−1
A
−1
(P
−1
)
−1
=P
−1
A
−1
P
soBis nonsingular.
Note thatA=PBP
−1
.So,ifBis nonsingular,
A
−1
=(PBP
−1
)
−1
=(P
−1
)
−1
B
−1
P
−1
=PB
−1
P
−1
soAis nonsingular.
(d) By part (c),A
−1
=(P
−1
)
−1
B
−1
(P
−1
),provingthatA
−1
is similar toB
−1
.
(e)B+I
=P
−1
AP+I =P
−1
AP+P
−1
IP=P
−1
(A+I )P.
(f) Exercise 28(c) in Section 1.5 states thattrace(AB) = trace(BA)for any two×matricesA
andB. In this particular case,
trace(B) = trace(P
−1
(AP)) = trace((AP)P
−1
)=trace(A(PP
−1
))
= trace(AI
) = trace(A)
(g) IfBis diagonalizable, then there is a matrixQsuch thatD=Q
−1
BQis diagonal. Thus,
D=Q
−1
(P
−1
AP)Q=(Q
−1
P
−1
)A(PQ)=(PQ)
−1
A(PQ),andsoAis diagonalizable. Since
Ais similar toBif and only ifBis similar toA(see Exercise 13(d) in Section 3.3), an analogous
argument shows thatAis diagonalizable=⇒Bis diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 50
Answers to Exercises Chapter 3 Review
(13)A(X+Y)=AX+AY=X+Y=(X+Y). This, and the given fact thatX+Yis
nonzero, prove thatX+Yis an eigenvector forAcorresponding to the eigenvalue.
(14) (a)
A()=
3
−
2
−10−8=(+2)(+1)(−4);
Eigenvalues:
1=−2, 2=−1, 3=4;
Eigenspaces:
−2={[−133]|∈R}, −1={[−277]|∈R},
4={[−31011]|∈R};
P=
⎡
⎣
−1−2−3
3710
3711
⎤
⎦;D=
⎡
⎣
−200
0−10
004
⎤
⎦
(b)
A()=
3
+
2
−21−45 = (+3)
2
(−5);
Eigenvalues:
1=5, 2=−3;
Eigenspaces:
5={[−144]|∈R}; −3={[−210] +[201]| ∈R};
P=
⎡
⎣
−1−22
410
401
⎤
⎦;D=
⎡
⎣
500
0−30
00 −3
⎤
⎦
(15) (a)
A()=
3
−1=(−1)(
2
++1).=1is the only eigenvalue, having algebraic multiplicity1.
Thus, at most1fundamental eigenvector will be produced, which is insufficient for diagonalization
by Step 4 of the Diagonalization Method.
(b)
A()=
4
+6
3
+9
2
=
2
(+3)
2
. Even though the eigenvalue=−3has algebraic multiplicity
2,only1fundamental eigenvector is produced forbecause(−3I
4−A)has rank3.Infact,we
get only3fundamental eigenvectors overall, which is insufficient for diagonalization by Step 4 of
the Diagonalization Method.
(16)A
13
=
⎡
⎣
−9565941 9565942 4782976
−12754588 12754589 6377300
3188648−3188648−1594325
⎤
⎦
(17) (a)
1=2, 2=−1, 3=3
(b)
2={[1−211]|∈R}, −1={[1001] +[37−32]| ∈R},
3={[28−43]|∈R}
(c)|A|=6
(18) (a) F
(b) F
(c) F
(d) F
(e) T
(f) T
(g) T
(h) T
(i) F
(j) T
(k) F
(l) F
(m) T
(n) F
(o) F
(p) F
(q) F
(r) T
(s) T
(t) F
(u) F
(v) T
(w) T
(x) T
(y) F
(z) F
Copyrightc°2016 Elsevier Ltd. All rights reserved. 51
(13)A(X+Y)=AX+AY=X+Y=(X+Y). This, and the given fact thatX+Yis
nonzero, prove thatX+Yis an eigenvector forAcorresponding to the eigenvalue.
(14) (a)
A()=
3
−
2
−10−8=(+2)(+1)(−4);
Eigenvalues:
1=−2, 2=−1, 3=4;
Eigenspaces:
−2={[−133]|∈R}, −1={[−277]|∈R},
4={[−31011]|∈R};
P=
⎡
⎣
−1−2−3
3710
3711
⎤
⎦;D=
⎡
⎣
−200
0−10
004
⎤
⎦
(b)
A()=
3
+
2
−21−45 = (+3)
2
(−5);
Eigenvalues:
1=5, 2=−3;
Eigenspaces:
5={[−144]|∈R}; −3={[−210] +[201]| ∈R};
P=
⎡
⎣
−1−22
410
401
⎤
⎦;D=
⎡
⎣
500
0−30
00 −3
⎤
⎦
(15) (a)
A()=
3
−1=(−1)(
2
++1).=1is the only eigenvalue, having algebraic multiplicity1.
Thus, at most1fundamental eigenvector will be produced, which is insufficient for diagonalization
by Step 4 of the Diagonalization Method.
(b)
A()=
4
+6
3
+9
2
=
2
(+3)
2
. Even though the eigenvalue=−3has algebraic multiplicity
2,only1fundamental eigenvector is produced forbecause(−3I
4−A)has rank3.Infact,we
get only3fundamental eigenvectors overall, which is insufficient for diagonalization by Step 4 of
the Diagonalization Method.
(16)A
13
=
⎡
⎣
−9565941 9565942 4782976
−12754588 12754589 6377300
3188648−3188648−1594325
⎤
⎦
(17) (a)
1=2, 2=−1, 3=3
(b)
2={[1−211]|∈R}, −1={[1001] +[37−32]| ∈R},
3={[28−43]|∈R}
(c)|A|=6
(18) (a) F
(b) F
(c) F
(d) F
(e) T
(f) T
(g) T
(h) T
(i) F
(j) T
(k) F
(l) F
(m) T
(n) F
(o) F
(p) F
(q) F
(r) T
(s) T
(t) F
(u) F
(v) T
(w) T
(x) T
(y) F
(z) F
Copyrightc°2016 Elsevier Ltd. All rights reserved. 51
Answers to Exercises Section 4.1
Chapter 4
Section 4.1
(1) Property (2):⊕(⊕)=(⊕)⊕
Property (5):¯(⊕)=(¯)⊕(¯)
Property (6):(+)¯=(¯)⊕(¯)
Property (7):()¯=¯(¯)
(2) LetW={[132]|∈R}.First,Wis closed under addition because
[132] +[132] = (+)[13
2]
which has the correct form. Similarly,([132]) = ()[132], proving closure under scalar multipli-
cation. Now, Properties (1), (2), (5), (6), (7), and (8) are true for all vectors inR
3
by Theorem 1.3,
and hence are true inW. Property (3) is satisfied because[000] = 0[132]∈W,and
[000] +[132] =[132] + [000] = [000]
Finally, Property (4) follows from the equation([132]) + ((−)[132]) = [000].
(3) LetW={f∈P
3|f(2) = 0}.First,Wis closed under addition since the sum of polynomials of degree
≤3has degree≤3, and because iffg∈W,then(f+g)(2) =f(2) +g(2) = 0 + 0 = 0,andso
(f+g)∈W. Similarly, iff∈Wand∈R,thenfhas the proper degree, and(f)(2) =f(2) =0=
0, thus establishing closure under scalar multiplication. Now Properties (2), (5), (6), (7), and (8) are
true for all real-valued functions onR, as shown in Example 6 in Section 4.1 of the text. Property (1)
holds because for every∈R,f()andg()are both real numbers, and sof()+g()=g()+f()
by the commutative law of addition for real numbers, which meansf+g=g+
f. Also, Property
(3) holds because the zero functionzis a polynomial of degree0,andz(2) = 0soz∈W. Finally,
Property (4) is true since−fis the additive inverse off,−fhas the correct degree, and(−f)(2) =
−f(2) =−0=0so−f∈W.
(4) Clearly the operation⊕is commutative. Also,⊕is associative because
(x⊕y)⊕z=(
3
+
3
)
13
⊕z=(((
3
+
3
)
13
)
3
+
3
)
13
=((
3
+
3
)+
3
)
13
=(
3
+(
3
+
3
))
13
=(
3
+((
3
+
3
)
13
)
3
)
13
=x⊕(
3
+
3
)
13
=x⊕(y⊕z)
Clearly, the real number0acts as the additive identity, and, for any real number, the additive inverse
ofis−,because(
3
+(−)
3
)
13
=0
13
=0, the additive identity.
Thefirst distributive law holds because
¯(x⊕y)= ¯(
3
+
3
)
13
=
13
(
3
+
3
)
13
=((
3
+
3
))
13
=(
3
+
3
)
13
=((
13
)
3
+(
13
)
3
)
13
=(
13
)⊕(
13
)
=(¯x)⊕(¯y)
Similarly, the other distributive law holds because
(+)¯x=(+)
13
=(+)
13
(
3
)
13
=(
3
+
3
)
13
=((
13
)
3
+(
13
)
3
)
13
=(
13
)⊕(
13
)
=(¯x)⊕(¯x)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 52
Chapter 4
Section 4.1
(1) Property (2):⊕(⊕)=(⊕)⊕
Property (5):¯(⊕)=(¯)⊕(¯)
Property (6):(+)¯=(¯)⊕(¯)
Property (7):()¯=¯(¯)
(2) LetW={[132]|∈R}.First,Wis closed under addition because
[132] +[132] = (+)[13
2]
which has the correct form. Similarly,([132]) = ()[132], proving closure under scalar multipli-
cation. Now, Properties (1), (2), (5), (6), (7), and (8) are true for all vectors inR
3
by Theorem 1.3,
and hence are true inW. Property (3) is satisfied because[000] = 0[132]∈W,and
[000] +[132] =[132] + [000] = [000]
Finally, Property (4) follows from the equation([132]) + ((−)[132]) = [000].
(3) LetW={f∈P
3|f(2) = 0}.First,Wis closed under addition since the sum of polynomials of degree
≤3has degree≤3, and because iffg∈W,then(f+g)(2) =f(2) +g(2) = 0 + 0 = 0,andso
(f+g)∈W. Similarly, iff∈Wand∈R,thenfhas the proper degree, and(f)(2) =f(2) =0=
0, thus establishing closure under scalar multiplication. Now Properties (2), (5), (6), (7), and (8) are
true for all real-valued functions onR, as shown in Example 6 in Section 4.1 of the text. Property (1)
holds because for every∈R,f()andg()are both real numbers, and sof()+g()=g()+f()
by the commutative law of addition for real numbers, which meansf+g=g+
f. Also, Property
(3) holds because the zero functionzis a polynomial of degree0,andz(2) = 0soz∈W. Finally,
Property (4) is true since−fis the additive inverse off,−fhas the correct degree, and(−f)(2) =
−f(2) =−0=0so−f∈W.
(4) Clearly the operation⊕is commutative. Also,⊕is associative because
(x⊕y)⊕z=(
3
+
3
)
13
⊕z=(((
3
+
3
)
13
)
3
+
3
)
13
=((
3
+
3
)+
3
)
13
=(
3
+(
3
+
3
))
13
=(
3
+((
3
+
3
)
13
)
3
)
13
=x⊕(
3
+
3
)
13
=x⊕(y⊕z)
Clearly, the real number0acts as the additive identity, and, for any real number, the additive inverse
ofis−,because(
3
+(−)
3
)
13
=0
13
=0, the additive identity.
Thefirst distributive law holds because
¯(x⊕y)= ¯(
3
+
3
)
13
=
13
(
3
+
3
)
13
=((
3
+
3
))
13
=(
3
+
3
)
13
=((
13
)
3
+(
13
)
3
)
13
=(
13
)⊕(
13
)
=(¯x)⊕(¯y)
Similarly, the other distributive law holds because
(+)¯x=(+)
13
=(+)
13
(
3
)
13
=(
3
+
3
)
13
=((
13
)
3
+(
13
)
3
)
13
=(
13
)⊕(
13
)
=(¯x)⊕(¯x)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 52
Answers to Exercises Section 4.1
Associativity of scalar multiplication holds since
()¯x=()
13
=
13
13
=
13
(
13
)=¯(
13
)=¯(¯x)
Finally,1¯x=1
13
=1=x.
(5) The set is not closed under addition. For example,
∙
11
11
¸
+
∙
12
24
¸
=
∙
23
35
¸
.
(6) The set does not contain the zero vector.
(7) Property (8) is not satisfied. For example,1¯5=06 =5.
(8) Properties (2), (3), and (6) are not satisfied. Property (4) makes no sense without Property (3). The
following is a counterexample for Property (2):
3⊕(4⊕5) = 3⊕18 = 42but(3⊕4)⊕5=14⊕5=38
(9) Properties (3) and (6) are not satisfied. Property (4) makes no sense without Property (3). The
following is a counterexample for Property (6):
(1 + 2)¯[34] = 3¯[34] = [912]but(1¯[34])⊕(2¯[34]) = [34]⊕[68] = [90]
(10) Not a vector space. Property (6) is not satisfied. For example,(1 + 2)¯3=3¯3=9, but
(1¯3)
⊕(2¯4) = 3⊕8=(3
5
+8
5
)
15
= 33011
15
≈801183
(11) Suppose0
1and0 2are both zero vectors. Then,0 1=01+02=02.
(12) Zero vector=[2−3];additiveinverseof[ ]=[4−−6−]
(13) (a) Add−vto both sides.
(b)v=v=⇒v+(−(v)) =v+(−(v)) =⇒v+(−1)(v)=0(by Theorem 4.1, part (3),
and Property (4))=⇒v+(−)v=0(by Property (7))=⇒(−)v=0(by Property (6))
=⇒−=0(by Theorem 4.1, part (4))=⇒=.
(c) Multiply both sides by the scalar
1
and use Property (7).
(14) For the Closure Properties, and Properties (2), (5), (6), (7), (8), mimic the steps in Example 6 in
Section 4.1 in the text, restricting the proof to polynomial functions only. For Property (1), mimic
the step for Property (1) in the answer to Exercise 3 above. Also, Property (3) holds because the zero
polynomialz(of degree zero) inPserves as an additive identity sincez+p=p+z=pfor any
polynomialp. Finally, Property (4) is true since, for any polynomialp,−pis a polynomial that has
the propertyp+(−p)=(−p)+p=z(the additive identity), and so−p∈Pserves as the additive
inverse ofp.
(15) For the Closure Properties, and Properties (2), (5), (6), (7), (8), mimic the steps in Example 6 in
Section 4.1 in the text, generalizing the proofsto allow the domain of the functions to be the set
rather thanR. For Property (1), mimic the step for Property (1) in the answer to Exercise 3 above.
Also, Property (3) holds because the zero functionzwith domainserves as an additive identity
sincez+f=f+z=ffor any functionf∈V. Finally, Property (4) is true since, for any function
f∈V,−fis a function with domainthat has the propertyf+(−f)=(−f)+f=z(the additive
identity), and so−f∈Vserves as the additive inverse off.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 53
Associativity of scalar multiplication holds since
()¯x=()
13
=
13
13
=
13
(
13
)=¯(
13
)=¯(¯x)
Finally,1¯x=1
13
=1=x.
(5) The set is not closed under addition. For example,
∙
11
11
¸
+
∙
12
24
¸
=
∙
23
35
¸
.
(6) The set does not contain the zero vector.
(7) Property (8) is not satisfied. For example,1¯5=06 =5.
(8) Properties (2), (3), and (6) are not satisfied. Property (4) makes no sense without Property (3). The
following is a counterexample for Property (2):
3⊕(4⊕5) = 3⊕18 = 42but(3⊕4)⊕5=14⊕5=38
(9) Properties (3) and (6) are not satisfied. Property (4) makes no sense without Property (3). The
following is a counterexample for Property (6):
(1 + 2)¯[34] = 3¯[34] = [912]but(1¯[34])⊕(2¯[34]) = [34]⊕[68] = [90]
(10) Not a vector space. Property (6) is not satisfied. For example,(1 + 2)¯3=3¯3=9, but
(1¯3)
⊕(2¯4) = 3⊕8=(3
5
+8
5
)
15
= 33011
15
≈801183
(11) Suppose0
1and0 2are both zero vectors. Then,0 1=01+02=02.
(12) Zero vector=[2−3];additiveinverseof[ ]=[4−−6−]
(13) (a) Add−vto both sides.
(b)v=v=⇒v+(−(v)) =v+(−(v)) =⇒v+(−1)(v)=0(by Theorem 4.1, part (3),
and Property (4))=⇒v+(−)v=0(by Property (7))=⇒(−)v=0(by Property (6))
=⇒−=0(by Theorem 4.1, part (4))=⇒=.
(c) Multiply both sides by the scalar
1
and use Property (7).
(14) For the Closure Properties, and Properties (2), (5), (6), (7), (8), mimic the steps in Example 6 in
Section 4.1 in the text, restricting the proof to polynomial functions only. For Property (1), mimic
the step for Property (1) in the answer to Exercise 3 above. Also, Property (3) holds because the zero
polynomialz(of degree zero) inPserves as an additive identity sincez+p=p+z=pfor any
polynomialp. Finally, Property (4) is true since, for any polynomialp,−pis a polynomial that has
the propertyp+(−p)=(−p)+p=z(the additive identity), and so−p∈Pserves as the additive
inverse ofp.
(15) For the Closure Properties, and Properties (2), (5), (6), (7), (8), mimic the steps in Example 6 in
Section 4.1 in the text, generalizing the proofsto allow the domain of the functions to be the set
rather thanR. For Property (1), mimic the step for Property (1) in the answer to Exercise 3 above.
Also, Property (3) holds because the zero functionzwith domainserves as an additive identity
sincez+f=f+z=ffor any functionf∈V. Finally, Property (4) is true since, for any function
f∈V,−fis a function with domainthat has the propertyf+(−f)=(−f)+f=z(the additive
identity), and so−f∈Vserves as the additive inverse off.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 53
Answers to Exercises Section 4.2
(16) Base Step:=1.Then 1v1∈VsinceVis closed under scalar multiplication.
Inductive Step: Assume
1v1+···+ v∈Vifv 1v ∈V,
1∈RProve 1v1+···+ v++1v+1∈V
ifv
1v v+1∈V, 1+1∈RBut,
1v1+···+ v++1v+1=( 1v1+···+ v)+ +1v+1
=w+ +1v+1 (by the inductive hypothesis).
Also,
+1v+1∈VsinceVis closed under scalar multiplication. Finally,w+ +1v+1∈VsinceV
is closed under addition.
(17)0v=0v+0=0v+(0v+(−(0v))) = (0v+0v)+(−(0v)) = (0 + 0)v+(−(0v)) = 0v+(−(0v)) =0.
(18) LetVbe a nontrivial vector space, and letv∈Vwithv6 =0. Then the set={v|∈R}is
asubsetofV,becauseVis closed under scalar multiplication. Also, if6 =thenv6 =v,since
v=v⇒v−v=0⇒(−)v=
0⇒=(by part (4) of Theorem 4.1). Hence, the elements of
are distinct, makingan infinite set, sinceRis infinite. Thus,Vhas an infinite number of elements
because it has an infinite subset.
(19) (a) F (b) F (c) T (d) T (e) F (f) T (g) T
Section 4.2
(1) In what follows,Vrepresents the given subset.
(a) Not a subspace; no zero vector.
(b) Not a subspace; not closed under either operation.
Counterexample for scalar multiplication:2[11] = [22]∈V
(c) Subspace. Nonempty:[00]∈V
The vectors inVare precisely those of the form[12],whereis a scalar.
Addition:Vis closed under addition because
1[12] + 2[12] = ( 1+2)[12], which has the
required form since
1+2is a scalar.
Scalar multiplication:Vis closed under scalar multiplication because([12]) = ()[12]which
has the required form sinceis a scalar.
(d) Not a subspace; not closed under addition.
Counterexample:[10] + [01] = [11]∈V
(e) Not a subspace; no zero vector.
(f) Subspace. Nonempty:[00]∈V.
The vectors inVare precisely those of the form[0] =[10]
Addition:Vis closed under addition since
1[10]+ 2[10] = ( 1+2)[10], which has the required
form since
1+2is a scalar.
Scalar multiplication:Vis closed under scalar multiplication since([10]) = ()[10],which
has the required form sinceis a scalar.
(g) Not a subspace; not closed under addition.
Counterexample:[11] + [1−1] = [20]∈V.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 54
(16) Base Step:=1.Then 1v1∈VsinceVis closed under scalar multiplication.
Inductive Step: Assume
1v1+···+ v∈Vifv 1v ∈V,
1∈RProve 1v1+···+ v++1v+1∈V
ifv
1v v+1∈V, 1+1∈RBut,
1v1+···+ v++1v+1=( 1v1+···+ v)+ +1v+1
=w+ +1v+1 (by the inductive hypothesis).
Also,
+1v+1∈VsinceVis closed under scalar multiplication. Finally,w+ +1v+1∈VsinceV
is closed under addition.
(17)0v=0v+0=0v+(0v+(−(0v))) = (0v+0v)+(−(0v)) = (0 + 0)v+(−(0v)) = 0v+(−(0v)) =0.
(18) LetVbe a nontrivial vector space, and letv∈Vwithv6 =0. Then the set={v|∈R}is
asubsetofV,becauseVis closed under scalar multiplication. Also, if6 =thenv6 =v,since
v=v⇒v−v=0⇒(−)v=
0⇒=(by part (4) of Theorem 4.1). Hence, the elements of
are distinct, makingan infinite set, sinceRis infinite. Thus,Vhas an infinite number of elements
because it has an infinite subset.
(19) (a) F (b) F (c) T (d) T (e) F (f) T (g) T
Section 4.2
(1) In what follows,Vrepresents the given subset.
(a) Not a subspace; no zero vector.
(b) Not a subspace; not closed under either operation.
Counterexample for scalar multiplication:2[11] = [22]∈V
(c) Subspace. Nonempty:[00]∈V
The vectors inVare precisely those of the form[12],whereis a scalar.
Addition:Vis closed under addition because
1[12] + 2[12] = ( 1+2)[12], which has the
required form since
1+2is a scalar.
Scalar multiplication:Vis closed under scalar multiplication because([12]) = ()[12]which
has the required form sinceis a scalar.
(d) Not a subspace; not closed under addition.
Counterexample:[10] + [01] = [11]∈V
(e) Not a subspace; no zero vector.
(f) Subspace. Nonempty:[00]∈V.
The vectors inVare precisely those of the form[0] =[10]
Addition:Vis closed under addition since
1[10]+ 2[10] = ( 1+2)[10], which has the required
form since
1+2is a scalar.
Scalar multiplication:Vis closed under scalar multiplication since([10]) = ()[10],which
has the required form sinceis a scalar.
(g) Not a subspace; not closed under addition.
Counterexample:[11] + [1−1] = [20]∈V.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 54
Answers to Exercises Section 4.2
(h) Subspace. Nonempty:[00]∈Vsince0=−3(0).
The vectors inVare precisely those of the form[−3]=[1−3]
Addition:Vis closed under addition since
1[1−3] + 2[1−3] = ( 1+2)[1−3], which has the
required form since
1+2is a scalar.
Scalar multiplication:Vis closed under scalar multiplication since([1−3]) = ()[1−3]which
has the required form sinceis a scalar.
(i) Not a subspace; no zero vector since06 =7(0)−5.
(j) Not a subspace; not closed under either operation.
Counterexample for addition:[11] + [24] = [35]∈V,since56 =3
2
.
(k) Not a subspace; not closed under either operation.
Counterexample for scalar multiplication:2[0−4] = [0−8]∈V,since−82(0)−5.
(l) Not a subspace; not closed under either operation.
Counterexample for scalar multiplication:2[0750] = [150]∈V.
(2) In what follows,Vrepresents the given subset.
(a) Subspace. Nonempty:O
22∈V
The matrices inVare precisely those of the form
∙
−
0
¸
=
∙
1−1
00
¸
+
∙
00
10
¸
Addition:Vis closed under matrix addition since
µ
1
∙
1−1
00
¸
+
1
∙
00
10
¸¶
+
µ
2
∙
1−1
00
¸
+
2
∙
00
10
¸¶
=(
1+2)
∙
1−1
00
¸
+( 1+2)
∙
00
10
¸
which is of the required form since
1+2and 1+2are scalars.
Scalar multiplication:Vis closed under scalar multiplication because
µ
∙
1−1
00
¸
+
∙
00
10
¸¶
=()
∙
1−1
00
¸
+()
∙
00
10
¸
which is of the required form sinceandare scalars.
(b) Not a subspace; not closed under addition.
Counterexample:
∙
11
00
¸
+
∙
00
11
¸
=
∙
11
11
¸
∈V.
(c) Subspace. Nonempty:O
22∈V.
The matrices inVare precisely those of the form
∙
¸
=
∙
10
00
¸
+
∙
01
10
¸
+
∙
00
01
¸
Addition:Vis closed under matrix addition since
µ
1
∙
10
00
¸
+
1
∙
01
10
¸
+
1
∙
00
01
¸¶
+
µ
2
∙
10
00
¸
+
2
∙
01
10
¸
+
2
∙
00
01
¸¶
=(
1+2)
∙
10
00
¸
+( 1+2)
∙
01
10
¸
+( 1+2)
∙
00
01
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 55
(h) Subspace. Nonempty:[00]∈Vsince0=−3(0).
The vectors inVare precisely those of the form[−3]=[1−3]
Addition:Vis closed under addition since
1[1−3] + 2[1−3] = ( 1+2)[1−3], which has the
required form since
1+2is a scalar.
Scalar multiplication:Vis closed under scalar multiplication since([1−3]) = ()[1−3]which
has the required form sinceis a scalar.
(i) Not a subspace; no zero vector since06 =7(0)−5.
(j) Not a subspace; not closed under either operation.
Counterexample for addition:[11] + [24] = [35]∈V,since56 =3
2
.
(k) Not a subspace; not closed under either operation.
Counterexample for scalar multiplication:2[0−4] = [0−8]∈V,since−82(0)−5.
(l) Not a subspace; not closed under either operation.
Counterexample for scalar multiplication:2[0750] = [150]∈V.
(2) In what follows,Vrepresents the given subset.
(a) Subspace. Nonempty:O
22∈V
The matrices inVare precisely those of the form
∙
−
0
¸
=
∙
1−1
00
¸
+
∙
00
10
¸
Addition:Vis closed under matrix addition since
µ
1
∙
1−1
00
¸
+
1
∙
00
10
¸¶
+
µ
2
∙
1−1
00
¸
+
2
∙
00
10
¸¶
=(
1+2)
∙
1−1
00
¸
+( 1+2)
∙
00
10
¸
which is of the required form since
1+2and 1+2are scalars.
Scalar multiplication:Vis closed under scalar multiplication because
µ
∙
1−1
00
¸
+
∙
00
10
¸¶
=()
∙
1−1
00
¸
+()
∙
00
10
¸
which is of the required form sinceandare scalars.
(b) Not a subspace; not closed under addition.
Counterexample:
∙
11
00
¸
+
∙
00
11
¸
=
∙
11
11
¸
∈V.
(c) Subspace. Nonempty:O
22∈V.
The matrices inVare precisely those of the form
∙
¸
=
∙
10
00
¸
+
∙
01
10
¸
+
∙
00
01
¸
Addition:Vis closed under matrix addition since
µ
1
∙
10
00
¸
+
1
∙
01
10
¸
+
1
∙
00
01
¸¶
+
µ
2
∙
10
00
¸
+
2
∙
01
10
¸
+
2
∙
00
01
¸¶
=(
1+2)
∙
10
00
¸
+( 1+2)
∙
01
10
¸
+( 1+2)
∙
00
01
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 55
Answers to Exercises Section 4.2
which is of the required form, since 1+21+2and 1+2are scalars.
Scalar multiplication:Vis closed under scalar multiplication because
µ
∙
10
00
¸
+
∙
01
10
¸
+
∙
00
01
¸¶
=(
1)
∙
10
00
¸
+( 1)
∙
01
10
¸
+( 1)
∙
00
01
¸
which is of the required form, since
1,1,and 1are scalars.
(d) Not a subspace; no zero vector sinceO
22is singular.
(e) Subspace. Nonempty:O
22∈V.
A typical element ofVhas the form
∙
¸
,where=−(++). Therefore, the matrices
inVare precisely those of the form
∙
−(++)
¸
=
∙
10
0−1
¸
+
∙
01
0−1
¸
+
∙
00
1−1
¸
Addition:Vis closed under addition because
µ
1
∙
10
0−1
¸
+
1
∙
01
0−1
¸
+
1
∙
00
1−1
¸¶
+
µ
2
∙
10
0−1
¸
+
2
∙
01
0−1
¸
+
2
∙
00
1−1
¸¶
=(
1+2)
∙
10
0−1
¸
+( 1+2)
∙
01
0−1
¸
+( 1+2)
∙
00
1−1
¸
which is of the required form, since
1+21+2and 1+2are scalars.
Scalar multiplication:Vis closed under scalar multiplication since
µ
∙
10
0−1
¸
+
∙
01
0−1
¸
+
∙
00
1−1
¸¶
=()
∙
10
0−1
¸
+()
∙
01
0−1
¸
+()
∙
00
1−1
¸
which is of the required form, since,,andare scalars.
(f) Subspace. Nonempty:O
22∈V.
The matrices inVare precisely those of the form
∙
−
¸
=
∙
10
0−1
¸
+
∙
01
00
¸
+
∙
00
10
¸
Addition:Vis closed under addition since
µ
1
∙
10
0−1
¸
+
1
∙
01
00
¸
+
1
∙
00
10
¸¶
+
µ
2
∙
10
0−1
¸
+
2
∙
01
00
¸
+
2
∙
00
10
¸¶
=(
1+2)
∙
10
0−1
¸
+( 1+2)
∙
01
00
¸
+( 1+2)
∙
00
10
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 56
which is of the required form, since 1+21+2and 1+2are scalars.
Scalar multiplication:Vis closed under scalar multiplication because
µ
∙
10
00
¸
+
∙
01
10
¸
+
∙
00
01
¸¶
=(
1)
∙
10
00
¸
+( 1)
∙
01
10
¸
+( 1)
∙
00
01
¸
which is of the required form, since
1,1,and 1are scalars.
(d) Not a subspace; no zero vector sinceO
22is singular.
(e) Subspace. Nonempty:O
22∈V.
A typical element ofVhas the form
∙
¸
,where=−(++). Therefore, the matrices
inVare precisely those of the form
∙
−(++)
¸
=
∙
10
0−1
¸
+
∙
01
0−1
¸
+
∙
00
1−1
¸
Addition:Vis closed under addition because
µ
1
∙
10
0−1
¸
+
1
∙
01
0−1
¸
+
1
∙
00
1−1
¸¶
+
µ
2
∙
10
0−1
¸
+
2
∙
01
0−1
¸
+
2
∙
00
1−1
¸¶
=(
1+2)
∙
10
0−1
¸
+( 1+2)
∙
01
0−1
¸
+( 1+2)
∙
00
1−1
¸
which is of the required form, since
1+21+2and 1+2are scalars.
Scalar multiplication:Vis closed under scalar multiplication since
µ
∙
10
0−1
¸
+
∙
01
0−1
¸
+
∙
00
1−1
¸¶
=()
∙
10
0−1
¸
+()
∙
01
0−1
¸
+()
∙
00
1−1
¸
which is of the required form, since,,andare scalars.
(f) Subspace. Nonempty:O
22∈V.
The matrices inVare precisely those of the form
∙
−
¸
=
∙
10
0−1
¸
+
∙
01
00
¸
+
∙
00
10
¸
Addition:Vis closed under addition since
µ
1
∙
10
0−1
¸
+
1
∙
01
00
¸
+
1
∙
00
10
¸¶
+
µ
2
∙
10
0−1
¸
+
2
∙
01
00
¸
+
2
∙
00
10
¸¶
=(
1+2)
∙
10
0−1
¸
+( 1+2)
∙
01
00
¸
+( 1+2)
∙
00
10
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 56
Answers to Exercises Section 4.2
which is of the required form, since 1+21+2and 1+2are scalars.
Scalar multiplication:Vis closed under scalar multiplication since
µ
∙
10
0−1
¸
+
∙
01
00
¸
+
∙
00
10
¸¶
=()
∙
10
0−1
¸
+()
∙
01
00
¸
+()
∙
00
10
¸
which is of the required form, since,,andare scalars.
(g) Subspace. LetB=
∙
13
−2−6
¸
.Nonempty:O
22∈VbecauseO 22B=O 22.
Addition: IfAC∈V,then(A+C)B=AB+CB=O
22+O22=O 22. Hence,(A+C)∈V.
Scalar multiplication: IfA∈V,then(A)B=(AB)=O
22=O 22,andso(A)∈V.
(h) Not a subspace; not closed under addition.
Counterexample:
∙
11
00
¸
+
∙
00
11
¸
=
∙
11
11
¸
∈V.
(3) In what follows,Vrepresents the given subset, andzrepresents the zero polynomial.
(a) Subspace. Nonempty: Clearly,z∈V.
Addition:
(
5
+
4
+
3
+
2
++)+(
5
+
4
+
3
+
2
++)
=(+)
5
+(+)
4
+(+)
3
+(+)
2
+(+)+(+)∈V;
Scalar multiplication:
(
5
+
4
+
3
+
2
++)
=()
5
+()
4
+()
3
+()
2
+()+()∈V
(b) Subspace. Nonempty:z∈V. Supposepq∈V.
Addition:(p+q)(3) =p(3) +q(3) = 0 + 0 = 0. Hence(p+q)∈V.
Scalar multiplication:(p)(3) =p(3) =(0) = 0.Thus(p)∈V.
(c) Subspace. Nonempty:z∈V.
Addition:
(
5
+
4
+
3
+
2
+−(++++))
+(
5
+
4
+
3
+
2
+−(++++))
=(+)
5
+(+)
4
+(+)
3
+(+)
2
+(+)
+(−(++++)−(++++))
=(+)
5
+(+)
4
+(+)
3
+(+)
2
+(+)
+(−((+)+(+)+(+)+(+)+(+)))
which is inV
Scalar multiplication:
(
5
+
4
+
3
+
2
+−(++++))
=()
5
+()
4
+()
3
+()
2
+()−(()+()+()+()+())
which is inV.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 57
which is of the required form, since 1+21+2and 1+2are scalars.
Scalar multiplication:Vis closed under scalar multiplication since
µ
∙
10
0−1
¸
+
∙
01
00
¸
+
∙
00
10
¸¶
=()
∙
10
0−1
¸
+()
∙
01
00
¸
+()
∙
00
10
¸
which is of the required form, since,,andare scalars.
(g) Subspace. LetB=
∙
13
−2−6
¸
.Nonempty:O
22∈VbecauseO 22B=O 22.
Addition: IfAC∈V,then(A+C)B=AB+CB=O
22+O22=O 22. Hence,(A+C)∈V.
Scalar multiplication: IfA∈V,then(A)B=(AB)=O
22=O 22,andso(A)∈V.
(h) Not a subspace; not closed under addition.
Counterexample:
∙
11
00
¸
+
∙
00
11
¸
=
∙
11
11
¸
∈V.
(3) In what follows,Vrepresents the given subset, andzrepresents the zero polynomial.
(a) Subspace. Nonempty: Clearly,z∈V.
Addition:
(
5
+
4
+
3
+
2
++)+(
5
+
4
+
3
+
2
++)
=(+)
5
+(+)
4
+(+)
3
+(+)
2
+(+)+(+)∈V;
Scalar multiplication:
(
5
+
4
+
3
+
2
++)
=()
5
+()
4
+()
3
+()
2
+()+()∈V
(b) Subspace. Nonempty:z∈V. Supposepq∈V.
Addition:(p+q)(3) =p(3) +q(3) = 0 + 0 = 0. Hence(p+q)∈V.
Scalar multiplication:(p)(3) =p(3) =(0) = 0.Thus(p)∈V.
(c) Subspace. Nonempty:z∈V.
Addition:
(
5
+
4
+
3
+
2
+−(++++))
+(
5
+
4
+
3
+
2
+−(++++))
=(+)
5
+(+)
4
+(+)
3
+(+)
2
+(+)
+(−(++++)−(++++))
=(+)
5
+(+)
4
+(+)
3
+(+)
2
+(+)
+(−((+)+(+)+(+)+(+)+(+)))
which is inV
Scalar multiplication:
(
5
+
4
+
3
+
2
+−(++++))
=()
5
+()
4
+()
3
+()
2
+()−(()+()+()+()+())
which is inV.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 57
Answers to Exercises Section 4.2
(d) Subspace. Nonempty:z∈V. Supposepq∈V.
Addition:(p+q)(3) =p(3) +q(3) =p(5) +q(5) = (p+q)(5). Hence(p+q)∈V.
Scalar multiplication:(p)(3) =p(3) =p(5) = (p)(5).Thus(p)∈V.
(e) Not a subspace;z∈V.
(f) Not a subspace; not closed under scalar multiplication.
Counterexample:−
2
has a relative maximum at=0, but(−1)(−
2
)=
2
does not
(it has a relative minimum instead).
(g) Subspace. Nonempty:z∈V. Supposepq∈V.
Addition:(p+q)
0
(4) =p
0
(4) +q
0
(4) = 0 + 0 = 0. Hence(p+q)∈V.
Scalar multiplication:(p)
0
(4) =p
0
(4) =(0) = 0.Thus(p)∈V.
(h) Not a subspace;z∈V.
(4) LetVbe the given set of vectors. Setting===0shows that0∈V. HenceVis nonempty. Note
that the vectors inVhave the form[ 0−2+]=[11001] +[0100−2] +[00011]
For closure under addition, note that
(
1[11001] + 1[0100−2] + 1[00011])
+(
2[11001] + 2[0100−2] + 2[00011])
=(
1+2)[11001] + ( 1+2)[0100−2] + ( 1+2)[00011]
which is clearly another vector of this same form, since
1+2,1+2,and 1+2are scalars. Similarly,
for closure under scalar multiplication, note that
([11001] +[0100−2] +[00011])
=()[11001] + ()[0100−2] + ()[00011]
which also has the correct form since,,andare scalars. ThusVis a subspace by Theorem 4.2.
(5) LetVbe the given set of vectors. Setting===0shows that0∈V. HenceVis nonempty. Note
that the vectors inVhave the form[2−3
−5 4− ]=[21101] +[−300−10] +
[0−5041]For closure under addition, note that
(
1[21101] + 1[−300−10] + 1[0−5041])
+(
2[21101] + 2[−300−10] + 2[0−5041])
=(
1+2)[21101] + ( 1+2)[−300−10] + ( 1+2)[0−5041]
which is clearly another vector of this same form, since
1+2,1+2,and 1+2are scalars. Similarly,
for closure under scalar multiplication,
([21101] +[−300−10] +[0−5041])
=()[21101] + ()[−300−10] + ()[0−5041]
which also has the correct form since,,andare scalars. ThusVis a subspace by Theorem 4.2.
(6) (a) LetV={x∈R
3
|x·[1−14] = 0}. Clearly[000]∈Vsince[000]·[1−14] = 0. HenceVis
nonempty. Next, ifxy∈V,then
(x+y)·[1−14] =x·[1−14] +y·[1−14] = 0 + 0 = 0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 58
(d) Subspace. Nonempty:z∈V. Supposepq∈V.
Addition:(p+q)(3) =p(3) +q(3) =p(5) +q(5) = (p+q)(5). Hence(p+q)∈V.
Scalar multiplication:(p)(3) =p(3) =p(5) = (p)(5).Thus(p)∈V.
(e) Not a subspace;z∈V.
(f) Not a subspace; not closed under scalar multiplication.
Counterexample:−
2
has a relative maximum at=0, but(−1)(−
2
)=
2
does not
(it has a relative minimum instead).
(g) Subspace. Nonempty:z∈V. Supposepq∈V.
Addition:(p+q)
0
(4) =p
0
(4) +q
0
(4) = 0 + 0 = 0. Hence(p+q)∈V.
Scalar multiplication:(p)
0
(4) =p
0
(4) =(0) = 0.Thus(p)∈V.
(h) Not a subspace;z∈V.
(4) LetVbe the given set of vectors. Setting===0shows that0∈V. HenceVis nonempty. Note
that the vectors inVhave the form[ 0−2+]=[11001] +[0100−2] +[00011]
For closure under addition, note that
(
1[11001] + 1[0100−2] + 1[00011])
+(
2[11001] + 2[0100−2] + 2[00011])
=(
1+2)[11001] + ( 1+2)[0100−2] + ( 1+2)[00011]
which is clearly another vector of this same form, since
1+2,1+2,and 1+2are scalars. Similarly,
for closure under scalar multiplication, note that
([11001] +[0100−2] +[00011])
=()[11001] + ()[0100−2] + ()[00011]
which also has the correct form since,,andare scalars. ThusVis a subspace by Theorem 4.2.
(5) LetVbe the given set of vectors. Setting===0shows that0∈V. HenceVis nonempty. Note
that the vectors inVhave the form[2−3
−5 4− ]=[21101] +[−300−10] +
[0−5041]For closure under addition, note that
(
1[21101] + 1[−300−10] + 1[0−5041])
+(
2[21101] + 2[−300−10] + 2[0−5041])
=(
1+2)[21101] + ( 1+2)[−300−10] + ( 1+2)[0−5041]
which is clearly another vector of this same form, since
1+2,1+2,and 1+2are scalars. Similarly,
for closure under scalar multiplication,
([21101] +[−300−10] +[0−5041])
=()[21101] + ()[−300−10] + ()[0−5041]
which also has the correct form since,,andare scalars. ThusVis a subspace by Theorem 4.2.
(6) (a) LetV={x∈R
3
|x·[1−14] = 0}. Clearly[000]∈Vsince[000]·[1−14] = 0. HenceVis
nonempty. Next, ifxy∈V,then
(x+y)·[1−14] =x·[1−14] +y·[1−14] = 0 + 0 = 0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 58
Answers to Exercises Section 4.2
Thus,(x+y)∈V,andsoVis closed under addition. Also, ifx∈Vand∈R,then
(x)·[1−14] =(x·[1−14]) =(0) = 0
Therefore(x)∈V,soVis closed under scalar multiplication. Hence, by Theorem 4.2,Vis a
subspace ofR
3
.
(b) Plane, since it contains the nonparallel vectors[041]and[110].
(7) (a) The zero functionz()=0is continuous, and so the set is nonempty. Also, from calculus, we
know that the sum of continuous functions is continuous, as is any scalar multiple of a continuous
function. Hence, both closure properties hold. Apply Theorem 4.2.
(b) Replace the word “continuous” everywhere in part (a) with the word “differentiable.”
(c) LetVbe the given set. The zero functionz()=0satisfiesz(
1
2
)=0,andsoVis nonempty.
Furthermore, iffg∈Vand∈R,then
(f+g)(
1
2
)=f(
1
2
)+g(
1
2
)=0+0=0
and
(f)(
1
2
)=f(
1
2
)=0=0
Hence, both closure properties hold. Now use Theorem 4.2.
(d) LetVbe the given set. The zero functionz()=0is continuous and satisfies
R
1
0
z()=0,
and soVis nonempty. Also, from calculus, we know that the sum of continuous functions is
continuous, as is any scalar multiple of a continuous function. Furthermore, iffg∈Vand∈R,
then
Z
1
0
(f+g)()=
Z
1
0
(f()+g())=
Z
1
0
f()+
Z
1
0
g()=0+0=0
and
Z
1
0
(f)()=
Z
1
0
f()=
Z
1
0
f()=0=0
Hence, both closure properties hold. Finally, apply Theorem 4.2.
(8) The zero functionz()=0is differentiable and satisfies3(z)−2z=0,andsoVis nonempty.
Also, from calculus, we know that the sum of differentiable functions is differentiable, as is any scalar
multiple of a differentiable function. Furthermore, iffg∈Vand∈R,then
3(f+g)
0
−2(f+g)=(3f
0
−2f)+(3g
0
−2g)=0+0=0
and
3(f)
0
−2(f)=3f
0
−2f=(3f
0
−2f)=0=0
Hence, both closure properties hold. Now use Theorem 4.2.
(9) LetVbe the given set. The zero functionz()=0is twice-differentiable and satisfiesz
00
+2z
0
−9z=0,
and soVis nonempty. From calculus, sums and scalar multiples of twice-differentiable functions are
twice-differentiable. Furthermore, iffg∈Vand∈R,then
(f+g)
00
+2(f+g)
0
−9(f+g)=(f
00
+2f
0
−9f)+(g
00
+2g
0
−9g)=0+0=0
and
(f)
00
+2(f)
0
−9(f)=f
00
+2f
0
−9f=(f
00
+2f
0
−9f)=0=0
Hence, both closure properties hold. Theorem 4.2finishes the proof.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 59
Thus,(x+y)∈V,andsoVis closed under addition. Also, ifx∈Vand∈R,then
(x)·[1−14] =(x·[1−14]) =(0) = 0
Therefore(x)∈V,soVis closed under scalar multiplication. Hence, by Theorem 4.2,Vis a
subspace ofR
3
.
(b) Plane, since it contains the nonparallel vectors[041]and[110].
(7) (a) The zero functionz()=0is continuous, and so the set is nonempty. Also, from calculus, we
know that the sum of continuous functions is continuous, as is any scalar multiple of a continuous
function. Hence, both closure properties hold. Apply Theorem 4.2.
(b) Replace the word “continuous” everywhere in part (a) with the word “differentiable.”
(c) LetVbe the given set. The zero functionz()=0satisfiesz(
1
2
)=0,andsoVis nonempty.
Furthermore, iffg∈Vand∈R,then
(f+g)(
1
2
)=f(
1
2
)+g(
1
2
)=0+0=0
and
(f)(
1
2
)=f(
1
2
)=0=0
Hence, both closure properties hold. Now use Theorem 4.2.
(d) LetVbe the given set. The zero functionz()=0is continuous and satisfies
R
1
0
z()=0,
and soVis nonempty. Also, from calculus, we know that the sum of continuous functions is
continuous, as is any scalar multiple of a continuous function. Furthermore, iffg∈Vand∈R,
then
Z
1
0
(f+g)()=
Z
1
0
(f()+g())=
Z
1
0
f()+
Z
1
0
g()=0+0=0
and
Z
1
0
(f)()=
Z
1
0
f()=
Z
1
0
f()=0=0
Hence, both closure properties hold. Finally, apply Theorem 4.2.
(8) The zero functionz()=0is differentiable and satisfies3(z)−2z=0,andsoVis nonempty.
Also, from calculus, we know that the sum of differentiable functions is differentiable, as is any scalar
multiple of a differentiable function. Furthermore, iffg∈Vand∈R,then
3(f+g)
0
−2(f+g)=(3f
0
−2f)+(3g
0
−2g)=0+0=0
and
3(f)
0
−2(f)=3f
0
−2f=(3f
0
−2f)=0=0
Hence, both closure properties hold. Now use Theorem 4.2.
(9) LetVbe the given set. The zero functionz()=0is twice-differentiable and satisfiesz
00
+2z
0
−9z=0,
and soVis nonempty. From calculus, sums and scalar multiples of twice-differentiable functions are
twice-differentiable. Furthermore, iffg∈Vand∈R,then
(f+g)
00
+2(f+g)
0
−9(f+g)=(f
00
+2f
0
−9f)+(g
00
+2g
0
−9g)=0+0=0
and
(f)
00
+2(f)
0
−9(f)=f
00
+2f
0
−9f=(f
00
+2f
0
−9f)=0=0
Hence, both closure properties hold. Theorem 4.2finishes the proof.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 59
Answers to Exercises Section 4.2
(10) The given subset does not contain the zero function.
(11) First note thatAO=OA,andsoO∈W.ThusWis nonempty. Next, letB
1B2∈W.Then
A(B
1+B2)=AB 1+AB 2=B1A+B 2A=(B 1+B2)A
and
A(B
1)=(AB 1)=(B 1A)=(B 1)A
(12) (a) Closure under addition is a required property of every vector space.
(b) LetAbe a singular×matrix. Then|A|=0.Butthen|A|=
|A|=0,soAis also
singular.
(c) All8properties hold.
(d) For=2,
∙
10
00
¸
+
∙
00
01
¸
=
∙
10
01
¸
.
(e) No; if|A|6 =0and=0,then|A|=|O
|=0.
(13) (a) Since the given line goes through the origin, it must have either the form=,or=0.Inthe
first case, all vectors on the line have the form[ ]=[1].Noticethat
1[1]+ 2[1]=
(
1+2)[1], which also lies on=,and([1]) = ()[1], which also lies on=.
Thus, both closure properties hold in this case. On the other hand, if the line has the form=0,
then all vectors on the line have the form[0]and these vectors clearly form a subspace ofR
2
.
(b) The given subset does not contain the zero vector.
(14) LetAbe an×matrix, and letVbe the set of solutions of the homogeneous systemAX=0.Now,
Vis nonempty since the zero-vector is a solution ofAX=0For closure under addition, notice that
ifX
1andX 2are solutions ofAX=0,thenA(X 1+X2)=AX 1+AX 2=0+0=0,soX 1+X2is
also a solution ofAX=0. For closure under scalar multiplication, notice that ifX
1is a solution of
AX=0,thenA(X
1)=(AX
1
)=0=0,soX 1is also a solution ofAX=0Thus, by Theorem
4.2,Vis a subspace ofR
.
(15)={0}, the trivial subspace ofR
.
(16) Leta∈Vwitha6 =0,andletb∈R.Then(
)a=b∈V. (We have used bold-faced variables when
we are considering objects as vectors rather than scalars. However,aandhave the same value as
real numbers; similarly forband.)
(17) The given subset does not contain the zero vector.
(18) Note that0∈W
1and0∈W 2. Hence0∈W 1∩W2,andsoW 1∩W2is nonempty. Letxy∈W 1∩W2.
Thenx+y∈W
1sinceW 1is a subspace, andx+y∈W 2sinceW 2is a subspace. Hence,x+y∈W 1∩W2.
Similarly, if∈R,thenx∈W
1andx∈W 2sinceW 1andW 2are subspaces. Thusx∈W 1∩W2.
Now apply Theorem 4.2.
(19) IfWis a subspace, thenw
1w2∈Wby closure under scalar multiplication.
Hencew
1+w 2∈Wby closure under addition.
Conversely, setting=0shows thatw
1∈Wfor allw 1∈W. This establishes closure under scalar
multiplication. Use=1and=1to get closure under addition.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 60
(10) The given subset does not contain the zero function.
(11) First note thatAO=OA,andsoO∈W.ThusWis nonempty. Next, letB
1B2∈W.Then
A(B
1+B2)=AB 1+AB 2=B1A+B 2A=(B 1+B2)A
and
A(B
1)=(AB 1)=(B 1A)=(B 1)A
(12) (a) Closure under addition is a required property of every vector space.
(b) LetAbe a singular×matrix. Then|A|=0.Butthen|A|=
|A|=0,soAis also
singular.
(c) All8properties hold.
(d) For=2,
∙
10
00
¸
+
∙
00
01
¸
=
∙
10
01
¸
.
(e) No; if|A|6 =0and=0,then|A|=|O
|=0.
(13) (a) Since the given line goes through the origin, it must have either the form=,or=0.Inthe
first case, all vectors on the line have the form[ ]=[1].Noticethat
1[1]+ 2[1]=
(
1+2)[1], which also lies on=,and([1]) = ()[1], which also lies on=.
Thus, both closure properties hold in this case. On the other hand, if the line has the form=0,
then all vectors on the line have the form[0]and these vectors clearly form a subspace ofR
2
.
(b) The given subset does not contain the zero vector.
(14) LetAbe an×matrix, and letVbe the set of solutions of the homogeneous systemAX=0.Now,
Vis nonempty since the zero-vector is a solution ofAX=0For closure under addition, notice that
ifX
1andX 2are solutions ofAX=0,thenA(X 1+X2)=AX 1+AX 2=0+0=0,soX 1+X2is
also a solution ofAX=0. For closure under scalar multiplication, notice that ifX
1is a solution of
AX=0,thenA(X
1)=(AX
1
)=0=0,soX 1is also a solution ofAX=0Thus, by Theorem
4.2,Vis a subspace ofR
.
(15)={0}, the trivial subspace ofR
.
(16) Leta∈Vwitha6 =0,andletb∈R.Then(
)a=b∈V. (We have used bold-faced variables when
we are considering objects as vectors rather than scalars. However,aandhave the same value as
real numbers; similarly forband.)
(17) The given subset does not contain the zero vector.
(18) Note that0∈W
1and0∈W 2. Hence0∈W 1∩W2,andsoW 1∩W2is nonempty. Letxy∈W 1∩W2.
Thenx+y∈W
1sinceW 1is a subspace, andx+y∈W 2sinceW 2is a subspace. Hence,x+y∈W 1∩W2.
Similarly, if∈R,thenx∈W
1andx∈W 2sinceW 1andW 2are subspaces. Thusx∈W 1∩W2.
Now apply Theorem 4.2.
(19) IfWis a subspace, thenw
1w2∈Wby closure under scalar multiplication.
Hencew
1+w 2∈Wby closure under addition.
Conversely, setting=0shows thatw
1∈Wfor allw 1∈W. This establishes closure under scalar
multiplication. Use=1and=1to get closure under addition.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 60
Answers to Exercises Section 4.3
(20) (a) Suppose thatWis a subspace of a vector spaceV. We give a proof by induction on.
Base Step:If=1, then we must show that ifv
1∈Wand 1is a scalar, then 1v1∈W
But this is certainly true since the subspaceWis closed under scalar multiplication.
Inductive Step:Assume that the theorem is true for any linear combination ofvectors
inW. We must prove the theorem holds for a linear combination of+1vectors. Suppose
v
1v2v v+1are vectors inW,and 12+1are scalars. We must show that
1v1+2v2+···+ v++1v+1∈W. However, by the inductive hypothesis, we know that
1v1+2v2+···+ v∈W.Also, +1v+1∈W,sinceWis closed under scalar multiplication.
But sinceWis also closed under addition, the sum of any two vectors inWis again inW,so
(
1v1+2v2+···+ v)+( +1v+1)∈W.
(b) Supposew
1w2∈WThenw 1+w2=1w 1+1w 2is afinite linear combination of vectors ofW,
sow
1+w2∈W.Also, 1w1(for 1∈R)isafinite linear combination inW,so 1w1∈W.Now
apply Theorem 4.2.
(21) Apply Theorems 4.4 and 4.3.
(22) (a) F (b) T (c) F (d) T (e) T (f) F (g) T (h) T
Section 4.3
(1) (a){[ −+]| ∈R}
(b){[1
1
3
−
2
3
]|∈R}
(c){[ −]| ∈R}
(d) Span()=R
3
(e){[ −2++]| ∈R}
(f){[ 2+ +]| ∈R}
(2) (a){
3
+
2
+−(++)| ∈R}
(b){
3
+
2
+| ∈R} (c){
3
−+| ∈R}
(3) (a)
½∙
−−−
¸¯
¯
¯
¯
∈R
¾
(b)
½∙
−−8+3
¸¯
¯
¯
¯
∈R
¾
(c)
½∙
¸¯
¯
¯
¯
∈R
¾
=M
22
(4) (a)W=row space ofA=row space of
⎡
⎣
1100
1010
0111
⎤
⎦
={[1100] +[1010] +[0111]| ∈R}
(b)B=
⎡
⎢
⎢
⎢
⎣
100 −
1
2
010
1
2
001
1
2
⎤
⎥
⎥
⎥
⎦
(c) Row space ofB={[100−
1
2
]+[010
1
2
]+[001
1
2
]| ∈R}
={[ −
1
2
+
1
2
+
1
2
]| ∈R}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 61
(20) (a) Suppose thatWis a subspace of a vector spaceV. We give a proof by induction on.
Base Step:If=1, then we must show that ifv
1∈Wand 1is a scalar, then 1v1∈W
But this is certainly true since the subspaceWis closed under scalar multiplication.
Inductive Step:Assume that the theorem is true for any linear combination ofvectors
inW. We must prove the theorem holds for a linear combination of+1vectors. Suppose
v
1v2v v+1are vectors inW,and 12+1are scalars. We must show that
1v1+2v2+···+ v++1v+1∈W. However, by the inductive hypothesis, we know that
1v1+2v2+···+ v∈W.Also, +1v+1∈W,sinceWis closed under scalar multiplication.
But sinceWis also closed under addition, the sum of any two vectors inWis again inW,so
(
1v1+2v2+···+ v)+( +1v+1)∈W.
(b) Supposew
1w2∈WThenw 1+w2=1w 1+1w 2is afinite linear combination of vectors ofW,
sow
1+w2∈W.Also, 1w1(for 1∈R)isafinite linear combination inW,so 1w1∈W.Now
apply Theorem 4.2.
(21) Apply Theorems 4.4 and 4.3.
(22) (a) F (b) T (c) F (d) T (e) T (f) F (g) T (h) T
Section 4.3
(1) (a){[ −+]| ∈R}
(b){[1
1
3
−
2
3
]|∈R}
(c){[ −]| ∈R}
(d) Span()=R
3
(e){[ −2++]| ∈R}
(f){[ 2+ +]| ∈R}
(2) (a){
3
+
2
+−(++)| ∈R}
(b){
3
+
2
+| ∈R} (c){
3
−+| ∈R}
(3) (a)
½∙
−−−
¸¯
¯
¯
¯
∈R
¾
(b)
½∙
−−8+3
¸¯
¯
¯
¯
∈R
¾
(c)
½∙
¸¯
¯
¯
¯
∈R
¾
=M
22
(4) (a)W=row space ofA=row space of
⎡
⎣
1100
1010
0111
⎤
⎦
={[1100] +[1010] +[0111]| ∈R}
(b)B=
⎡
⎢
⎢
⎢
⎣
100 −
1
2
010
1
2
001
1
2
⎤
⎥
⎥
⎥
⎦
(c) Row space ofB={[100−
1
2
]+[010
1
2
]+[001
1
2
]| ∈R}
={[ −
1
2
+
1
2
+
1
2
]| ∈R}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 61
Answers to Exercises Section 4.3
(5) (a)W=row space ofA=row space of
⎡
⎣
2 1 034
31 −142
−4−1 700
⎤
⎦
={[21034] +[31−142] +[−4−1700]}
(b)B=
⎡
⎣
100 2 −2
010 −18
001 1 0
⎤
⎦
(c) Row space ofB={[ 2−+−2+8]| ∈R}
(6)spansR
3
by the Simplified Span Method since
⎡
⎣
13−1
27−3
48−7
⎤
⎦row reduces to
⎡
⎣
100
010
001
⎤
⎦.
(7)does not spanR
3
by the Simplified Span Method since
⎡
⎢
⎢
⎣
1−22
3−4−1
1−49
02 −7
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎣
10 −5
01−
7
2
00 0
00 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
.
Thus,span()={[5
7
2
]}so, for example,[001]is not in span().
(8)
2
++=(
2
++1)+(−)(+1)+(−)1.
(9) The set does not spanP
2by the Simplified Span Method since
⎡
⎣
14 −3
21 5
07−11
⎤
⎦row reduces to
⎡
⎢
⎢
⎣
10
23
7
01−
11
7
00 0
⎤
⎥
⎥
⎦
.
Thus,span()={−
23
7
2
+
11
7
+}so, for example,0
2
+0+1is not in the span of the set.
(10) (a)[−45−13] = 5[1−2−2]−3[3−51] + 0[−11−5].
(b)does not spanR
3
by the Simplified Span Method since
⎡
⎣
1−2−2
3−51
−11 −5
⎤
⎦row reduces to
⎡
⎣
1012
01 7
00 0
⎤
⎦.
Thus,span()={[−12−7 ]},so,forexample,[001]is not in span().
(11) One answer is
−1(
3
−2
2
+−3) + 2(2
3
−3
2
+2+5)−1(4
2
+−3) + 0(4
3
−7
2
+4−1)
(12) Let
1={[1100][1010][1001][0011]}. The Simplified Span Method shows that 1spans
R
4
. Thus, since 1⊆,also spansR
4
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 62
(5) (a)W=row space ofA=row space of
⎡
⎣
2 1 034
31 −142
−4−1 700
⎤
⎦
={[21034] +[31−142] +[−4−1700]}
(b)B=
⎡
⎣
100 2 −2
010 −18
001 1 0
⎤
⎦
(c) Row space ofB={[ 2−+−2+8]| ∈R}
(6)spansR
3
by the Simplified Span Method since
⎡
⎣
13−1
27−3
48−7
⎤
⎦row reduces to
⎡
⎣
100
010
001
⎤
⎦.
(7)does not spanR
3
by the Simplified Span Method since
⎡
⎢
⎢
⎣
1−22
3−4−1
1−49
02 −7
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎣
10 −5
01−
7
2
00 0
00 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
.
Thus,span()={[5
7
2
]}so, for example,[001]is not in span().
(8)
2
++=(
2
++1)+(−)(+1)+(−)1.
(9) The set does not spanP
2by the Simplified Span Method since
⎡
⎣
14 −3
21 5
07−11
⎤
⎦row reduces to
⎡
⎢
⎢
⎣
10
23
7
01−
11
7
00 0
⎤
⎥
⎥
⎦
.
Thus,span()={−
23
7
2
+
11
7
+}so, for example,0
2
+0+1is not in the span of the set.
(10) (a)[−45−13] = 5[1−2−2]−3[3−51] + 0[−11−5].
(b)does not spanR
3
by the Simplified Span Method since
⎡
⎣
1−2−2
3−51
−11 −5
⎤
⎦row reduces to
⎡
⎣
1012
01 7
00 0
⎤
⎦.
Thus,span()={[−12−7 ]},so,forexample,[001]is not in span().
(11) One answer is
−1(
3
−2
2
+−3) + 2(2
3
−3
2
+2+5)−1(4
2
+−3) + 0(4
3
−7
2
+4−1)
(12) Let
1={[1100][1010][1001][0011]}. The Simplified Span Method shows that 1spans
R
4
. Thus, since 1⊆,also spansR
4
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 62
Answers to Exercises Section 4.3
(13) Letb∈R.Thenb=(
)a∈span({a}). (We have used bold-faced variables when we are considering
objects as vectors rather than scalars. However,aandhave the same value as real numbers; similarly
forband.)
(14) Apply the Simplified Span Method to.Thematrix
⎡
⎣
01 01
10 10
−12 3−23
⎤
⎦represents the vectors in.
Then
⎡
⎣
1000
0101
0010
⎤
⎦is the corresponding reduced row echelon form matrix, which clearly indicates
the desired result.
(15) (a){[−320][405]}
(b)
1={[−
3
2
+
4
5
]}={[−
3
2
10] +[
4
5
01]}={
2
[−320] +
5
[405]}so the set in part (a)
spans
1.
(16) Span(
1)⊆span( 2), since 1v1+···+ v=(− 1)(−v 1)+···+(− )(−v )
Span(
2)⊆span( 1), since 1(−v1)+···+ (−v)=(− 1)v1+···+(− )v
(17) Ifu=v, then all vectors inspan() are scalar multiples ofv6 =0Hence,span() is a line through
the origin. Ifu6 =vthenspan() is the set of all linear combinations of the vectorsuvand since
these vectors point in different directions,span() is a plane through the origin.
(18) First, suppose thatspan()=R
3
.Letxbe a solution toAx=0.Thenu·x=v·x=w·x=0,since
these are the three entries ofAx. Becausespan()=R
3
,thereexist such thatu+v+w=x.
Hence,x·x=(u+v+w)·x=u·x+v·x+w·x=0. Therefore, by part (3) of Theorem
1.5,x=0.Thus,Ax=0has only the trivial solution. Theorem 2.7 and Corollary 3.6 then show that
|A|6 =0.
Next, suppose that|A|6 =0. Then Corollary 3.6 implies thatrank(A)=3. This means that the
reduced row echelon form forAhas three nonzero rows. This can only occur ifArow reduces toI
3.
Thus,Ais row equivalent toI
3. By Theorem 2.9,AandI 3havethesamerowspace.Hence,span()
=row space ofA=row space ofI
3=R
3
.
(19) Choose=max(degree(p
1),, degree(p )).
(20) This follows immediately from Theorem 1.15.
(21) Consider the set{Ψ
}of×matrices, where eachΨ has1as its( )entry, and every remaining
entry equal to0.Any×matrixAcan be expressed as a linear combination of the matrices in{Ψ
}
by letting the coefficient ofΨ
be,the( )entry ofA. Therefore, the set{Ψ }spansM But
since eachΨ
is upper triangular or lower triangular (or both),M is spanned byU ∪L
(22)
1⊆2⊆span( 2), by Theorem 4.5, part (1). Then, sincespan( 2)isasubspaceofVcontaining 1
(by Theorem 4.5, part (2)),span( 1)⊆span( 2) (by Theorem 4.5, part (3)).
(23) (a) Supposeis a subspace. Then⊆span() by Theorem 4.5, part (1). Also,is a subspace
containing,sospan()⊆by Theorem 4.5, part (3). Hence,=span().
Conversely, ifspan()=,thenis a subspace by Theorem 4.5, part (2).
(b)span({skew-symmetric3×3matrices})={skew-symmetric3×3matrices}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 63
(13) Letb∈R.Thenb=(
)a∈span({a}). (We have used bold-faced variables when we are considering
objects as vectors rather than scalars. However,aandhave the same value as real numbers; similarly
forband.)
(14) Apply the Simplified Span Method to.Thematrix
⎡
⎣
01 01
10 10
−12 3−23
⎤
⎦represents the vectors in.
Then
⎡
⎣
1000
0101
0010
⎤
⎦is the corresponding reduced row echelon form matrix, which clearly indicates
the desired result.
(15) (a){[−320][405]}
(b)
1={[−
3
2
+
4
5
]}={[−
3
2
10] +[
4
5
01]}={
2
[−320] +
5
[405]}so the set in part (a)
spans
1.
(16) Span(
1)⊆span( 2), since 1v1+···+ v=(− 1)(−v 1)+···+(− )(−v )
Span(
2)⊆span( 1), since 1(−v1)+···+ (−v)=(− 1)v1+···+(− )v
(17) Ifu=v, then all vectors inspan() are scalar multiples ofv6 =0Hence,span() is a line through
the origin. Ifu6 =vthenspan() is the set of all linear combinations of the vectorsuvand since
these vectors point in different directions,span() is a plane through the origin.
(18) First, suppose thatspan()=R
3
.Letxbe a solution toAx=0.Thenu·x=v·x=w·x=0,since
these are the three entries ofAx. Becausespan()=R
3
,thereexist such thatu+v+w=x.
Hence,x·x=(u+v+w)·x=u·x+v·x+w·x=0. Therefore, by part (3) of Theorem
1.5,x=0.Thus,Ax=0has only the trivial solution. Theorem 2.7 and Corollary 3.6 then show that
|A|6 =0.
Next, suppose that|A|6 =0. Then Corollary 3.6 implies thatrank(A)=3. This means that the
reduced row echelon form forAhas three nonzero rows. This can only occur ifArow reduces toI
3.
Thus,Ais row equivalent toI
3. By Theorem 2.9,AandI 3havethesamerowspace.Hence,span()
=row space ofA=row space ofI
3=R
3
.
(19) Choose=max(degree(p
1),, degree(p )).
(20) This follows immediately from Theorem 1.15.
(21) Consider the set{Ψ
}of×matrices, where eachΨ has1as its( )entry, and every remaining
entry equal to0.Any×matrixAcan be expressed as a linear combination of the matrices in{Ψ
}
by letting the coefficient ofΨ
be,the( )entry ofA. Therefore, the set{Ψ }spansM But
since eachΨ
is upper triangular or lower triangular (or both),M is spanned byU ∪L
(22)
1⊆2⊆span( 2), by Theorem 4.5, part (1). Then, sincespan( 2)isasubspaceofVcontaining 1
(by Theorem 4.5, part (2)),span( 1)⊆span( 2) (by Theorem 4.5, part (3)).
(23) (a) Supposeis a subspace. Then⊆span() by Theorem 4.5, part (1). Also,is a subspace
containing,sospan()⊆by Theorem 4.5, part (3). Hence,=span().
Conversely, ifspan()=,thenis a subspace by Theorem 4.5, part (2).
(b)span({skew-symmetric3×3matrices})={skew-symmetric3×3matrices}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 63
Answers to Exercises Section 4.3
(24) Ifspan( 1)=span( 2), then 1⊆span( 1)=span( 2)and 2⊆span( 2)=span( 1).
Conversely, sincespan(
1)andspan( 2) are subspaces, part (3) of Theorem 4.5 shows that
if
1⊆span( 2)and 2⊆span( 1), thenspan( 1)⊆span( 2)andspan( 2)⊆span( 1), and so
span(
1)=span( 2).
(25) (a) Clearly
1∩2⊆1and 1∩2⊆2. Corollary 4.6 then impliesspan( 1∩2)⊆span( 1)and
span(
1∩2)⊆span( 2). The desired result easily follows.
(b)
1={[100][010]}, 2={[010][001]}
(c)
1={[100][010]}, 2={[100][110]}
(26) (a) Clearly
1⊆1∪2and 2⊆1∪2. Corollary 4.6 then impliesspan( 1)⊆span( 1∪2)and
span(
2)⊆span( 1∪2). The desired result easily follows.
(b) If
1⊆2thenspan( 1)⊆span( 2), sospan( 1)∪span( 2)=span( 2). Also 1∪2=2
(c)
1={
5
},2={
4
}
(27) Step 3 of the Diagonalization Method creates a set={X
1X }of fundamental eigenvectors
where eachX
is the particular solution ofI −A=0obtained by letting theth independent
variable equal1while letting all other independent variables equal0.Now,letX∈
ThenXis a
solution ofI
−A=0SupposeXis obtained by setting theth independent variable equal to
for eachwith1≤≤ThenX=
1X1++ XHenceXis a linear combination of the
X
’s, and thusspans
(28) (a) Letv
1andv 2be two vectors inspan(). By the definition ofspan(), bothv 1andv 2can be
expressed asfinite linear combinations of vectors from.Thatis,therearefinite subsets
1=
{w
1w }and 2={x 1x }ofsuch thatv 1=1w1+···+ wandv 2=1x1+···+ x
for some real numbers 11
(b) The natural thing to do at this point would be to combine the expressions forv
1andv 2by adding
corresponding coefficients. However, each of the subsets{w
1w }or{x 1x }may contain
elements not found in the other. Therefore, we create a larger set
3=1∪2containing all of
the vectors in both subsets. We rename the elements of thefinite subset
3as{z 1z }.Then
v
1=1w1+···+ wcan be expressed as 1z1+···+ z,where =ifz=w ,and
=0ifz ∈{w 1w }Similarly,v 2=1x1+···+ xcan be expressed as 1z1+···+ z,
where
=ifz=x,and =0ifz ∈{x 1x }In this way, bothv 1andv 2can be
expressed as linear combinations of the vectors in
3.
(c) From parts (a) and (b), we have
v
1+v2=
X
=1
z+
X
=1
z=
X
=1
(+)z
a linear combination of the vectorsz
in the subset 3of.Thus,v 1+v2∈span().
(29) Ifcontains a nonzero vectorv, then span()containsv, for every scalar.Now,ifandare
different scalars, we havev6 =v(or else(−)v=0,andthenv=0by part (4) of Theorem 4.1).
That is, distinct scalar multiples ofvform distinct vectors. Because the number of scalars is infinite,
the set of scalar multiples ofvforms an infinite set of vectors, and sospan()isinfinite.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 64
(24) Ifspan( 1)=span( 2), then 1⊆span( 1)=span( 2)and 2⊆span( 2)=span( 1).
Conversely, sincespan(
1)andspan( 2) are subspaces, part (3) of Theorem 4.5 shows that
if
1⊆span( 2)and 2⊆span( 1), thenspan( 1)⊆span( 2)andspan( 2)⊆span( 1), and so
span(
1)=span( 2).
(25) (a) Clearly
1∩2⊆1and 1∩2⊆2. Corollary 4.6 then impliesspan( 1∩2)⊆span( 1)and
span(
1∩2)⊆span( 2). The desired result easily follows.
(b)
1={[100][010]}, 2={[010][001]}
(c)
1={[100][010]}, 2={[100][110]}
(26) (a) Clearly
1⊆1∪2and 2⊆1∪2. Corollary 4.6 then impliesspan( 1)⊆span( 1∪2)and
span(
2)⊆span( 1∪2). The desired result easily follows.
(b) If
1⊆2thenspan( 1)⊆span( 2), sospan( 1)∪span( 2)=span( 2). Also 1∪2=2
(c)
1={
5
},2={
4
}
(27) Step 3 of the Diagonalization Method creates a set={X
1X }of fundamental eigenvectors
where eachX
is the particular solution ofI −A=0obtained by letting theth independent
variable equal1while letting all other independent variables equal0.Now,letX∈
ThenXis a
solution ofI
−A=0SupposeXis obtained by setting theth independent variable equal to
for eachwith1≤≤ThenX=
1X1++ XHenceXis a linear combination of the
X
’s, and thusspans
(28) (a) Letv
1andv 2be two vectors inspan(). By the definition ofspan(), bothv 1andv 2can be
expressed asfinite linear combinations of vectors from.Thatis,therearefinite subsets
1=
{w
1w }and 2={x 1x }ofsuch thatv 1=1w1+···+ wandv 2=1x1+···+ x
for some real numbers 11
(b) The natural thing to do at this point would be to combine the expressions forv
1andv 2by adding
corresponding coefficients. However, each of the subsets{w
1w }or{x 1x }may contain
elements not found in the other. Therefore, we create a larger set
3=1∪2containing all of
the vectors in both subsets. We rename the elements of thefinite subset
3as{z 1z }.Then
v
1=1w1+···+ wcan be expressed as 1z1+···+ z,where =ifz=w ,and
=0ifz ∈{w 1w }Similarly,v 2=1x1+···+ xcan be expressed as 1z1+···+ z,
where
=ifz=x,and =0ifz ∈{x 1x }In this way, bothv 1andv 2can be
expressed as linear combinations of the vectors in
3.
(c) From parts (a) and (b), we have
v
1+v2=
X
=1
z+
X
=1
z=
X
=1
(+)z
a linear combination of the vectorsz
in the subset 3of.Thus,v 1+v2∈span().
(29) Ifcontains a nonzero vectorv, then span()containsv, for every scalar.Now,ifandare
different scalars, we havev6 =v(or else(−)v=0,andthenv=0by part (4) of Theorem 4.1).
That is, distinct scalar multiples ofvform distinct vectors. Because the number of scalars is infinite,
the set of scalar multiples ofvforms an infinite set of vectors, and sospan()isinfinite.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 64
Answers to Exercises Section 4.4
(30) (a) F (b) T (c) F (d) F (e) F (f) T
Section 4.4
(1) (a) Linearly independent ([011]is not the zero vector)
(b) Linearly independent (neither vector is a multiple of the other)
(c) Linearly dependent (each vector is a multiple of the other)
(d) Linearly dependent (contains[000])
(e) Linearly dependent (Theorem 4.7)
(2) Linearly independent: (b), (c), (f). The others are linearly dependent, and the appropriate reduced
row echelon form matrix for each is given.
(a)
⎡
⎣
10 1
01−1
00 0
⎤
⎦ (d)
⎡
⎣
103
012
000
⎤
⎦
(e)
⎡
⎢
⎢
⎢
⎣
10−
1
2
01
1
2
00 0
00 0
⎤
⎥
⎥
⎥
⎦
(3) Linearly independent: (a), (b). The others are linearly dependent, and the appropriate reduced row
echelon form matrix for each is given.
(c)
∙
460 −24
−2−25 9
¸
row reduces to
"
10 3
01−
3
5
#
(d)
⎡
⎣
1111
11 −1−1
1−11 −1
⎤
⎦row reduces to
⎡
⎣
100 −1
010 1
001 1
⎤
⎦
(4) (a) Linearly independent:
⎡
⎣
111
001
−110
⎤
⎦row reduces toI
3.
(b) Linearly independent:
⎡
⎢
⎢
⎣
−101
100
021
1−10
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎣
100
010
001
000
⎤
⎥
⎥
⎦
.
(c) Consider the polynomials inP
2as corresponding vectors inR
3
. Then, the set is linearly dependent
by Theorem 4.7.
(d) Linearly dependent:
⎡
⎢
⎢
⎣
31 0 1
00 0 0
21 1 1
10−5−10
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎣
100
5
2
010 −
13
2
001
5
2
000 0
⎤
⎥
⎥
⎥
⎥
⎦
, although row reduction really is not
necessary since the original matrix clearly has rank≤3due to the row of zeroes.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 65
(30) (a) F (b) T (c) F (d) F (e) F (f) T
Section 4.4
(1) (a) Linearly independent ([011]is not the zero vector)
(b) Linearly independent (neither vector is a multiple of the other)
(c) Linearly dependent (each vector is a multiple of the other)
(d) Linearly dependent (contains[000])
(e) Linearly dependent (Theorem 4.7)
(2) Linearly independent: (b), (c), (f). The others are linearly dependent, and the appropriate reduced
row echelon form matrix for each is given.
(a)
⎡
⎣
10 1
01−1
00 0
⎤
⎦ (d)
⎡
⎣
103
012
000
⎤
⎦
(e)
⎡
⎢
⎢
⎢
⎣
10−
1
2
01
1
2
00 0
00 0
⎤
⎥
⎥
⎥
⎦
(3) Linearly independent: (a), (b). The others are linearly dependent, and the appropriate reduced row
echelon form matrix for each is given.
(c)
∙
460 −24
−2−25 9
¸
row reduces to
"
10 3
01−
3
5
#
(d)
⎡
⎣
1111
11 −1−1
1−11 −1
⎤
⎦row reduces to
⎡
⎣
100 −1
010 1
001 1
⎤
⎦
(4) (a) Linearly independent:
⎡
⎣
111
001
−110
⎤
⎦row reduces toI
3.
(b) Linearly independent:
⎡
⎢
⎢
⎣
−101
100
021
1−10
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎣
100
010
001
000
⎤
⎥
⎥
⎦
.
(c) Consider the polynomials inP
2as corresponding vectors inR
3
. Then, the set is linearly dependent
by Theorem 4.7.
(d) Linearly dependent:
⎡
⎢
⎢
⎣
31 0 1
00 0 0
21 1 1
10−5−10
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎣
100
5
2
010 −
13
2
001
5
2
000 0
⎤
⎥
⎥
⎥
⎥
⎦
, although row reduction really is not
necessary since the original matrix clearly has rank≤3due to the row of zeroes.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 65
Answers to Exercises Section 4.4
(e) Linearly independent. See the remarks just before Example 13 in the textbook. No polynomial
inthesetcanbeexpressedasafinite linear combination of the others since no single power of
can be expressed as a combination of other powers of.
(f) Linearly independent. Use an argument similar to that in Example 13 in the textbook.
(5) Solving the appropriate linear system yields
21
∙
1−2
01
¸
+15
∙
32
−61
¸
−18
∙
4−1
−52
¸
+2
∙
3−3
00
¸
=
∙
00
00
¸
.
(6)
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1400
2217
−1−615
11 −12
302 −1
0126
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1000
0100
0010
0001
0000
0000
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
, so the set is linearly independent by the
Independence Test Method.
(7) (a)[−201]is not a scalar multiple of[1
10].
(b)[010][001]
(c) Any nonzero linear combination of[110]and[−201], other than[110]and[−201]them-
selves, will work. One possibility isu=[110] + [−201] = [−111].
(8) (a) Applying the Independence Test Method, we obtain a pivot in every column.
(b)[−203−4] =−1[2−105] + 1[1−120] + 1[−101
1].
(c) No, becauseis linearly independent (see Theorem 4.9).
(9) (a)
3
−4+8=2(2
3
−+3)−(3
3
+2−2).(Coefficients were obtained using Independence
Test Method.)
(b) See part (a). Also, solving appropriate systems yields:
4
3
+5−7=−1(2
3
−+3)+2(3
3
+2−2);
2
3
−+3=
2
3
(
3
−4+8)+
1
3
(4
3
+5−7);
3
3
+2−2=
1
3
(
3
−4+8)+
2
3
(4
3
+5−7)
(c) No polynomial inis a scalar multiple of any other polynomial in.
(10) Following the hint in the textbook, letAbe the matrix whose rows are the vectorsu,v,andw.By
the Independence Test Method,is linearly independent iffA
row reduces to a matrix with a pivot
in every column iffA
has rank3iff|A
|6 =0iff|A|6 =0.
(11) In each part, there are many different correct answers. Only one possibility is given here.
(a){e
1e2e3e4} (b){e 1e2e3e4} (c){1
2
3
}
(d)
½∙
100
000
¸
∙
010
000
¸
∙
001
000
¸
∙
000
100
¸¾
(e)
⎧
⎨
⎩
⎡
⎣
100
000
000
⎤
⎦
⎡
⎣
010
100
000
⎤
⎦
⎡
⎣
001
000
100
⎤
⎦
⎡
⎣
000
010
000
⎤
⎦
⎫
⎬
⎭
(Notice that each matrix is symmetric.)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 66
(e) Linearly independent. See the remarks just before Example 13 in the textbook. No polynomial
inthesetcanbeexpressedasafinite linear combination of the others since no single power of
can be expressed as a combination of other powers of.
(f) Linearly independent. Use an argument similar to that in Example 13 in the textbook.
(5) Solving the appropriate linear system yields
21
∙
1−2
01
¸
+15
∙
32
−61
¸
−18
∙
4−1
−52
¸
+2
∙
3−3
00
¸
=
∙
00
00
¸
.
(6)
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1400
2217
−1−615
11 −12
302 −1
0126
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1000
0100
0010
0001
0000
0000
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
, so the set is linearly independent by the
Independence Test Method.
(7) (a)[−201]is not a scalar multiple of[1
10].
(b)[010][001]
(c) Any nonzero linear combination of[110]and[−201], other than[110]and[−201]them-
selves, will work. One possibility isu=[110] + [−201] = [−111].
(8) (a) Applying the Independence Test Method, we obtain a pivot in every column.
(b)[−203−4] =−1[2−105] + 1[1−120] + 1[−101
1].
(c) No, becauseis linearly independent (see Theorem 4.9).
(9) (a)
3
−4+8=2(2
3
−+3)−(3
3
+2−2).(Coefficients were obtained using Independence
Test Method.)
(b) See part (a). Also, solving appropriate systems yields:
4
3
+5−7=−1(2
3
−+3)+2(3
3
+2−2);
2
3
−+3=
2
3
(
3
−4+8)+
1
3
(4
3
+5−7);
3
3
+2−2=
1
3
(
3
−4+8)+
2
3
(4
3
+5−7)
(c) No polynomial inis a scalar multiple of any other polynomial in.
(10) Following the hint in the textbook, letAbe the matrix whose rows are the vectorsu,v,andw.By
the Independence Test Method,is linearly independent iffA
row reduces to a matrix with a pivot
in every column iffA
has rank3iff|A
|6 =0iff|A|6 =0.
(11) In each part, there are many different correct answers. Only one possibility is given here.
(a){e
1e2e3e4} (b){e 1e2e3e4} (c){1
2
3
}
(d)
½∙
100
000
¸
∙
010
000
¸
∙
001
000
¸
∙
000
100
¸¾
(e)
⎧
⎨
⎩
⎡
⎣
100
000
000
⎤
⎦
⎡
⎣
010
100
000
⎤
⎦
⎡
⎣
001
000
100
⎤
⎦
⎡
⎣
000
010
000
⎤
⎦
⎫
⎬
⎭
(Notice that each matrix is symmetric.)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 66
Answers to Exercises Section 4.4
(12) (a) Ifis linearly dependent, then 1v1+···+ v=0for some{v 1v }⊆and 1∈R,
with some
6 =0.Then
v
=−
1
v1−···−
−1
v−1−
+1
v+1−···−
v
So, ifw∈span(),thenw=v
+1w1+···+ w,forsome{w 1w }⊆−{v }and
1∈R. Hence,
w=−
1
v1−···−
−1
v−1−
+1
v+1−···−
v+1w1+···+ w∈span(−{v })
So,span()⊆span(−{v
}).Also,span(−{v })⊆span()by Corollary 4.6. Thus, we have
span(−{v
})=span()
Conversely, ifspan(−{v})=span()forsomev∈, then by Theorem 4.5, part (1), we have
v∈span()=span(−{v}),andsois linearly dependent by the Alternate Characterization
of linear dependence in Table 4.1.
(b) Supposev∈is redundant. Then,v∈span()=span(−{v}),andsovis a linear combination
of vectors in−{v}.
Conversely, supposevis a linear combination of vectors in−{v}.Thenv=
1v1+···+
v,forsomev 1v ∈−{v}. We need to show thatspan()=span(−{v}).Now,
span(−{v})⊆span()by Corollary 4.6. So, we need to show thatspan()⊆span(−{v}).
Supposew∈span().Thenw=v+
1w1+···+ w,forsome{w 1w }⊆−{v}and
1∈R. Hence,
w=
1v1+···+ v+1w1+···+ w∈span(−{v})
completing the proof.
(13) In each part, you can prove that the indicated vector is redundant by creating a matrix using the given
vectors as rows, and showing that the Simplified Span Method leads to the same reduced row echelon
form (except, perhaps, with an extra row of zeroes) with or without the indicated vector as one of the
rows.
(a) Letv=[0000]. Any linear combination of the other three vectors becomes a linear combination
of all four vectors when1[0000]is added to it.
(b) Letv=[00−60]Letbe the given set of three vectors. To prove thatv=[00−60]is
redundant, we need to show thatspan(−{v})=span(). We apply the Simplified Span Method
to both−{v}and.
Forspan(−{v}):
∙
1100
1110
¸
row reduces to
∙
1100
0010
¸
.
Forspan(
):
⎡
⎣
11 00
11 10
00−60
⎤
⎦row reduces to
⎡
⎣
1100
0010
0000
⎤
⎦.
Since the reduced row echelon form matrices are the same, except for the extra row of zeroes, the
two spans are equal, andv=[00−60]is a redundant vector.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 67
(12) (a) Ifis linearly dependent, then 1v1+···+ v=0for some{v 1v }⊆and 1∈R,
with some
6 =0.Then
v
=−
1
v1−···−
−1
v−1−
+1
v+1−···−
v
So, ifw∈span(),thenw=v
+1w1+···+ w,forsome{w 1w }⊆−{v }and
1∈R. Hence,
w=−
1
v1−···−
−1
v−1−
+1
v+1−···−
v+1w1+···+ w∈span(−{v })
So,span()⊆span(−{v
}).Also,span(−{v })⊆span()by Corollary 4.6. Thus, we have
span(−{v
})=span()
Conversely, ifspan(−{v})=span()forsomev∈, then by Theorem 4.5, part (1), we have
v∈span()=span(−{v}),andsois linearly dependent by the Alternate Characterization
of linear dependence in Table 4.1.
(b) Supposev∈is redundant. Then,v∈span()=span(−{v}),andsovis a linear combination
of vectors in−{v}.
Conversely, supposevis a linear combination of vectors in−{v}.Thenv=
1v1+···+
v,forsomev 1v ∈−{v}. We need to show thatspan()=span(−{v}).Now,
span(−{v})⊆span()by Corollary 4.6. So, we need to show thatspan()⊆span(−{v}).
Supposew∈span().Thenw=v+
1w1+···+ w,forsome{w 1w }⊆−{v}and
1∈R. Hence,
w=
1v1+···+ v+1w1+···+ w∈span(−{v})
completing the proof.
(13) In each part, you can prove that the indicated vector is redundant by creating a matrix using the given
vectors as rows, and showing that the Simplified Span Method leads to the same reduced row echelon
form (except, perhaps, with an extra row of zeroes) with or without the indicated vector as one of the
rows.
(a) Letv=[0000]. Any linear combination of the other three vectors becomes a linear combination
of all four vectors when1[0000]is added to it.
(b) Letv=[00−60]Letbe the given set of three vectors. To prove thatv=[00−60]is
redundant, we need to show thatspan(−{v})=span(). We apply the Simplified Span Method
to both−{v}and.
Forspan(−{v}):
∙
1100
1110
¸
row reduces to
∙
1100
0010
¸
.
Forspan(
):
⎡
⎣
11 00
11 10
00−60
⎤
⎦row reduces to
⎡
⎣
1100
0010
0000
⎤
⎦.
Since the reduced row echelon form matrices are the same, except for the extra row of zeroes, the
two spans are equal, andv=[00−60]is a redundant vector.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 67
Answers to Exercises Section 4.4
For an alternate approach to proving thatspan(−{v})=span(),first note that−{v}⊆,
and sospan(−{v})⊆span()by Corollary 4.6. Next,[1100]and[1110]are clearly in
span(−{v})by part (1) of Theorem 4.5. Butv∈span(−{v})as well, because
[00−60] = (6)[1100] + (−6)[1110]
Hence,is a subset of the subspacespan(−{v})
. Therefore, by part (3) of Theorem 4.5,
span()⊆span(−{v})
Thus, since we have proven subset inclusion in both directions,span(−{v})=span().
(c) Letbe the given set of vectors. Any of the given16vectors incanbeconsideredasthe
redundant vector. One clever way to see this is to consider the matrix
A=
⎡
⎢
⎢
⎣
1111
1−111
11 −11
111 −1
⎤
⎥
⎥
⎦
whose rows are4of the vectors in.Now,Arow reduces toI
4.Thus,thefourrowsofAspan
R
4
, by the Simplified Span Method. Hence, any of the remaining12vectors inare redundant
sincetheyareinthespanofthese4rows. Now, repeat this argument using−Ato show that the
four rows inAare also individually redundant. (Note thatAand−Aare row equivalent, so we
do not have to perform a second row reduction.)
(14) Assume
1is linearly independent. Suppose 1(v1)+ 2(v2)+···+ (v)=0Divide both
sides byand use the fact that{v
1v2v }is linearly independent to obtain 1=2=···=
=0Thus 2is linearly independent. Conversely, assume 2is linearly independent. Suppose
1v1+2v2+···+ v=0Multiply both sides byand use the fact that{v 1v2v }is
linearly independent to obtain
1=2=···= =0Thus 1is linearly independent.
(15) Notice thatf
0
()is not a scalar multiple off(). For if the terms off()are written in order
of descending degree, and thefirst two nonzero terms are, respectively,
and
, then the
corresponding terms off
0
()are
and
,whicharedifferent multiples of
and
,
respectively, since6 =.
(16) Theorem 4.5, part (3) shows thatspan()⊆W.Thusv6 ∈span(). Now if∪{v}is linearly
dependent, then there exists{w
1w }⊆so that 1w1+···+ w+v=0, with not all of
1equal to zero. Also,6 =0sinceis linearly independent. Hence,
v=−
1
w
1−···−
w
∈span()
a contradiction.
(17) (a) Suppose
1v1+···+ v=0.Then 1Av1+···+ Av=0=⇒ 1=···= =0,since
is linearly independent.
(b) The converse to the statement is: If={v
1v }is a linearly independent subset ofR
,
then={Av
1Av }is a linearly independent subset ofR
withAv 1Av distinct.
We can construct specific counterexamples that contradict either (or both) of the conclusions of
the converse. For a specific counterexample in whichAv
1Av are distinct butis linearly
Copyrightc°2016 Elsevier Ltd. All rights reserved. 68
For an alternate approach to proving thatspan(−{v})=span(),first note that−{v}⊆,
and sospan(−{v})⊆span()by Corollary 4.6. Next,[1100]and[1110]are clearly in
span(−{v})by part (1) of Theorem 4.5. Butv∈span(−{v})as well, because
[00−60] = (6)[1100] + (−6)[1110]
Hence,is a subset of the subspacespan(−{v})
. Therefore, by part (3) of Theorem 4.5,
span()⊆span(−{v})
Thus, since we have proven subset inclusion in both directions,span(−{v})=span().
(c) Letbe the given set of vectors. Any of the given16vectors incanbeconsideredasthe
redundant vector. One clever way to see this is to consider the matrix
A=
⎡
⎢
⎢
⎣
1111
1−111
11 −11
111 −1
⎤
⎥
⎥
⎦
whose rows are4of the vectors in.Now,Arow reduces toI
4.Thus,thefourrowsofAspan
R
4
, by the Simplified Span Method. Hence, any of the remaining12vectors inare redundant
sincetheyareinthespanofthese4rows. Now, repeat this argument using−Ato show that the
four rows inAare also individually redundant. (Note thatAand−Aare row equivalent, so we
do not have to perform a second row reduction.)
(14) Assume
1is linearly independent. Suppose 1(v1)+ 2(v2)+···+ (v)=0Divide both
sides byand use the fact that{v
1v2v }is linearly independent to obtain 1=2=···=
=0Thus 2is linearly independent. Conversely, assume 2is linearly independent. Suppose
1v1+2v2+···+ v=0Multiply both sides byand use the fact that{v 1v2v }is
linearly independent to obtain
1=2=···= =0Thus 1is linearly independent.
(15) Notice thatf
0
()is not a scalar multiple off(). For if the terms off()are written in order
of descending degree, and thefirst two nonzero terms are, respectively,
and
, then the
corresponding terms off
0
()are
and
,whicharedifferent multiples of
and
,
respectively, since6 =.
(16) Theorem 4.5, part (3) shows thatspan()⊆W.Thusv6 ∈span(). Now if∪{v}is linearly
dependent, then there exists{w
1w }⊆so that 1w1+···+ w+v=0, with not all of
1equal to zero. Also,6 =0sinceis linearly independent. Hence,
v=−
1
w
1−···−
w
∈span()
a contradiction.
(17) (a) Suppose
1v1+···+ v=0.Then 1Av1+···+ Av=0=⇒ 1=···= =0,since
is linearly independent.
(b) The converse to the statement is: If={v
1v }is a linearly independent subset ofR
,
then={Av
1Av }is a linearly independent subset ofR
withAv 1Av distinct.
We can construct specific counterexamples that contradict either (or both) of the conclusions of
the converse. For a specific counterexample in whichAv
1Av are distinct butis linearly
Copyrightc°2016 Elsevier Ltd. All rights reserved. 68
Answers to Exercises Section 4.4
dependent, letA=
∙
11
11
¸
and={[10][12]}.Then={[11][33]}.Notethatis
linearly independent, but the vectorsAv
1andAv 2are distinct andis linearly dependent. For
aspecific counterexample in whichAv
1Av are not distinct butis linearly independent,
useA=
∙
11
11
¸
and={[12][21]}.Then={[33]}.NotethatAv
1andAv 2both equal
[33], and that bothandare linearly independent. For afinal counterexample in which both
parts of the conclusion are false, letA=O
22and={e 1e2}.ThenAv 1andAv 2both equal
0,is linearly independent, but={0}is linearly dependent.
(c) The converse to the statement is: If={v
1v }is a linearly independent subset ofR
,
then={Av
1Av }is a linearly independent subset ofR
withAv 1Av distinct.
To prove thatAv
1Av are distinct, we assumeAv =Av ,with6 =, and then multiply
both sides byA
−1
to obtainv =v, a contradiction. To prove{Av 1Av }is linearly
independent, suppose
1Av1+···+ Av=0.Then 1A
−1
Av1+···+ A
−1
Av=0=⇒
1v1+···+ v=0=⇒ 1=···= =0,sinceis linearly independent.
(18) Case 1:is empty. This case is obvious, since the empty set is the only subset of.
Case 2: Suppose={v
1v }is afinite nonempty linearly independent set inV,andlet 1
be a subset of.If 1is empty, it is linearly independent by definition. Suppose 1is nonempty.
After renumbering the subscripts if necessary, we can assume
1={v 1v },with≤.If
1v1+···+ v=0,then 1v1+···+ v+0v +1+···0v =0,andso 1=···= =0,since
is linearly independent. Hence,
1is linearly independent.
Case 3: Supposeis an infinite linearly independent subset ofV.Thenanyfinite subset ofis linearly
independent by the definition of linear independence for infinite subsets. If
1is an infinite subset,
then anyfinite subset
2of1is also afinite subset of. Hence 2is linearly independent because
is. Thus,
1is linearly independent, by definition.
(19) First, suppose thatv
16 =0and that for every,with2≤≤,wehavev 6 ∈span({v 1v −1}).It
is enough to show thatis linearly independent by assuming thatis linearlydependentand showing
that this leads to a contradiction. Ifis linearly dependent, then there real numbers
1,not
all zero, such that
0=
1v1+···+ v+···+ v
Supposeis the largest subscript such that
6 =0.Then
0=
1v1+···+ v
If=1,then
1v1=0with 16 =0.Thisimpliesthatv 1=0, a contradiction. If≥2,thensolving
forv
yields
v
=
µ
−
1
¶
v
1+···+
µ
−
−1
¶
v
−1
wherev
is expressed as a linear combination ofv 1v −1, contradicting the fact that
v
6∈span({v 1v −1}).
Conversely, supposeis linearly independent. Notice thatv
16 =0since anyfinite subset ofV
containing0is linearly dependent. We must prove that for eachwith1≤≤,
v
6∈span({v 1v −1}). Now, by Corollary 4.6,span({v 1v −1})⊆span(−{v })since
{v
1v −1}⊆−{v }.But,bythefirst “boxed” alternate characterization of linear independence
following Example 10 in the textbook,v
6 ∈span(−{v }). Therefore,v 6 ∈span({v 1v −1}).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 69
dependent, letA=
∙
11
11
¸
and={[10][12]}.Then={[11][33]}.Notethatis
linearly independent, but the vectorsAv
1andAv 2are distinct andis linearly dependent. For
aspecific counterexample in whichAv
1Av are not distinct butis linearly independent,
useA=
∙
11
11
¸
and={[12][21]}.Then={[33]}.NotethatAv
1andAv 2both equal
[33], and that bothandare linearly independent. For afinal counterexample in which both
parts of the conclusion are false, letA=O
22and={e 1e2}.ThenAv 1andAv 2both equal
0,is linearly independent, but={0}is linearly dependent.
(c) The converse to the statement is: If={v
1v }is a linearly independent subset ofR
,
then={Av
1Av }is a linearly independent subset ofR
withAv 1Av distinct.
To prove thatAv
1Av are distinct, we assumeAv =Av ,with6 =, and then multiply
both sides byA
−1
to obtainv =v, a contradiction. To prove{Av 1Av }is linearly
independent, suppose
1Av1+···+ Av=0.Then 1A
−1
Av1+···+ A
−1
Av=0=⇒
1v1+···+ v=0=⇒ 1=···= =0,sinceis linearly independent.
(18) Case 1:is empty. This case is obvious, since the empty set is the only subset of.
Case 2: Suppose={v
1v }is afinite nonempty linearly independent set inV,andlet 1
be a subset of.If 1is empty, it is linearly independent by definition. Suppose 1is nonempty.
After renumbering the subscripts if necessary, we can assume
1={v 1v },with≤.If
1v1+···+ v=0,then 1v1+···+ v+0v +1+···0v =0,andso 1=···= =0,since
is linearly independent. Hence,
1is linearly independent.
Case 3: Supposeis an infinite linearly independent subset ofV.Thenanyfinite subset ofis linearly
independent by the definition of linear independence for infinite subsets. If
1is an infinite subset,
then anyfinite subset
2of1is also afinite subset of. Hence 2is linearly independent because
is. Thus,
1is linearly independent, by definition.
(19) First, suppose thatv
16 =0and that for every,with2≤≤,wehavev 6 ∈span({v 1v −1}).It
is enough to show thatis linearly independent by assuming thatis linearlydependentand showing
that this leads to a contradiction. Ifis linearly dependent, then there real numbers
1,not
all zero, such that
0=
1v1+···+ v+···+ v
Supposeis the largest subscript such that
6 =0.Then
0=
1v1+···+ v
If=1,then
1v1=0with 16 =0.Thisimpliesthatv 1=0, a contradiction. If≥2,thensolving
forv
yields
v
=
µ
−
1
¶
v
1+···+
µ
−
−1
¶
v
−1
wherev
is expressed as a linear combination ofv 1v −1, contradicting the fact that
v
6∈span({v 1v −1}).
Conversely, supposeis linearly independent. Notice thatv
16 =0since anyfinite subset ofV
containing0is linearly dependent. We must prove that for eachwith1≤≤,
v
6∈span({v 1v −1}). Now, by Corollary 4.6,span({v 1v −1})⊆span(−{v })since
{v
1v −1}⊆−{v }.But,bythefirst “boxed” alternate characterization of linear independence
following Example 10 in the textbook,v
6 ∈span(−{v }). Therefore,v 6 ∈span({v 1v −1}).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 69
Answers to Exercises Section 4.4
(20) (a) Reversing the order of the elements gives{727236
2
−3012
3
−15
2
+43
4
−5
3
+4−8}.
Each polynomial in the set now has a higher degree than those preceding it, and hence is not
a linear combination of those preceding polynomials. So, by Exercise 19 this set is linearly
independent.
(b) Reversing the order of the elements gives
©
f
()
f
(2)
f
(1)
f
ª
.Buteachf
()
6 =+1f
(+1)
+
···+
f
()
, for any +1, because all polynomials on the right side of the equation have
lower degrees thanf
()
. Thus, by Exercise 19 this set is linearly independent.
(21) (a) Ifis linearly independent, then the zero vector has a unique expression by Theorem 4.9.
Conversely, supposev∈span()has a unique expression of the formv=
1v1+···+ v,where
v
1v are in. To show thatis linearly independent, it is enough to show that anyfinite
subset
1={w 1w }ofis linearly independent. Suppose 1w1+···+ w=0.Now
1w1+···+ w+1v1+···+ v=0+v=v.Ifw =vfor some, then the uniqueness
of the expression forvimplies that
+=, yielding =0.If,instead,w does not equal
anyv
,then =0, again by the uniqueness assumption. Thus each must equal zero. Hence,
1is linearly independent by the definition of linear independence.
(b)is linearly dependent iffevery vectorvinspan() can be expressed in more than one way as a
linear combination of vectors in(ignoring zero coefficients).
(22) Follow the hint in the text. The fundamental eigenvectors are found by solving the homogeneous system
(I
−A)v=0.Eachv has a “1” in the position corresponding to theth independent variable for
the system and a “0” in the position corresponding to every other independent variable. Hence, any
linear combination
1v1+···+ vis an-vector having the value in the position corresponding
to theth independent variable for the system, for each. So if this linear combination equals0,then
each
must equal0, proving linear independence.
(23) (a) Since∪{v}is linearly dependent, by the definition of linear independence, there is somefinite
subset{t
1t }ofsuch that 1t1+···+ t+v=0with not all coefficients equal to
zero. But if=0,then
1t1+···+ t=0with not all =0, contradicting the fact thatis
linearly independent. Hence,6 =0and thusv=(−
1
)t1+···+(−
)tHence,v∈span().
(b) The statement in part (b) is merely the contrapositive of the statement in part (a).
(24) Suppose thatis linearly independent, and supposev∈span()can be expressed both as
v=
1u1+···+ uandv= 1v1+···+ vfor distinctu 1u ∈and distinct
v
1v ∈, where these expressions differ in at least one nonzero term. Since theu ’s might not
be distinct from thev
’s, we consider the set={u 1u }∪{v 1v }and label the distinct
vectors inas{w
1w }Then we can expressv= 1u1+···+ uasv= 1w1+···+ w
and alsov=
1v1+···+ vasv= 1w1+···+ wchoosing the scalars 1≤≤
with
=ifw=u,=0otherwise, and =ifw=v,=0otherwise. Since the
original linear combinations forvare distinct, we know that
6 =for some.Now,v−v=
0=(
1−1)w1+···+( −)wSince{w 1w }⊆a linearly independent set, each
−=0for everywith1≤≤. But this is a contradiction since 6 =
Conversely, assume every vector inspan() can be uniquely expressed as a linear combination of
elements ofSince0∈span(), there is exactly one linear combination of elements ofthat equals0.
Now, if{v
1v }is anyfinite subset of,wehave0=0v 1+···+0v . Because this representation
is unique, it means that in any linear combination of{v
1v }that equals0, the only possible
coefficients are zeroes. Thus, by definition,is linearly independent.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 70
(20) (a) Reversing the order of the elements gives{727236
2
−3012
3
−15
2
+43
4
−5
3
+4−8}.
Each polynomial in the set now has a higher degree than those preceding it, and hence is not
a linear combination of those preceding polynomials. So, by Exercise 19 this set is linearly
independent.
(b) Reversing the order of the elements gives
©
f
()
f
(2)
f
(1)
f
ª
.Buteachf
()
6 =+1f
(+1)
+
···+
f
()
, for any +1, because all polynomials on the right side of the equation have
lower degrees thanf
()
. Thus, by Exercise 19 this set is linearly independent.
(21) (a) Ifis linearly independent, then the zero vector has a unique expression by Theorem 4.9.
Conversely, supposev∈span()has a unique expression of the formv=
1v1+···+ v,where
v
1v are in. To show thatis linearly independent, it is enough to show that anyfinite
subset
1={w 1w }ofis linearly independent. Suppose 1w1+···+ w=0.Now
1w1+···+ w+1v1+···+ v=0+v=v.Ifw =vfor some, then the uniqueness
of the expression forvimplies that
+=, yielding =0.If,instead,w does not equal
anyv
,then =0, again by the uniqueness assumption. Thus each must equal zero. Hence,
1is linearly independent by the definition of linear independence.
(b)is linearly dependent iffevery vectorvinspan() can be expressed in more than one way as a
linear combination of vectors in(ignoring zero coefficients).
(22) Follow the hint in the text. The fundamental eigenvectors are found by solving the homogeneous system
(I
−A)v=0.Eachv has a “1” in the position corresponding to theth independent variable for
the system and a “0” in the position corresponding to every other independent variable. Hence, any
linear combination
1v1+···+ vis an-vector having the value in the position corresponding
to theth independent variable for the system, for each. So if this linear combination equals0,then
each
must equal0, proving linear independence.
(23) (a) Since∪{v}is linearly dependent, by the definition of linear independence, there is somefinite
subset{t
1t }ofsuch that 1t1+···+ t+v=0with not all coefficients equal to
zero. But if=0,then
1t1+···+ t=0with not all =0, contradicting the fact thatis
linearly independent. Hence,6 =0and thusv=(−
1
)t1+···+(−
)tHence,v∈span().
(b) The statement in part (b) is merely the contrapositive of the statement in part (a).
(24) Suppose thatis linearly independent, and supposev∈span()can be expressed both as
v=
1u1+···+ uandv= 1v1+···+ vfor distinctu 1u ∈and distinct
v
1v ∈, where these expressions differ in at least one nonzero term. Since theu ’s might not
be distinct from thev
’s, we consider the set={u 1u }∪{v 1v }and label the distinct
vectors inas{w
1w }Then we can expressv= 1u1+···+ uasv= 1w1+···+ w
and alsov=
1v1+···+ vasv= 1w1+···+ wchoosing the scalars 1≤≤
with
=ifw=u,=0otherwise, and =ifw=v,=0otherwise. Since the
original linear combinations forvare distinct, we know that
6 =for some.Now,v−v=
0=(
1−1)w1+···+( −)wSince{w 1w }⊆a linearly independent set, each
−=0for everywith1≤≤. But this is a contradiction since 6 =
Conversely, assume every vector inspan() can be uniquely expressed as a linear combination of
elements ofSince0∈span(), there is exactly one linear combination of elements ofthat equals0.
Now, if{v
1v }is anyfinite subset of,wehave0=0v 1+···+0v . Because this representation
is unique, it means that in any linear combination of{v
1v }that equals0, the only possible
coefficients are zeroes. Thus, by definition,is linearly independent.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 70
Answers to Exercises Section 4.5
(25) (a)F (b)T (c)T (d)F (e)T (f)T (g)F (h)T (i)T
Section 4.5
(1) Foreachpart,rowreducethematrixwhoserowsarethegivenvectors. Theresultwillbe I 4. Hence,
by the Simplified Span Method, the set of vectors spansR
4
. Row reducing the matrix whose columns
are the given vectors also results inI
4. Therefore, by the Independence Test Method, the set of vectors
is linearly independent. (After thefirst row reduction, you could just apply part (1) of Theorem 4.12,
andthenskipthesecondrowreduction.)
(2) Follow the procedure in the answer to Exercise 1.
(3) Follow the procedure in the answer to Exercise 1, except that row reduction results inI
5instead ofI 4.
(4) (a) Not a basis:||dim(R
4
). (linearly independent but does not span)
(b) Not a basis:||dim(R
4
). (linearly dependent and does not span)
(c) Basis: Follow the procedure in the answer to Exercise 1.
(d) Not a basis:
⎡
⎢
⎢
⎣
13 2 −1
−20 6 −10
0610 −12
210−331
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎣
100 −4
010 3
001 −3
000 0
⎤
⎥
⎥
⎦
, and so does not span.
⎡
⎢
⎢
⎣
1−202
30610
2610 −3
−1−10−12 31
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎣
1020
0110
0001
0000
⎤
⎥
⎥
⎦
, and so is linearly dependent.
(e) Not a basis:||dim(R
4
). (linearly dependent but spans)
(5) (a)Wis nonempty, since0∈WLetX
1X2∈W∈RThenA(X 1+X 2)=AX 1+AX 2=
0+0=0Also,A(X
1)=(AX 1)=0=0HenceWis closed under addition and scalar
multiplication, and so is a subspace by Theorem 4.2.
(b)Arow reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎣
10
1
5
2
5
1
01
2
5
−
1
5
−1
00 0 0 0
00 0 0 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
.
Thus, a basis forWis{[−1−2500][−21050][−11001]}.
(c) dim(W)=3;rank(A)=2;3+2=5
(6) (a)Wis nonempty, since0∈WLetX
1X2∈W∈RThenA(X 1+X 2)=AX 1+AX 2=
0+0=0Also,A(X
1)=(AX 1)=0=0HenceWis closed under addition and scalar
multiplication, and so is a subspace by Theorem 4.2.
(b)Arow reduces to
⎡
⎢
⎢
⎢
⎢
⎣
10−20
01 30
00 01
00 00
00 00
⎤
⎥
⎥
⎥
⎥
⎦
.
Thus, a basis forWis{[2−310]}.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 71
(25) (a)F (b)T (c)T (d)F (e)T (f)T (g)F (h)T (i)T
Section 4.5
(1) Foreachpart,rowreducethematrixwhoserowsarethegivenvectors. Theresultwillbe I 4. Hence,
by the Simplified Span Method, the set of vectors spansR
4
. Row reducing the matrix whose columns
are the given vectors also results inI
4. Therefore, by the Independence Test Method, the set of vectors
is linearly independent. (After thefirst row reduction, you could just apply part (1) of Theorem 4.12,
andthenskipthesecondrowreduction.)
(2) Follow the procedure in the answer to Exercise 1.
(3) Follow the procedure in the answer to Exercise 1, except that row reduction results inI
5instead ofI 4.
(4) (a) Not a basis:||dim(R
4
). (linearly independent but does not span)
(b) Not a basis:||dim(R
4
). (linearly dependent and does not span)
(c) Basis: Follow the procedure in the answer to Exercise 1.
(d) Not a basis:
⎡
⎢
⎢
⎣
13 2 −1
−20 6 −10
0610 −12
210−331
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎣
100 −4
010 3
001 −3
000 0
⎤
⎥
⎥
⎦
, and so does not span.
⎡
⎢
⎢
⎣
1−202
30610
2610 −3
−1−10−12 31
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎣
1020
0110
0001
0000
⎤
⎥
⎥
⎦
, and so is linearly dependent.
(e) Not a basis:||dim(R
4
). (linearly dependent but spans)
(5) (a)Wis nonempty, since0∈WLetX
1X2∈W∈RThenA(X 1+X 2)=AX 1+AX 2=
0+0=0Also,A(X
1)=(AX 1)=0=0HenceWis closed under addition and scalar
multiplication, and so is a subspace by Theorem 4.2.
(b)Arow reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎣
10
1
5
2
5
1
01
2
5
−
1
5
−1
00 0 0 0
00 0 0 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
.
Thus, a basis forWis{[−1−2500][−21050][−11001]}.
(c) dim(W)=3;rank(A)=2;3+2=5
(6) (a)Wis nonempty, since0∈WLetX
1X2∈W∈RThenA(X 1+X 2)=AX 1+AX 2=
0+0=0Also,A(X
1)=(AX 1)=0=0HenceWis closed under addition and scalar
multiplication, and so is a subspace by Theorem 4.2.
(b)Arow reduces to
⎡
⎢
⎢
⎢
⎢
⎣
10−20
01 30
00 01
00 00
00 00
⎤
⎥
⎥
⎥
⎥
⎦
.
Thus, a basis forWis{[2−310]}.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 71
Answers to Exercises Section 4.5
(c) dim(W)=1;rank(A)=3;1+3=4
(7) These vectors are linearly independent by part (b) of Exercise 20 in Section 4.4. The given set has
+1distinct elements, anddim(P
)=+1. Hence, by part (2) of Theorem 4.12, the set is a basis
forP
.
(8) (a) Let={I
2AA
2
A
3
A
4
}.Ifis linearly independent, thenis a basis for
W= span()⊆M
22,andsincedim(W)=5and dim(M 22)=4, this contradicts Theorem 4.13.
Henceis linearly dependent. Now use the definition of linear dependence.
(b) Generalize the proof in part (a).
(9) (a)is easily seen to be linearly independentusing Exercise 19, Section 4.4. To showspansV,
note that everyf∈Vcan be expressed asf=(−2)g,forsomeg∈P
4.
(b)5
(c){(−2)(−3)(−2)(−3)
2
(−2)(−3)
3
(−2)(−3)}
(d)4
(10) (a)[−111−1]is not a scalar multiple of[230−1].Also,
[141−2] = 1[230−1] + 1[−111−1],and[32−10] = 1[230−1]−1[−111−1].
(b) We illustrate two approaches:
First approach: We check thatandboth span the same subspace ofR
4
by using the
Simplified Span Method and showing that bothandproduce the same bases for their spans.
For:
⎡
⎢
⎢
⎣
14 1 −2
−11 1 −1
32−10
23 0 −1
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎣
10−
3
5
2
5
01
2
5
−
3
5
0000
0000
⎤
⎥
⎥
⎥
⎦
.
For:
∙
230 −1
−111 −1
¸
row reduces to
"
10−
3
5
2
5
01
2
5
−
3
5
#
.
Since both row reduced matrices have identical nonzero rows,span()andspan()have identical
bases (the set of nonzero rows of the row reduced matrices). Hence, they must be the same
subspaces. Also, since the basis produced forspan()contains two vectors,dim(span()) = 2.
Finally, becauseis a linearly independent subset ofspan()with||= dim(span()), part (2)
of Theorem 4.12 implies thatis a basis forspan().
Second approach: In part (a) we showed that no larger subset ofcontainingis linearly
independent; that is, each of the remaining vectors incan be expressed as a linear combination of
the vectors in. Therefore,⊆span(). Then, part (3) of Theorem 4.5 implies thatspan()⊆
span(). We also know that⊆, and so Corollary 4.6 implies thatspan()⊆span().
Therefore,span()=span().Sincewas shown to be linearly independent in part (a),is a
basis forspan(). Hence,dim(span()) =||
=2.
(c) No; dim(span())=26 =4=dim(R
4
)
(11) (a) Solving an appropriate homogeneous system shows thatis linearly independent. Also,
−1=
1
12
(
3
−
2
+2+1)+
1
12
(2
3
+4−7)−
1
12
(3
3
−
2
−6+6)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 72
(c) dim(W)=1;rank(A)=3;1+3=4
(7) These vectors are linearly independent by part (b) of Exercise 20 in Section 4.4. The given set has
+1distinct elements, anddim(P
)=+1. Hence, by part (2) of Theorem 4.12, the set is a basis
forP
.
(8) (a) Let={I
2AA
2
A
3
A
4
}.Ifis linearly independent, thenis a basis for
W= span()⊆M
22,andsincedim(W)=5and dim(M 22)=4, this contradicts Theorem 4.13.
Henceis linearly dependent. Now use the definition of linear dependence.
(b) Generalize the proof in part (a).
(9) (a)is easily seen to be linearly independentusing Exercise 19, Section 4.4. To showspansV,
note that everyf∈Vcan be expressed asf=(−2)g,forsomeg∈P
4.
(b)5
(c){(−2)(−3)(−2)(−3)
2
(−2)(−3)
3
(−2)(−3)}
(d)4
(10) (a)[−111−1]is not a scalar multiple of[230−1].Also,
[141−2] = 1[230−1] + 1[−111−1],and[32−10] = 1[230−1]−1[−111−1].
(b) We illustrate two approaches:
First approach: We check thatandboth span the same subspace ofR
4
by using the
Simplified Span Method and showing that bothandproduce the same bases for their spans.
For:
⎡
⎢
⎢
⎣
14 1 −2
−11 1 −1
32−10
23 0 −1
⎤
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎣
10−
3
5
2
5
01
2
5
−
3
5
0000
0000
⎤
⎥
⎥
⎥
⎦
.
For:
∙
230 −1
−111 −1
¸
row reduces to
"
10−
3
5
2
5
01
2
5
−
3
5
#
.
Since both row reduced matrices have identical nonzero rows,span()andspan()have identical
bases (the set of nonzero rows of the row reduced matrices). Hence, they must be the same
subspaces. Also, since the basis produced forspan()contains two vectors,dim(span()) = 2.
Finally, becauseis a linearly independent subset ofspan()with||= dim(span()), part (2)
of Theorem 4.12 implies thatis a basis forspan().
Second approach: In part (a) we showed that no larger subset ofcontainingis linearly
independent; that is, each of the remaining vectors incan be expressed as a linear combination of
the vectors in. Therefore,⊆span(). Then, part (3) of Theorem 4.5 implies thatspan()⊆
span(). We also know that⊆, and so Corollary 4.6 implies thatspan()⊆span().
Therefore,span()=span().Sincewas shown to be linearly independent in part (a),is a
basis forspan(). Hence,dim(span()) =||
=2.
(c) No; dim(span())=26 =4=dim(R
4
)
(11) (a) Solving an appropriate homogeneous system shows thatis linearly independent. Also,
−1=
1
12
(
3
−
2
+2+1)+
1
12
(2
3
+4−7)−
1
12
(3
3
−
2
−6+6)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 72
Answers to Exercises Section 4.5
and
3
−3
2
−22+34=1(
3
−
2
+2+1)−3(2
3
+4−7) + 2(3
3
−
2
−6+6)
(b) We illustrate two approaches:
First approach: We check thatandboth span the same subspace ofP
3by using the
Simplified Span Method and showing that bothandproduce the same bases for their spans.
For:
⎡
⎢
⎢
⎢
⎢
⎣
1−121
00 1 −1
20 4 −7
1−3−22 34
3−1−66
⎤
⎥
⎥
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎣
100 −
3
2
010 −
9
2
001 −1
000 0
000 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
.
For:
⎡
⎣
1−121
204 −7
3−1−66
⎤
⎦row reduces to
⎡
⎢
⎣
100 −
3
2
010 −
9
2
001 −1
⎤
⎥
⎦.
Since both row reduced matrices have identical nonzero rows,span()andspan()have identical
bases (the set of nonzero rows of the row reduced matrices). Hence, they must be the same
subspaces. Also, since the basis produced forspan()contains three vectors,dim(span()) = 3.
Finally, becauseis a linearly independent subset ofspan()with||= dim(span()), part (2)
of Theorem 4.12 implies thatis a basis forspan().
Second approach: In part (a) we showed that no larger subset ofcontainingis linearly
independent; that is, each of the remaining vectors incan be expressed as a linear combination of
the vectors in. Therefore,⊆span(). Then, part (3) of Theorem 4.5 implies thatspan()⊆
span(). We also know that⊆, and so Corollary 4.6 implies thatspan()⊆span().
Therefore,span()=span().Sincewas shown to be linearly independent in part (a),is a
basis forspan(). Hence,dim(span(
)) =||=3.
(c) No; dim(span())=36 =4=dim(P
3)
(12) (a) LetV=R
3
,andlet={[100][200][300]}.
(b) LetV=R
3
,andlet={[100][200][300]}.
(13) IfspansV,thenis a basis by part (1) of Theorem 4.12. Ifis linearly independent, thenis a
basis by part (2) of Theorem 4.12.
(14) By part (1) of Theorem 4.12, ifspansV,thenis a basis forV,sois linearly independent.
Similarly, by part (2) of Theorem 4.12, ifis linearly independent, thenis a basis forV,sospans
V.
(15) (a) Suppose={v
1v }.
Linear independence of
1:1Av1+···+ Av=0
=⇒A
−1
(1Av1+···+ Av)=A
−1
0=⇒ 1v1+···+ v=0
=⇒
1=···= =0,sinceis linearly independent.
1is a basis by part (2) of Theorem 4.12.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 73
and
3
−3
2
−22+34=1(
3
−
2
+2+1)−3(2
3
+4−7) + 2(3
3
−
2
−6+6)
(b) We illustrate two approaches:
First approach: We check thatandboth span the same subspace ofP
3by using the
Simplified Span Method and showing that bothandproduce the same bases for their spans.
For:
⎡
⎢
⎢
⎢
⎢
⎣
1−121
00 1 −1
20 4 −7
1−3−22 34
3−1−66
⎤
⎥
⎥
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎣
100 −
3
2
010 −
9
2
001 −1
000 0
000 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
.
For:
⎡
⎣
1−121
204 −7
3−1−66
⎤
⎦row reduces to
⎡
⎢
⎣
100 −
3
2
010 −
9
2
001 −1
⎤
⎥
⎦.
Since both row reduced matrices have identical nonzero rows,span()andspan()have identical
bases (the set of nonzero rows of the row reduced matrices). Hence, they must be the same
subspaces. Also, since the basis produced forspan()contains three vectors,dim(span()) = 3.
Finally, becauseis a linearly independent subset ofspan()with||= dim(span()), part (2)
of Theorem 4.12 implies thatis a basis forspan().
Second approach: In part (a) we showed that no larger subset ofcontainingis linearly
independent; that is, each of the remaining vectors incan be expressed as a linear combination of
the vectors in. Therefore,⊆span(). Then, part (3) of Theorem 4.5 implies thatspan()⊆
span(). We also know that⊆, and so Corollary 4.6 implies thatspan()⊆span().
Therefore,span()=span().Sincewas shown to be linearly independent in part (a),is a
basis forspan(). Hence,dim(span(
)) =||=3.
(c) No; dim(span())=36 =4=dim(P
3)
(12) (a) LetV=R
3
,andlet={[100][200][300]}.
(b) LetV=R
3
,andlet={[100][200][300]}.
(13) IfspansV,thenis a basis by part (1) of Theorem 4.12. Ifis linearly independent, thenis a
basis by part (2) of Theorem 4.12.
(14) By part (1) of Theorem 4.12, ifspansV,thenis a basis forV,sois linearly independent.
Similarly, by part (2) of Theorem 4.12, ifis linearly independent, thenis a basis forV,sospans
V.
(15) (a) Suppose={v
1v }.
Linear independence of
1:1Av1+···+ Av=0
=⇒A
−1
(1Av1+···+ Av)=A
−1
0=⇒ 1v1+···+ v=0
=⇒
1=···= =0,sinceis linearly independent.
1is a basis by part (2) of Theorem 4.12.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 73
Answers to Exercises Section 4.5
(b) Similar to part (a).
(c) Note thatAe
=(th column ofA).
(d) Note thate
A=(th row ofA).
(16) The dimension of a proper nontrivial subspace must be1or2, by Theorem 4.13. If the basis for the
subspace contains one [two] element(s), then the subspace is a line [plane] through the origin as in
Exercise 17 of Section 4.3.
(17) (a) If={f
1f },let=max(degree(f 1), degree(f )). Then⊆P .
(b) Apply Theorem 4.5, part (3). (c)
+1
6 ∈span().
(18) (a) SupposeVisfinite dimensional. SinceVhas an infinite linearly independent subset, part (2) of
Theorem 4.12 is contradicted.
(b){1
2
}is a linearly independent subset ofP, and hence, of any vector space containingP.
Apply part (a).
(19) Supposeis a basis forVand thatis a subset ofVcontainingwith6 =.Letv∈with
v∈.Thenv∈span()becausespansV.Thus,vis a linear combination of vectors in,which
means thatvis a linear combination of vectors inother thanvitself. Hence,is linearly dependent
by Theorem 4.8.
(20) Supposeis a basis forV.Supposethatis a subset ofwith6 =.Letv∈withv∈.
Nowis linearly independent, so by the Alternate Characterization of linear independence in Table
4.1,v∈span(−{v}).But⊆−{v}, implying
span()⊆span(−{v})(Corollary 4.6). Hence
v∈span().Therefore,does not spanV.
(21) LetVbe afinite dimensional vector space and letWbe a subspace ofV. Consider the setof
nonnegative integers defined by
={|asetexists with⊆W||=andlinearly independent}
(a) The empty set is linearly independent and is a subset ofW. Hence0∈. Therefore,is not
the empty set.
(b) SinceVandWshare the same operations, every linearly independent subsetofWis also a
linearly independent subset ofV(by the definition of linear independence). Hence, using part (2)
of Theorem 4.12 oninVshows that=||≤dim(V). Therefore,contains no numbers
larger than
dim(V),andsohasafinite number of elements.
(c) Becauseis afinite nonempty set of numbers, it has a largest number. Therefore, because
∈, the definition of the setimplies that there must exist some setsuch that⊆W,
||=,andis linearly independent. Let={v
1v }be such a set.
Now, we are given that⊆W,andsospan()⊆Wby part (3) of Theorem 4.5.
(d) Supposew∈Wwithw∈span().Thenw∈,andso∪{w}is a subset ofWcontaining
+1vectors. But sinceis the largest element of,+1is not in. Hence, the set∪{w}
must fail to satisfy at least one of the conditions in the description of.Since∪{w}⊆W,we
must have that∪{w}is not linearly independent. Therefore,∪{w}is linearly dependent.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 74
(b) Similar to part (a).
(c) Note thatAe
=(th column ofA).
(d) Note thate
A=(th row ofA).
(16) The dimension of a proper nontrivial subspace must be1or2, by Theorem 4.13. If the basis for the
subspace contains one [two] element(s), then the subspace is a line [plane] through the origin as in
Exercise 17 of Section 4.3.
(17) (a) If={f
1f },let=max(degree(f 1), degree(f )). Then⊆P .
(b) Apply Theorem 4.5, part (3). (c)
+1
6 ∈span().
(18) (a) SupposeVisfinite dimensional. SinceVhas an infinite linearly independent subset, part (2) of
Theorem 4.12 is contradicted.
(b){1
2
}is a linearly independent subset ofP, and hence, of any vector space containingP.
Apply part (a).
(19) Supposeis a basis forVand thatis a subset ofVcontainingwith6 =.Letv∈with
v∈.Thenv∈span()becausespansV.Thus,vis a linear combination of vectors in,which
means thatvis a linear combination of vectors inother thanvitself. Hence,is linearly dependent
by Theorem 4.8.
(20) Supposeis a basis forV.Supposethatis a subset ofwith6 =.Letv∈withv∈.
Nowis linearly independent, so by the Alternate Characterization of linear independence in Table
4.1,v∈span(−{v}).But⊆−{v}, implying
span()⊆span(−{v})(Corollary 4.6). Hence
v∈span().Therefore,does not spanV.
(21) LetVbe afinite dimensional vector space and letWbe a subspace ofV. Consider the setof
nonnegative integers defined by
={|asetexists with⊆W||=andlinearly independent}
(a) The empty set is linearly independent and is a subset ofW. Hence0∈. Therefore,is not
the empty set.
(b) SinceVandWshare the same operations, every linearly independent subsetofWis also a
linearly independent subset ofV(by the definition of linear independence). Hence, using part (2)
of Theorem 4.12 oninVshows that=||≤dim(V). Therefore,contains no numbers
larger than
dim(V),andsohasafinite number of elements.
(c) Becauseis afinite nonempty set of numbers, it has a largest number. Therefore, because
∈, the definition of the setimplies that there must exist some setsuch that⊆W,
||=,andis linearly independent. Let={v
1v }be such a set.
Now, we are given that⊆W,andsospan()⊆Wby part (3) of Theorem 4.5.
(d) Supposew∈Wwithw∈span().Thenw∈,andso∪{w}is a subset ofWcontaining
+1vectors. But sinceis the largest element of,+1is not in. Hence, the set∪{w}
must fail to satisfy at least one of the conditions in the description of.Since∪{w}⊆W,we
must have that∪{w}is not linearly independent. Therefore,∪{w}is linearly dependent.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 74
Answers to Exercises Section 4.5
(e) Supposew∈W, butw∈span()as in part (d). Because∪{w}={v 1v w}is linearly
dependent (by part (d)), there exist scalars
1such that 1v1+···+ v+w=0
with not all coefficients equal to zero. But if=0,then
1v1+···+ v=0with not all
=0, contradicting the fact thatis linearly independent. Hence,6 =0and thus
w=(−
1
)v
1+···+(−
)v
(f) We assumed thatw∈span(). However, the conclusion of part (e) shows thatw∈span().
Since both clearly can not be true, we have a contradiction. Hence, our assumption that there is
aw∈Wwithw∈span()is false. That is, there are no vectors inWoutside ofspan().Thus,
W⊆span().
(g) From the conclusions of parts (c) and (f),W= span().Now,is a basis forW, since it is
linearly independent by assumption (part (c)), and spansW. Hence,Wisfinite dimensional
becauseis afinite basis (containingelements). Also, using part (b),
dim(W)=||=≤dim(V)
(h) Supposedim(W)=dim(V),andletbe a basis forW.Thenis a linearly independent subset
ofV(see the discussion in part (b)) with||=dim(W)=dim(V). Part (2) of Theorem 4.12
then shows thatis a basis forV. Hence,W= span()=V.
(i) The converse of part (h) is “IfW=V,thendim(W)=dim(V).” This is obviously true, since the
dimension of afinite dimensional vector space is unique.
(22) We need tofind afinite basis forV.Ifitself is linearly independent, thenis afinite basis, and we
arefinished. Ifis linearly dependent, then part (a) of Exercise 12 in Section 4.4 shows that there is
a redundant vectorv∈such thatspan(−{v})=span()=V.Let
1=−{v}.
If
1is linearly independent, we arefinished, since it is afinite basis forV. Otherwise, repeat the
above process, producing a subset
2of1withspan( 2)=V. Continue in a similar manner until a
linearly independent subset
is produced withspan( )=V.Then is afinite basis forV.This
will take afinite number of steps because there are at most||vectors overall that can be removed.
(This solution can be expressed more formally using induction.)
(23) Let={v
1v −1}⊆R
be a basis forV.LetAbe the(−1)×matrix whose rows are the
vectors in, and consider the homogeneous systemAX=0. Now, by Corollary 2.3, the system has
a nontrivial solutionx. Therefore,Ax=0.Thatis,x·v
=0for each1≤≤−1.Now,suppose
v∈V. Then there are
1−1∈Rsuch thatv= 1v1+···+ −1v−1. Hence,
x·v=x·
1v1+···+x· −1v−1=0+···+0=0
ThereforeV⊆{v∈R
|x·v=0}.LetW={v∈R
|x·v=0}.So,V⊆W.Now,itiseasyto
see thatWis a subspace ofR
.(Clearly0∈Wsincex·0=0. HenceWis nonempty. Next, if
w
1w2∈W,then
x·(w
1+w2)=x·w 1+x·w 2=0+0=0
Thus,(w
1+w2)∈W,andsoWis closed under addition. Also, ifw∈Wand∈R,then
x·(w)=(x·w)=(0) = 0
Therefore(w)∈W,andWis closed under scalar multiplication. Hence, by Theorem 4.2,Wis a
subspace ofR
.) IfV=W, we are done. Suppose, then, thatV6 =W. Then, by Theorem 4.13,
−1=dim(V)dim(W)≤dim(R
)=
Hence,dim(W)=,andsoW=R
, by Theorem 4.13. Butx∈W,sincex·x6 =0, becausexis
Copyrightc°2016 Elsevier Ltd. All rights reserved. 75
(e) Supposew∈W, butw∈span()as in part (d). Because∪{w}={v 1v w}is linearly
dependent (by part (d)), there exist scalars
1such that 1v1+···+ v+w=0
with not all coefficients equal to zero. But if=0,then
1v1+···+ v=0with not all
=0, contradicting the fact thatis linearly independent. Hence,6 =0and thus
w=(−
1
)v
1+···+(−
)v
(f) We assumed thatw∈span(). However, the conclusion of part (e) shows thatw∈span().
Since both clearly can not be true, we have a contradiction. Hence, our assumption that there is
aw∈Wwithw∈span()is false. That is, there are no vectors inWoutside ofspan().Thus,
W⊆span().
(g) From the conclusions of parts (c) and (f),W= span().Now,is a basis forW, since it is
linearly independent by assumption (part (c)), and spansW. Hence,Wisfinite dimensional
becauseis afinite basis (containingelements). Also, using part (b),
dim(W)=||=≤dim(V)
(h) Supposedim(W)=dim(V),andletbe a basis forW.Thenis a linearly independent subset
ofV(see the discussion in part (b)) with||=dim(W)=dim(V). Part (2) of Theorem 4.12
then shows thatis a basis forV. Hence,W= span()=V.
(i) The converse of part (h) is “IfW=V,thendim(W)=dim(V).” This is obviously true, since the
dimension of afinite dimensional vector space is unique.
(22) We need tofind afinite basis forV.Ifitself is linearly independent, thenis afinite basis, and we
arefinished. Ifis linearly dependent, then part (a) of Exercise 12 in Section 4.4 shows that there is
a redundant vectorv∈such thatspan(−{v})=span()=V.Let
1=−{v}.
If
1is linearly independent, we arefinished, since it is afinite basis forV. Otherwise, repeat the
above process, producing a subset
2of1withspan( 2)=V. Continue in a similar manner until a
linearly independent subset
is produced withspan( )=V.Then is afinite basis forV.This
will take afinite number of steps because there are at most||vectors overall that can be removed.
(This solution can be expressed more formally using induction.)
(23) Let={v
1v −1}⊆R
be a basis forV.LetAbe the(−1)×matrix whose rows are the
vectors in, and consider the homogeneous systemAX=0. Now, by Corollary 2.3, the system has
a nontrivial solutionx. Therefore,Ax=0.Thatis,x·v
=0for each1≤≤−1.Now,suppose
v∈V. Then there are
1−1∈Rsuch thatv= 1v1+···+ −1v−1. Hence,
x·v=x·
1v1+···+x· −1v−1=0+···+0=0
ThereforeV⊆{v∈R
|x·v=0}.LetW={v∈R
|x·v=0}.So,V⊆W.Now,itiseasyto
see thatWis a subspace ofR
.(Clearly0∈Wsincex·0=0. HenceWis nonempty. Next, if
w
1w2∈W,then
x·(w
1+w2)=x·w 1+x·w 2=0+0=0
Thus,(w
1+w2)∈W,andsoWis closed under addition. Also, ifw∈Wand∈R,then
x·(w)=(x·w)=(0) = 0
Therefore(w)∈W,andWis closed under scalar multiplication. Hence, by Theorem 4.2,Wis a
subspace ofR
.) IfV=W, we are done. Suppose, then, thatV6 =W. Then, by Theorem 4.13,
−1=dim(V)dim(W)≤dim(R
)=
Hence,dim(W)=,andsoW=R
, by Theorem 4.13. Butx∈W,sincex·x6 =0, becausexis
Copyrightc°2016 Elsevier Ltd. All rights reserved. 75
Answers to Exercises Section 4.6
nontrivial. This contradiction completes the proof.
(24) (a)T (b)F (c)F (d)F (e)F (f)T (g)F (h)F (i)T
Section 4.6
Problems 1 through 12 ask you to construct a basis forsome vector space. The answers here are those you
will most likely get using the techniques in the textbook. However, other answers are possible.
(1) (a){[1002−2][01001][001−10]}
(b){[10−
1
5
6
5
3
5
][01−
1
5
−
9
5
−
2
5
]}
(c){e
1e2e3e4e5}
(d){[100−2−
13
4
][0103
9
2
][0010−
1
4
]}
(2){
3
−3
2
−1}
(3)
⎧
⎨
⎩
⎡
⎣
10
4
3
1
3
20
⎤
⎦
⎡
⎣
01
−
1
3
−
1
3
00
⎤
⎦
⎡
⎣
00
00
01
⎤
⎦
⎫
⎬
⎭
(4) (a){[13−2][214][01−1]}
(b){[14−2][2−85][0−72]}
(c){[3−22][12−1][3−27]}
(d){[310][2−17][15
7]}
(5) (a){
3
−8
2
+13
3
−2
2
+4
3
+2−10
3
−20
2
−+12}
(b){−2
3
++23
3
−
2
+4+68
3
+
2
+6+10}
(6) (a){[31−2][62−3]}
(b){[471][100]}
(c){e
1e3}
(d){[1−30][011]}
(7) (a){
3
2
}
(b){1
2
}
(c){
3
+
2
1}
(d){8
3
8
3
+
2
8
3
+8
3
+1}
(8) (a) One answer is{Ψ
|1≤ ≤3},whereΨ is the3×3matrix with( )entry=1and all
other entries0.
(b)
⎧
⎨
⎩
⎡
⎣
111
111
111
⎤
⎦,
⎡
⎣
−111
111
111
⎤
⎦,
⎡
⎣
1−11
111
111
⎤
⎦,
⎡
⎣
11−1
11 1
11 1
⎤
⎦,
⎡
⎣
111
−111
111
⎤
⎦,
⎡
⎣
111
1−11
111
⎤
⎦,
⎡
⎣
11 1
11−1
11 1
⎤
⎦,
⎡
⎣
111
111
−111
⎤
⎦,
⎡
⎣
111
111
1−11
⎤
⎦
⎫
⎬
⎭
(c)
⎧
⎨
⎩
⎡
⎣
100
000
000
⎤
⎦
⎡
⎣
010
100
000
⎤
⎦
⎡
⎣
001
000
100
⎤
⎦
⎡
⎣
000
010
000
⎤
⎦
⎡
⎣
000
001
010
⎤
⎦
⎡
⎣
000
000
001
⎤
⎦
⎫
⎬
⎭
Copyrightc°2016 Elsevier Ltd. All rights reserved. 76
nontrivial. This contradiction completes the proof.
(24) (a)T (b)F (c)F (d)F (e)F (f)T (g)F (h)F (i)T
Section 4.6
Problems 1 through 12 ask you to construct a basis forsome vector space. The answers here are those you
will most likely get using the techniques in the textbook. However, other answers are possible.
(1) (a){[1002−2][01001][001−10]}
(b){[10−
1
5
6
5
3
5
][01−
1
5
−
9
5
−
2
5
]}
(c){e
1e2e3e4e5}
(d){[100−2−
13
4
][0103
9
2
][0010−
1
4
]}
(2){
3
−3
2
−1}
(3)
⎧
⎨
⎩
⎡
⎣
10
4
3
1
3
20
⎤
⎦
⎡
⎣
01
−
1
3
−
1
3
00
⎤
⎦
⎡
⎣
00
00
01
⎤
⎦
⎫
⎬
⎭
(4) (a){[13−2][214][01−1]}
(b){[14−2][2−85][0−72]}
(c){[3−22][12−1][3−27]}
(d){[310][2−17][15
7]}
(5) (a){
3
−8
2
+13
3
−2
2
+4
3
+2−10
3
−20
2
−+12}
(b){−2
3
++23
3
−
2
+4+68
3
+
2
+6+10}
(6) (a){[31−2][62−3]}
(b){[471][100]}
(c){e
1e3}
(d){[1−30][011]}
(7) (a){
3
2
}
(b){1
2
}
(c){
3
+
2
1}
(d){8
3
8
3
+
2
8
3
+8
3
+1}
(8) (a) One answer is{Ψ
|1≤ ≤3},whereΨ is the3×3matrix with( )entry=1and all
other entries0.
(b)
⎧
⎨
⎩
⎡
⎣
111
111
111
⎤
⎦,
⎡
⎣
−111
111
111
⎤
⎦,
⎡
⎣
1−11
111
111
⎤
⎦,
⎡
⎣
11−1
11 1
11 1
⎤
⎦,
⎡
⎣
111
−111
111
⎤
⎦,
⎡
⎣
111
1−11
111
⎤
⎦,
⎡
⎣
11 1
11−1
11 1
⎤
⎦,
⎡
⎣
111
111
−111
⎤
⎦,
⎡
⎣
111
111
1−11
⎤
⎦
⎫
⎬
⎭
(c)
⎧
⎨
⎩
⎡
⎣
100
000
000
⎤
⎦
⎡
⎣
010
100
000
⎤
⎦
⎡
⎣
001
000
100
⎤
⎦
⎡
⎣
000
010
000
⎤
⎦
⎡
⎣
000
001
010
⎤
⎦
⎡
⎣
000
000
001
⎤
⎦
⎫
⎬
⎭
Copyrightc°2016 Elsevier Ltd. All rights reserved. 76
Answers to Exercises Section 4.6
(d)
⎧
⎨
⎩
⎡
⎣
100
010
001
⎤
⎦,
⎡
⎣
110
010
001
⎤
⎦,
⎡
⎣
101
010
001
⎤
⎦,
⎡
⎣
100
110
001
⎤
⎦,
⎡
⎣
100
011
001
⎤
⎦,
⎡
⎣
100
010
101
⎤
⎦,
⎡
⎣
100
010
011
⎤
⎦,
⎡
⎣
100
020
001
⎤
⎦,
⎡
⎣
100
010
002
⎤
⎦
⎫
⎬
⎭
(9) LetVbe the subspace ofM
22consisting of all2×2symmetric matrices. Letbe the set of nonsingular
matrices inV,andletW=span()=span({nonsingular, symmetric2×2matrices}). Using the
strategy from Exercise 7, we reducetoabasisforWusing the Independence Test Method, even
thoughis infinite.
The strategy is to guess afinitesubsetofthat spansW. We then use the Independence Test
Method ontofind the desired basis. We try to pick vectors forwhose forms are as simple as
possible to make computation easier. In this case, we choose the set of all nonsingular symmetric2×2
matrices having only zeroes and ones as entries. That is,
=
½∙
10
01
¸
∙
11
10
¸
∙
01
10
¸
∙
01
11
¸¾
Now, before continuing, we must ensure that span()=W. That is, we must show every nonsin-
gular symmetric2×2matrix is in span(). In fact, we will show every symmetric2×2matrix is in
span()byfinding real numbers and
so that
∙
¸
=
∙
10
01
¸
+
∙
11
10
¸
+
∙
01
10
¸
+
∙
01
11
¸
Thus, we must prove that the system
⎧
⎪
⎪
⎨
⎪
⎪
⎩
+ =
++=
++=
+=
has solutions for andin terms of and.But=0= =−− =certainly
satisfies the system. Hence,V⊆span(). Since span()⊆Vwe have span(
)=V=W
We can now use the Independence Test Method on. We express the matrices inas corre-
sponding vectors inR
4
and create the matrix with these vectors as columns, as follows:
A=
⎡
⎢
⎢
⎣
1100
0111
0111
1001
⎤
⎥
⎥
⎦
which reduces toC=
⎡
⎢
⎢
⎣
100 1
010 −1
001 2
000 0
⎤
⎥
⎥
⎦
Then, the desired basis is
=
½∙
10
01
¸
∙
11
10
¸
∙
01
10
¸¾
the elements ofcorresponding to the pivot columns of.
(10) (a){[1−3014][221−31][10000][01000]
[00100]}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 77
(d)
⎧
⎨
⎩
⎡
⎣
100
010
001
⎤
⎦,
⎡
⎣
110
010
001
⎤
⎦,
⎡
⎣
101
010
001
⎤
⎦,
⎡
⎣
100
110
001
⎤
⎦,
⎡
⎣
100
011
001
⎤
⎦,
⎡
⎣
100
010
101
⎤
⎦,
⎡
⎣
100
010
011
⎤
⎦,
⎡
⎣
100
020
001
⎤
⎦,
⎡
⎣
100
010
002
⎤
⎦
⎫
⎬
⎭
(9) LetVbe the subspace ofM
22consisting of all2×2symmetric matrices. Letbe the set of nonsingular
matrices inV,andletW=span()=span({nonsingular, symmetric2×2matrices}). Using the
strategy from Exercise 7, we reducetoabasisforWusing the Independence Test Method, even
thoughis infinite.
The strategy is to guess afinitesubsetofthat spansW. We then use the Independence Test
Method ontofind the desired basis. We try to pick vectors forwhose forms are as simple as
possible to make computation easier. In this case, we choose the set of all nonsingular symmetric2×2
matrices having only zeroes and ones as entries. That is,
=
½∙
10
01
¸
∙
11
10
¸
∙
01
10
¸
∙
01
11
¸¾
Now, before continuing, we must ensure that span()=W. That is, we must show every nonsin-
gular symmetric2×2matrix is in span(). In fact, we will show every symmetric2×2matrix is in
span()byfinding real numbers and
so that
∙
¸
=
∙
10
01
¸
+
∙
11
10
¸
+
∙
01
10
¸
+
∙
01
11
¸
Thus, we must prove that the system
⎧
⎪
⎪
⎨
⎪
⎪
⎩
+ =
++=
++=
+=
has solutions for andin terms of and.But=0= =−− =certainly
satisfies the system. Hence,V⊆span(). Since span()⊆Vwe have span(
)=V=W
We can now use the Independence Test Method on. We express the matrices inas corre-
sponding vectors inR
4
and create the matrix with these vectors as columns, as follows:
A=
⎡
⎢
⎢
⎣
1100
0111
0111
1001
⎤
⎥
⎥
⎦
which reduces toC=
⎡
⎢
⎢
⎣
100 1
010 −1
001 2
000 0
⎤
⎥
⎥
⎦
Then, the desired basis is
=
½∙
10
01
¸
∙
11
10
¸
∙
01
10
¸¾
the elements ofcorresponding to the pivot columns of.
(10) (a){[1−3014][221−31][10000][01000]
[00100]}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 77
Answers to Exercises Section 4.6
(b){[11111][01111][00111][00100][00010]}
(c){[10−100][01−110][23−8−10][10000][00
001]}
(11) (a){
3
−
2
4
−3
3
+5
2
−
4
3
1}
(b){3−2
3
−6+4
4
2
}
(c){
4
−
3
+
2
−+1
3
−
2
+−1
2
−+1
2
}
(12) (a)
⎧
⎨
⎩
⎡
⎣
1−1
−11
00
⎤
⎦
⎡
⎣
00
1−1
−11
⎤
⎦
⎡
⎣
10
00
00
⎤
⎦,
⎡
⎣
01
00
00
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎡
⎣
00
00
10
⎤
⎦
⎫
⎬
⎭
(b)
⎧
⎨
⎩
⎡
⎣
0−2
10
−12
⎤
⎦
⎡
⎣
1−3
01
3−6
⎤
⎦
⎡
⎣
4−13
23
7−14
⎤
⎦,
⎡
⎣
01
00
00
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎡
⎣
00
00
10
⎤
⎦
⎫
⎬
⎭
(c)
⎧
⎨
⎩
⎡
⎣
3−1
2−6
51
⎤
⎦
⎡
⎣
−12
−42
00
⎤
⎦
⎡
⎣
62
−2−9
10 2
⎤
⎦
⎡
⎣
3−4
8−9
51
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎡
⎣
00
00
10
⎤
⎦
⎫
⎬
⎭
(13) (a)2 (b)8 (c)4 (d)3
(14) (a) A basis forU
is{Ψ |1≤≤≤},whereΨ is the×matrixhavingazeroinevery
entry, except for the( )th entry, which equals1. Since there are−+1entries in theth row
of an×matrix on or above the main diagonal, this basis contains
X
=1
(−+1)=
X
=1
−
X
=1
+
X
=1
1=
2
−
(+1)
2
+=
2
+
2
elements.
Similarly, a basis forL
is{Ψ |1≤≤≤}, and a basis for the symmetric×matrices
is{Ψ
+Ψ
|1≤≤≤}.
(b)(
2
−)2
(15) LetBbe the reduced row echelon form ofA. Each basis element of
Ais found by setting one
independent variable of the systemAX=Oequal to1and the remaining independent variables equal
to zero. (The values of the dependent variables are then determined from these values.) Since there is
one independent variable for each nonpivot column inB,dim(
A)=number of nonpivot columns of
B.But,clearly,rank(A)=number of pivot columns inB.
So, dim(
A)+rank(A)=total number of columns inB=.
(16) LetVandbe as given in the statement of the theorem. Let
={|asetexists with⊆,||=,andlinearly independent}.
The empty set is linearly independent and is a subset of. Hence0∈,andis nonempty.
Suppose∈. Then, every linearly independent subsetofis also a linearly independent subset
ofV. Hence, using part (2) of Theorem 4.12 onshows that=||≤dim(V).
Supposeis the largest number in, which must exist, sinceis nonempty and all its elements
are≤dim(V).Letbe a linearly independent subset ofwith||=, which exists because∈.
We want to show thatis a basis for
V=span().Now,is given as linearly independent. We must
show thatspansV.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 78
(b){[11111][01111][00111][00100][00010]}
(c){[10−100][01−110][23−8−10][10000][00
001]}
(11) (a){
3
−
2
4
−3
3
+5
2
−
4
3
1}
(b){3−2
3
−6+4
4
2
}
(c){
4
−
3
+
2
−+1
3
−
2
+−1
2
−+1
2
}
(12) (a)
⎧
⎨
⎩
⎡
⎣
1−1
−11
00
⎤
⎦
⎡
⎣
00
1−1
−11
⎤
⎦
⎡
⎣
10
00
00
⎤
⎦,
⎡
⎣
01
00
00
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎡
⎣
00
00
10
⎤
⎦
⎫
⎬
⎭
(b)
⎧
⎨
⎩
⎡
⎣
0−2
10
−12
⎤
⎦
⎡
⎣
1−3
01
3−6
⎤
⎦
⎡
⎣
4−13
23
7−14
⎤
⎦,
⎡
⎣
01
00
00
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎡
⎣
00
00
10
⎤
⎦
⎫
⎬
⎭
(c)
⎧
⎨
⎩
⎡
⎣
3−1
2−6
51
⎤
⎦
⎡
⎣
−12
−42
00
⎤
⎦
⎡
⎣
62
−2−9
10 2
⎤
⎦
⎡
⎣
3−4
8−9
51
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎡
⎣
00
00
10
⎤
⎦
⎫
⎬
⎭
(13) (a)2 (b)8 (c)4 (d)3
(14) (a) A basis forU
is{Ψ |1≤≤≤},whereΨ is the×matrixhavingazeroinevery
entry, except for the( )th entry, which equals1. Since there are−+1entries in theth row
of an×matrix on or above the main diagonal, this basis contains
X
=1
(−+1)=
X
=1
−
X
=1
+
X
=1
1=
2
−
(+1)
2
+=
2
+
2
elements.
Similarly, a basis forL
is{Ψ |1≤≤≤}, and a basis for the symmetric×matrices
is{Ψ
+Ψ
|1≤≤≤}.
(b)(
2
−)2
(15) LetBbe the reduced row echelon form ofA. Each basis element of
Ais found by setting one
independent variable of the systemAX=Oequal to1and the remaining independent variables equal
to zero. (The values of the dependent variables are then determined from these values.) Since there is
one independent variable for each nonpivot column inB,dim(
A)=number of nonpivot columns of
B.But,clearly,rank(A)=number of pivot columns inB.
So, dim(
A)+rank(A)=total number of columns inB=.
(16) LetVandbe as given in the statement of the theorem. Let
={|asetexists with⊆,||=,andlinearly independent}.
The empty set is linearly independent and is a subset of. Hence0∈,andis nonempty.
Suppose∈. Then, every linearly independent subsetofis also a linearly independent subset
ofV. Hence, using part (2) of Theorem 4.12 onshows that=||≤dim(V).
Supposeis the largest number in, which must exist, sinceis nonempty and all its elements
are≤dim(V).Letbe a linearly independent subset ofwith||=, which exists because∈.
We want to show thatis a basis for
V=span().Now,is given as linearly independent. We must
show thatspansV.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 78
Answers to Exercises Section 4.6
Now,⊆,andsospan()⊆span()(by Corollary 4.6). We need to show thatspan()⊆
span(). To do this, we will prove that⊆span(), for then, parts (2) and (3) of Theorem 4.5 will
give us thatspan()⊆span().
We prove that⊆span(): Suppose={v
1v }andv∈.Ifv∈, then, clearly
v∈span(),andwearefinished. So, suppose thatv∈.Then∪{v}is a subset ofcontaining
+1elements. Therefore, sinceis the largest number in,∪{v}must be linearly dependent.
Hence, there exist scalars
1such that 1v1+···+ v+v=0with not all coefficients
equal to zero. But if=0,then
1v1+···+ v=0with not all =0, contradicting the fact that
is linearly independent. Hence,6 =0and thus
v=(−
1
)v
1+···+(−
)v
This shows thatv∈span(), completing the proof thatspansV. Hence, the subsetofis the
desired basis forV,finishing the proof of Theorem 4.14.
(17) (a) Let
0
be a basis forW.Expand
0
to a basisforV(using Theorem 4.15).
(b) No; consider the subspaceWofR
3
given byW={[00]|∈R}. No subset of
={[110][1−10][001]}(a basis forR
3
)isabasisforW.
(c) Yes; considerY=span(
0
).
(18) (a) Let
1be a basis forW.Expand 1toabasisforV(using Theorem 4.15). Let 2=− 1,
and letW
0
=span( 2). Suppose 1={v 1v }and 2={u 1u }.Sinceis a basis for
V,allv∈Vcan be written asv=w+w
0
withw= 1v1+···+ v∈Wand
w
0
=1u1+···+ u∈W
0
.
For uniqueness, supposev=w
1+w
0
1
=w2+w
0
2
,withw 1w2∈Wandw
0
1
w
0
2
∈W
0
.Then
w
1−w2=w
0
2
−w
0
1
∈W∩W
0
={0}. Hence,w 1=w2andw
0
1
=w
0
2
.
(b) InR
3
,considerW={[ 0]| ∈R}. We could let
W
0
={[00]|∈R}orW
0
={[0]|∈R}.
(19) (a) LetAbe the matrix whose rows are the-vectors in.LetCbe the reduced row echelon form
matrix forA. If the nonzero rows ofCform the standard basis forR
then
span()=span({e
1e })=R
Conversely, if span()=R
then there must be at leastnonzero rows ofCby part (1) of
Theorem 4.12. But this means there are at leastrows with pivots, and hence allcolumns have
pivots. Thus the nonzero rows ofCaree
1e which form the standard basis forR
.
(b) LetAandCbe as in the answer to part (a). Note thatAandCare×matrices. Now, since
||=,is a basis forR
⇐⇒span()=R
(by Theorem 4.12)⇐⇒the rows ofCform the
standard basis forR
(from part (a))⇐⇒C=I ⇐⇒|A|6 =0
(20) LetAbe an×matrix and let⊆R
be the set of therows ofA.
(a) SupposeCis the reduced row echelon form ofA. Then the Simplified Span Method shows that
the nonzero rows ofCform a basis forspan(). But the number of such rows isrank(A). Hence,
dim(span()) = rank(A).
(b) LetDbe the reduced row echelon form ofA
. By the Independence Test Method, the pivot
columns ofDcorrespond to the columns ofA
which make up a basis forspan(). Hence, the
number of pivot columns ofDequalsdim(span()). However, the number of pivot columns ofD
equals the number of nonzero rows ofD(each pivot column shares the single pivot element with
a unique pivot row, and vice-versa), which equalsrank(A
).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 79
Now,⊆,andsospan()⊆span()(by Corollary 4.6). We need to show thatspan()⊆
span(). To do this, we will prove that⊆span(), for then, parts (2) and (3) of Theorem 4.5 will
give us thatspan()⊆span().
We prove that⊆span(): Suppose={v
1v }andv∈.Ifv∈, then, clearly
v∈span(),andwearefinished. So, suppose thatv∈.Then∪{v}is a subset ofcontaining
+1elements. Therefore, sinceis the largest number in,∪{v}must be linearly dependent.
Hence, there exist scalars
1such that 1v1+···+ v+v=0with not all coefficients
equal to zero. But if=0,then
1v1+···+ v=0with not all =0, contradicting the fact that
is linearly independent. Hence,6 =0and thus
v=(−
1
)v
1+···+(−
)v
This shows thatv∈span(), completing the proof thatspansV. Hence, the subsetofis the
desired basis forV,finishing the proof of Theorem 4.14.
(17) (a) Let
0
be a basis forW.Expand
0
to a basisforV(using Theorem 4.15).
(b) No; consider the subspaceWofR
3
given byW={[00]|∈R}. No subset of
={[110][1−10][001]}(a basis forR
3
)isabasisforW.
(c) Yes; considerY=span(
0
).
(18) (a) Let
1be a basis forW.Expand 1toabasisforV(using Theorem 4.15). Let 2=− 1,
and letW
0
=span( 2). Suppose 1={v 1v }and 2={u 1u }.Sinceis a basis for
V,allv∈Vcan be written asv=w+w
0
withw= 1v1+···+ v∈Wand
w
0
=1u1+···+ u∈W
0
.
For uniqueness, supposev=w
1+w
0
1
=w2+w
0
2
,withw 1w2∈Wandw
0
1
w
0
2
∈W
0
.Then
w
1−w2=w
0
2
−w
0
1
∈W∩W
0
={0}. Hence,w 1=w2andw
0
1
=w
0
2
.
(b) InR
3
,considerW={[ 0]| ∈R}. We could let
W
0
={[00]|∈R}orW
0
={[0]|∈R}.
(19) (a) LetAbe the matrix whose rows are the-vectors in.LetCbe the reduced row echelon form
matrix forA. If the nonzero rows ofCform the standard basis forR
then
span()=span({e
1e })=R
Conversely, if span()=R
then there must be at leastnonzero rows ofCby part (1) of
Theorem 4.12. But this means there are at leastrows with pivots, and hence allcolumns have
pivots. Thus the nonzero rows ofCaree
1e which form the standard basis forR
.
(b) LetAandCbe as in the answer to part (a). Note thatAandCare×matrices. Now, since
||=,is a basis forR
⇐⇒span()=R
(by Theorem 4.12)⇐⇒the rows ofCform the
standard basis forR
(from part (a))⇐⇒C=I ⇐⇒|A|6 =0
(20) LetAbe an×matrix and let⊆R
be the set of therows ofA.
(a) SupposeCis the reduced row echelon form ofA. Then the Simplified Span Method shows that
the nonzero rows ofCform a basis forspan(). But the number of such rows isrank(A). Hence,
dim(span()) = rank(A).
(b) LetDbe the reduced row echelon form ofA
. By the Independence Test Method, the pivot
columns ofDcorrespond to the columns ofA
which make up a basis forspan(). Hence, the
number of pivot columns ofDequalsdim(span()). However, the number of pivot columns ofD
equals the number of nonzero rows ofD(each pivot column shares the single pivot element with
a unique pivot row, and vice-versa), which equalsrank(A
).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 79
Answers to Exercises Section 4.7
(c) Both ranks equaldim(span()), and so they must be equal.
(21) Use the fact thatsin(
+
)=(cos )sin
+(sin )cos
Note that theth row ofAis then
(cos
)x1+(sin )x2Hence each row ofAis a linear combination of{x 1x2}and so
dim(span({rows ofA}))≤2. Hence,Ahas less thanpivots when row reduced, and so|A|=0.
(22) (a) T (b) T (c) F (d) T (e) F (f) F (g) F
Section 4.7
(1) (a)[v] =[7−1−5]
(b)[v]
=[3−2−1−2]
(c)[v]
=[−24−5]
(d)[v]
=[1032−1]
(e)[v]
=[4−53]
(f)[v]
=[−323]
(g)[v]
=[−14−2]
(h)[v]
=[2−31]
(i)[v]
=[52−6]
(j)[v]
=[5−2]
(2) (a)
⎡
⎣
−102 20 3
67−13−2
36−7−1
⎤
⎦
(b)
⎡
⎣
−2−6−1
361
−6−7−1
⎤
⎦
(c)
⎡
⎣
20−30−69
24−24−80
−91131
⎤
⎦
(d)
⎡
⎢
⎢
⎣
−1−42 −9
4513
02 −31
−4−13 13−15
⎤
⎥
⎥
⎦
(e)
∙
32
−2−3
¸
(f)
⎡
⎣
612
112
−1−1−3
⎤
⎦
(g)
⎡
⎣
−3−24
23 −5
213
⎤
⎦
(3)[(26)]
=(22)(see Figure 11)
(4) (a)P=
∙
13 31
−18−43
¸
,Q=
∙
−11−8
29 21
¸
,T=
∙
13
−1−4
¸
(b)P=
⎡
⎣
010
001
100
⎤
⎦,Q=
⎡
⎣
010
100
001
⎤
⎦,T=
⎡
⎣
001
010
100
⎤
⎦
(c)P=
⎡
⎣
2813
−6−25−43
11 45 76
⎤
⎦,Q=
⎡
⎣
−24−21
30 3 −1
139 13−5
⎤
⎦,T=
⎡
⎣
−25−97−150
31 120 185
145 562 868
⎤
⎦
(d)P=
⎡
⎢
⎢
⎣
−3−45−77−38
−41−250−420−205
19 113 191 93
−4−22−37−18
⎤
⎥
⎥
⎦
,Q=
⎡
⎢
⎢
⎣
1013
012 −1
1266
3−16 36
⎤
⎥
⎥
⎦
,T=
⎡
⎢
⎢
⎣
4231
1−2−1−1
5172
2131
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 80
(c) Both ranks equaldim(span()), and so they must be equal.
(21) Use the fact thatsin(
+
)=(cos )sin
+(sin )cos
Note that theth row ofAis then
(cos
)x1+(sin )x2Hence each row ofAis a linear combination of{x 1x2}and so
dim(span({rows ofA}))≤2. Hence,Ahas less thanpivots when row reduced, and so|A|=0.
(22) (a) T (b) T (c) F (d) T (e) F (f) F (g) F
Section 4.7
(1) (a)[v] =[7−1−5]
(b)[v]
=[3−2−1−2]
(c)[v]
=[−24−5]
(d)[v]
=[1032−1]
(e)[v]
=[4−53]
(f)[v]
=[−323]
(g)[v]
=[−14−2]
(h)[v]
=[2−31]
(i)[v]
=[52−6]
(j)[v]
=[5−2]
(2) (a)
⎡
⎣
−102 20 3
67−13−2
36−7−1
⎤
⎦
(b)
⎡
⎣
−2−6−1
361
−6−7−1
⎤
⎦
(c)
⎡
⎣
20−30−69
24−24−80
−91131
⎤
⎦
(d)
⎡
⎢
⎢
⎣
−1−42 −9
4513
02 −31
−4−13 13−15
⎤
⎥
⎥
⎦
(e)
∙
32
−2−3
¸
(f)
⎡
⎣
612
112
−1−1−3
⎤
⎦
(g)
⎡
⎣
−3−24
23 −5
213
⎤
⎦
(3)[(26)]
=(22)(see Figure 11)
(4) (a)P=
∙
13 31
−18−43
¸
,Q=
∙
−11−8
29 21
¸
,T=
∙
13
−1−4
¸
(b)P=
⎡
⎣
010
001
100
⎤
⎦,Q=
⎡
⎣
010
100
001
⎤
⎦,T=
⎡
⎣
001
010
100
⎤
⎦
(c)P=
⎡
⎣
2813
−6−25−43
11 45 76
⎤
⎦,Q=
⎡
⎣
−24−21
30 3 −1
139 13−5
⎤
⎦,T=
⎡
⎣
−25−97−150
31 120 185
145 562 868
⎤
⎦
(d)P=
⎡
⎢
⎢
⎣
−3−45−77−38
−41−250−420−205
19 113 191 93
−4−22−37−18
⎤
⎥
⎥
⎦
,Q=
⎡
⎢
⎢
⎣
1013
012 −1
1266
3−16 36
⎤
⎥
⎥
⎦
,T=
⎡
⎢
⎢
⎣
4231
1−2−1−1
5172
2131
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 80
Answers to Exercises Section 4.7
Figure 11:(26)in-coordinates
(5) (a)=([1−40−20][00140][00001]);P=
⎡
⎣
163
153
132
⎤
⎦;
Q=P
−1
=
⎡
⎣
1−33
1−10
−23 −1
⎤
⎦;[v] =[174−13];[v] =[2−23]
(b)=([1017021][01305][00012]);P=
⎡
⎣
131
−5−14−4
023
⎤
⎦;
Q=P
−1
=
⎡
⎣
−34−72
15 3 −1
−10−21
⎤
⎦;[v] =[5−23];[v] =[2−95]
(c)=([1000][0100][0010][0001]);
Copyrightc°2016 Elsevier Ltd. All rights reserved. 81
Figure 11:(26)in-coordinates
(5) (a)=([1−40−20][00140][00001]);P=
⎡
⎣
163
153
132
⎤
⎦;
Q=P
−1
=
⎡
⎣
1−33
1−10
−23 −1
⎤
⎦;[v] =[174−13];[v] =[2−23]
(b)=([1017021][01305][00012]);P=
⎡
⎣
131
−5−14−4
023
⎤
⎦;
Q=P
−1
=
⎡
⎣
−34−72
15 3 −1
−10−21
⎤
⎦;[v] =[5−23];[v] =[2−95]
(c)=([1000][0100][0010][0001]);
Copyrightc°2016 Elsevier Ltd. All rights reserved. 81
Answers to Exercises Section 4.7
P=
⎡
⎢
⎢
⎣
36 −4−2
−17 −30
4−331
6−242
⎤
⎥
⎥
⎦
;Q=P
−1
=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
1−4−12 7
−2927 −
31
2
−52267 −
77
2
5−23−71 41
⎤
⎥
⎥
⎥
⎥
⎥
⎦
;
[v]
=[21−37];[v] =[1014312]
(6) For each,v
=0v 1+0v 2+···+1v +···+0v and so[v ]=[0010] =e
(7) (a) Transition matrices to
1,2,3,4,and 5, respectively:
⎡
⎣
010
001
100
⎤
⎦,
⎡
⎣
001
100
010
⎤
⎦,
⎡
⎣
100
001
010
⎤
⎦,
⎡
⎣
010
100
001
⎤
⎦,
⎡
⎣
001
010
100
⎤
⎦
(b) Let=(v
1v )and=(v 1v ),where 12··· is a rearrangement of the
numbers12···.LetPbe the transition matrix fromto. The goal is to show, for
1≤≤, that theth row ofPequalse
.Thisistrueiff(for1≤≤)the th column ofP
equalse
.Nowthe th column ofPis, by definition,[v ]. But sincev is theth vector in
,[v
]=e, completing the proof.
(8) (a) Following the Transition Matrix Method, it is necessary to row reduce
⎡
⎢
⎢
⎣
First Second th
vector vector···vector
in in in
¯
¯
¯
¯
¯
¯
¯
¯
I
⎤
⎥
⎥
⎦
The algorithm from Section 2.4 for calculating inverses shows that the result obtained in the last
columnsisthedesiredinverse.
(b) Following the Transition Matrix Method, it is necessary to row reduce
⎡
⎢
⎢
⎣
I
¯
¯
¯
¯
¯
¯
¯
¯
First Second th
vector vector···vector
in in in
⎤
⎥
⎥
⎦
Clearly this is already in row reduced form, and so the result is as desired.
(9) Letbe the standard basis forR
.Exercise8(b)showsthatPis the transition matrix fromto
, and Exercise 8(a) shows thatQ
−1
is the transition matrix fromto. Theorem 4.18finishes the
proof.
(10)=([−14264167][−532463][−246111290])
(11) (a) Easily verified with straightforward computations.
(b)[v]
=[12−3];Av=[1425];D[v] =[Av] =[2−23]
(12) (a)
1=2,v 1=[215]; 2=−1,v 2=[19231]; 3=−3,v 3=[6−25]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 82
P=
⎡
⎢
⎢
⎣
36 −4−2
−17 −30
4−331
6−242
⎤
⎥
⎥
⎦
;Q=P
−1
=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
1−4−12 7
−2927 −
31
2
−52267 −
77
2
5−23−71 41
⎤
⎥
⎥
⎥
⎥
⎥
⎦
;
[v]
=[21−37];[v] =[1014312]
(6) For each,v
=0v 1+0v 2+···+1v +···+0v and so[v ]=[0010] =e
(7) (a) Transition matrices to
1,2,3,4,and 5, respectively:
⎡
⎣
010
001
100
⎤
⎦,
⎡
⎣
001
100
010
⎤
⎦,
⎡
⎣
100
001
010
⎤
⎦,
⎡
⎣
010
100
001
⎤
⎦,
⎡
⎣
001
010
100
⎤
⎦
(b) Let=(v
1v )and=(v 1v ),where 12··· is a rearrangement of the
numbers12···.LetPbe the transition matrix fromto. The goal is to show, for
1≤≤, that theth row ofPequalse
.Thisistrueiff(for1≤≤)the th column ofP
equalse
.Nowthe th column ofPis, by definition,[v ]. But sincev is theth vector in
,[v
]=e, completing the proof.
(8) (a) Following the Transition Matrix Method, it is necessary to row reduce
⎡
⎢
⎢
⎣
First Second th
vector vector···vector
in in in
¯
¯
¯
¯
¯
¯
¯
¯
I
⎤
⎥
⎥
⎦
The algorithm from Section 2.4 for calculating inverses shows that the result obtained in the last
columnsisthedesiredinverse.
(b) Following the Transition Matrix Method, it is necessary to row reduce
⎡
⎢
⎢
⎣
I
¯
¯
¯
¯
¯
¯
¯
¯
First Second th
vector vector···vector
in in in
⎤
⎥
⎥
⎦
Clearly this is already in row reduced form, and so the result is as desired.
(9) Letbe the standard basis forR
.Exercise8(b)showsthatPis the transition matrix fromto
, and Exercise 8(a) shows thatQ
−1
is the transition matrix fromto. Theorem 4.18finishes the
proof.
(10)=([−14264167][−532463][−246111290])
(11) (a) Easily verified with straightforward computations.
(b)[v]
=[12−3];Av=[1425];D[v] =[Av] =[2−23]
(12) (a)
1=2,v 1=[215]; 2=−1,v 2=[19231]; 3=−3,v 3=[6−25]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 82
Answers to Exercises Section 4.7
(b)D=
⎡
⎣
200
0−10
00 −3
⎤
⎦ (c)
⎡
⎣
219 6
12 −2
531 5
⎤
⎦
(13) LetV,,andthe
’s andw’s be as given in the statement of the theorem.
Proof of Part (1): Suppose that[w
1]=[1]and[w 2]=[1]. Then,
w
1=1v1+···+ vandw 2=1v1+···+ v
Hence,
w
1+w2=1v1+···+ v+1v1+···+ v=(1+1)v1+···+( +)v
implying
[w
1+w2]=[( 1+1)( +)]
which equals[w
1]+[w 2].
Proof of Part (2): Suppose that[w
1]=[1]. Then,w 1=1v1+···+ v. Hence,
1w1=1(1v1+···+ v)= 11v1+···+ 1v
implying[
1w1]=[ 111], which equals 1[1]= 1[w].
Proof of Part (3): Use induction on.
Base Step(=1): Thisisjustpart(2).
Inductive Step: Assume true for any linear combination ofvectors, and prove true for a linear
combination of+1vectors. Now,
[
1w1+···+ w++1w+1]=[ 1w1+···+ w]+[+1w+1](by part (1))
=[
1w1+···+ w]++1[w+1](by part (2))
=
1[w1]+···+ [w]++1[w+1]
(by the inductive hypothesis).
(14) Letv∈V. Then, using Theorem 4.17,(QP)[v]
=Q[v] =[v] . Hence, by Theorem 4.17,QPis
the (unique) transition matrix fromto.
(15) Let,,V,andPbe as given in the statement of the theorem. LetQbe the transition matrix from
to. Then, by Theorem 4.18,QPis the transition matrix fromto. But, clearly, since for all
v∈V,I[v]
=[v] , Theorem 4.17 implies thatImust be the transition matrix fromto. Hence,
QP=I,finishing the proof of the theorem.
(16) Let,V,andPbe as given in the statement of the problem. Let=(v
1v ),wherev is the
vector inVsuch that[v
]=th column ofP
−1
.Thenis a basis forVbecause it haslinearly
independent vectors, since the columns ofP
−1
are linearly independent. By definition,P
−1
is the
transition matrix fromto. Hence, by Theorem 4.19,Pis the transition matrix fromto.
(17) (a) F (b) T (c) T (d) F (e) F (f) T (g) F
Copyrightc°2016 Elsevier Ltd. All rights reserved. 83
(b)D=
⎡
⎣
200
0−10
00 −3
⎤
⎦ (c)
⎡
⎣
219 6
12 −2
531 5
⎤
⎦
(13) LetV,,andthe
’s andw’s be as given in the statement of the theorem.
Proof of Part (1): Suppose that[w
1]=[1]and[w 2]=[1]. Then,
w
1=1v1+···+ vandw 2=1v1+···+ v
Hence,
w
1+w2=1v1+···+ v+1v1+···+ v=(1+1)v1+···+( +)v
implying
[w
1+w2]=[( 1+1)( +)]
which equals[w
1]+[w 2].
Proof of Part (2): Suppose that[w
1]=[1]. Then,w 1=1v1+···+ v. Hence,
1w1=1(1v1+···+ v)= 11v1+···+ 1v
implying[
1w1]=[ 111], which equals 1[1]= 1[w].
Proof of Part (3): Use induction on.
Base Step(=1): Thisisjustpart(2).
Inductive Step: Assume true for any linear combination ofvectors, and prove true for a linear
combination of+1vectors. Now,
[
1w1+···+ w++1w+1]=[ 1w1+···+ w]+[+1w+1](by part (1))
=[
1w1+···+ w]++1[w+1](by part (2))
=
1[w1]+···+ [w]++1[w+1]
(by the inductive hypothesis).
(14) Letv∈V. Then, using Theorem 4.17,(QP)[v]
=Q[v] =[v] . Hence, by Theorem 4.17,QPis
the (unique) transition matrix fromto.
(15) Let,,V,andPbe as given in the statement of the theorem. LetQbe the transition matrix from
to. Then, by Theorem 4.18,QPis the transition matrix fromto. But, clearly, since for all
v∈V,I[v]
=[v] , Theorem 4.17 implies thatImust be the transition matrix fromto. Hence,
QP=I,finishing the proof of the theorem.
(16) Let,V,andPbe as given in the statement of the problem. Let=(v
1v ),wherev is the
vector inVsuch that[v
]=th column ofP
−1
.Thenis a basis forVbecause it haslinearly
independent vectors, since the columns ofP
−1
are linearly independent. By definition,P
−1
is the
transition matrix fromto. Hence, by Theorem 4.19,Pis the transition matrix fromto.
(17) (a) F (b) T (c) T (d) F (e) F (f) T (g) F
Copyrightc°2016 Elsevier Ltd. All rights reserved. 83
Answers to Exercises Chapter 4 Review
Chapter 4 Review Exercises
(1) This is a vector space. To prove this, modify the proof in Example 7 in Section 4.1.
(2) Zero vector=[−45];additiveinverseof[ ]is[−−8−+ 10]
(3) (a), (b), (d), (f) and (g) are not subspaces; (c) and (e) are subspaces
(4) (a)span()={[ 5−3+]| ∈R}6 =R
4
(b) Basis={[1005][010−3][0011]};dim(span()) = 3
(5) (a)span()={
3
+
2
+(−3+2)+| ∈R}6 =P 3
(b) Basis={
3
−3
2
+21};dim(span()) = 3
(6) (a)span()=
½∙
4−3
−2+−
¸¯
¯
¯
¯
∈R
¾
6=M
23
(b) Basis=
½∙
104
0−20
¸
∙
01−3
01 0
¸
∙
000
1−10
¸
∙
000
001
¸¾
;dim(span()) = 4
(7) (a)is linearly independent
(b)itself is a basis forspan().spansR
3
.
(c) No, by Theorem 4.9
(8) (a)is linearly dependent;
3
−2
2
−+2=3(−5
3
+2
2
+5−2) + 8(2
3
−
2
−2+1)
(b) The subset
{−5
3
+2
2
+5−22
3
−
2
−2+1−2
3
+2
2
+3−5}
ofis a basis forspan().does not spanP
3.
(c) Yes, by part (b) of Exercise 21 in Section 4.4. One such alternate linear combination is
−5
¡
−5
3
+2
2
+5−2
¢
−5
¡
2
3
−
2
−2+1
¢
+1
¡
3
−2
2
−+2
¢
−1
¡
−2
3
+2
2
+3−5
¢
(9) (a) Linearly dependent;
⎡
⎣
46
34
25
⎤
⎦=−
⎡
⎣
−10−14
−10−8
−6−12
⎤
⎦−2
⎡
⎣
712
57
310
⎤
⎦+
⎡
⎣
816
310
213
⎤
⎦
(b)
⎧
⎨
⎩
⎡
⎣
40
11−2
6−1
⎤
⎦
⎡
⎣
−10−14
−10−8
−6−12
⎤
⎦
⎡
⎣
712
57
310
⎤
⎦
⎡
⎣
816
310
213
⎤
⎦
⎡
⎣
611
47
39
⎤
⎦
⎫
⎬
⎭
;does not spanM
32
(c) Yes, by part (b) of Exercise 21 in Section 4.4. One such alternate linear combination is
2
⎡
⎣
40
11−2
6−1
⎤
⎦+
⎡
⎣
−10−14
−10−8
−6−12
⎤
⎦−
⎡
⎣
712
57
310
⎤
⎦−
⎡
⎣
816
310
213
⎤
⎦+
⎡
⎣
46
34
25
⎤
⎦+
⎡
⎣
611
47
39
⎤
⎦
(10)v=1v.Also,sincev∈span(),v=
1v1+···+ v,forsome 1.Theseare2different
ways to expressvas a linear combination of vectors in.
(11) Use the same technique as in Example 13 in Section 4.4.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 84
Chapter 4 Review Exercises
(1) This is a vector space. To prove this, modify the proof in Example 7 in Section 4.1.
(2) Zero vector=[−45];additiveinverseof[ ]is[−−8−+ 10]
(3) (a), (b), (d), (f) and (g) are not subspaces; (c) and (e) are subspaces
(4) (a)span()={[ 5−3+]| ∈R}6 =R
4
(b) Basis={[1005][010−3][0011]};dim(span()) = 3
(5) (a)span()={
3
+
2
+(−3+2)+| ∈R}6 =P 3
(b) Basis={
3
−3
2
+21};dim(span()) = 3
(6) (a)span()=
½∙
4−3
−2+−
¸¯
¯
¯
¯
∈R
¾
6=M
23
(b) Basis=
½∙
104
0−20
¸
∙
01−3
01 0
¸
∙
000
1−10
¸
∙
000
001
¸¾
;dim(span()) = 4
(7) (a)is linearly independent
(b)itself is a basis forspan().spansR
3
.
(c) No, by Theorem 4.9
(8) (a)is linearly dependent;
3
−2
2
−+2=3(−5
3
+2
2
+5−2) + 8(2
3
−
2
−2+1)
(b) The subset
{−5
3
+2
2
+5−22
3
−
2
−2+1−2
3
+2
2
+3−5}
ofis a basis forspan().does not spanP
3.
(c) Yes, by part (b) of Exercise 21 in Section 4.4. One such alternate linear combination is
−5
¡
−5
3
+2
2
+5−2
¢
−5
¡
2
3
−
2
−2+1
¢
+1
¡
3
−2
2
−+2
¢
−1
¡
−2
3
+2
2
+3−5
¢
(9) (a) Linearly dependent;
⎡
⎣
46
34
25
⎤
⎦=−
⎡
⎣
−10−14
−10−8
−6−12
⎤
⎦−2
⎡
⎣
712
57
310
⎤
⎦+
⎡
⎣
816
310
213
⎤
⎦
(b)
⎧
⎨
⎩
⎡
⎣
40
11−2
6−1
⎤
⎦
⎡
⎣
−10−14
−10−8
−6−12
⎤
⎦
⎡
⎣
712
57
310
⎤
⎦
⎡
⎣
816
310
213
⎤
⎦
⎡
⎣
611
47
39
⎤
⎦
⎫
⎬
⎭
;does not spanM
32
(c) Yes, by part (b) of Exercise 21 in Section 4.4. One such alternate linear combination is
2
⎡
⎣
40
11−2
6−1
⎤
⎦+
⎡
⎣
−10−14
−10−8
−6−12
⎤
⎦−
⎡
⎣
712
57
310
⎤
⎦−
⎡
⎣
816
310
213
⎤
⎦+
⎡
⎣
46
34
25
⎤
⎦+
⎡
⎣
611
47
39
⎤
⎦
(10)v=1v.Also,sincev∈span(),v=
1v1+···+ v,forsome 1.Theseare2different
ways to expressvas a linear combination of vectors in.
(11) Use the same technique as in Example 13 in Section 4.4.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 84
Answers to Exercises Chapter 4 Review
(12) (a) The matrix whose rows are the given vectors row reduces toI 4,sotheSimplified Span Method
shows that the given set spansR
4
. Since the set has4vectors anddim(R
4
)=4, part (1) of
Theorem 4.12 shows that the given set is a basis forR
4
.
(b) Similar to part (a). The matrix
⎡
⎣
2213
10 3
4116
⎤
⎦row reduces toI
3
(c) Similar to part (a). The matrix
⎡
⎢
⎢
⎣
1503
6−143
7−47 −1
−37 −24
⎤
⎥
⎥
⎦
row reduces toI
4
(13) (a)Wnonempty:0∈WbecauseA0=O.
Closure under addition: IfX
1X2∈W,thenA(X 1+X2)=AX 1+AX 2=O+O=O.
Closure under scalar multiplication: IfX∈W,thenA(X)=AX=O=O.
(b) Basis={[3100][−2011]}
(c)dim(W)=2,rank(A)=2;2+2=4
(14) (a) First, we use direct computation to check that every polynomial inis inV:
Ifp()=
3
−3,thenp
0
()=3
2
−3,andsop
0
(1) = 3−3=0.
Ifp()=
2
−2,thenp
0
()=2−2,andsop
0
(1) = 2−2=0.
Ifp()=1,thenp
0
()=0,andsop
0
(1) = 0.
Thus, every polynomial inis actually inV.
Next, we convert the polynomials into4-vectors, and use the Independence Test Method. Now,
⎡
⎢
⎢
⎣
100
010
−3−20
001
⎤
⎥
⎥
⎦
clearly row reduces to
⎡
⎢
⎢
⎣
100
010
001
000
⎤
⎥
⎥
⎦
,sois linearly independent. Finally, since
the polynomial∈V,dim(V)dim(P
3)=4by Theorem 4.13. But, part (2) of Theorem 4.12
shows that||≤dim(V). Hence,3=||≤dim(V)dim(P
3)=4,andsodim(V)=3. Part (2)
of Theorem 4.12 then implies thatis a basis forV.
(b)={1
3
−3
2
+3}is a basis forWanddim(W)=2.
(15)={[2−301][4304][1021]}
(16)={
2
−2
3
−2
3
−2
2
+1}
(17){[21−12][1−22−4][0100][0010]}
(18)
⎧
⎨
⎩
⎡
⎣
3−1
02
10
⎤
⎦
⎡
⎣
−12
0−1
30
⎤
⎦
⎡
⎣
2−1
01
40
⎤
⎦
⎡
⎣
10
00
00
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎡
⎣
00
00
01
⎤
⎦
⎫
⎬
⎭
(19)
{
4
+3
2
+1
3
−2
2
−1+3}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 85
(12) (a) The matrix whose rows are the given vectors row reduces toI 4,sotheSimplified Span Method
shows that the given set spansR
4
. Since the set has4vectors anddim(R
4
)=4, part (1) of
Theorem 4.12 shows that the given set is a basis forR
4
.
(b) Similar to part (a). The matrix
⎡
⎣
2213
10 3
4116
⎤
⎦row reduces toI
3
(c) Similar to part (a). The matrix
⎡
⎢
⎢
⎣
1503
6−143
7−47 −1
−37 −24
⎤
⎥
⎥
⎦
row reduces toI
4
(13) (a)Wnonempty:0∈WbecauseA0=O.
Closure under addition: IfX
1X2∈W,thenA(X 1+X2)=AX 1+AX 2=O+O=O.
Closure under scalar multiplication: IfX∈W,thenA(X)=AX=O=O.
(b) Basis={[3100][−2011]}
(c)dim(W)=2,rank(A)=2;2+2=4
(14) (a) First, we use direct computation to check that every polynomial inis inV:
Ifp()=
3
−3,thenp
0
()=3
2
−3,andsop
0
(1) = 3−3=0.
Ifp()=
2
−2,thenp
0
()=2−2,andsop
0
(1) = 2−2=0.
Ifp()=1,thenp
0
()=0,andsop
0
(1) = 0.
Thus, every polynomial inis actually inV.
Next, we convert the polynomials into4-vectors, and use the Independence Test Method. Now,
⎡
⎢
⎢
⎣
100
010
−3−20
001
⎤
⎥
⎥
⎦
clearly row reduces to
⎡
⎢
⎢
⎣
100
010
001
000
⎤
⎥
⎥
⎦
,sois linearly independent. Finally, since
the polynomial∈V,dim(V)dim(P
3)=4by Theorem 4.13. But, part (2) of Theorem 4.12
shows that||≤dim(V). Hence,3=||≤dim(V)dim(P
3)=4,andsodim(V)=3. Part (2)
of Theorem 4.12 then implies thatis a basis forV.
(b)={1
3
−3
2
+3}is a basis forWanddim(W)=2.
(15)={[2−301][4304][1021]}
(16)={
2
−2
3
−2
3
−2
2
+1}
(17){[21−12][1−22−4][0100][0010]}
(18)
⎧
⎨
⎩
⎡
⎣
3−1
02
10
⎤
⎦
⎡
⎣
−12
0−1
30
⎤
⎦
⎡
⎣
2−1
01
40
⎤
⎦
⎡
⎣
10
00
00
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎡
⎣
00
00
01
⎤
⎦
⎫
⎬
⎭
(19)
{
4
+3
2
+1
3
−2
2
−1+3}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 85
Answers to Exercises Chapter 4 Review
(20) (a)[v] =[−3−1−2] (b)[v] =[42−3] (c)[v] =[−35−1]
(21) (a)[v]
=[27−626];P=
⎡
⎣
42 5
−10 1
31−2
⎤
⎦;[v] =[14−217]
(b)[v]
=[−4−13];P=
⎡
⎣
41−2
12 0
−11 1
⎤
⎦;[v] =[−23−66]
(c)[v]
=[4−21−3];P=
⎡
⎢
⎢
⎣
62 2 −1
50 1 −1
−31−10
−52 0 1
⎤
⎥
⎥
⎦
;[v] =[2524−15−27]
(22) (a)P=
⎡
⎢
⎢
⎣
0 010
1 011
1−111
0 101
⎤
⎥
⎥
⎦
(b)Q=
⎡
⎢
⎢
⎣
0101
1011
1101
0010
⎤
⎥
⎥
⎦
(c)R=QP=
⎡
⎢
⎢
⎣
1 112
1 022
1 122
1−111
⎤
⎥
⎥
⎦
(d)R
−1
=
⎡
⎢
⎢
⎣
0−322
0−110
−1010
12 −2−1
⎤
⎥
⎥
⎦
(23) (a)
1=0,v 1=[4413]; 2=2,v 2=[−320]v 3=[304]
(b)D=
⎡
⎣
000
020
002
⎤
⎦ (c)
⎡
⎣
4−33
420
13 0 4
⎤
⎦
(24) (a)={[100−7][0106][0014]}
(b)P=
⎡
⎣
1−13
−2−1−5
112
⎤
⎦
(c)[v]
=[25−2−10];(Note:P
−1
=
⎡
⎣
358
−1−1−1
−1−2−3
⎤
⎦)
(d)v=[−32345]
(25) By part (b) of Exercise 8 in Section 4.7, the matrixCis the transition matrix from-coordinates to
standard coordinates. Hence, by Theorem 4.18,CPis the transition matrix from-coordinates to
standard coordinates. However, again by part (b) ofExercise 8 in Section 4.7, this transition matrix
is the matrix whose columns are the vectors in.
(26) (a) T
(b) T
(c) F
(d) T
(e) F
(f) T
(g) T
(h) F
(i) F
(j) T
(k) F
(l) F
(m) T
(n) T
(o) F
(p) F
(q) T
(r) F
(s) T
(t) T
(u) T
(v) T
(w) F
(x) T
(y) T
(z) F
Copyrightc°2016 Elsevier Ltd. All rights reserved. 86
(20) (a)[v] =[−3−1−2] (b)[v] =[42−3] (c)[v] =[−35−1]
(21) (a)[v]
=[27−626];P=
⎡
⎣
42 5
−10 1
31−2
⎤
⎦;[v] =[14−217]
(b)[v]
=[−4−13];P=
⎡
⎣
41−2
12 0
−11 1
⎤
⎦;[v] =[−23−66]
(c)[v]
=[4−21−3];P=
⎡
⎢
⎢
⎣
62 2 −1
50 1 −1
−31−10
−52 0 1
⎤
⎥
⎥
⎦
;[v] =[2524−15−27]
(22) (a)P=
⎡
⎢
⎢
⎣
0 010
1 011
1−111
0 101
⎤
⎥
⎥
⎦
(b)Q=
⎡
⎢
⎢
⎣
0101
1011
1101
0010
⎤
⎥
⎥
⎦
(c)R=QP=
⎡
⎢
⎢
⎣
1 112
1 022
1 122
1−111
⎤
⎥
⎥
⎦
(d)R
−1
=
⎡
⎢
⎢
⎣
0−322
0−110
−1010
12 −2−1
⎤
⎥
⎥
⎦
(23) (a)
1=0,v 1=[4413]; 2=2,v 2=[−320]v 3=[304]
(b)D=
⎡
⎣
000
020
002
⎤
⎦ (c)
⎡
⎣
4−33
420
13 0 4
⎤
⎦
(24) (a)={[100−7][0106][0014]}
(b)P=
⎡
⎣
1−13
−2−1−5
112
⎤
⎦
(c)[v]
=[25−2−10];(Note:P
−1
=
⎡
⎣
358
−1−1−1
−1−2−3
⎤
⎦)
(d)v=[−32345]
(25) By part (b) of Exercise 8 in Section 4.7, the matrixCis the transition matrix from-coordinates to
standard coordinates. Hence, by Theorem 4.18,CPis the transition matrix from-coordinates to
standard coordinates. However, again by part (b) ofExercise 8 in Section 4.7, this transition matrix
is the matrix whose columns are the vectors in.
(26) (a) T
(b) T
(c) F
(d) T
(e) F
(f) T
(g) T
(h) F
(i) F
(j) T
(k) F
(l) F
(m) T
(n) T
(o) F
(p) F
(q) T
(r) F
(s) T
(t) T
(u) T
(v) T
(w) F
(x) T
(y) T
(z) F
Copyrightc°2016 Elsevier Ltd. All rights reserved. 86
Answers to Exercises Section 5.1
Chapter 5
Section 5.1
(1) (a) Linear operator:
([ ]+[ ]) =([+ +])
=[3(+)−4(+)−(+)+2(+)]
=[(3−4)+(3−4)(−+2)+(−+2)]
=[3−4−+2]+[3−4−+2]=([ ]) +([ ])
Also,
([ ]) =([ ]) = [3()
−4()−()+2()]
=[(3−4)(−+2)]
=[3−4−+2]=([ ])
(b) Not a linear transformation:([0000]) = [2−10−3]6 =0.
(c) Linear operator:
([ ]+[ ]) =([+ + +])
=[+ ++]
=[ ]+[ ]=([ ]) +([
])
Also,
([ ]) =([ ]) = [ ]
=[ ]=([ ])
(d) Linear operator:
µ∙
¸
+
∙
¸¶
=
µ∙
++
++
¸¶
=
∙
(+)−2(+)+(+)3( +)
−4(+)( +)+(+)−3(+)
¸
=
∙
(−2
+)+(−2+)3 +3
−4−4 (+−3)+(+−3)
¸
=
∙
−2+ 3
−4 +−3
¸
+
∙
−2+ 3
−4 +−3
¸
=
µ∙
¸¶
+
µ∙
¸¶
Copyrightc°2016 Elsevier Ltd. All rights reserved. 87
Chapter 5
Section 5.1
(1) (a) Linear operator:
([ ]+[ ]) =([+ +])
=[3(+)−4(+)−(+)+2(+)]
=[(3−4)+(3−4)(−+2)+(−+2)]
=[3−4−+2]+[3−4−+2]=([ ]) +([ ])
Also,
([ ]) =([ ]) = [3()
−4()−()+2()]
=[(3−4)(−+2)]
=[3−4−+2]=([ ])
(b) Not a linear transformation:([0000]) = [2−10−3]6 =0.
(c) Linear operator:
([ ]+[ ]) =([+ + +])
=[+ ++]
=[ ]+[ ]=([ ]) +([
])
Also,
([ ]) =([ ]) = [ ]
=[ ]=([ ])
(d) Linear operator:
µ∙
¸
+
∙
¸¶
=
µ∙
++
++
¸¶
=
∙
(+)−2(+)+(+)3( +)
−4(+)( +)+(+)−3(+)
¸
=
∙
(−2
+)+(−2+)3 +3
−4−4 (+−3)+(+−3)
¸
=
∙
−2+ 3
−4 +−3
¸
+
∙
−2+ 3
−4 +−3
¸
=
µ∙
¸¶
+
µ∙
¸¶
Copyrightc°2016 Elsevier Ltd. All rights reserved. 87
Answers to Exercises Section 5.1
Also,
µ
∙
¸¶
=
µ∙
¸¶
=
∙
()−2()+()3( )
−4()( )+()−3()
¸
=
∙
(−2+) (3)
(−4) (+−3)
¸
=
∙
−2+ 3
−4 +−3
¸
=
µ∙
¸¶
(e) Not a linear transformation:
µ
2
∙
12
01
¸¶
=
µ∙
24
02
¸¶
=4,but2
µ∙
12
01
¸¶
=
2(1) = 2.
(f) Not a linear transformation:(8
3
)=2
2
,but8(
3
)=8(
2
).
(g) Not a linear transformation:(0)=[101]6 =0.
(h) Linear transformation:
µ
(
3
+
2
++)
+(
3
+
2
++)
¶
=((+)
3
+(+)
2
+(+)+(+))
=(+)+(+)+(+)+(+)
=(+++)+(+++)
=(
3
+
2
++)+(
3
+
2
++)
Also,
((
3
+
2
++)) =(()
3
+()
2
+()+())
=()+()+()+()
=(+++)
=(
3
+
2
++)
(i) Not a linear transformation:((−1)[0100]) =([0−100]) = 1, but(−1)([0100]) =
(−1)(1) =−1.
(j) Not a linear transformation:(
2
+(+1)) =(
2
++1) = 1,but(
2
)+(+1) = 0+0 = 0.
(k) Linear transformation:
⎛
⎝
⎡
⎣
⎤
⎦+
⎡
⎣
⎤
⎦
⎞
⎠=
⎛
⎝
⎡
⎣
++
++
++
⎤
⎦
⎞
⎠
=(+)
4
−(+)
2
+(+)
=(
4
−
2
+)+(
4
−
2
+)
=
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠+
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠
Copyrightc°2016 Elsevier Ltd. All rights reserved. 88
Also,
µ
∙
¸¶
=
µ∙
¸¶
=
∙
()−2()+()3( )
−4()( )+()−3()
¸
=
∙
(−2+) (3)
(−4) (+−3)
¸
=
∙
−2+ 3
−4 +−3
¸
=
µ∙
¸¶
(e) Not a linear transformation:
µ
2
∙
12
01
¸¶
=
µ∙
24
02
¸¶
=4,but2
µ∙
12
01
¸¶
=
2(1) = 2.
(f) Not a linear transformation:(8
3
)=2
2
,but8(
3
)=8(
2
).
(g) Not a linear transformation:(0)=[101]6 =0.
(h) Linear transformation:
µ
(
3
+
2
++)
+(
3
+
2
++)
¶
=((+)
3
+(+)
2
+(+)+(+))
=(+)+(+)+(+)+(+)
=(+++)+(+++)
=(
3
+
2
++)+(
3
+
2
++)
Also,
((
3
+
2
++)) =(()
3
+()
2
+()+())
=()+()+()+()
=(+++)
=(
3
+
2
++)
(i) Not a linear transformation:((−1)[0100]) =([0−100]) = 1, but(−1)([0100]) =
(−1)(1) =−1.
(j) Not a linear transformation:(
2
+(+1)) =(
2
++1) = 1,but(
2
)+(+1) = 0+0 = 0.
(k) Linear transformation:
⎛
⎝
⎡
⎣
⎤
⎦+
⎡
⎣
⎤
⎦
⎞
⎠=
⎛
⎝
⎡
⎣
++
++
++
⎤
⎦
⎞
⎠
=(+)
4
−(+)
2
+(+)
=(
4
−
2
+)+(
4
−
2
+)
=
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠+
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠
Copyrightc°2016 Elsevier Ltd. All rights reserved. 88
Answers to Exercises Section 5.1
Also,
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠=
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠
=()
4
−()
2
+()
=(
4
−
2
+)
=
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠
(l) Not a linear transformation:([30] + [04]) =([34]) =
√
3
2
+4
2
=5,but([30]) +([04])
=
√
3
2
+0
2
+
√
0
2
+4
2
=3+4=7 .
(2) (a)(v+w)=v+w=(v)+(w);also,(v)=v=(v).
(b)(v+w)=0=0+0=(v)+(w);also,(v)=0=0=(v).
(3)(v+w)=(v+w)=v+w=(v)+(w);
(v)=v=(v)=(v).
(4) (a) For addition:
([ ]+[ ]) =([+ + +])
=[−(+)+ +]
=[(−)+(−)+ +]
=[− ]+[− ]=([ ]) +([ ])
For scalar multiplication:
([ ]) =([ ]) = [−()]
=[(−)]=
[− ]=([ ])
(b)([ ]) = [− ]. It is a linear operator.
(c) Through the-axis:([ ]) = [− ]; Through the-axis:([ ]) = [−]; both are linear
operators.
(5) First,is a linear operator because
([
123]+[ 123]) =([ 1+12+23+3])
=[
1+12+20]
=[
120] + [ 120] =([ 123]) +([ 123])and
([
123]) =([ 123]) = [ 120] =[ 120] =([ 123])
Also,is a linear operator because
([
1234]+[ 1234]) =([ 1+12+23+34+4])
=[0
2+204+4]
=[0
204]+[0 204]
=([
1234]) +([ 1234])and
Copyrightc°2016 Elsevier Ltd. All rights reserved. 89
Also,
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠=
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠
=()
4
−()
2
+()
=(
4
−
2
+)
=
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠
(l) Not a linear transformation:([30] + [04]) =([34]) =
√
3
2
+4
2
=5,but([30]) +([04])
=
√
3
2
+0
2
+
√
0
2
+4
2
=3+4=7 .
(2) (a)(v+w)=v+w=(v)+(w);also,(v)=v=(v).
(b)(v+w)=0=0+0=(v)+(w);also,(v)=0=0=(v).
(3)(v+w)=(v+w)=v+w=(v)+(w);
(v)=v=(v)=(v).
(4) (a) For addition:
([ ]+[ ]) =([+ + +])
=[−(+)+ +]
=[(−)+(−)+ +]
=[− ]+[− ]=([ ]) +([ ])
For scalar multiplication:
([ ]) =([ ]) = [−()]
=[(−)]=
[− ]=([ ])
(b)([ ]) = [− ]. It is a linear operator.
(c) Through the-axis:([ ]) = [− ]; Through the-axis:([ ]) = [−]; both are linear
operators.
(5) First,is a linear operator because
([
123]+[ 123]) =([ 1+12+23+3])
=[
1+12+20]
=[
120] + [ 120] =([ 123]) +([ 123])and
([
123]) =([ 123]) = [ 120] =[ 120] =([ 123])
Also,is a linear operator because
([
1234]+[ 1234]) =([ 1+12+23+34+4])
=[0
2+204+4]
=[0
204]+[0 204]
=([
1234]) +([ 1234])and
Copyrightc°2016 Elsevier Ltd. All rights reserved. 89
Answers to Exercises Section 5.1
([ 1234]) =([ 1234]) = [0 204]
=[0
204]=([ 1234])
(6) For addition:
µ
[
12]
+[
12]
¶
=([
1+12+2++])
=
+
=([ 12]) +([ 12])
For scalar multiplication:
([
12]) =([ 12])
=
=([ 12])
(7) For addition:
(y+z)= proj
x(y+z)=
x·(y+z)
kxk
2
x=
(x·y)+(x·z)
kxk
2
x
=
(x·y)
kxk
2
x+
(x·z)
kxk
2
x=(y)+(z)
For scalar multiplication:
(y)=
x·(y)
kxk
2
x=
µ
x·y
kxk
2
¶
x=(y)
(8)(y
1+y2)=x·(y 1+y2)=(x·y 1)+(x·y 2)=(y 1)+(y 2);
(y
1)=x·(y 1)=(x·y 1)=(y 1)
(9) Follow the hint in the text, and use the sum of angle identities for sine and cosine.
(10) (a) Use Example 10.
(b) Use the hint in Exercise 9.
(c)
⎡
⎣
sin0cos
01 0
cos0−sin
⎤
⎦
(11) The proofs follow the proof in Example 10, but with the matrixAreplaced by the specific matrices of
this exercise.
(12)(A+B) = trace(A+B)=
P
=1
(+)=
P
=1
+
P
=1
=trace(A)+trace(B)=(A)+(B);
(A)=trace(A)=
P
=1
=
P
=1
=trace(A)=(A).
(13)(A
1+A2)=(A 1+A2)+(A 1+A2)
=(A 1+A
1
)+(A 2+A
2
)=(A 1)+(A 2);
(A)=A+(A)
=(A+A
)=(A)The proof foris done similarly.
(14) (a)(p
1+p2)=
R
(p 1+p2)=
R
p 1+
R
p 2=(p 1)+(p 2)(since the constant of integration
is assumed to be zero for all these indefinite integrals);
(p)=
R
(p)=
R
p=(p)(since the constant of integration is assumed to be zero for
these indefinite integrals).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 90
([ 1234]) =([ 1234]) = [0 204]
=[0
204]=([ 1234])
(6) For addition:
µ
[
12]
+[
12]
¶
=([
1+12+2++])
=
+
=([ 12]) +([ 12])
For scalar multiplication:
([
12]) =([ 12])
=
=([ 12])
(7) For addition:
(y+z)= proj
x(y+z)=
x·(y+z)
kxk
2
x=
(x·y)+(x·z)
kxk
2
x
=
(x·y)
kxk
2
x+
(x·z)
kxk
2
x=(y)+(z)
For scalar multiplication:
(y)=
x·(y)
kxk
2
x=
µ
x·y
kxk
2
¶
x=(y)
(8)(y
1+y2)=x·(y 1+y2)=(x·y 1)+(x·y 2)=(y 1)+(y 2);
(y
1)=x·(y 1)=(x·y 1)=(y 1)
(9) Follow the hint in the text, and use the sum of angle identities for sine and cosine.
(10) (a) Use Example 10.
(b) Use the hint in Exercise 9.
(c)
⎡
⎣
sin0cos
01 0
cos0−sin
⎤
⎦
(11) The proofs follow the proof in Example 10, but with the matrixAreplaced by the specific matrices of
this exercise.
(12)(A+B) = trace(A+B)=
P
=1
(+)=
P
=1
+
P
=1
=trace(A)+trace(B)=(A)+(B);
(A)=trace(A)=
P
=1
=
P
=1
=trace(A)=(A).
(13)(A
1+A2)=(A 1+A2)+(A 1+A2)
=(A 1+A
1
)+(A 2+A
2
)=(A 1)+(A 2);
(A)=A+(A)
=(A+A
)=(A)The proof foris done similarly.
(14) (a)(p
1+p2)=
R
(p 1+p2)=
R
p 1+
R
p 2=(p 1)+(p 2)(since the constant of integration
is assumed to be zero for all these indefinite integrals);
(p)=
R
(p)=
R
p=(p)(since the constant of integration is assumed to be zero for
these indefinite integrals).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 90
Answers to Exercises Section 5.1
(b)(p 1+p2)=
R
(p1+p2)=
R
p1+
R
p2=(p 1)+(p 2);
(p)=
R
(p)=
R
p=(p)
(15) Base Step:=1: An argument similar to that in Example 3 in Section 5.1 shows that:V−→V
given by()=
0
is a linear operator.
Inductive Step: Assume that
:V−→Vgiven by ()=
()
is a linear operator. We need to
show that
+1:V−→Vgiven by +1()=
(+1)
is a linear operator. But since +1=◦
Theorem 5.2 assures us that
+1is a linear operator.
(16)(A
1+A2)=B(A 1+A2)=BA 1+BA 2=(A 1)+(A 2);
(A)=B(A)=(BA)=(A)
(17)(A
1+A2)=B
−1
(A1+A2)B=B
−1
A1B+B
−1
A2B=(A 1)+(A 2);
(A)=B
−1
(A)B=(B
−1
AB)=(A)
(18) (a)(p+q)=(p+q)()=p()+q()=(p)+(q).
Similarly,(p)=(p)()=(p()) =(p).
(b) For all∈R,((p+q)()) = (p+q)(+)=p(+)+q(+)=(p()) +(q(
)).
Also,((p)()) = (p)(+)=(p(+)) =(p()).
(19) Letp=
+···+ 1+ 0and letq=
+···+ 1+ 0.Then
(p+q)= (
+···+ 1+ 0+
+···+ 1+ 0)
=((
+)
+···+( 1+1)+( 0+0))
=(
+)A
+···+( 1+1)A+( 0+0)I
=( A
+···+ 1A+ 0I)+( A
+···+ 1A+ 0I)
=(p)+(q)
Similarly,
(p)= ((
+···+ 1+ 0))
=(
+···+ 1+ 0)
=
A
+···+ 1A+ 0I
=( A
+···+ 1A+ 0I)
=(p)
(20)(⊕)=()=ln()=ln()+ln()=()+();
(¯)=(
)=ln(
)=ln()=()
(21)(0)=0+x=x6 =0, contradicting Theorem 5.1, part (1).
(22)(0)=A0+y=y6 =0, contradicting Theorem 5.1, part (1).
(23)
⎛
⎝
⎡
⎣
100
010
001
⎤
⎦
⎞
⎠=1, but
⎛
⎝
⎡
⎣
100
010
000
⎤
⎦
⎞
⎠+
⎛
⎝
⎡
⎣
000
000
001
⎤
⎦
⎞
⎠=0+0=0 . (Note: For any
1,(2I
)=2
, but2(I ) = 2(1) = 2.)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 91
(b)(p 1+p2)=
R
(p1+p2)=
R
p1+
R
p2=(p 1)+(p 2);
(p)=
R
(p)=
R
p=(p)
(15) Base Step:=1: An argument similar to that in Example 3 in Section 5.1 shows that:V−→V
given by()=
0
is a linear operator.
Inductive Step: Assume that
:V−→Vgiven by ()=
()
is a linear operator. We need to
show that
+1:V−→Vgiven by +1()=
(+1)
is a linear operator. But since +1=◦
Theorem 5.2 assures us that
+1is a linear operator.
(16)(A
1+A2)=B(A 1+A2)=BA 1+BA 2=(A 1)+(A 2);
(A)=B(A)=(BA)=(A)
(17)(A
1+A2)=B
−1
(A1+A2)B=B
−1
A1B+B
−1
A2B=(A 1)+(A 2);
(A)=B
−1
(A)B=(B
−1
AB)=(A)
(18) (a)(p+q)=(p+q)()=p()+q()=(p)+(q).
Similarly,(p)=(p)()=(p()) =(p).
(b) For all∈R,((p+q)()) = (p+q)(+)=p(+)+q(+)=(p()) +(q(
)).
Also,((p)()) = (p)(+)=(p(+)) =(p()).
(19) Letp=
+···+ 1+ 0and letq=
+···+ 1+ 0.Then
(p+q)= (
+···+ 1+ 0+
+···+ 1+ 0)
=((
+)
+···+( 1+1)+( 0+0))
=(
+)A
+···+( 1+1)A+( 0+0)I
=( A
+···+ 1A+ 0I)+( A
+···+ 1A+ 0I)
=(p)+(q)
Similarly,
(p)= ((
+···+ 1+ 0))
=(
+···+ 1+ 0)
=
A
+···+ 1A+ 0I
=( A
+···+ 1A+ 0I)
=(p)
(20)(⊕)=()=ln()=ln()+ln()=()+();
(¯)=(
)=ln(
)=ln()=()
(21)(0)=0+x=x6 =0, contradicting Theorem 5.1, part (1).
(22)(0)=A0+y=y6 =0, contradicting Theorem 5.1, part (1).
(23)
⎛
⎝
⎡
⎣
100
010
001
⎤
⎦
⎞
⎠=1, but
⎛
⎝
⎡
⎣
100
010
000
⎤
⎦
⎞
⎠+
⎛
⎝
⎡
⎣
000
000
001
⎤
⎦
⎞
⎠=0+0=0 . (Note: For any
1,(2I
)=2
, but2(I ) = 2(1) = 2.)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 91
Answers to Exercises Section 5.1
(24) 2(v+w)= 1(2(v+w)) = 2 1(v+w)=2( 1(v)+ 1(w)) = 2 1(v)+2 1(w)= 2(v)+ 2(w);
2(v)= 1(2v)=2 1(v)=(2 1(v)) = 2(v).
(25)([−324]) = [12−1114];
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠=
⎡
⎣
−230
1−2−1
013
⎤
⎦
⎡
⎣
⎤
⎦=
⎡
⎣
−2+3
−2−
+3
⎤
⎦
(26)(i)=
7
5
i−
11
5
j;(j)=−
2
5
i−
4
5
j
(27)(x−y)=(x+(−y)) =(x)+((−1)y)=(x)+(−1)(y)=(x)−(y).
(28) Follow the hint in the text. Letting==0proves property (1) of a linear transformation. Letting
=0proves property (2).
(29) Suppose part (3) of Theorem 5.1 is true for some.Weproveittruefor+1.
(
1v1+···+ v++1v+1)= ( 1v1+···+ v)+( +1v+1)
(by property (1) of a linear transformation)
=(
1(v1)+···+ (v)) +( +1v+1)
(by the inductive hypothesis)
=
1(v1)+···+ (v)+ +1(v+1)
(by property (2) of a linear transformation),
and we are done.
(30) (a)
1v1+···+ v=0V=⇒( 1v1+···+ v)=(0 V)
=⇒
1(v1)+···+ (v)=0 W=⇒ 1=2=···= =0.
(b) Consider the zero linear transformation.
(31)(
2◦1)(v)= 2(1(v)) = 2(1(v)) = 2(1(v)) =( 2◦1)(v).
(32) Let=(1).Then()=(1) =(1) ==.
(33)0
W∈W
0
,so0 V∈
−1
({0W})⊆
−1
(W
0
). Hence,
−1
(W
0
)is nonempty.
Also,xy∈
−1
(W
0
)=⇒(x)(y)∈W
0
=⇒(x)+(y)∈W
0
=⇒(x+y)∈W
0
=⇒x+y∈
−1
(W
0
)
Finally,x∈
−1
(W
0
)=⇒(x)∈W
0
=⇒(x)∈W
0
(for any∈R)
=⇒(x)∈W
0
=⇒x∈
−1
(W
0
).
Hence
−1
(W
0
)is a subspace by Theorem 4.2.
(34) (a) For
1⊕2:
1⊕2(x+y)= 1(x+y)+ 2(x+y)
=(
1(x)+ 1(y)) + ( 2(x)+ 2(y))
=(
1(x)+ 2(x)) + ( 1(y)+ 2(y))
=
1⊕2(x)+ 1⊕2(y)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 92
(24) 2(v+w)= 1(2(v+w)) = 2 1(v+w)=2( 1(v)+ 1(w)) = 2 1(v)+2 1(w)= 2(v)+ 2(w);
2(v)= 1(2v)=2 1(v)=(2 1(v)) = 2(v).
(25)([−324]) = [12−1114];
⎛
⎝
⎡
⎣
⎤
⎦
⎞
⎠=
⎡
⎣
−230
1−2−1
013
⎤
⎦
⎡
⎣
⎤
⎦=
⎡
⎣
−2+3
−2−
+3
⎤
⎦
(26)(i)=
7
5
i−
11
5
j;(j)=−
2
5
i−
4
5
j
(27)(x−y)=(x+(−y)) =(x)+((−1)y)=(x)+(−1)(y)=(x)−(y).
(28) Follow the hint in the text. Letting==0proves property (1) of a linear transformation. Letting
=0proves property (2).
(29) Suppose part (3) of Theorem 5.1 is true for some.Weproveittruefor+1.
(
1v1+···+ v++1v+1)= ( 1v1+···+ v)+( +1v+1)
(by property (1) of a linear transformation)
=(
1(v1)+···+ (v)) +( +1v+1)
(by the inductive hypothesis)
=
1(v1)+···+ (v)+ +1(v+1)
(by property (2) of a linear transformation),
and we are done.
(30) (a)
1v1+···+ v=0V=⇒( 1v1+···+ v)=(0 V)
=⇒
1(v1)+···+ (v)=0 W=⇒ 1=2=···= =0.
(b) Consider the zero linear transformation.
(31)(
2◦1)(v)= 2(1(v)) = 2(1(v)) = 2(1(v)) =( 2◦1)(v).
(32) Let=(1).Then()=(1) =(1) ==.
(33)0
W∈W
0
,so0 V∈
−1
({0W})⊆
−1
(W
0
). Hence,
−1
(W
0
)is nonempty.
Also,xy∈
−1
(W
0
)=⇒(x)(y)∈W
0
=⇒(x)+(y)∈W
0
=⇒(x+y)∈W
0
=⇒x+y∈
−1
(W
0
)
Finally,x∈
−1
(W
0
)=⇒(x)∈W
0
=⇒(x)∈W
0
(for any∈R)
=⇒(x)∈W
0
=⇒x∈
−1
(W
0
).
Hence
−1
(W
0
)is a subspace by Theorem 4.2.
(34) (a) For
1⊕2:
1⊕2(x+y)= 1(x+y)+ 2(x+y)
=(
1(x)+ 1(y)) + ( 2(x)+ 2(y))
=(
1(x)+ 2(x)) + ( 1(y)+ 2(y))
=
1⊕2(x)+ 1⊕2(y)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 92
Answers to Exercises Section 5.2
Also,
1⊕2(x)= 1(x)+ 2(x)
=
1(x)+ 2(x)
=(
1(x)+ 2(x))
=
1⊕2(x)
For¯
1:
¯
1(x+y)= ( 1(x+y))
=(
1(x)+ 1(y))
=
1(x)+ 2(y)
=¯
1(x)+¯ 1()
Also,
¯
1(x)= 1(x)
=(
1(x))
=(
1(x))
=(¯
1(x))
(b) Use Theorem 4.2: the existence of the zero linear transformation shows that the set is nonempty,
while part (a) proves closure under addition and scalar multiplication.
(35) Define a lineparametrically by{x+y|∈R},forsomefixedxy∈R
2
.Since(x+y)=
(x)+(y),mapsto the set{(x)+(y)|∈R}.If(x)6 =0this set represents a line;
otherwise, it represents a point.
(36) (a) F (b) T (c) F (d) F (e) T (f) F (g) T (h) T
Section 5.2
(1) For each operator, check that theth column of the given matrix (for1≤≤3) is the image ofe .
This is easily done by inspection.
(2) (a)
⎡
⎣
−64 −1
−23 −5
3−17
⎤
⎦
(b)
∙
3−51 −2
51 −28
¸
(c)
⎡
⎣
4−133
13 −15
−2−75 −1
⎤
⎦
(d)
⎡
⎢
⎢
⎣
−30 −20
0−104
04 −13
−6−102
⎤
⎥
⎥
⎦
(3) (a)
∙
−47 128−288
−18 51−104
¸
(b)
⎡
⎣
−35
2−1
4−2
⎤
⎦
(c)
⎡
⎣
22 14
62 39
68 43
⎤
⎦
(d)
⎡
⎣
−32 16 24 15
−10 7 12 6
−121 63 96 58
⎤
⎦
(e)
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
560
−11−26−6
−14−19−1
63 −2
−111
11 13 0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 93
Also,
1⊕2(x)= 1(x)+ 2(x)
=
1(x)+ 2(x)
=(
1(x)+ 2(x))
=
1⊕2(x)
For¯
1:
¯
1(x+y)= ( 1(x+y))
=(
1(x)+ 1(y))
=
1(x)+ 2(y)
=¯
1(x)+¯ 1()
Also,
¯
1(x)= 1(x)
=(
1(x))
=(
1(x))
=(¯
1(x))
(b) Use Theorem 4.2: the existence of the zero linear transformation shows that the set is nonempty,
while part (a) proves closure under addition and scalar multiplication.
(35) Define a lineparametrically by{x+y|∈R},forsomefixedxy∈R
2
.Since(x+y)=
(x)+(y),mapsto the set{(x)+(y)|∈R}.If(x)6 =0this set represents a line;
otherwise, it represents a point.
(36) (a) F (b) T (c) F (d) F (e) T (f) F (g) T (h) T
Section 5.2
(1) For each operator, check that theth column of the given matrix (for1≤≤3) is the image ofe .
This is easily done by inspection.
(2) (a)
⎡
⎣
−64 −1
−23 −5
3−17
⎤
⎦
(b)
∙
3−51 −2
51 −28
¸
(c)
⎡
⎣
4−133
13 −15
−2−75 −1
⎤
⎦
(d)
⎡
⎢
⎢
⎣
−30 −20
0−104
04 −13
−6−102
⎤
⎥
⎥
⎦
(3) (a)
∙
−47 128−288
−18 51−104
¸
(b)
⎡
⎣
−35
2−1
4−2
⎤
⎦
(c)
⎡
⎣
22 14
62 39
68 43
⎤
⎦
(d)
⎡
⎣
−32 16 24 15
−10 7 12 6
−121 63 96 58
⎤
⎦
(e)
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
560
−11−26−6
−14−19−1
63 −2
−111
11 13 0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 93
Answers to Exercises Section 5.2
(4) (a)
⎡
⎣
−202−32−43
−146−23−31
83 14 18
⎤
⎦
(b)
∙
21 7 21 16
−51−13−51−38
¸
(c)
⎡
⎣
98−91 60−76
28−25 17−21
135−125 82−104
⎤
⎦
(5) See the answer for Exercise 3(d).
(6) (a)
∙
67−123
37−68
¸
(b)
⎡
⎣
−7210
5−2−9
−618
⎤
⎦
(c)
⎡
⎢
⎢
⎣
8−99 −14
8−85 −7
6−6−16
1−1−25
⎤
⎥
⎥
⎦
(7) (a)
⎡
⎣
3000
0200
0010
⎤
⎦;12
2
−10+6
(b)
⎡
⎢
⎢
⎢
⎢
⎣
1
3
00
0
1
2
0
001
000
⎤
⎥
⎥
⎥
⎥
⎦
;
2
3
3
−
1
2
2
+5
(8) (a)
"
√
3
2
−
1
2
1
2
√
3
2
#
(b)
"
1
2
√
3−9−
13
2
25
2
1
2
√
3+9
#
(9) (a)
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
100000
000100
010000
000010
001000
000001
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(b)
1
2
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
100 −100
100100
0110 −10
011010
0−10 0 −11
0−1001 −1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(10)
⎡
⎣
−12 12−2
−46 −2
−10−37
⎤
⎦
(11) (a) Matrix for
1:
⎡
⎢
⎢
⎣
1−1−1
023
130
−201
⎤
⎥
⎥
⎦
; matrix for 2:
∙
02−23
10−11
¸
(b) Matrix for
2◦1:
∙
−8−29
−2−40
¸
(c) Easily verified with straightforward computations.
(12) (a)A
2
represents a rotation through the angletwice, which corresponds to a rotation through the
angle2.
(b) Generalize the argument in part (a).
(13) (a)I
(b)O (c)I
(d) The×matrix whose columns aree 2e3e e1, respectively
(e) The×matrix whose columns aree
,e1,e2,,e −1, respectively
Copyrightc°2016 Elsevier Ltd. All rights reserved. 94
(4) (a)
⎡
⎣
−202−32−43
−146−23−31
83 14 18
⎤
⎦
(b)
∙
21 7 21 16
−51−13−51−38
¸
(c)
⎡
⎣
98−91 60−76
28−25 17−21
135−125 82−104
⎤
⎦
(5) See the answer for Exercise 3(d).
(6) (a)
∙
67−123
37−68
¸
(b)
⎡
⎣
−7210
5−2−9
−618
⎤
⎦
(c)
⎡
⎢
⎢
⎣
8−99 −14
8−85 −7
6−6−16
1−1−25
⎤
⎥
⎥
⎦
(7) (a)
⎡
⎣
3000
0200
0010
⎤
⎦;12
2
−10+6
(b)
⎡
⎢
⎢
⎢
⎢
⎣
1
3
00
0
1
2
0
001
000
⎤
⎥
⎥
⎥
⎥
⎦
;
2
3
3
−
1
2
2
+5
(8) (a)
"
√
3
2
−
1
2
1
2
√
3
2
#
(b)
"
1
2
√
3−9−
13
2
25
2
1
2
√
3+9
#
(9) (a)
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
100000
000100
010000
000010
001000
000001
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(b)
1
2
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
100 −100
100100
0110 −10
011010
0−10 0 −11
0−1001 −1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(10)
⎡
⎣
−12 12−2
−46 −2
−10−37
⎤
⎦
(11) (a) Matrix for
1:
⎡
⎢
⎢
⎣
1−1−1
023
130
−201
⎤
⎥
⎥
⎦
; matrix for 2:
∙
02−23
10−11
¸
(b) Matrix for
2◦1:
∙
−8−29
−2−40
¸
(c) Easily verified with straightforward computations.
(12) (a)A
2
represents a rotation through the angletwice, which corresponds to a rotation through the
angle2.
(b) Generalize the argument in part (a).
(13) (a)I
(b)O (c)I
(d) The×matrix whose columns aree 2e3e e1, respectively
(e) The×matrix whose columns aree
,e1,e2,,e −1, respectively
Copyrightc°2016 Elsevier Ltd. All rights reserved. 94
Answers to Exercises Section 5.2
(14) LetAbe the matrix for(with respect to the standard bases). ThenAis a1×matrix (an-vector).
If we letx=A,thenAy=x·yfor ally∈R
(15)[(
2◦1)(v)]=[ 2(1(v))]=A [1(v)]=A (A[v])=(A A)[v]. Now apply
the uniqueness condition in Theorem 5.5.
(16) (a)
⎡
⎣
0−4−13
−65 6
2−2−3
⎤
⎦ (b)
⎡
⎣
200
0−10
001
⎤
⎦
(c) The vectors inare eigenvectors for the matrix from part (a) corresponding to the eigenvalues,
2,−1,and1, respectively.
(17) Easily verified with straightforward computations.
(18) (a)
A
()=(−1)
2
(b) 1={[210] +[201]};basisfor 1={[210][201]};
0={[−122]};basisfor 0={[−122]};
=([210][201][−122])
(Note: The remaining answers will vary if the vectors inare ordered differently.)
(c) Using the basisordered as([210][201][−122]),theanswerisP=
⎡
⎣
22−1
10 2
01 2
⎤
⎦,with
P
−1
=
1
9
⎡
⎣
25 −4
2−45
−122
⎤
⎦. Another possible answer, using=([−122][210][201]),is
P=
⎡
⎣
−122
210
201
⎤
⎦,withP
−1
=
1
9
⎡
⎣
−122
254
2−45
⎤
⎦
(d) Using=([210][201][−122])producesA
=
⎡
⎣
100
010
000
⎤
⎦.
Using=([−122][210][201])yieldsA
=P
−1
AP=
⎡
⎣
000
010
001
⎤
⎦instead.
(e)is a projection onto the plane formed by[210]and[201]while[−122]is orthogonal to
this plane.
(19)[(v)]
=
1
[(v)] =
1
A[v]=
1
A([v] )=A [v]. By Theorem 5.5,A =A .
(20) Find the matrix forwith respect to, and then apply Theorem 5.6.
(21) LetA=
∙
10
0−1
¸
andP=
1
√
1+
2
∙
1−
1
¸
.ThenArepresents a reflection through the-axis,
andPrepresents a counterclockwise rotation putting the-axis on the line=(using Example 9,
Section 5.1). Thus, the desired matrix isPAP
−1
.
An alternate solution is obtained byfinding the images of the standard basis vectors using Exercise
21 in Section 1.2. A vector in the direction of the line=is[1]. Then for any vectorx,
p=proj
[1]x=
[1]·x
1+
2
Copyrightc°2016 Elsevier Ltd. All rights reserved. 95
(14) LetAbe the matrix for(with respect to the standard bases). ThenAis a1×matrix (an-vector).
If we letx=A,thenAy=x·yfor ally∈R
(15)[(
2◦1)(v)]=[ 2(1(v))]=A [1(v)]=A (A[v])=(A A)[v]. Now apply
the uniqueness condition in Theorem 5.5.
(16) (a)
⎡
⎣
0−4−13
−65 6
2−2−3
⎤
⎦ (b)
⎡
⎣
200
0−10
001
⎤
⎦
(c) The vectors inare eigenvectors for the matrix from part (a) corresponding to the eigenvalues,
2,−1,and1, respectively.
(17) Easily verified with straightforward computations.
(18) (a)
A
()=(−1)
2
(b) 1={[210] +[201]};basisfor 1={[210][201]};
0={[−122]};basisfor 0={[−122]};
=([210][201][−122])
(Note: The remaining answers will vary if the vectors inare ordered differently.)
(c) Using the basisordered as([210][201][−122]),theanswerisP=
⎡
⎣
22−1
10 2
01 2
⎤
⎦,with
P
−1
=
1
9
⎡
⎣
25 −4
2−45
−122
⎤
⎦. Another possible answer, using=([−122][210][201]),is
P=
⎡
⎣
−122
210
201
⎤
⎦,withP
−1
=
1
9
⎡
⎣
−122
254
2−45
⎤
⎦
(d) Using=([210][201][−122])producesA
=
⎡
⎣
100
010
000
⎤
⎦.
Using=([−122][210][201])yieldsA
=P
−1
AP=
⎡
⎣
000
010
001
⎤
⎦instead.
(e)is a projection onto the plane formed by[210]and[201]while[−122]is orthogonal to
this plane.
(19)[(v)]
=
1
[(v)] =
1
A[v]=
1
A([v] )=A [v]. By Theorem 5.5,A =A .
(20) Find the matrix forwith respect to, and then apply Theorem 5.6.
(21) LetA=
∙
10
0−1
¸
andP=
1
√
1+
2
∙
1−
1
¸
.ThenArepresents a reflection through the-axis,
andPrepresents a counterclockwise rotation putting the-axis on the line=(using Example 9,
Section 5.1). Thus, the desired matrix isPAP
−1
.
An alternate solution is obtained byfinding the images of the standard basis vectors using Exercise
21 in Section 1.2. A vector in the direction of the line=is[1]. Then for any vectorx,
p=proj
[1]x=
[1]·x
1+
2
Copyrightc°2016 Elsevier Ltd. All rights reserved. 95
Answers to Exercises Section 5.3
and so by Exercise 21 in Section 1.2, the reflection ofxthrough=is
2p−x=2
µ
[1]·x
1+
2
¶
[1]−x
Therefore,
([10]) = 2
µ
[1]·[10]
1+
2
¶
[1]−[10]which simplifies to
∙
1−
2
1+
2
2
1+
2
¸
thefirst column of the desired matrixSimilarly,
([01]) = 2
µ
[1]·[01]
1+
2
¶
[1]−[01]which simplifies to
∙
2
1+
2
2
−1
1+
2
¸
the second column of the desired matrix
(22) Let{v
1v }be a basis forY. Extendthistoabasis={v 1v }forV. By Theorem 5.4,
there is a unique linear transformation
0
:V−→Wsuch that
0
(v)=
½
(v
) if≤
0 if
Clearly,
0
agrees withonY.
(23) Let:V−→Wbe a linear transformation such that(v
1)=w 1(v 2)=w 2,(v )=w with
{v
1v2v }abasisforV.Ifv∈V,thenv= 1v1+2v2+···+ vfor unique 12∈R
(by Theorem 4.9). But then
(v)= (
1v1+2v2+···+ v)
=
1(v1)+ 2(v2)+···+ (v)
=
1w1+2w2+···+ w
Hence(v)is determined uniquely, for eachv∈V,andsois uniquely determined.
(24) (a)T (b)T (c)F (d)F (e)T (f)F (g)T (h)T (i)T (j)F
Section 5.3
(1) (a) Yes, because([1−23]) = [000]
(b) No, because([2−14]) = [5−2−1]
(c) No; the system
⎧
⎨
⎩
5
1+ 2− 3=2
−3
1 + 3=−1
1− 2− 3=4
has no solutions.
Note:
⎡
⎣
51 −1
−301
1−1−1
¯
¯
¯
¯
¯
¯
2
−1
4
⎤
⎦row reduces to
⎡
⎢
⎣
10−
1
3
01
2
3
00 0
¯
¯
¯
¯
¯
¯
¯
1
3
1
3
4
⎤
⎥
⎦,
wherewehavenotrow reduced beyond the augmentation bar.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 96
and so by Exercise 21 in Section 1.2, the reflection ofxthrough=is
2p−x=2
µ
[1]·x
1+
2
¶
[1]−x
Therefore,
([10]) = 2
µ
[1]·[10]
1+
2
¶
[1]−[10]which simplifies to
∙
1−
2
1+
2
2
1+
2
¸
thefirst column of the desired matrixSimilarly,
([01]) = 2
µ
[1]·[01]
1+
2
¶
[1]−[01]which simplifies to
∙
2
1+
2
2
−1
1+
2
¸
the second column of the desired matrix
(22) Let{v
1v }be a basis forY. Extendthistoabasis={v 1v }forV. By Theorem 5.4,
there is a unique linear transformation
0
:V−→Wsuch that
0
(v)=
½
(v
) if≤
0 if
Clearly,
0
agrees withonY.
(23) Let:V−→Wbe a linear transformation such that(v
1)=w 1(v 2)=w 2,(v )=w with
{v
1v2v }abasisforV.Ifv∈V,thenv= 1v1+2v2+···+ vfor unique 12∈R
(by Theorem 4.9). But then
(v)= (
1v1+2v2+···+ v)
=
1(v1)+ 2(v2)+···+ (v)
=
1w1+2w2+···+ w
Hence(v)is determined uniquely, for eachv∈V,andsois uniquely determined.
(24) (a)T (b)T (c)F (d)F (e)T (f)F (g)T (h)T (i)T (j)F
Section 5.3
(1) (a) Yes, because([1−23]) = [000]
(b) No, because([2−14]) = [5−2−1]
(c) No; the system
⎧
⎨
⎩
5
1+ 2− 3=2
−3
1 + 3=−1
1− 2− 3=4
has no solutions.
Note:
⎡
⎣
51 −1
−301
1−1−1
¯
¯
¯
¯
¯
¯
2
−1
4
⎤
⎦row reduces to
⎡
⎢
⎣
10−
1
3
01
2
3
00 0
¯
¯
¯
¯
¯
¯
¯
1
3
1
3
4
⎤
⎥
⎦,
wherewehavenotrow reduced beyond the augmentation bar.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 96
Answers to Exercises Section 5.3
(d) Yes, because([−440]) = [−1612−8]
(2) (a) No, since(
3
−5
2
+3−6) = 6
3
+4−96 =0
(b) Yes, because(4
3
−4
2
)=0
(c) Yes, because, for example,(
3
+4+3)=8
3
−−1
(d) No, since every element ofrange()hasazerocoefficient for its
2
term.
(3) In each part, we give the reduced row echelon form for the matrix for the linear transformation as well.
(a)
⎡
⎣
10 2
01−3
00 0
⎤
⎦;dim(ker())=1,basisforker()={[−231]},
dim(range())=2,basisforrange()={[1−23][−13−3]}
(b)
⎡
⎢
⎢
⎣
10 1
01−2
00 0
00 0
⎤
⎥
⎥
⎦
;dim(ker())=1,basisforker()={[−12
1]},
dim(range())=2,basisforrange()={[47−23][−21−1−2]}
(c)
∙
10 5
01−2
¸
;dim(ker())=1,basisforker()={[−521]},
dim(range())=2,basisforrange()={[32][21]}
(d)
⎡
⎢
⎢
⎢
⎢
⎣
10−11
01 3
−2
00 0 0
00 0 0
00 0 0
⎤
⎥
⎥
⎥
⎥
⎦
;dim(ker())=2,basisforker()={[1−310][−1201]},
dim(range())=2,basisforrange()={[−14−4−634][−8−12−72]}
(4) Students can generally solve these problems by inspection.
(a)dim(ker())=2,basisforker()={
[100],[001]},
dim(range())=1,basisforrange()={[01]}
(b)dim(ker())=0,basisforker()={},
dim(range())=2,basisforrange()={[110][011]}
(c)dim(ker())=2,basisforker()=
½∙
10
00
¸
∙
00
01
¸¾
,
dim(range
())=2,basisforrange()=
⎧
⎨
⎩
⎡
⎣
01
00
00
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎫
⎬
⎭
(d)dim(ker())=2,basisforker()={
4
3
},
dim(range())=3,basisforrange()={
2
1}
(e)dim(ker())=0,basisforker()={},
dim(range())=3,basisforrange()={
3
2
}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 97
(d) Yes, because([−440]) = [−1612−8]
(2) (a) No, since(
3
−5
2
+3−6) = 6
3
+4−96 =0
(b) Yes, because(4
3
−4
2
)=0
(c) Yes, because, for example,(
3
+4+3)=8
3
−−1
(d) No, since every element ofrange()hasazerocoefficient for its
2
term.
(3) In each part, we give the reduced row echelon form for the matrix for the linear transformation as well.
(a)
⎡
⎣
10 2
01−3
00 0
⎤
⎦;dim(ker())=1,basisforker()={[−231]},
dim(range())=2,basisforrange()={[1−23][−13−3]}
(b)
⎡
⎢
⎢
⎣
10 1
01−2
00 0
00 0
⎤
⎥
⎥
⎦
;dim(ker())=1,basisforker()={[−12
1]},
dim(range())=2,basisforrange()={[47−23][−21−1−2]}
(c)
∙
10 5
01−2
¸
;dim(ker())=1,basisforker()={[−521]},
dim(range())=2,basisforrange()={[32][21]}
(d)
⎡
⎢
⎢
⎢
⎢
⎣
10−11
01 3
−2
00 0 0
00 0 0
00 0 0
⎤
⎥
⎥
⎥
⎥
⎦
;dim(ker())=2,basisforker()={[1−310][−1201]},
dim(range())=2,basisforrange()={[−14−4−634][−8−12−72]}
(4) Students can generally solve these problems by inspection.
(a)dim(ker())=2,basisforker()={
[100],[001]},
dim(range())=1,basisforrange()={[01]}
(b)dim(ker())=0,basisforker()={},
dim(range())=2,basisforrange()={[110][011]}
(c)dim(ker())=2,basisforker()=
½∙
10
00
¸
∙
00
01
¸¾
,
dim(range
())=2,basisforrange()=
⎧
⎨
⎩
⎡
⎣
01
00
00
⎤
⎦
⎡
⎣
00
10
00
⎤
⎦
⎫
⎬
⎭
(d)dim(ker())=2,basisforker()={
4
3
},
dim(range())=3,basisforrange()={
2
1}
(e)dim(ker())=0,basisforker()={},
dim(range())=3,basisforrange()={
3
2
}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 97
Answers to Exercises Section 5.3
(f)dim(ker())=1,basisforker()={[011]},
dim(range())=2,basisforrange()={[101][00−1]}
(A simpler basis forrange()={[100][001]}.)
(g)dim(ker())=0,basisforker()={}(empty set),
dim(range())=4,basisforrange()=standard basis forM
22
(h)dim(ker())=6,basisforker()
=
⎧
⎨
⎩
⎡
⎣
100
000
000
⎤
⎦,
⎡
⎣
010
100
000
⎤
⎦,
⎡
⎣
001
000
100
⎤
⎦,
⎡
⎣
000
010
000
⎤
⎦,
⎡
⎣
000
001
010
⎤
⎦,
⎡
⎣
000
000
001
⎤
⎦
⎫
⎬
⎭
,
dim(range())=3,basisforrange()=
⎧
⎨
⎩
⎡
⎣
010
−100
000
⎤
⎦,
⎡
⎣
001
000
−100
⎤
⎦
,
⎡
⎣
000
001
0−10
⎤
⎦
⎫
⎬
⎭
(i)dim(ker())=1,basisforker()={
2
−2+1},
dim(range())=2,basisforrange()={[12][11]}
(A simpler basis forrange()=standard basis forR
2
.)
(j)dim(ker())=2,basisforker()={(+1)(−1)
2
(+1)(−1)},
dim(range())=3,basisforrange()={e
1e2e3}
(5) (a)ker()=V,range()={0
W} (b)ker()={0 V},range()=V
(6)(A+B)=trace(A+B)=
P
3
=1
(+)=
P
3
=1
+
P
3
=1
= trace(A)+trace(B)=(A)+(B);
(A) = trace(A)=
P
3
=1
=
P
3
=1
=(trace(A)) =(A);
ker()=
⎧
⎨
⎩
⎡
⎣
−−
⎤
⎦
¯
¯
¯
¯
¯
¯
∈R
⎫
⎬
⎭
,
dim(ker())=8,range()=R,dim(range())=1
(7)range()=V,ker()={0
V}
(8)ker()={0},range()={
4
+
3
+
2
},dim(ker())=0,dim(range())=3
(9)ker()={+| ∈R},range()=P
2,dim(ker())=2,dim(range())=3
(10) When≤,ker()=all polynomials of degree less than,dim(ker())=,range()=P
−,and
dim(range())=−+1.When,ker()=P
,dim(ker())=+1,range()={0},and
dim(range())=0.
(11) Sincerange()=R,dim(range())=1. By the Dimension Theorem,dim(ker())=.Sincethe
given set is clearly a linearly independent subset ofker() containingdistinct elements, it must be
abasisforker() by part (2) of Theorem 4.12.
(12)ker()={[000]},range()=R
(Note: Every vectorXis in therangesince(A
−1
X)=A(A
−1
X)=X.)
(13)ker()={0
V}iffdim(ker())=0iffdim(range())=dim(V)−0=dim(V)iffrange()=V.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 98
(f)dim(ker())=1,basisforker()={[011]},
dim(range())=2,basisforrange()={[101][00−1]}
(A simpler basis forrange()={[100][001]}.)
(g)dim(ker())=0,basisforker()={}(empty set),
dim(range())=4,basisforrange()=standard basis forM
22
(h)dim(ker())=6,basisforker()
=
⎧
⎨
⎩
⎡
⎣
100
000
000
⎤
⎦,
⎡
⎣
010
100
000
⎤
⎦,
⎡
⎣
001
000
100
⎤
⎦,
⎡
⎣
000
010
000
⎤
⎦,
⎡
⎣
000
001
010
⎤
⎦,
⎡
⎣
000
000
001
⎤
⎦
⎫
⎬
⎭
,
dim(range())=3,basisforrange()=
⎧
⎨
⎩
⎡
⎣
010
−100
000
⎤
⎦,
⎡
⎣
001
000
−100
⎤
⎦
,
⎡
⎣
000
001
0−10
⎤
⎦
⎫
⎬
⎭
(i)dim(ker())=1,basisforker()={
2
−2+1},
dim(range())=2,basisforrange()={[12][11]}
(A simpler basis forrange()=standard basis forR
2
.)
(j)dim(ker())=2,basisforker()={(+1)(−1)
2
(+1)(−1)},
dim(range())=3,basisforrange()={e
1e2e3}
(5) (a)ker()=V,range()={0
W} (b)ker()={0 V},range()=V
(6)(A+B)=trace(A+B)=
P
3
=1
(+)=
P
3
=1
+
P
3
=1
= trace(A)+trace(B)=(A)+(B);
(A) = trace(A)=
P
3
=1
=
P
3
=1
=(trace(A)) =(A);
ker()=
⎧
⎨
⎩
⎡
⎣
−−
⎤
⎦
¯
¯
¯
¯
¯
¯
∈R
⎫
⎬
⎭
,
dim(ker())=8,range()=R,dim(range())=1
(7)range()=V,ker()={0
V}
(8)ker()={0},range()={
4
+
3
+
2
},dim(ker())=0,dim(range())=3
(9)ker()={+| ∈R},range()=P
2,dim(ker())=2,dim(range())=3
(10) When≤,ker()=all polynomials of degree less than,dim(ker())=,range()=P
−,and
dim(range())=−+1.When,ker()=P
,dim(ker())=+1,range()={0},and
dim(range())=0.
(11) Sincerange()=R,dim(range())=1. By the Dimension Theorem,dim(ker())=.Sincethe
given set is clearly a linearly independent subset ofker() containingdistinct elements, it must be
abasisforker() by part (2) of Theorem 4.12.
(12)ker()={[000]},range()=R
(Note: Every vectorXis in therangesince(A
−1
X)=A(A
−1
X)=X.)
(13)ker()={0
V}iffdim(ker())=0iffdim(range())=dim(V)−0=dim(V)iffrange()=V.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 98
Answers to Exercises Section 5.3
(14) Forker():0 V∈ker(), soker()6 ={}.Ifv 1v2∈ker(), then
(v
1+v2)=(v 1)+(v 2)=0 V+0V=0V
and sov
1+v2∈ker(). Similarly,(v 1)=0 V,andsov 1∈ker().
Forrange():0
W∈range()since0 W=(0 V).Ifw 1w2∈range(), then there existv 1v2∈V
such that(v
1)=w 1and(v 2)=w 2. Hence,
(v
1+v2)=(v 1)+(v 2)=w 1+w2
Thusw
1+w2∈range(). Similarly,(v 1)=(v 1)=w 1,sow 1∈range().
(15) (a) Ifv∈ker(
1), then( 2◦1)(v)= 2(1(v)) = 2(0W)=0 X.
(b) Ifx∈range(
2◦1), then there is avsuch that( 2◦1)(v)=x. Hence 2(1(v)) =x,sox
∈range(
2).
(c)dim(range(
2◦1))=dim(V)−dim(ker( 2◦1))≤dim(V)−dim(ker( 1)) (by part (a))=
dim(range(
1))
(16) Consider
µ∙
¸¶
=
∙
1−1
1−1
¸∙
¸
. Then,ker()= range()={[ ]|∈R}.
(17) (a) LetPbe the transition matrix from the standard basis to,andletQbe the transition matrix
from the standard basis to. Note that bothPandQare nonsingular by Theorem 4.19. Then,
by Theorem 5.6,B=QAP
−1
. Hence,rank(B)=rank(QAP
−1
)=rank(AP
−1
)(by part (a) of
Review Exercise 16 in Chapter 2)=rank(A)(by part (b) of Review Exercise 16 in Chapter 2).
(b) The comments prior to Theorem 5.9 in the textshow that parts (a) and (b) of the theorem are
true ifAis the matrix forwith respect to the standard bases. However, part (a) of this exercise
shows that ifBis the matrix forwith respect to any bases,rank(B)=rank(A). Hence,rank(B)
can be substituted for any occurrence ofrank(A)in the statement of Theorem 5.9. Thus, parts
(a) and (b) of Theorem 5.9 hold when the matrix forwith respect to any bases is used. Part
(c) of Theorem 5.9 follows immediately from parts (a) and (b).
(18) (a) By Theorem 4.15, there are vectorsq
1q inVsuch that={k 1k q1q }is a
basis forV. Hence,dim(V)=+.
(b) Ifv∈Vthenv=
1k1+···+ k+1q1+···+ qfor some 11∈RHence,
(v)= (
1k1+···+ k+1q1+···+ q)
=
1(k1)+···+ (k)+ 1(q1)+···+ (q)
=
10+···+ 0+ 1(q1)+···+ (q)
=
1(q1)+···+ (q)
(c) By part (b), any element ofrange()is a linear combination of(q
1)(q ),sorange()is
spanned by{(q
1)(q )}. Hence, by part (1) of Theorem 4.12,range()isfinite dimensional
anddim(range())≤.
(d) Suppose
1(q1)+···+ (q)=0 W.Then( 1q1+···+ q)=0 W,
and so
1q1+···+ q∈ker().
(e) Every element ofker()can be expressed as a linear combination of the basis vectorsk
1k
forker().
Copyrightc°2016 Elsevier Ltd. All rights reserved. 99
(14) Forker():0 V∈ker(), soker()6 ={}.Ifv 1v2∈ker(), then
(v
1+v2)=(v 1)+(v 2)=0 V+0V=0V
and sov
1+v2∈ker(). Similarly,(v 1)=0 V,andsov 1∈ker().
Forrange():0
W∈range()since0 W=(0 V).Ifw 1w2∈range(), then there existv 1v2∈V
such that(v
1)=w 1and(v 2)=w 2. Hence,
(v
1+v2)=(v 1)+(v 2)=w 1+w2
Thusw
1+w2∈range(). Similarly,(v 1)=(v 1)=w 1,sow 1∈range().
(15) (a) Ifv∈ker(
1), then( 2◦1)(v)= 2(1(v)) = 2(0W)=0 X.
(b) Ifx∈range(
2◦1), then there is avsuch that( 2◦1)(v)=x. Hence 2(1(v)) =x,sox
∈range(
2).
(c)dim(range(
2◦1))=dim(V)−dim(ker( 2◦1))≤dim(V)−dim(ker( 1)) (by part (a))=
dim(range(
1))
(16) Consider
µ∙
¸¶
=
∙
1−1
1−1
¸∙
¸
. Then,ker()= range()={[ ]|∈R}.
(17) (a) LetPbe the transition matrix from the standard basis to,andletQbe the transition matrix
from the standard basis to. Note that bothPandQare nonsingular by Theorem 4.19. Then,
by Theorem 5.6,B=QAP
−1
. Hence,rank(B)=rank(QAP
−1
)=rank(AP
−1
)(by part (a) of
Review Exercise 16 in Chapter 2)=rank(A)(by part (b) of Review Exercise 16 in Chapter 2).
(b) The comments prior to Theorem 5.9 in the textshow that parts (a) and (b) of the theorem are
true ifAis the matrix forwith respect to the standard bases. However, part (a) of this exercise
shows that ifBis the matrix forwith respect to any bases,rank(B)=rank(A). Hence,rank(B)
can be substituted for any occurrence ofrank(A)in the statement of Theorem 5.9. Thus, parts
(a) and (b) of Theorem 5.9 hold when the matrix forwith respect to any bases is used. Part
(c) of Theorem 5.9 follows immediately from parts (a) and (b).
(18) (a) By Theorem 4.15, there are vectorsq
1q inVsuch that={k 1k q1q }is a
basis forV. Hence,dim(V)=+.
(b) Ifv∈Vthenv=
1k1+···+ k+1q1+···+ qfor some 11∈RHence,
(v)= (
1k1+···+ k+1q1+···+ q)
=
1(k1)+···+ (k)+ 1(q1)+···+ (q)
=
10+···+ 0+ 1(q1)+···+ (q)
=
1(q1)+···+ (q)
(c) By part (b), any element ofrange()is a linear combination of(q
1)(q ),sorange()is
spanned by{(q
1)(q )}. Hence, by part (1) of Theorem 4.12,range()isfinite dimensional
anddim(range())≤.
(d) Suppose
1(q1)+···+ (q)=0 W.Then( 1q1+···+ q)=0 W,
and so
1q1+···+ q∈ker().
(e) Every element ofker()can be expressed as a linear combination of the basis vectorsk
1k
forker().
Copyrightc°2016 Elsevier Ltd. All rights reserved. 99
Answers to Exercises Section 5.4
(f) The fact that 1q1+···+ q=1k1+···+ kimplies that
1k1+···+ k−1q1−···− q=0W
But since={k
1k q1q }is linearly independent,
1=···= =1=···= =0
by the definition of linear independence.
(g) In part (d) we assumed that
1(q1)+···+ (q)=0 W. This led to the conclusion in part (f)
that
1=···= =0. Hence{(q 1)(q )}is linearly independent by definition.
(h) Part (c) proved that{(q
1)(q )}spansrange()and part (g) shows that{(q 1)(q )}
is linearly independent. Hence,{(q
1)(q )}is a basis forrange().
(i) We assumed thatdim(ker()) =. Part (h) shows thatdim(range()) =.
Therefore,dim(ker()) + dim(range()) =+=dim(V)(by part (a)).
(19) From Theorem 4.13, we havedim(ker())≤dim(V). The Dimension Theorem states thatdim(range())
isfinite anddim(ker()) + dim(range()) = dim(V).Sincedim(ker())is nonnegative, it follows that
dim(range())≤dim(V).
(20) (a) F (b) F (c) T (d) F (e) T (f) F (g) F (h) F
Section 5.4
(1) (a) Not one-to-one, because([100]) =([000]) = [0000];
not onto, because[0001]is not inrange()
(b) Not one-to-one, because([1−11]) = [00];onto, because([0]) = [ ]
(c) One-to-one, because([ ]) = [000]implies that[2 ++−]=[000],
which gives===0;
onto, because every vector[ ]can be expressed as[2 ++−],
where
=
2
,=−,and=−
2
+
(d) Not one-to-one, because(1) = 0;onto, because(
3
+
2
+)=
2
++
(e) One-to-one, because(
2
++)=0implies that+=+=+=0,whichgives=
and hence===0;
onto, because every polynomial
2
++can be expressed as(+)
2
+(+)+(+),
where=(−+)2,=(+−)2,and=(−++)2
(f) One-to-one, because
µ∙
¸¶
=
∙
00
00
¸
=⇒
∙
+
−
¸
=
∙
00
00
¸
=⇒==+=−=0 =⇒====0;
onto, because
Ã"
+
2
−
2
#!
=
∙
¸
(g) Not one-to-one, because
µ∙
01 0
10−1
¸¶
=
µ∙
000
000
¸¶
=
∙
00
00
¸
;
onto, because every2×2matrix
∙
¸
can be expressed as
∙
−
2+
¸
,
where=,=−,=2,=,and=0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 100
(f) The fact that 1q1+···+ q=1k1+···+ kimplies that
1k1+···+ k−1q1−···− q=0W
But since={k
1k q1q }is linearly independent,
1=···= =1=···= =0
by the definition of linear independence.
(g) In part (d) we assumed that
1(q1)+···+ (q)=0 W. This led to the conclusion in part (f)
that
1=···= =0. Hence{(q 1)(q )}is linearly independent by definition.
(h) Part (c) proved that{(q
1)(q )}spansrange()and part (g) shows that{(q 1)(q )}
is linearly independent. Hence,{(q
1)(q )}is a basis forrange().
(i) We assumed thatdim(ker()) =. Part (h) shows thatdim(range()) =.
Therefore,dim(ker()) + dim(range()) =+=dim(V)(by part (a)).
(19) From Theorem 4.13, we havedim(ker())≤dim(V). The Dimension Theorem states thatdim(range())
isfinite anddim(ker()) + dim(range()) = dim(V).Sincedim(ker())is nonnegative, it follows that
dim(range())≤dim(V).
(20) (a) F (b) F (c) T (d) F (e) T (f) F (g) F (h) F
Section 5.4
(1) (a) Not one-to-one, because([100]) =([000]) = [0000];
not onto, because[0001]is not inrange()
(b) Not one-to-one, because([1−11]) = [00];onto, because([0]) = [ ]
(c) One-to-one, because([ ]) = [000]implies that[2 ++−]=[000],
which gives===0;
onto, because every vector[ ]can be expressed as[2 ++−],
where
=
2
,=−,and=−
2
+
(d) Not one-to-one, because(1) = 0;onto, because(
3
+
2
+)=
2
++
(e) One-to-one, because(
2
++)=0implies that+=+=+=0,whichgives=
and hence===0;
onto, because every polynomial
2
++can be expressed as(+)
2
+(+)+(+),
where=(−+)2,=(+−)2,and=(−++)2
(f) One-to-one, because
µ∙
¸¶
=
∙
00
00
¸
=⇒
∙
+
−
¸
=
∙
00
00
¸
=⇒==+=−=0 =⇒====0;
onto, because
Ã"
+
2
−
2
#!
=
∙
¸
(g) Not one-to-one, because
µ∙
01 0
10−1
¸¶
=
µ∙
000
000
¸¶
=
∙
00
00
¸
;
onto, because every2×2matrix
∙
¸
can be expressed as
∙
−
2+
¸
,
where=,=−,=2,=,and=0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 100
Answers to Exercises Section 5.4
(h) One-to-one, because(
2
++)=
∙
00
00
¸
implies that+=−=−3=0,which
gives===0;
not onto, because
∙
01
00
¸
is not inrange()
(2) (a) One-to-one; onto;
the matrix row reduces toI
2,whichmeansthatdim(ker())=0anddim(range())=2.
(b) One-to-one; not onto;
the matrix row reduces to
⎡
⎣
10
01
00
⎤
⎦, which means thatdim(ker())=0anddim(range())=2.
(c) Not one-to-one; not onto;
the matrix row reduces to
⎡
⎢
⎢
⎣
10−
2
5
01−
6
5
00 0
⎤
⎥
⎥
⎦
,whichmeansthatdim(ker())=1
anddim(range())=2.
(d) Not one-to-one; onto;
the matrix row reduces to
⎡
⎣
100 −2
010 3
001 −1
⎤
⎦,whichmeansthatdim(ker())=1
anddim(range())=3.
(3) (a) One-to-one; onto;
the matrix row reduces toI
3which means thatdim(ker())=0anddim(range())=3.
(b) Not one-to-one; not onto;
the matrix row reduces to
⎡
⎢
⎢
⎣
100 −2
010 −2
001 1
000 0
⎤
⎥
⎥
⎦
,whichmeansthatdim(ker())=1
anddim(range())=3.
(c) Not one-to-one; not onto;
the matrix row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎣
10−
10
11
19
11
01
3
11
−
9
11
00 0 0
00 0 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
, which means thatdim(ker())=2
anddim(range())=2.
(4) (a) Let:R
→R
.Thendim(range())=−dim(ker())≤,sois not onto.
(b) Let:R
→R
.Thendim(ker())=−dim(range())≥−0,sois not one-to-one.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 101
(h) One-to-one, because(
2
++)=
∙
00
00
¸
implies that+=−=−3=0,which
gives===0;
not onto, because
∙
01
00
¸
is not inrange()
(2) (a) One-to-one; onto;
the matrix row reduces toI
2,whichmeansthatdim(ker())=0anddim(range())=2.
(b) One-to-one; not onto;
the matrix row reduces to
⎡
⎣
10
01
00
⎤
⎦, which means thatdim(ker())=0anddim(range())=2.
(c) Not one-to-one; not onto;
the matrix row reduces to
⎡
⎢
⎢
⎣
10−
2
5
01−
6
5
00 0
⎤
⎥
⎥
⎦
,whichmeansthatdim(ker())=1
anddim(range())=2.
(d) Not one-to-one; onto;
the matrix row reduces to
⎡
⎣
100 −2
010 3
001 −1
⎤
⎦,whichmeansthatdim(ker())=1
anddim(range())=3.
(3) (a) One-to-one; onto;
the matrix row reduces toI
3which means thatdim(ker())=0anddim(range())=3.
(b) Not one-to-one; not onto;
the matrix row reduces to
⎡
⎢
⎢
⎣
100 −2
010 −2
001 1
000 0
⎤
⎥
⎥
⎦
,whichmeansthatdim(ker())=1
anddim(range())=3.
(c) Not one-to-one; not onto;
the matrix row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎣
10−
10
11
19
11
01
3
11
−
9
11
00 0 0
00 0 0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
, which means thatdim(ker())=2
anddim(range())=2.
(4) (a) Let:R
→R
.Thendim(range())=−dim(ker())≤,sois not onto.
(b) Let:R
→R
.Thendim(ker())=−dim(range())≥−0,sois not one-to-one.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 101
Answers to Exercises Section 5.4
(5) (a)(I )=AI −IA=A−A=O . HenceI ∈ker(),andsois not one-to-one by part (1)
of Theorem 5.12.
(b) Use part (a) of this exercise, part (2) of Theorem 5.12, and the Dimension Theorem.
(6) If(A)=O
3,thenA
=−A,soAis skew-symmetric. But sinceAis also upper triangular,A=O 3.
Thus,ker()={0}andis one-to-one. By the Dimension Theorem,dim(range())=dim(U
3)=6
(sincedim(ker())=0). Hence,range()6 =M
33,andis not onto.
(7) (a) No, by Corollary 5.13, becausedim(R
6
)=dim(P 5).
(b) No, by Corollary 5.13, becausedim(M
22)=dim(P 3).
(8) (a) Clearly, multiplying a nonzero polynomial byproduces another nonzero polynomial. Hence,
ker()={0
P},andis one-to-one. But, every nonzero element ofrange()has degree at least
1,sincemultiplies by. Hence, the constant polynomial1is not inrange(),andis not onto.
(b) Corollary 5.13 requires that the domain and codomain of the linear transformation befinite
dimensional. However,Pis infinite dimensional.
(9) (a) Consider:P
2→R
3
given by(p)=[p( 1)p( 2)p( 3)]Now,dim(P 2)=dim(R
3
)=3
Hence, by Corollary 5.13, ifis either one-to-one or onto, it has the other property as well.
We will show thatis one-to-one using part (1) of Theorem 5.12. Ifp∈ker(),then(p)=0,
and sop(
1)=p( 2)=p( 3)=0. Hencepis a polynomial of degree≤2touching the-axis at
=
1,= 2,and= 3.Sincethegraphofpmust be either a parabola or a line, it cannot
touch the-axis at three distinct points unless its graph is the line=0.Thatis,p=0inP
2.
Therefore,ker()={0},andis one-to-one.
Now, by Corollary 5.13,is onto. Thus, given any3-vector[ ],thereissomep∈P
2such
thatp(
1)=,p( 2)=,andp( 3)=.
(b) From part (a),is an isomorphism. Hence, any[ ]∈R
3
has a unique pre-image underin
P
2.
(c) Generalize parts (a) and (b) using:P
→R
+1
given by(p)=[p( 1)p( )p( +1)]
Note that any polynomial inker()has+1distinct roots, and so must be trivial. Henceis one-
to-one, and therefore by Corollary 5.13,is onto. Thus, given any(+1)-vector[
1+1],
there is a uniquep∈P
such thatp( 1)= 1p( )= ,andp( +1)= +1.
(10) Supposeis not one-to-one. Thenker() is nontrivial. Letv∈ker()withv6 =0Then={v}is
linearly independent. But()={0
W}, which is linearly dependent, a contradiction.
(11) (a) Supposew∈(span()). Then there is a vectorv∈span()such that(v)=w.Thereare
also vectorsv
1v ∈and scalars 1such that 1v1+···+ v=v. Hence,
w=(v)=(
1v1+···+ v)
=(
1v1)+···+( v)
=
1(v1)+···+ (v)
which shows thatw∈span(()). Hence,(span())⊆span(()).
Next,⊆span()(by Theorem 4.5)
=⇒()⊆(span())
=⇒span(())⊆(span())(by part (3) of Theorem 4.5, since(span())is a subspace
ofWby part (1) of Theorem 5.3). Therefore,(span()) = span(()).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 102
(5) (a)(I )=AI −IA=A−A=O . HenceI ∈ker(),andsois not one-to-one by part (1)
of Theorem 5.12.
(b) Use part (a) of this exercise, part (2) of Theorem 5.12, and the Dimension Theorem.
(6) If(A)=O
3,thenA
=−A,soAis skew-symmetric. But sinceAis also upper triangular,A=O 3.
Thus,ker()={0}andis one-to-one. By the Dimension Theorem,dim(range())=dim(U
3)=6
(sincedim(ker())=0). Hence,range()6 =M
33,andis not onto.
(7) (a) No, by Corollary 5.13, becausedim(R
6
)=dim(P 5).
(b) No, by Corollary 5.13, becausedim(M
22)=dim(P 3).
(8) (a) Clearly, multiplying a nonzero polynomial byproduces another nonzero polynomial. Hence,
ker()={0
P},andis one-to-one. But, every nonzero element ofrange()has degree at least
1,sincemultiplies by. Hence, the constant polynomial1is not inrange(),andis not onto.
(b) Corollary 5.13 requires that the domain and codomain of the linear transformation befinite
dimensional. However,Pis infinite dimensional.
(9) (a) Consider:P
2→R
3
given by(p)=[p( 1)p( 2)p( 3)]Now,dim(P 2)=dim(R
3
)=3
Hence, by Corollary 5.13, ifis either one-to-one or onto, it has the other property as well.
We will show thatis one-to-one using part (1) of Theorem 5.12. Ifp∈ker(),then(p)=0,
and sop(
1)=p( 2)=p( 3)=0. Hencepis a polynomial of degree≤2touching the-axis at
=
1,= 2,and= 3.Sincethegraphofpmust be either a parabola or a line, it cannot
touch the-axis at three distinct points unless its graph is the line=0.Thatis,p=0inP
2.
Therefore,ker()={0},andis one-to-one.
Now, by Corollary 5.13,is onto. Thus, given any3-vector[ ],thereissomep∈P
2such
thatp(
1)=,p( 2)=,andp( 3)=.
(b) From part (a),is an isomorphism. Hence, any[ ]∈R
3
has a unique pre-image underin
P
2.
(c) Generalize parts (a) and (b) using:P
→R
+1
given by(p)=[p( 1)p( )p( +1)]
Note that any polynomial inker()has+1distinct roots, and so must be trivial. Henceis one-
to-one, and therefore by Corollary 5.13,is onto. Thus, given any(+1)-vector[
1+1],
there is a uniquep∈P
such thatp( 1)= 1p( )= ,andp( +1)= +1.
(10) Supposeis not one-to-one. Thenker() is nontrivial. Letv∈ker()withv6 =0Then={v}is
linearly independent. But()={0
W}, which is linearly dependent, a contradiction.
(11) (a) Supposew∈(span()). Then there is a vectorv∈span()such that(v)=w.Thereare
also vectorsv
1v ∈and scalars 1such that 1v1+···+ v=v. Hence,
w=(v)=(
1v1+···+ v)
=(
1v1)+···+( v)
=
1(v1)+···+ (v)
which shows thatw∈span(()). Hence,(span())⊆span(()).
Next,⊆span()(by Theorem 4.5)
=⇒()⊆(span())
=⇒span(())⊆(span())(by part (3) of Theorem 4.5, since(span())is a subspace
ofWby part (1) of Theorem 5.3). Therefore,(span()) = span(()).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 102
Answers to Exercises Section 5.5
(b) By part (a),()spans(span()) =(V)=range().
(c)W= span(()) =(span())(by part (a))⊆(V)(becausespan()⊆V)= range().Thus,
is onto.
(12) (a)F (b)F (c)T (d)T (e)T (f)T (g)T (h)F (i)F
Section 5.5
(1) Ineachpart,letArepresent the given matrix for 1and letBrepresent the given matrix for 2.By
Theorem 5.16,
1is an isomorphism if and only ifAis nonsingular, and 2is an isomorphism if and
only ifBis nonsingular. In each part, we will give|A|and|B|to show thatAandBare nonsingular.
(a)|A|=1,|B|=3,
−1
1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
001
0−10
1−20
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
−1
2
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
10 0
00−
1
3
21 0
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,(
2◦1)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
0−21
14 −2
030
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
(
2◦1)
−1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=(
−1
1
◦
−1
2
)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎢
⎢
⎣
21 0
00
1
3
10
2
3
⎤
⎥
⎥
⎦
⎡
⎣
1
2
3
⎤
⎦
(b)|A|=−1,|B|=−1,
−1
1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
0−21
010
1−84
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
−1
2
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
010
101
203
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,(
2◦1)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
−110
−401
100
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
(
2◦1)
−1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=(
−1
1
◦
−1
2
)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
001
101
014
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
(c)|A|=−1,|B|=1,
−1
1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
2−4−1
7−13−3
5−10−3
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
−1
2
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
10 −1
31 −3
−1−22
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
(
2◦1)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
29−6−4
21−5−2
38−8−5
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
(
2◦1)
−1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=(
−1
1
◦
−1
2
)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
−9−28
−29−726
−22−419
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 103
(b) By part (a),()spans(span()) =(V)=range().
(c)W= span(()) =(span())(by part (a))⊆(V)(becausespan()⊆V)= range().Thus,
is onto.
(12) (a)F (b)F (c)T (d)T (e)T (f)T (g)T (h)F (i)F
Section 5.5
(1) Ineachpart,letArepresent the given matrix for 1and letBrepresent the given matrix for 2.By
Theorem 5.16,
1is an isomorphism if and only ifAis nonsingular, and 2is an isomorphism if and
only ifBis nonsingular. In each part, we will give|A|and|B|to show thatAandBare nonsingular.
(a)|A|=1,|B|=3,
−1
1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
001
0−10
1−20
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
−1
2
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
10 0
00−
1
3
21 0
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,(
2◦1)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
0−21
14 −2
030
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
(
2◦1)
−1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=(
−1
1
◦
−1
2
)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎢
⎢
⎣
21 0
00
1
3
10
2
3
⎤
⎥
⎥
⎦
⎡
⎣
1
2
3
⎤
⎦
(b)|A|=−1,|B|=−1,
−1
1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
0−21
010
1−84
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
−1
2
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
010
101
203
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,(
2◦1)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
−110
−401
100
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
(
2◦1)
−1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=(
−1
1
◦
−1
2
)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
001
101
014
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
(c)|A|=−1,|B|=1,
−1
1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
2−4−1
7−13−3
5−10−3
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
−1
2
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
10 −1
31 −3
−1−22
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
(
2◦1)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
29−6−4
21−5−2
38−8−5
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦,
(
2◦1)
−1
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=(
−1
1
◦
−1
2
)
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
−9−28
−29−726
−22−419
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 103
Answers to Exercises Section 5.5
(2)is clearly a linear operator, andis invertible (
−1
=). By Theorem 5.15,is an isomorphism.
(3) (a)
1is a linear operator:
1(B+C)= A(B+C)=AB+AC= 1(B)+ 1(C);
1(B)= A(B)=(AB)= 1(B)
Note that
1is invertible (
−1
1
(B)=A
−1
B). Use Theorem 5.15.
(b)
2is a linear operator:
2(B+C)= A(B+C)A
−1
=(AB+AC)A
−1
=ABA
−1
+ACA
−1
=2(B)+ 2(C);
2(B)= A(B)A
−1
=(ABA
−1
)= 2(B)
Note that
2is invertible (
−1
2
(B)=A
−1
BA). Use Theorem 5.15.
(4)is a linear operator:
(p+q)=( p+q)+(p+q)
0
=p+q+p
0
+q
0
=(p+p
0
)+(q+q
0
)=(p)+();
(p)=( p)+(p)
0
=p+p
0
=(p+p
0
)=(p)
Now, if(p)=0,thenp+p
0
=0=⇒p=−p
0
But ifpis not a constant polynomial, thenpand
p
0
have different degrees, a contradiction. Hence,p
0
=0and sop=0.Thus,ker()={0}andis
one-to-one. Then, by Corollary 5.13,is an isomorphism.
(5) (a)
∙
01
10
¸
(b) Use Theorem 5.16 (since the matrix in part (a) is nonsingular).
(c) IfAis the matrix forin part (a), thenA
2
=I2,so=
−1
.
(d) Performing the reflectiontwice in succession gives the identity mapping.
(6) The change of basis is performed by multiplying by a nonsingular transition matrix. Use Theorem
5.16. (Note that multiplication by a matrix is a linear transformation.)
(7)(◦
1)(v)=(◦ 2)(v)=⇒(
−1
◦◦ 1)(v)=(
−1
◦◦ 2)(v)=⇒ 1(v)= 2(v)
(8) Supposedim(V)=0anddim(W)=0.
SupposeA
is nonsingular, thenA is square,=,A
−1
exists, andA
−1
A=A A
−1
=
I
.Letbe the linear operator fromWtoVwhose matrix with respect toandisA
−1
. Then,
for allv∈V,[(◦)(v)]
=A
−1
A[v]=I[v]=[v] . Therefore,(◦)(v)=vfor allv∈V,
since coordinatizations are unique.
Similarly, for allw∈W,[(◦)(w)]
=A A
−1
[w]=I[w]=[w] . Hence(◦)(w)=w
for allw∈W. Therefore,acts as an inverse for. Hence,is an isomorphism by Theorem 5.15.
(9) In all parts, use Corollary 5.20. In part (c), the answer to the question is yes.
(10) Ifis an isomorphism,(◦)
−1
=
−1
◦
−1
,so◦is an isomorphism by Theorem 5.15.
Conversely, if◦is an isomorphism, it is easy to proveis one-to-one and onto directly. Or, let
=◦(◦)
−1
and=(◦)
−1
◦. Then clearly,◦=(identity operator onV)=◦.But
=◦=◦◦=◦=,andso==
−1
. Henceis an isomorphism by Theorem 5.15.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 104
(2)is clearly a linear operator, andis invertible (
−1
=). By Theorem 5.15,is an isomorphism.
(3) (a)
1is a linear operator:
1(B+C)= A(B+C)=AB+AC= 1(B)+ 1(C);
1(B)= A(B)=(AB)= 1(B)
Note that
1is invertible (
−1
1
(B)=A
−1
B). Use Theorem 5.15.
(b)
2is a linear operator:
2(B+C)= A(B+C)A
−1
=(AB+AC)A
−1
=ABA
−1
+ACA
−1
=2(B)+ 2(C);
2(B)= A(B)A
−1
=(ABA
−1
)= 2(B)
Note that
2is invertible (
−1
2
(B)=A
−1
BA). Use Theorem 5.15.
(4)is a linear operator:
(p+q)=( p+q)+(p+q)
0
=p+q+p
0
+q
0
=(p+p
0
)+(q+q
0
)=(p)+();
(p)=( p)+(p)
0
=p+p
0
=(p+p
0
)=(p)
Now, if(p)=0,thenp+p
0
=0=⇒p=−p
0
But ifpis not a constant polynomial, thenpand
p
0
have different degrees, a contradiction. Hence,p
0
=0and sop=0.Thus,ker()={0}andis
one-to-one. Then, by Corollary 5.13,is an isomorphism.
(5) (a)
∙
01
10
¸
(b) Use Theorem 5.16 (since the matrix in part (a) is nonsingular).
(c) IfAis the matrix forin part (a), thenA
2
=I2,so=
−1
.
(d) Performing the reflectiontwice in succession gives the identity mapping.
(6) The change of basis is performed by multiplying by a nonsingular transition matrix. Use Theorem
5.16. (Note that multiplication by a matrix is a linear transformation.)
(7)(◦
1)(v)=(◦ 2)(v)=⇒(
−1
◦◦ 1)(v)=(
−1
◦◦ 2)(v)=⇒ 1(v)= 2(v)
(8) Supposedim(V)=0anddim(W)=0.
SupposeA
is nonsingular, thenA is square,=,A
−1
exists, andA
−1
A=A A
−1
=
I
.Letbe the linear operator fromWtoVwhose matrix with respect toandisA
−1
. Then,
for allv∈V,[(◦)(v)]
=A
−1
A[v]=I[v]=[v] . Therefore,(◦)(v)=vfor allv∈V,
since coordinatizations are unique.
Similarly, for allw∈W,[(◦)(w)]
=A A
−1
[w]=I[w]=[w] . Hence(◦)(w)=w
for allw∈W. Therefore,acts as an inverse for. Hence,is an isomorphism by Theorem 5.15.
(9) In all parts, use Corollary 5.20. In part (c), the answer to the question is yes.
(10) Ifis an isomorphism,(◦)
−1
=
−1
◦
−1
,so◦is an isomorphism by Theorem 5.15.
Conversely, if◦is an isomorphism, it is easy to proveis one-to-one and onto directly. Or, let
=◦(◦)
−1
and=(◦)
−1
◦. Then clearly,◦=(identity operator onV)=◦.But
=◦=◦◦=◦=,andso==
−1
. Henceis an isomorphism by Theorem 5.15.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 104
Answers to Exercises Section 5.5
(11) (a) Letv∈Vwithv6 =0 V.Since(◦)(v)=0 V,either(v)=0 Vor((v)) =0 V,with
(v)6 =0
V. Hence,ker()6 ={0 V}, since one of the two vectorsvor(v)is nonzero and is in
ker().
(b) Proof by contrapositive: The goal is to prove that ifis an isomorphism, then◦6 =oris
the identity transformation. Assumeis an isomorphism and◦=(see “IfThenor
Proofs” in Section 1.3). Thenis the identity transformation because
(◦)(v)=(v)=⇒(
−1
◦◦)(v)=(
−1
◦)(v)=⇒(v)=v
(12) Range()=column space ofA(see comments just before the Range Method in Section 5.3 of the
textbook).
Now,is an isomorphism iffis onto (by Corollary 5.13)
iffrange()=R
iffspan({columns ofA})=R
iffthecolumns ofAare linearly independent (by part (2) of Theorem 4.12).
(13) (a) Use Theorem 5.14.
(b) By Exercise 11(c) in Section 5.4,is onto. Now, ifis not one-to-one, thenker()is nontrivial.
Supposev∈ker()withv6 =0If={v
1v }then there are 1∈Rsuch thatv=
1v1+···+ vwhere not every =0(sincev6 =0). Then
0
W=(v)=( 1v1+···+ v)= 1(v1)+···+ (v
)
But since each(v
)is distinct, and some 6 =0this contradicts the linear independence of
(). Henceis one-to-one.
(c)(e
1)=[31](e 2)=[52]and(e 3)=[31]Hence,()={[31][52]}(because identical
elements of a set are not listed more than once). This set is clearly a basis forR
2
.is not an
isomorphism or else we would contradict Theorem 5.18, sincedim(R
3
)6 =dim(R
2
).
(d) Part (c) is not a counterexample to part (b) because in part (c) the images of the vectors in
are not distinct.
(14) Let=(v
1v )letv∈V,andlet 1∈Rsuch thatv= 1v1+···+ v
Then[v]
=[1]But
(v)=(
1v1+···+ v)= 1(v1)+···+ (v
)
From Exercise 13(a),{(v
1) (v
)}is a basis for()=W(sinceis onto).
Hence,[(v)]
()=[1]as well.
(15) (Note: This exercise will be used as part of the proof of the Dimension Theorem, so we do not use the
Dimension Theorem in this proof. Notice that we can use both Theorem 5.14 and part (a) of Exercise
11 because they were previously proven without invoking the Dimension Theorem.)
BecauseYis a subspace of thefinite dimensional vector spaceV,Yis alsofinite dimensional (Theo-
rem 4.13). Suppose{y
1y }is a basis forY. Becauseis one-to-one, the vectors(y 1)(y )
are distinct. By part (1) of Theorem 5.14,{(y
1)(y )}=({y 1y })is linearly indepen-
dent. Part (a) of Exercise 11 in Section 5.4 shows that
span(({y
1y })) =(span({y 1y })) =(Y)
Therefore,{(y
1)(y )}is a-element basis for(Y). Hence,
dim((Y)) ==dim(Y)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 105
(11) (a) Letv∈Vwithv6 =0 V.Since(◦)(v)=0 V,either(v)=0 Vor((v)) =0 V,with
(v)6 =0
V. Hence,ker()6 ={0 V}, since one of the two vectorsvor(v)is nonzero and is in
ker().
(b) Proof by contrapositive: The goal is to prove that ifis an isomorphism, then◦6 =oris
the identity transformation. Assumeis an isomorphism and◦=(see “IfThenor
Proofs” in Section 1.3). Thenis the identity transformation because
(◦)(v)=(v)=⇒(
−1
◦◦)(v)=(
−1
◦)(v)=⇒(v)=v
(12) Range()=column space ofA(see comments just before the Range Method in Section 5.3 of the
textbook).
Now,is an isomorphism iffis onto (by Corollary 5.13)
iffrange()=R
iffspan({columns ofA})=R
iffthecolumns ofAare linearly independent (by part (2) of Theorem 4.12).
(13) (a) Use Theorem 5.14.
(b) By Exercise 11(c) in Section 5.4,is onto. Now, ifis not one-to-one, thenker()is nontrivial.
Supposev∈ker()withv6 =0If={v
1v }then there are 1∈Rsuch thatv=
1v1+···+ vwhere not every =0(sincev6 =0). Then
0
W=(v)=( 1v1+···+ v)= 1(v1)+···+ (v
)
But since each(v
)is distinct, and some 6 =0this contradicts the linear independence of
(). Henceis one-to-one.
(c)(e
1)=[31](e 2)=[52]and(e 3)=[31]Hence,()={[31][52]}(because identical
elements of a set are not listed more than once). This set is clearly a basis forR
2
.is not an
isomorphism or else we would contradict Theorem 5.18, sincedim(R
3
)6 =dim(R
2
).
(d) Part (c) is not a counterexample to part (b) because in part (c) the images of the vectors in
are not distinct.
(14) Let=(v
1v )letv∈V,andlet 1∈Rsuch thatv= 1v1+···+ v
Then[v]
=[1]But
(v)=(
1v1+···+ v)= 1(v1)+···+ (v
)
From Exercise 13(a),{(v
1) (v
)}is a basis for()=W(sinceis onto).
Hence,[(v)]
()=[1]as well.
(15) (Note: This exercise will be used as part of the proof of the Dimension Theorem, so we do not use the
Dimension Theorem in this proof. Notice that we can use both Theorem 5.14 and part (a) of Exercise
11 because they were previously proven without invoking the Dimension Theorem.)
BecauseYis a subspace of thefinite dimensional vector spaceV,Yis alsofinite dimensional (Theo-
rem 4.13). Suppose{y
1y }is a basis forY. Becauseis one-to-one, the vectors(y 1)(y )
are distinct. By part (1) of Theorem 5.14,{(y
1)(y )}=({y 1y })is linearly indepen-
dent. Part (a) of Exercise 11 in Section 5.4 shows that
span(({y
1y })) =(span({y 1y })) =(Y)
Therefore,{(y
1)(y )}is a-element basis for(Y). Hence,
dim((Y)) ==dim(Y)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 105
Answers to Exercises Section 5.5
(16) (a) Letv∈ker().Then
(
2◦)(v)= 2((v)) = 2(0W)=0 Y
and sov∈ker(
2◦). Hence,ker()⊆ker( 2◦).
Next, supposev∈ker(
2◦).Then
0
Y=( 2◦)(v)= 2((v))
which implies that(v)∈ker(
2).But 2is one-to-one, and soker( 2)={0 W}. Therefore,
(v)=0
W, implying thatv∈ker(). Hence,ker( 2◦)⊆ker().
Therefore,ker()=ker(
2◦).
(b) Supposew∈range(◦
1)Then there is somex∈Xsuch that(◦ 1)(x)=wimplying
(
1(x)) =wHencewis the image underof 1(x)and sow∈range().Thus,range(◦ 1)
⊆range(). Similarly, ifw∈range()there is somev∈Vsuch that(v)=wSince
−1
1
exists,(◦ 1)(
−1
1
(v)) =wand sow∈range(◦ 1)Thus,range()⊆range(◦ 1)finishing
the proof.
(c) Letv∈
1(ker(◦ 1)).Thus,thereisanx∈ker(◦ 1)such that 1(x)=v. Then,
(v)=(
1(x)) = (◦ 1)(x)=0 W
Thus,v∈ker(). Therefore,
1(ker(◦ 1))⊆ker().
Now suppose thatv∈ker().Then(v)=0
W.Thus,(◦ 1)(
−1
1
(v)) =0 Was well.
Hence,
−1
1
(v)∈ker(◦ 1). Therefore,
v=
1(
−1
1
(v))∈ 1(ker(◦ 1))
implyingker()⊆
1(ker(◦ 1)), completing the proof.
(d) From part (c),dim(ker()) = dim(
1(ker(◦ 1))). Apply Exercise 15 using 1asand
Y=ker(◦
1)to show that
dim(
1(ker(◦ 1))) = dim(ker(◦ 1))
(e) Supposey∈range(
2◦). Hence, there is av∈Vsuch that( 2◦)(v)=y.Thus,
y=
2((v)). Clearly,(v)∈range(),andsoy∈ 2(range()).Thisprovesthat
range(
2◦)⊆ 2(range())
Next, suppose thaty∈
2(range()). Then there is somew∈range()such that 2(w)=y.
Sincew∈range(),thereisav∈Vsuch that(v)=w. Therefore,
(
2◦)(v)= 2((v)) = 2(w)=y
This establishes that
2(range())⊆range( 2◦),finishing the proof.
(f) From part (e),
dim(range(
2◦)) = dim( 2(range()))
Apply Exercise 15 using
2asandY= range()to show that
dim(
2(range())) = dim(range())
Copyrightc°2016 Elsevier Ltd. All rights reserved. 106
(16) (a) Letv∈ker().Then
(
2◦)(v)= 2((v)) = 2(0W)=0 Y
and sov∈ker(
2◦). Hence,ker()⊆ker( 2◦).
Next, supposev∈ker(
2◦).Then
0
Y=( 2◦)(v)= 2((v))
which implies that(v)∈ker(
2).But 2is one-to-one, and soker( 2)={0 W}. Therefore,
(v)=0
W, implying thatv∈ker(). Hence,ker( 2◦)⊆ker().
Therefore,ker()=ker(
2◦).
(b) Supposew∈range(◦
1)Then there is somex∈Xsuch that(◦ 1)(x)=wimplying
(
1(x)) =wHencewis the image underof 1(x)and sow∈range().Thus,range(◦ 1)
⊆range(). Similarly, ifw∈range()there is somev∈Vsuch that(v)=wSince
−1
1
exists,(◦ 1)(
−1
1
(v)) =wand sow∈range(◦ 1)Thus,range()⊆range(◦ 1)finishing
the proof.
(c) Letv∈
1(ker(◦ 1)).Thus,thereisanx∈ker(◦ 1)such that 1(x)=v. Then,
(v)=(
1(x)) = (◦ 1)(x)=0 W
Thus,v∈ker(). Therefore,
1(ker(◦ 1))⊆ker().
Now suppose thatv∈ker().Then(v)=0
W.Thus,(◦ 1)(
−1
1
(v)) =0 Was well.
Hence,
−1
1
(v)∈ker(◦ 1). Therefore,
v=
1(
−1
1
(v))∈ 1(ker(◦ 1))
implyingker()⊆
1(ker(◦ 1)), completing the proof.
(d) From part (c),dim(ker()) = dim(
1(ker(◦ 1))). Apply Exercise 15 using 1asand
Y=ker(◦
1)to show that
dim(
1(ker(◦ 1))) = dim(ker(◦ 1))
(e) Supposey∈range(
2◦). Hence, there is av∈Vsuch that( 2◦)(v)=y.Thus,
y=
2((v)). Clearly,(v)∈range(),andsoy∈ 2(range()).Thisprovesthat
range(
2◦)⊆ 2(range())
Next, suppose thaty∈
2(range()). Then there is somew∈range()such that 2(w)=y.
Sincew∈range(),thereisav∈Vsuch that(v)=w. Therefore,
(
2◦)(v)= 2((v)) = 2(w)=y
This establishes that
2(range())⊆range( 2◦),finishing the proof.
(f) From part (e),
dim(range(
2◦)) = dim( 2(range()))
Apply Exercise 15 using
2asandY= range()to show that
dim(
2(range())) = dim(range())
Copyrightc°2016 Elsevier Ltd. All rights reserved. 106
Answers to Exercises Section 5.5
(17) (a) Part (c) of Exercise 16 states that 1(ker (◦ 1)) = ker (). Substituting= 2◦and
1=
−1
1
yields
−1
1
¡
ker
¡
2◦◦
−1
1
¢¢
=ker(
2◦)
Now=
2◦◦
−1
1
. Hence,
−1
1
(ker ()) = ker ( 2◦).
(b) With
2=2and=, part (a) of Exercise 16 shows thatker ( 2◦)=ker().This,
combined with part (a) of this exercise, shows that
−1
1
(ker ()) = ker ().
(c) From part (b),
dim(ker ()) = dim(
−1
1
(ker ()))
Apply Exercise 15 withreplaced by
−1
1
andY=ker()to show that
dim(
−1
1
(ker ())) = dim(ker())
completing the proof.
(d) Part (e) of Exercise 16 states thatrange (
2◦)= 2(range()). Substituting◦
−1
1
forand
2for 2yields
range
¡
2◦◦
−1
1
¢
=
2
¡
range(◦
−1
1
)
¢
Apply
−1
2
to both sides to produce
−1
2
(range
¡
2◦◦
−1
1
¢
)=range(◦
−1
1
)
Using=
2◦◦
−1
1
gives
−1
2
(range ()) = range(◦
−1
1
), the desired result.
(e) Part (b) of Exercise 16 states thatrange (◦
1)=range(). Substitutingforand
−1
1
for
1producesrange(◦
−1
1
)=range(). Combining this with
−1
2
(range ()) = range(◦
−1
1
)
from part (d) of this exercise yields
−1
2
(range()) = range().
(f) From part (e),dim(
−1
2
(range())) = dim(range()).ApplyExercise15withreplaced by
−1
2
andY= range()to show that
dim(
−1
2
(range())) = dim(range())
completing the proof.
(18) (a) A nonsingular2×2matrix must have at least one of itsfirst column entries nonzero. Then, use
an appropriate equation from the two given in the problem.
(b)
∙
0
01
¸∙
¸
=
∙
¸
, a contraction (if≤1) or dilation (if≥1)alongthe-coordinate.
Similarly,
∙
10
0
¸∙
¸
=
∙
¸
, a contraction (if≤1) or dilation (if≥1)alongthe
-coordinate.
(c) By the hint, multiplying by
∙
0
01
¸
first performs a dilation/contraction by part (a) followed by
areflection about the-axis. Using
∙
10
0
¸
=
∙
10
0−1
¸∙
10
0−
¸
shows that multiplying
Copyrightc
°2016 Elsevier Ltd. All rights reserved. 107
(17) (a) Part (c) of Exercise 16 states that 1(ker (◦ 1)) = ker (). Substituting= 2◦and
1=
−1
1
yields
−1
1
¡
ker
¡
2◦◦
−1
1
¢¢
=ker(
2◦)
Now=
2◦◦
−1
1
. Hence,
−1
1
(ker ()) = ker ( 2◦).
(b) With
2=2and=, part (a) of Exercise 16 shows thatker ( 2◦)=ker().This,
combined with part (a) of this exercise, shows that
−1
1
(ker ()) = ker ().
(c) From part (b),
dim(ker ()) = dim(
−1
1
(ker ()))
Apply Exercise 15 withreplaced by
−1
1
andY=ker()to show that
dim(
−1
1
(ker ())) = dim(ker())
completing the proof.
(d) Part (e) of Exercise 16 states thatrange (
2◦)= 2(range()). Substituting◦
−1
1
forand
2for 2yields
range
¡
2◦◦
−1
1
¢
=
2
¡
range(◦
−1
1
)
¢
Apply
−1
2
to both sides to produce
−1
2
(range
¡
2◦◦
−1
1
¢
)=range(◦
−1
1
)
Using=
2◦◦
−1
1
gives
−1
2
(range ()) = range(◦
−1
1
), the desired result.
(e) Part (b) of Exercise 16 states thatrange (◦
1)=range(). Substitutingforand
−1
1
for
1producesrange(◦
−1
1
)=range(). Combining this with
−1
2
(range ()) = range(◦
−1
1
)
from part (d) of this exercise yields
−1
2
(range()) = range().
(f) From part (e),dim(
−1
2
(range())) = dim(range()).ApplyExercise15withreplaced by
−1
2
andY= range()to show that
dim(
−1
2
(range())) = dim(range())
completing the proof.
(18) (a) A nonsingular2×2matrix must have at least one of itsfirst column entries nonzero. Then, use
an appropriate equation from the two given in the problem.
(b)
∙
0
01
¸∙
¸
=
∙
¸
, a contraction (if≤1) or dilation (if≥1)alongthe-coordinate.
Similarly,
∙
10
0
¸∙
¸
=
∙
¸
, a contraction (if≤1) or dilation (if≥1)alongthe
-coordinate.
(c) By the hint, multiplying by
∙
0
01
¸
first performs a dilation/contraction by part (a) followed by
areflection about the-axis. Using
∙
10
0
¸
=
∙
10
0−1
¸∙
10
0−
¸
shows that multiplying
Copyrightc
°2016 Elsevier Ltd. All rights reserved. 107
Answers to Exercises Section 5.6
by
∙
10
0
¸
first performs a dilation/contraction by part (a) followed by a reflection about the
-axis.
(d) These matrices are defined to represent shears in Exercise 11 in Section 5.1.
(e)
∙
01
10
¸∙
¸
=
∙
¸
; that is, this matrix multiplication switches the coordinates of a vector
inR
2
. That is precisely the result of a reflection through the line=.
(f) This is merely a summary of parts (a) through (e).
(19)
"
√
2
2
−
√
2
2
√
2
2
√
2
2
#
=
" √
2
2
0
01
#"
10
√
2
2
1
#
∙
10
0
√
2
¸∙
1−1
01
¸
,
where the matrices on the right side are, respectively, a contraction along the-coordinate, a shear in
the-direction, a dilation along the-coordinate, and a shear in the-direction.
(20) (a) T (b) T (c) F (d) F (e) F (f) T (g) T (h) T
Section 5.6
(1) (a)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=2 {[10]} 2 1
(b)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=3 {[14]} 1 1
2=2 {[01]} 1 1
(c)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=1 {[211]} 1 1
2=−1 {[−101]} 1 1
3=2 {[111]} 1 1
(d)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=2 {[540][302]} 2 2
2=3 {[01−1]} 1 1
(e)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=−1 {[−123]} 2 1
2=0 {[−113]} 1 1
(f)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=0 {[6−114]} 2 1
2=2 {[7105]} 2 1
(2) (a)=(e 1e2e3e4);=([1100][0011][1−100][001−1]);
A=
⎡
⎢
⎢
⎣
0100
1000
0001
0010
⎤
⎥
⎥
⎦
;D=
⎡
⎢
⎢
⎣
10 0 0
01 0 0
00−10
00 0 −1
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
10 1 0
10−10
01 0 1
01 0 −1
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 108
by
∙
10
0
¸
first performs a dilation/contraction by part (a) followed by a reflection about the
-axis.
(d) These matrices are defined to represent shears in Exercise 11 in Section 5.1.
(e)
∙
01
10
¸∙
¸
=
∙
¸
; that is, this matrix multiplication switches the coordinates of a vector
inR
2
. That is precisely the result of a reflection through the line=.
(f) This is merely a summary of parts (a) through (e).
(19)
"
√
2
2
−
√
2
2
√
2
2
√
2
2
#
=
" √
2
2
0
01
#"
10
√
2
2
1
#
∙
10
0
√
2
¸∙
1−1
01
¸
,
where the matrices on the right side are, respectively, a contraction along the-coordinate, a shear in
the-direction, a dilation along the-coordinate, and a shear in the-direction.
(20) (a) T (b) T (c) F (d) F (e) F (f) T (g) T (h) T
Section 5.6
(1) (a)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=2 {[10]} 2 1
(b)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=3 {[14]} 1 1
2=2 {[01]} 1 1
(c)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=1 {[211]} 1 1
2=−1 {[−101]} 1 1
3=2 {[111]} 1 1
(d)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=2 {[540][302]} 2 2
2=3 {[01−1]} 1 1
(e)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=−1 {[−123]} 2 1
2=0 {[−113]} 1 1
(f)
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=0 {[6−114]} 2 1
2=2 {[7105]} 2 1
(2) (a)=(e 1e2e3e4);=([1100][0011][1−100][001−1]);
A=
⎡
⎢
⎢
⎣
0100
1000
0001
0010
⎤
⎥
⎥
⎦
;D=
⎡
⎢
⎢
⎣
10 0 0
01 0 0
00−10
00 0 −1
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
10 1 0
10−10
01 0 1
01 0 −1
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 108
Answers to Exercises Section 5.6
(b)=(
2
1);=(
2
−2+1−+11);
A=
⎡
⎣
200
−210
0−10
⎤
⎦;D=
⎡
⎣
200
010
000
⎤
⎦;P=
⎡
⎣
100
−2−10
111
⎤
⎦
(c) Not diagonalizable;=(
2
1);A=
⎡
⎣
−100
2−30
01 −3
⎤
⎦;
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=−1{2
2
+2+1} 1 1
2=−3 {1} 2 1
(d)=(
2
1);=(2
2
−8+91);
A=
⎡
⎣
−100
−12−40
18 0 −5
⎤
⎦;D=
⎡
⎣
−100
0−40
00 −5
⎤
⎦;P=
⎡
⎣
200
−810
901
⎤
⎦
(e) Not diagonalizable; no eigenvalues;A=
"
1
2
−
√
3
2
√
3
2
1
2
#
(f)=
µ∙
10
00
¸
∙
01
00
¸
∙
00
10
¸
∙
00
01
¸¶
;
=
µ∙
10
00
¸
∙
00
01
¸
∙
01
10
¸
∙
01
−10
¸¶
;
A=
⎡
⎢
⎢
⎣
1000
0010
0100
0001
⎤
⎥
⎥
⎦
;D=
⎡
⎢
⎢
⎣
100 0
010 0
001 0
000 −1
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
100 0
001 1
001 −1
010 0
⎤
⎥
⎥
⎦
(g)=
µ∙
10
00
¸
∙
01
00
¸
∙
00
10
¸
∙
00
01
¸¶
;
=
µ∙
10
00
¸
∙
01
10
¸
∙
00
01
¸
∙
0−1
10
¸¶
;
A=
⎡
⎢
⎢
⎣
0000
01 −10
0−110
0000
⎤
⎥
⎥
⎦
;D=
⎡
⎢
⎢
⎣
0000
0000
0000
0002
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
100 0
010 −1
010 1
001 0
⎤
⎥
⎥
⎦
(h)=
µ∙
10
00
¸
∙
01
00
¸
∙
00
10
¸
∙
00
01
¸¶
;
=
µ∙
30
50
¸
∙
03
05
¸
∙
10
20
¸
∙
01
02
¸¶
;
A=
⎡
⎢
⎢
⎣
−4 030
0−403
−10 070
0−1007
⎤
⎥
⎥
⎦
;D=
⎡
⎢
⎢
⎣
1000
0100
0020
0002
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
3010
0301
5020
0502
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 109
(b)=(
2
1);=(
2
−2+1−+11);
A=
⎡
⎣
200
−210
0−10
⎤
⎦;D=
⎡
⎣
200
010
000
⎤
⎦;P=
⎡
⎣
100
−2−10
111
⎤
⎦
(c) Not diagonalizable;=(
2
1);A=
⎡
⎣
−100
2−30
01 −3
⎤
⎦;
EigenvalueBasis for Alg. Mult.Geom. Mult.
1=−1{2
2
+2+1} 1 1
2=−3 {1} 2 1
(d)=(
2
1);=(2
2
−8+91);
A=
⎡
⎣
−100
−12−40
18 0 −5
⎤
⎦;D=
⎡
⎣
−100
0−40
00 −5
⎤
⎦;P=
⎡
⎣
200
−810
901
⎤
⎦
(e) Not diagonalizable; no eigenvalues;A=
"
1
2
−
√
3
2
√
3
2
1
2
#
(f)=
µ∙
10
00
¸
∙
01
00
¸
∙
00
10
¸
∙
00
01
¸¶
;
=
µ∙
10
00
¸
∙
00
01
¸
∙
01
10
¸
∙
01
−10
¸¶
;
A=
⎡
⎢
⎢
⎣
1000
0010
0100
0001
⎤
⎥
⎥
⎦
;D=
⎡
⎢
⎢
⎣
100 0
010 0
001 0
000 −1
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
100 0
001 1
001 −1
010 0
⎤
⎥
⎥
⎦
(g)=
µ∙
10
00
¸
∙
01
00
¸
∙
00
10
¸
∙
00
01
¸¶
;
=
µ∙
10
00
¸
∙
01
10
¸
∙
00
01
¸
∙
0−1
10
¸¶
;
A=
⎡
⎢
⎢
⎣
0000
01 −10
0−110
0000
⎤
⎥
⎥
⎦
;D=
⎡
⎢
⎢
⎣
0000
0000
0000
0002
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
100 0
010 −1
010 1
001 0
⎤
⎥
⎥
⎦
(h)=
µ∙
10
00
¸
∙
01
00
¸
∙
00
10
¸
∙
00
01
¸¶
;
=
µ∙
30
50
¸
∙
03
05
¸
∙
10
20
¸
∙
01
02
¸¶
;
A=
⎡
⎢
⎢
⎣
−4 030
0−403
−10 070
0−1007
⎤
⎥
⎥
⎦
;D=
⎡
⎢
⎢
⎣
1000
0100
0020
0002
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
3010
0301
5020
0502
⎤
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 109
Answers to Exercises Section 5.6
(3) (a) ()=
¯
¯
¯
¯
¯
¯
¯
¯
(−5)−20 −1
2( −1) 0 1
−4 −4(−3)−2
−16 0 8 ( +5)
¯
¯
¯
¯
¯
¯
¯
¯
=(−3)
¯
¯
¯
¯
¯
¯
(−5)−2 −1
2( −1) 1
−16 0 +5
¯
¯
¯
¯
¯
¯
−8
¯
¯
¯
¯
¯
¯
(−5)−2−1
2( −1) 1
−4 −4−2
¯
¯
¯
¯
¯
¯
=(
−3)
µ
−16
¯
¯
¯
¯
−2−1
(−1) 1
¯
¯
¯
¯
+(+5)
¯
¯
¯
¯
(−5)−2
2( −1)
¯
¯
¯
¯
¶
−8
µ
(−5)
¯
¯
¯
¯
(−1) 1
−4−2
¯
¯
¯
¯
−2
¯
¯
¯
¯
−2−1
−4−2
¯
¯
¯
¯
−4
¯
¯
¯
¯
−2−1
(−1) 1
¯
¯
¯
¯
¶
=(−3)(−16(
−3) + (+5)(
2
−6+9))−8((−5)(−2+6)−2(0)−4(−3))
=(−3)(−16(−3) + (+5)(−3)
2
)−8((−2)(−5)(−3)−4(−3))
=(−3)
2
(−16 + (+5)(−3)) +16(−3)((−5) + 2)
=(−3)
2
(
2
+2−31) + 16(−3)
2
=(−3)
2
(
2
+2−15) = (−3)
3
(+5).
You can check that this polynomial expands as claimed.
(b)
⎡
⎢
⎢
⎣
−10−20 −1
2−601
−4−4−8−2
−16080
⎤
⎥
⎥
⎦
becomes
⎡
⎢
⎢
⎢
⎢
⎣
100
1
8
010 −
1
8
001
1
4
000 0
⎤
⎥
⎥
⎥
⎥
⎦
when put into reduced row echelon form.
(4) For both parts, the only eigenvalue is=1;
1={1}.
(5) SinceAis upper triangular, so isI
−A.Thus,byTheorem3.2,|I −A|equals the product of the
main diagonal entries ofI
−A,whichis(− )
,since 11=22=···= ThusAhas exactly one
eigenvaluewith algebraic multiplicity. HenceAis diagonalizable⇐⇒has geometric multiplicity
⇐⇒
=R
⇐⇒Av=vfor allv∈R
⇐⇒Av=vfor allv∈{e 1e }⇐⇒A=I
(using Exercise 14(b) in Section 1.5).
(6) In both cases, the given linear operator is diagonalizable, but there are fewer thandistinct eigenvalues.
This occurs because at least one of the eigenvalues has geometric multiplicity greater than1.
(7) (a)
⎡
⎣
11−1
01 0
00 1
⎤
⎦;eigenvalue=1;basis for
1={[100][011]};has algebraic multiplicity3
and geometric multiplicity2
(b)
⎡
⎣
100
010
000
⎤
⎦;eigenvalues
1=1 2=0;basis for 1={[100][010]}; 1has algebraic
multiplicity2and geometric multiplicity2
(8) (a) Zero is not an eigenvalue foriffker()={0}. Now apply part (1) of Theorem 5.12 and
Corollary 5.13.
(b)(v)=v=⇒
−1
((v)) =
−1
(v)=⇒v=
−1
(v)=⇒
1
v=
−1
(v)(since6 =0by
part (a)).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 110
(3) (a) ()=
¯
¯
¯
¯
¯
¯
¯
¯
(−5)−20 −1
2( −1) 0 1
−4 −4(−3)−2
−16 0 8 ( +5)
¯
¯
¯
¯
¯
¯
¯
¯
=(−3)
¯
¯
¯
¯
¯
¯
(−5)−2 −1
2( −1) 1
−16 0 +5
¯
¯
¯
¯
¯
¯
−8
¯
¯
¯
¯
¯
¯
(−5)−2−1
2( −1) 1
−4 −4−2
¯
¯
¯
¯
¯
¯
=(
−3)
µ
−16
¯
¯
¯
¯
−2−1
(−1) 1
¯
¯
¯
¯
+(+5)
¯
¯
¯
¯
(−5)−2
2( −1)
¯
¯
¯
¯
¶
−8
µ
(−5)
¯
¯
¯
¯
(−1) 1
−4−2
¯
¯
¯
¯
−2
¯
¯
¯
¯
−2−1
−4−2
¯
¯
¯
¯
−4
¯
¯
¯
¯
−2−1
(−1) 1
¯
¯
¯
¯
¶
=(−3)(−16(
−3) + (+5)(
2
−6+9))−8((−5)(−2+6)−2(0)−4(−3))
=(−3)(−16(−3) + (+5)(−3)
2
)−8((−2)(−5)(−3)−4(−3))
=(−3)
2
(−16 + (+5)(−3)) +16(−3)((−5) + 2)
=(−3)
2
(
2
+2−31) + 16(−3)
2
=(−3)
2
(
2
+2−15) = (−3)
3
(+5).
You can check that this polynomial expands as claimed.
(b)
⎡
⎢
⎢
⎣
−10−20 −1
2−601
−4−4−8−2
−16080
⎤
⎥
⎥
⎦
becomes
⎡
⎢
⎢
⎢
⎢
⎣
100
1
8
010 −
1
8
001
1
4
000 0
⎤
⎥
⎥
⎥
⎥
⎦
when put into reduced row echelon form.
(4) For both parts, the only eigenvalue is=1;
1={1}.
(5) SinceAis upper triangular, so isI
−A.Thus,byTheorem3.2,|I −A|equals the product of the
main diagonal entries ofI
−A,whichis(− )
,since 11=22=···= ThusAhas exactly one
eigenvaluewith algebraic multiplicity. HenceAis diagonalizable⇐⇒has geometric multiplicity
⇐⇒
=R
⇐⇒Av=vfor allv∈R
⇐⇒Av=vfor allv∈{e 1e }⇐⇒A=I
(using Exercise 14(b) in Section 1.5).
(6) In both cases, the given linear operator is diagonalizable, but there are fewer thandistinct eigenvalues.
This occurs because at least one of the eigenvalues has geometric multiplicity greater than1.
(7) (a)
⎡
⎣
11−1
01 0
00 1
⎤
⎦;eigenvalue=1;basis for
1={[100][011]};has algebraic multiplicity3
and geometric multiplicity2
(b)
⎡
⎣
100
010
000
⎤
⎦;eigenvalues
1=1 2=0;basis for 1={[100][010]}; 1has algebraic
multiplicity2and geometric multiplicity2
(8) (a) Zero is not an eigenvalue foriffker()={0}. Now apply part (1) of Theorem 5.12 and
Corollary 5.13.
(b)(v)=v=⇒
−1
((v)) =
−1
(v)=⇒v=
−1
(v)=⇒
1
v=
−1
(v)(since6 =0by
part (a)).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 110
Answers to Exercises Section 5.6
(9) LetPbe such thatP
−1
APis diagonal. Let=(v 1v ),where[v ]=th column ofP. Then,
by definition,Pis the transition matrix fromto. Then, for allv∈V,
P
−1
AP[v] =P
−1
A[v] =P
−1
[(v)] =[(v)]
Hence, as in part (a),P
−1
APis the matrix forwith respect to.ButP
−1
APis diagonal.
(10) IfPis the matrix with columnsv
1v ,thenP
−1
AP=Dis a diagonal matrix with 1as
the main diagonal entries ofD. Clearly|D|=
12··· .But
|D|=|P
−1
AP|=|P
−1
||A||P|=|A||P
−1
||P|=|A|
(11) Clearly≥
P
=1
(algebraic multiplicity of )≥
P
=1
(geometric multiplicity of ), by Theorem 5.26.
(12) Proof by contrapositive: If
()has a nonreal root, then the sum of the algebraic multiplicities of the
eigenvalues ofis less than. Apply Theorem 5.28.
(13) (a)A(Bv)=(AB)v=(BA)v=B(Av)=B(v)=(Bv).
(b) The eigenspaces
forAare one-dimensional. Letv∈ ,withv6 =0.ThenBv∈ ,bypart
(a). HenceBv=vfor some∈R,andsovis an eigenvector forB.Repeatingthisforeach
eigenspace ofAshows thatBhaslinearly independent eigenvectors. Now apply Theorem 5.22.
(14) (a) Supposev
1andv 2are eigenvectors forAcorresponding to 1and 2, respectively. LetK 1be
the2×2matrix[v
10](where the vectors representcolumns). Similarly, letK 2=[0v 1],
K
3=[v 20],andK 4=[0v 2]. Then a little thought shows that{K 1K2}is a basis for 1
in
M
22,and{K 3K4}is a basis for 2inM 22.
(b) Generalize the construction in part (a).
(15) Supposev∈
1∩2Thenv∈ 1=⇒v∈ 1
=⇒(v)= 1v
Similarly,v∈
2=⇒v∈ 2=⇒(v)= 2v
Hence,
1v= 2vand so( 1−2)v=0Since 16 =2v=0
But0can not be an element of any basis, and so
1∩2is empty.
(16) (a)
is a subspace, hence closed.
(b) First, substituteu
for
P
=1
vin the given double sum equation. This proves
P
=1
u=0.
Now, the set of all nonzerou
’s is linearly independent by Theorem 5.23, since they are eigenvectors
corresponding to distinct eigenvalues. But then, the nonzero terms in
P
=1
uwould give a
nontrivial linear combination from a linearly independent set equal to the zero vector. This
contradiction shows that all of theu
’s must equal0.
(c) Using part (b) and the definition ofu
,0=
P
=1
vfor each.But{v 1v }is linearly
independent, since it is a basis for
. Hence, for each, 1=···= =0.
(d) Apply the definition of linear independence to.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 111
(9) LetPbe such thatP
−1
APis diagonal. Let=(v 1v ),where[v ]=th column ofP. Then,
by definition,Pis the transition matrix fromto. Then, for allv∈V,
P
−1
AP[v] =P
−1
A[v] =P
−1
[(v)] =[(v)]
Hence, as in part (a),P
−1
APis the matrix forwith respect to.ButP
−1
APis diagonal.
(10) IfPis the matrix with columnsv
1v ,thenP
−1
AP=Dis a diagonal matrix with 1as
the main diagonal entries ofD. Clearly|D|=
12··· .But
|D|=|P
−1
AP|=|P
−1
||A||P|=|A||P
−1
||P|=|A|
(11) Clearly≥
P
=1
(algebraic multiplicity of )≥
P
=1
(geometric multiplicity of ), by Theorem 5.26.
(12) Proof by contrapositive: If
()has a nonreal root, then the sum of the algebraic multiplicities of the
eigenvalues ofis less than. Apply Theorem 5.28.
(13) (a)A(Bv)=(AB)v=(BA)v=B(Av)=B(v)=(Bv).
(b) The eigenspaces
forAare one-dimensional. Letv∈ ,withv6 =0.ThenBv∈ ,bypart
(a). HenceBv=vfor some∈R,andsovis an eigenvector forB.Repeatingthisforeach
eigenspace ofAshows thatBhaslinearly independent eigenvectors. Now apply Theorem 5.22.
(14) (a) Supposev
1andv 2are eigenvectors forAcorresponding to 1and 2, respectively. LetK 1be
the2×2matrix[v
10](where the vectors representcolumns). Similarly, letK 2=[0v 1],
K
3=[v 20],andK 4=[0v 2]. Then a little thought shows that{K 1K2}is a basis for 1
in
M
22,and{K 3K4}is a basis for 2inM 22.
(b) Generalize the construction in part (a).
(15) Supposev∈
1∩2Thenv∈ 1=⇒v∈ 1
=⇒(v)= 1v
Similarly,v∈
2=⇒v∈ 2=⇒(v)= 2v
Hence,
1v= 2vand so( 1−2)v=0Since 16 =2v=0
But0can not be an element of any basis, and so
1∩2is empty.
(16) (a)
is a subspace, hence closed.
(b) First, substituteu
for
P
=1
vin the given double sum equation. This proves
P
=1
u=0.
Now, the set of all nonzerou
’s is linearly independent by Theorem 5.23, since they are eigenvectors
corresponding to distinct eigenvalues. But then, the nonzero terms in
P
=1
uwould give a
nontrivial linear combination from a linearly independent set equal to the zero vector. This
contradiction shows that all of theu
’s must equal0.
(c) Using part (b) and the definition ofu
,0=
P
=1
vfor each.But{v 1v }is linearly
independent, since it is a basis for
. Hence, for each, 1=···= =0.
(d) Apply the definition of linear independence to.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 111
Answers to Exercises Chapter 5 Review
(17) Example 7 states that A()=
3
−4
2
++6.So,
A(A)= A
3
−4A
2
+A+6I 3
=
⎡
⎣
−17 22 76
−374 214 1178
54−27−163
⎤
⎦−4
⎡
⎣
52 −4
−106 62 334
18−9−53
⎤
⎦
+
⎡
⎣
31−14−92
−50 28 158
18−9−55
⎤
⎦+6
⎡
⎣
100
010
001
⎤
⎦
=
⎡
⎣
000
000
000
⎤
⎦
verifying the Cayley-Hamilton Theorem for this matrix.
(18) (a)F (b)T (c)T (d)F (e)T (f)T (g)T (h)F (i)F (j)T
Chapter 5 Review Exercises
(1) (a) Not a linear transformation:([000])6 =[000]
(b) Linear transformation:
((
3
+
2
++)+(
3
+
2
++))
=((+)
3
+(+)
2
+(+)+(+))
=
⎡
⎣
4(+)−(+)3( +)−(+)
2(+)+3(+)4( +)
5(+)+(+)+2(+)2(+)−3(+)
⎤
⎦
=
⎡
⎣
4− 3−
2+3 4
5++22−3
⎤
⎦+
⎡
⎣
4− 3−
2+3 4
5++22−3
⎤
⎦
=(
3
+
2
++)+(
3
+
2
++)
Also,
((
3
+
2
++)) =(()
3
+()
2
+()+())
=
⎡
⎣
4()−()3( )−()
2()+3()4( )
5()+()+2()2()−3()
⎤
⎦
=
⎡
⎣
4− 3−
2+3 4
5++22−3
⎤
⎦
=((
3
+
2
++))
(c) Not a linear transformation:([10] + [01])6 =([10]) +([01])since([11]) = [2−3]and
([10]) +([01]) = [00] + [00] = [00].
(2)[15983232]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 112
(17) Example 7 states that A()=
3
−4
2
++6.So,
A(A)= A
3
−4A
2
+A+6I 3
=
⎡
⎣
−17 22 76
−374 214 1178
54−27−163
⎤
⎦−4
⎡
⎣
52 −4
−106 62 334
18−9−53
⎤
⎦
+
⎡
⎣
31−14−92
−50 28 158
18−9−55
⎤
⎦+6
⎡
⎣
100
010
001
⎤
⎦
=
⎡
⎣
000
000
000
⎤
⎦
verifying the Cayley-Hamilton Theorem for this matrix.
(18) (a)F (b)T (c)T (d)F (e)T (f)T (g)T (h)F (i)F (j)T
Chapter 5 Review Exercises
(1) (a) Not a linear transformation:([000])6 =[000]
(b) Linear transformation:
((
3
+
2
++)+(
3
+
2
++))
=((+)
3
+(+)
2
+(+)+(+))
=
⎡
⎣
4(+)−(+)3( +)−(+)
2(+)+3(+)4( +)
5(+)+(+)+2(+)2(+)−3(+)
⎤
⎦
=
⎡
⎣
4− 3−
2+3 4
5++22−3
⎤
⎦+
⎡
⎣
4− 3−
2+3 4
5++22−3
⎤
⎦
=(
3
+
2
++)+(
3
+
2
++)
Also,
((
3
+
2
++)) =(()
3
+()
2
+()+())
=
⎡
⎣
4()−()3( )−()
2()+3()4( )
5()+()+2()2()−3()
⎤
⎦
=
⎡
⎣
4− 3−
2+3 4
5++22−3
⎤
⎦
=((
3
+
2
++))
(c) Not a linear transformation:([10] + [01])6 =([10]) +([01])since([11]) = [2−3]and
([10]) +([01]) = [00] + [00] = [00].
(2)[15983232]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 112
Answers to Exercises Chapter 5 Review
(3)(A 1)+(A 2)=CA 1B
−1
+CA 2B
−1
=C(A 1B
−1
+A2B
−1
)=C(A 1+A2)B
−1
=(A 1+A2);
(A)=C(A)B
−1
=CAB
−1
=(A).
(4)([62−7]) = [201044];
([ ]) =
⎡
⎣
−35 −4
2−10
43 −2
⎤
⎦
⎡
⎣
⎤
⎦=[−3+5−42−4+3−2]
(5) (a) Use Theorem 5.2 and part (1) of Theorem 5.3.
(b) Use Theorem 5.2 and part (2) of Theorem 5.3.
(6) (a)A
=
∙
29 32−2
43 42−6
¸
(b)A
=
⎡
⎣
113−58 51 −58
566−291 255 −283
−1648 847 −742 823
⎤
⎦
(7) (a)A
=
⎡
⎣
21−30
13 0 −4
00 1 −2
⎤
⎦;A =
⎡
⎣
−151 99 163 16
171−113−186−17
238−158−260−25
⎤
⎦
(b)A
=
⎡
⎢
⎢
⎣
6−1−1
032
20 −4
1−51
⎤
⎥
⎥
⎦
;A =
⎡
⎢
⎢
⎣
115−45 59
374−146 190
−46 15 −25
−271 108 −137
⎤
⎥
⎥
⎦
(8)
⎡
⎣
0−10
000
001
⎤
⎦
(9) (a)
A
()=
3
−
2
−+1=(+1)(−1)
2
(b) Fundamental eigenvectors for 1=1:{[210][203]};for 2=−1:{[3−62]}
(c)=([210][203][3−62])
(d)A
=
⎡
⎣
10 0
01 0
00−1
⎤
⎦.(Note:P=
⎡
⎣
22 3
10−6
03 2
⎤
⎦andP
−1
=
1
41
⎡
⎣
18 5 −12
−2415
3−6−2
⎤
⎦)
(e) This is a reflection through the plane spanned by{[210][203]}. The equation for this plane
is3−6+2=0.
(10) (a) Basis forker()={[2−310][−3401]};basisforrange()={[3221][1134]}
(b)2+2=4
(c)[−1826−42]
∈ker()because([−1826−42]) = [−6−21020]6 =0;
[−1826−62]∈ker()because([−1826−62]) =0
(d)[83−11−23]∈range(); row reduction shows[83−11−23] = 5[3221]−7[1134],and
so,([5−7
00]) = [83−11−23]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 113
(3)(A 1)+(A 2)=CA 1B
−1
+CA 2B
−1
=C(A 1B
−1
+A2B
−1
)=C(A 1+A2)B
−1
=(A 1+A2);
(A)=C(A)B
−1
=CAB
−1
=(A).
(4)([62−7]) = [201044];
([ ]) =
⎡
⎣
−35 −4
2−10
43 −2
⎤
⎦
⎡
⎣
⎤
⎦=[−3+5−42−4+3−2]
(5) (a) Use Theorem 5.2 and part (1) of Theorem 5.3.
(b) Use Theorem 5.2 and part (2) of Theorem 5.3.
(6) (a)A
=
∙
29 32−2
43 42−6
¸
(b)A
=
⎡
⎣
113−58 51 −58
566−291 255 −283
−1648 847 −742 823
⎤
⎦
(7) (a)A
=
⎡
⎣
21−30
13 0 −4
00 1 −2
⎤
⎦;A =
⎡
⎣
−151 99 163 16
171−113−186−17
238−158−260−25
⎤
⎦
(b)A
=
⎡
⎢
⎢
⎣
6−1−1
032
20 −4
1−51
⎤
⎥
⎥
⎦
;A =
⎡
⎢
⎢
⎣
115−45 59
374−146 190
−46 15 −25
−271 108 −137
⎤
⎥
⎥
⎦
(8)
⎡
⎣
0−10
000
001
⎤
⎦
(9) (a)
A
()=
3
−
2
−+1=(+1)(−1)
2
(b) Fundamental eigenvectors for 1=1:{[210][203]};for 2=−1:{[3−62]}
(c)=([210][203][3−62])
(d)A
=
⎡
⎣
10 0
01 0
00−1
⎤
⎦.(Note:P=
⎡
⎣
22 3
10−6
03 2
⎤
⎦andP
−1
=
1
41
⎡
⎣
18 5 −12
−2415
3−6−2
⎤
⎦)
(e) This is a reflection through the plane spanned by{[210][203]}. The equation for this plane
is3−6+2=0.
(10) (a) Basis forker()={[2−310][−3401]};basisforrange()={[3221][1134]}
(b)2+2=4
(c)[−1826−42]
∈ker()because([−1826−42]) = [−6−21020]6 =0;
[−1826−62]∈ker()because([−1826−62]) =0
(d)[83−11−23]∈range(); row reduction shows[83−11−23] = 5[3221]−7[1134],and
so,([5−7
00]) = [83−11−23]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 113
Answers to Exercises Chapter 5 Review
(11) Matrix forwith respect to standard bases:
⎡
⎢
⎢
⎣
10 000 −1
01−200 0
00 010 −3
00 000 0
⎤
⎥
⎥
⎦
;
basis forker()=
½∙
021
000
¸
∙
000
010
¸
∙
100
301
¸¾
;
basis forrange()={
3
2
};
dim(ker()) + dim(range())=3+3=6=dim( M
32)
(12) (a) Ifv∈ker(
1),then 1(v)=0 W.Thus,( 2◦1)(v)= 2(1(v)) = 2(0W)=0 X,andso
v∈ker(
2◦1). Therefore,ker( 1)⊆ker( 2◦1). Hence,dim(ker( 1))≤dim(ker( 2◦1)).
(b) Let
1be the projection onto the-axis ( 1([ ]) = [0])and 2be the projection onto the
-axis (
2([ ]) = [0]). A basis forker( 1)is{[01]},whileker( 2◦1)=R
2
.
(13) (a) By part (2) of Theorem 5.9 applied toand,
dim(ker())−dim(ker()) = (−rank(A))−(−rank(A
))
=(−)−(rank(A)−rank(A
)) =−
by Corollary 5.11.
(b) Sinceis onto,dim(range()) ==rank(A)(by part (1) of Theorem 5.9)=rank(A
)
(by Corollary 5.11)= dim(range()). Thus, by part (3) of Theorem 5.9,dim(ker()) +
dim(range()) =, implyingdim(ker()) +=,andsodim(ker()) = 0,andis
one-to-one (by part (1) of Theorem 5.12).
(c) Converse:one-to-one=⇒onto.Thisistrue.
one-to-one=⇒dim(ker()) = 0
=⇒dim(ker()) =−(by part (a))
=⇒(−)+dim(range()) =(by part (3) of Theorem 5.9)
=⇒dim(range()) =
=⇒is onto (by Theorem 4.13, and part (2) of Theorem 5.12).
(14) (a)is not one-to-one because(
3
−+1)=O 22. Corollary 5.13 then shows thatis not onto.
(b)ker()has{
3
−+1}as a basis, sodim(ker()) = 1.Thus,dim(range()) = 3by the Dimension
Theorem.
(15) In each part, letArepresent the given matrix forwith respect to the standard bases.
(a) The reduced row echelon form ofAisI
3. Therefore,ker()={0},andsodim(ker())=0and
is one-to-one. By Corollary 5.13,is also onto. Hencedim(range()) = 3.
(b) The reduced row echelon form ofAis
⎡
⎣
10 40
01−10
00 01
⎤
⎦.
dim(ker()) = 4−rank(A)=1.is not one-to-one.
dim(range()) = rank(A)=3.is onto.
(16) (a)dim(ker()) = dim(P
3)−dim(range())≥4−dim(R
3
)=1,socannot be one-to-one.
(b)dim(range()) = dim(P
2)−dim(ker())≤dim(P 2)=3dim(M 22),socannot be onto.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 114
(11) Matrix forwith respect to standard bases:
⎡
⎢
⎢
⎣
10 000 −1
01−200 0
00 010 −3
00 000 0
⎤
⎥
⎥
⎦
;
basis forker()=
½∙
021
000
¸
∙
000
010
¸
∙
100
301
¸¾
;
basis forrange()={
3
2
};
dim(ker()) + dim(range())=3+3=6=dim( M
32)
(12) (a) Ifv∈ker(
1),then 1(v)=0 W.Thus,( 2◦1)(v)= 2(1(v)) = 2(0W)=0 X,andso
v∈ker(
2◦1). Therefore,ker( 1)⊆ker( 2◦1). Hence,dim(ker( 1))≤dim(ker( 2◦1)).
(b) Let
1be the projection onto the-axis ( 1([ ]) = [0])and 2be the projection onto the
-axis (
2([ ]) = [0]). A basis forker( 1)is{[01]},whileker( 2◦1)=R
2
.
(13) (a) By part (2) of Theorem 5.9 applied toand,
dim(ker())−dim(ker()) = (−rank(A))−(−rank(A
))
=(−)−(rank(A)−rank(A
)) =−
by Corollary 5.11.
(b) Sinceis onto,dim(range()) ==rank(A)(by part (1) of Theorem 5.9)=rank(A
)
(by Corollary 5.11)= dim(range()). Thus, by part (3) of Theorem 5.9,dim(ker()) +
dim(range()) =, implyingdim(ker()) +=,andsodim(ker()) = 0,andis
one-to-one (by part (1) of Theorem 5.12).
(c) Converse:one-to-one=⇒onto.Thisistrue.
one-to-one=⇒dim(ker()) = 0
=⇒dim(ker()) =−(by part (a))
=⇒(−)+dim(range()) =(by part (3) of Theorem 5.9)
=⇒dim(range()) =
=⇒is onto (by Theorem 4.13, and part (2) of Theorem 5.12).
(14) (a)is not one-to-one because(
3
−+1)=O 22. Corollary 5.13 then shows thatis not onto.
(b)ker()has{
3
−+1}as a basis, sodim(ker()) = 1.Thus,dim(range()) = 3by the Dimension
Theorem.
(15) In each part, letArepresent the given matrix forwith respect to the standard bases.
(a) The reduced row echelon form ofAisI
3. Therefore,ker()={0},andsodim(ker())=0and
is one-to-one. By Corollary 5.13,is also onto. Hencedim(range()) = 3.
(b) The reduced row echelon form ofAis
⎡
⎣
10 40
01−10
00 01
⎤
⎦.
dim(ker()) = 4−rank(A)=1.is not one-to-one.
dim(range()) = rank(A)=3.is onto.
(16) (a)dim(ker()) = dim(P
3)−dim(range())≥4−dim(R
3
)=1,socannot be one-to-one.
(b)dim(range()) = dim(P
2)−dim(ker())≤dim(P 2)=3dim(M 22),socannot be onto.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 114
Answers to Exercises Chapter 5 Review
(17) (a)(v 1)=(v 2)=(v 2)=⇒v 1=v 2,becauseis one-to-one. In this case, the set
{(v
1)(v 2)}is linearly dependent, since(v 1)is a nonzero multiple of(v 2). Part (1) of
Theorem 5.14 then implies that{v
1v2}must be linearly dependent, which is correct, sincev 1is
a nonzero multiple ofv
2.
(b) Consider the contrapositive: Foronto andw∈W,if{v
1v2}spansV,then(v 1+v 2)=w
for some ∈R.
Proof of the contrapositive: Supposeis onto,w∈W,and{v
1v2}spansV. By part (2) of
Theorem 5.14,{(v
1)(v 2)}spansW,sothereexist ∈Rsuch that
w=(v
1)+(v 2)=(v 1+v 2)
(18) (a)
1and 2are isomorphisms (by Theorem 5.16) because their (given) matrices are nonsingular.
Theirinversesaregiveninpart(b).
(b) Matrix for
2◦1=
⎡
⎢
⎢
⎣
81 71 −15 18
107 77 −31 19
69 45 −23 11
−29−36−1−9
⎤
⎥
⎥
⎦
;
matrix for
−1
1
:
1
5
⎡
⎢
⎢
⎣
2−10 19 11
05 −10−5
3−15 26 14
−415 −23−17
⎤
⎥
⎥
⎦
;
matrix for
−1
2
:
1
2
⎡
⎢
⎢
⎣
−826 −30 2
10−35 41 −4
10−30 34 −2
−14 49 −57 6
⎤
⎥
⎥
⎦
(c) Both computations yield
1
10
⎡
⎢
⎢
⎣
−80 371 −451 72
20−120 150 −30
−110 509 −619 98
190−772 922 −124
⎤
⎥
⎥
⎦
(19) (a) The determinant of the shear matrix given in Table 5.1 is1, so this matrix is nonsingular. There-
fore, the given mapping is an isomorphism by Theorem 5.16.
(b) The inverse isomorphism is
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
10−
01−
00 1
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦=
⎡
⎣
1−3
2−3
3
⎤
⎦.This
represents a shear in the-direction with factor−.
(20) (a)(B
AB)
=B
A
(B
)
=B
AB,sinceAis symmetric.
(b) SupposeAC∈W.NotethatifB
AB=B
CB,thenA=CbecauseBis nonsingular. Thus
is one-to-one, and sois onto by Corollary 5.13.
An alternate approach is as follows: SupposeC∈W.IfA=(B
)
−1
CB
−1
then(A)=C
Thusis onto, and sois one-to-one by Corollary 5.13.
(21) (a)(
4
+
3
+
2
)=4
3
+(12+3)
2
+(6+2)+2. Clearlyker()={0}. Apply part
(1) of Theorem 5.12.
(b) No.dim(W)=36 =dim(P
3)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 115
(17) (a)(v 1)=(v 2)=(v 2)=⇒v 1=v 2,becauseis one-to-one. In this case, the set
{(v
1)(v 2)}is linearly dependent, since(v 1)is a nonzero multiple of(v 2). Part (1) of
Theorem 5.14 then implies that{v
1v2}must be linearly dependent, which is correct, sincev 1is
a nonzero multiple ofv
2.
(b) Consider the contrapositive: Foronto andw∈W,if{v
1v2}spansV,then(v 1+v 2)=w
for some ∈R.
Proof of the contrapositive: Supposeis onto,w∈W,and{v
1v2}spansV. By part (2) of
Theorem 5.14,{(v
1)(v 2)}spansW,sothereexist ∈Rsuch that
w=(v
1)+(v 2)=(v 1+v 2)
(18) (a)
1and 2are isomorphisms (by Theorem 5.16) because their (given) matrices are nonsingular.
Theirinversesaregiveninpart(b).
(b) Matrix for
2◦1=
⎡
⎢
⎢
⎣
81 71 −15 18
107 77 −31 19
69 45 −23 11
−29−36−1−9
⎤
⎥
⎥
⎦
;
matrix for
−1
1
:
1
5
⎡
⎢
⎢
⎣
2−10 19 11
05 −10−5
3−15 26 14
−415 −23−17
⎤
⎥
⎥
⎦
;
matrix for
−1
2
:
1
2
⎡
⎢
⎢
⎣
−826 −30 2
10−35 41 −4
10−30 34 −2
−14 49 −57 6
⎤
⎥
⎥
⎦
(c) Both computations yield
1
10
⎡
⎢
⎢
⎣
−80 371 −451 72
20−120 150 −30
−110 509 −619 98
190−772 922 −124
⎤
⎥
⎥
⎦
(19) (a) The determinant of the shear matrix given in Table 5.1 is1, so this matrix is nonsingular. There-
fore, the given mapping is an isomorphism by Theorem 5.16.
(b) The inverse isomorphism is
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
10−
01−
00 1
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦=
⎡
⎣
1−3
2−3
3
⎤
⎦.This
represents a shear in the-direction with factor−.
(20) (a)(B
AB)
=B
A
(B
)
=B
AB,sinceAis symmetric.
(b) SupposeAC∈W.NotethatifB
AB=B
CB,thenA=CbecauseBis nonsingular. Thus
is one-to-one, and sois onto by Corollary 5.13.
An alternate approach is as follows: SupposeC∈W.IfA=(B
)
−1
CB
−1
then(A)=C
Thusis onto, and sois one-to-one by Corollary 5.13.
(21) (a)(
4
+
3
+
2
)=4
3
+(12+3)
2
+(6+2)+2. Clearlyker()={0}. Apply part
(1) of Theorem 5.12.
(b) No.dim(W)=36 =dim(P
3)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 115
Answers to Exercises Chapter 5 Review
(c) The polynomial∈range()since it does not have the correct form for an image under(as
stated in part (a)).
(22) In parts (a), (b), and (c), letArepresent the given matrix.
(a) (i)
A()=
3
−3
2
−+3=(−1)(+1)(−3);
eigenvalues for:
1=1, 2=−1,and 3=3;
basis for
1:{[−134]};basisfor −1:{[−145]};basisfor 3:{[−62027]}
(ii) All algebraic and geometric multiplicities equal1;is diagonalizable.
(iii)={[−134][−145][−62027]};
D=
⎡
⎣
100
0−10
003
⎤
⎦;P=
⎡
⎣
−1−1−6
3420
4527
⎤
⎦(Note thatP
−1
=
⎡
⎣
−83 −4
13 −2
1−11
⎤
⎦).
(b) (i)
A()=
3
+5
2
+20+28=(+2)(
2
+3+ 14);
eigenvalue for:=−2since
2
+3+14has no real roots; basis for −2:{[011]}
(ii) For=−2: algebraic multiplicity=geometric multiplicity=1;is not diagonalizable.
(c) (i)
A()=
3
−5
2
+3+9=(+1)(−3)
2
;
eigenvalues for:
1=−1,and 2=3;
basis for
−1:{[133]};basisfor 3:{[150][3025]}
(ii) For
1=−1: algebraic multiplicity=geometric multiplicity=1;
For
2=3: algebraic multiplicity=geometric multiplicity=2;is diagonalizable.
(iii)={[133][150][3025]};
D=
⎡
⎣
−100
030
003
⎤
⎦;P=
⎡
⎣
11 3
35 0
3025
⎤
⎦(Note thatP
−1
=
1
5
⎡
⎣
125−25−15
−75 16 9
−15 3 2
⎤
⎦).
(d) Matrix forwith respect to standard coordinates:A=
⎡
⎢
⎢
⎣
1000
−3000
0−2−10
00 −1−2
⎤
⎥
⎥
⎦
.
(i)
A()=
4
+2
3
−
2
−2=(−1)(+1)(+2);
eigenvalues for:
1=1, 2=0, 3=−1,and 4=−2;
basis for
1:{−
3
+3
2
−3+1}({[−13−31]}in standard coordinates);
basis for
0:{
2
−2+1}({[01−21]}in standard coordinates);
basis for
−1:{−+1}({[00−11]}in standard coordinates);
basis for
−2:{1}({[0001]}in standard coordinates)
(ii) All algebraic and geometric multiplicities equal1;is diagonalizable.
(iii)={−
3
+3
2
−3+1
2
−2+1−+11}
({[−13−31][01−21][00−11][0001]}in standard coordinates);
D=
⎡
⎢
⎢
⎣
1000
0000
00−10
00 0 −2
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
−1000
3100
−3−2−10
1111
⎤
⎥
⎥
⎦
(Note thatP
−1
=P).
(23) Basis for
1:{[ ][ ]};basisfor −1:{[− − −]}.
basis of eigenvectors:{[ ][ ][− − −]};
Copyrightc°2016 Elsevier Ltd. All rights reserved. 116
(c) The polynomial∈range()since it does not have the correct form for an image under(as
stated in part (a)).
(22) In parts (a), (b), and (c), letArepresent the given matrix.
(a) (i)
A()=
3
−3
2
−+3=(−1)(+1)(−3);
eigenvalues for:
1=1, 2=−1,and 3=3;
basis for
1:{[−134]};basisfor −1:{[−145]};basisfor 3:{[−62027]}
(ii) All algebraic and geometric multiplicities equal1;is diagonalizable.
(iii)={[−134][−145][−62027]};
D=
⎡
⎣
100
0−10
003
⎤
⎦;P=
⎡
⎣
−1−1−6
3420
4527
⎤
⎦(Note thatP
−1
=
⎡
⎣
−83 −4
13 −2
1−11
⎤
⎦).
(b) (i)
A()=
3
+5
2
+20+28=(+2)(
2
+3+ 14);
eigenvalue for:=−2since
2
+3+14has no real roots; basis for −2:{[011]}
(ii) For=−2: algebraic multiplicity=geometric multiplicity=1;is not diagonalizable.
(c) (i)
A()=
3
−5
2
+3+9=(+1)(−3)
2
;
eigenvalues for:
1=−1,and 2=3;
basis for
−1:{[133]};basisfor 3:{[150][3025]}
(ii) For
1=−1: algebraic multiplicity=geometric multiplicity=1;
For
2=3: algebraic multiplicity=geometric multiplicity=2;is diagonalizable.
(iii)={[133][150][3025]};
D=
⎡
⎣
−100
030
003
⎤
⎦;P=
⎡
⎣
11 3
35 0
3025
⎤
⎦(Note thatP
−1
=
1
5
⎡
⎣
125−25−15
−75 16 9
−15 3 2
⎤
⎦).
(d) Matrix forwith respect to standard coordinates:A=
⎡
⎢
⎢
⎣
1000
−3000
0−2−10
00 −1−2
⎤
⎥
⎥
⎦
.
(i)
A()=
4
+2
3
−
2
−2=(−1)(+1)(+2);
eigenvalues for:
1=1, 2=0, 3=−1,and 4=−2;
basis for
1:{−
3
+3
2
−3+1}({[−13−31]}in standard coordinates);
basis for
0:{
2
−2+1}({[01−21]}in standard coordinates);
basis for
−1:{−+1}({[00−11]}in standard coordinates);
basis for
−2:{1}({[0001]}in standard coordinates)
(ii) All algebraic and geometric multiplicities equal1;is diagonalizable.
(iii)={−
3
+3
2
−3+1
2
−2+1−+11}
({[−13−31][01−21][00−11][0001]}in standard coordinates);
D=
⎡
⎢
⎢
⎣
1000
0000
00−10
00 0 −2
⎤
⎥
⎥
⎦
;P=
⎡
⎢
⎢
⎣
−1000
3100
−3−2−10
1111
⎤
⎥
⎥
⎦
(Note thatP
−1
=P).
(23) Basis for
1:{[ ][ ]};basisfor −1:{[− − −]}.
basis of eigenvectors:{[ ][ ][− − −]};
Copyrightc°2016 Elsevier Ltd. All rights reserved. 116
Answers to Exercises Chapter 5 Review
D=
⎡
⎣
10 0
01 0
00−1
⎤
⎦
(24)
A(A)=A
4
−4A
3
−18A
2
+ 108A−135I 4
=
⎡
⎢
⎢
⎣
649 216 −176−68
−568−135 176 68
1136 432 −271−136
−1088 0 544 625
⎤
⎥
⎥
⎦
−4
⎡
⎢
⎢
⎣
97 54 −819
−70−27 8 −19
140 108 11 38
304 0 −152−125
⎤
⎥
⎥
⎦
−18
⎡
⎢
⎢
⎣
37 12−8−2
−28−382
56 24−7−4
−32 01625
⎤
⎥
⎥
⎦
+ 108
⎡
⎢
⎢
⎣
52 0 1
−21 0 −1
44 3 2
16 0−8−5
⎤
⎥
⎥
⎦
−
⎡
⎢
⎢
⎣
135000
0 135 0 0
0 0 135 0
000135
⎤
⎥
⎥
⎦
=O
44
(25) (a) T
(b) F
(c) T
(d) T
(e) T
(f) F
(g) F
(h) T
(i) T
(j) F
(k) F
(l) T
(m) F
(n) T
(o) T
(p) F
(q) F
(r) T
(s) F
(t) T
(u) T
(v) F
(w) T
(x) T
(y) T
(z) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 117
D=
⎡
⎣
10 0
01 0
00−1
⎤
⎦
(24)
A(A)=A
4
−4A
3
−18A
2
+ 108A−135I 4
=
⎡
⎢
⎢
⎣
649 216 −176−68
−568−135 176 68
1136 432 −271−136
−1088 0 544 625
⎤
⎥
⎥
⎦
−4
⎡
⎢
⎢
⎣
97 54 −819
−70−27 8 −19
140 108 11 38
304 0 −152−125
⎤
⎥
⎥
⎦
−18
⎡
⎢
⎢
⎣
37 12−8−2
−28−382
56 24−7−4
−32 01625
⎤
⎥
⎥
⎦
+ 108
⎡
⎢
⎢
⎣
52 0 1
−21 0 −1
44 3 2
16 0−8−5
⎤
⎥
⎥
⎦
−
⎡
⎢
⎢
⎣
135000
0 135 0 0
0 0 135 0
000135
⎤
⎥
⎥
⎦
=O
44
(25) (a) T
(b) F
(c) T
(d) T
(e) T
(f) F
(g) F
(h) T
(i) T
(j) F
(k) F
(l) T
(m) F
(n) T
(o) T
(p) F
(q) F
(r) T
(s) F
(t) T
(u) T
(v) F
(w) T
(x) T
(y) T
(z) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 117
Answers to Exercises Section 6.1
Chapter 6
Section 6.1
(1) Orthogonal, not orthonormal: (a), (f);
Orthogonal and orthonormal: (b), (d), (e), (g);
Neither: (c)
(2) Orthogonal: (a), (d), (e);
Not orthogonal, columns not normalized: (b), (c)
(3) (a)[v]
=
h
2
√
3+3
2
3
√
3−2
2
i
(b)[v]
=[2−14] (c)[v]
=[3
13
√
3
3
5
√
6
3
4
√
2]
(4) (a){[5−12][5−3−14]}
(b){[2−131][−7−1013]}
(c){[210−1][−113−1][5−753]}
(d){[013−2][23−10][−3201]}
(e){[4−1−22][203−1][38−4−6]}
(5) (a){[22−3][13−46][032]}
(b){[1−43][254−3][034]}
(c){[1−31][2513][41−1]}
(d){[3
1−2][5−36][021]}
(e){[21−21][3−12−1][052−1]
[0012]}
(f){[210−3][0321][5−103]
[03−51]}
(6) Orthogonal basis forW={[−2
−14−21][4−30−21][−1−331538−19][333−2−7]}
(7) (a)[−133] (b)[331] (c)[511] (d)[432]
(8) (a)(
v)·( v)= (v·v)= 0=0,for6 =.
(b) No
(9) (a) Expressvandwas linear combinations of{u
1u }using Theorem 6.3. Then expand and
simplifyv·w.
(b) Letw=vin part (a).
(10) Follow the hint. Then use Exercise 9(b). Finally, drop some terms to get the inequality.
(11) (a)(A
−1
)
=(A
)
−1
=(A
−1
)
−1
,sinceAis orthogonal.
(b)AB(AB)
=AB(B
A
)=A(BB
)A
=AIA
=AA
=I. Hence(AB)
=(AB)
−1
.
(12)A=A
=⇒A
2
=AA
=⇒I =AA
=⇒A
=A
−1
.
Conversely,A
=A
−1
=⇒I =AA
=⇒A=A
2
A
=⇒A=I A
=⇒A=A
.
(13) SupposeAis both orthogonal and skew-symmetric. ThenA
2
=A(−A
)=−AA
=−I . Hence
|A|
2
=|−I |=(−1)
|I|(by Corollary 3.4)=−1,ifis odd, a contradiction.
(14)A(A+I
)
=A(A
+I)=AA
+A=I +A(sinceAis orthogonal). Hence|I +A|=
|A||(A+I
)
|=(−1)|A+I |,implying|A+I |=0.ThusA+I is singular.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 118
Chapter 6
Section 6.1
(1) Orthogonal, not orthonormal: (a), (f);
Orthogonal and orthonormal: (b), (d), (e), (g);
Neither: (c)
(2) Orthogonal: (a), (d), (e);
Not orthogonal, columns not normalized: (b), (c)
(3) (a)[v]
=
h
2
√
3+3
2
3
√
3−2
2
i
(b)[v]
=[2−14] (c)[v]
=[3
13
√
3
3
5
√
6
3
4
√
2]
(4) (a){[5−12][5−3−14]}
(b){[2−131][−7−1013]}
(c){[210−1][−113−1][5−753]}
(d){[013−2][23−10][−3201]}
(e){[4−1−22][203−1][38−4−6]}
(5) (a){[22−3][13−46][032]}
(b){[1−43][254−3][034]}
(c){[1−31][2513][41−1]}
(d){[3
1−2][5−36][021]}
(e){[21−21][3−12−1][052−1]
[0012]}
(f){[210−3][0321][5−103]
[03−51]}
(6) Orthogonal basis forW={[−2
−14−21][4−30−21][−1−331538−19][333−2−7]}
(7) (a)[−133] (b)[331] (c)[511] (d)[432]
(8) (a)(
v)·( v)= (v·v)= 0=0,for6 =.
(b) No
(9) (a) Expressvandwas linear combinations of{u
1u }using Theorem 6.3. Then expand and
simplifyv·w.
(b) Letw=vin part (a).
(10) Follow the hint. Then use Exercise 9(b). Finally, drop some terms to get the inequality.
(11) (a)(A
−1
)
=(A
)
−1
=(A
−1
)
−1
,sinceAis orthogonal.
(b)AB(AB)
=AB(B
A
)=A(BB
)A
=AIA
=AA
=I. Hence(AB)
=(AB)
−1
.
(12)A=A
=⇒A
2
=AA
=⇒I =AA
=⇒A
=A
−1
.
Conversely,A
=A
−1
=⇒I =AA
=⇒A=A
2
A
=⇒A=I A
=⇒A=A
.
(13) SupposeAis both orthogonal and skew-symmetric. ThenA
2
=A(−A
)=−AA
=−I . Hence
|A|
2
=|−I |=(−1)
|I|(by Corollary 3.4)=−1,ifis odd, a contradiction.
(14)A(A+I
)
=A(A
+I)=AA
+A=I +A(sinceAis orthogonal). Hence|I +A|=
|A||(A+I
)
|=(−1)|A+I |,implying|A+I |=0.ThusA+I is singular.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 118
Answers to Exercises Section 6.1
(15) Use part (2) of Theorem 6.7. To be a unit vector, thefirst column must be[±100]. To be a unit
vector orthogonal to thefirst column, the second column must be[0±10],etc.
(16) (a) By Theorem 6.5, we can expand the orthonormal set{u}to an orthonormal basis forR
.Form
the matrix using these basis vectors as rows. Then use part (1) of Theorem 6.7.
(b)
⎡
⎢
⎢
⎣
√
6
6
√
6
3
√
6
6
√
30
6
−
√
30
15
−
√
30
30
0
√
5
5
−
2
√
5
5
⎤
⎥
⎥
⎦
is one possible answer.
(17) Proof of the other half of part (1) of Theorem 6.7:( )entry ofAA
=(th row ofA)·(th row ofA)
=
½
1=
06 =
(since the rows ofAform an orthonormal set). HenceAA
=I,andAis orthogonal.
Proof of part (2):Ais orthogonal iffA
is orthogonal (by part (2) of Theorem 6.6) iffthe rows of
A
form an orthonormal basis forR
(by part (1) of Theorem 6.7) iffthe columns ofAform an
orthonormal basis forR
.
(18) (a)kvk
2
=v·v=Av·Av(by Theorem 6.9)=kAvk
2
. Then take square roots.
(b) Using part (a) and Theorem 6.9, we have
v·w
(kvk)(kwk)
=
(Av)·(Aw)
(kAvk)(kAwk)
and so the cosines of the appropriate angles are equal.
(19) A detailed proof of this can be found in the beginning of the proof of Theorem 6.19 in Appendix A.
(That portion of the proof uses no results beyond Section 6.1.)
(20) (a) LetQ
1andQ 2be the matrices whose columns are the vectors inand, respectively. Then
Q
1is orthogonal by part (2) of Theorem 6.7. Also,Q 1(respectively,Q 2) is the transition matrix
from(respectively,) to standard coordinates. By Theorem 4.19,Q
−1
2
is the transition matrix
from standard coordinates to. Theorem 4.18 then impliesP=Q
−1
2
Q1Hence,Q 2=Q 1P
−1
By parts (2) and (3) of Theorem 6.6,Q
2is orthogonal, since bothPandQ 1are orthogonal. Thus
is an orthonormal basis by part (2) of Theorem 6.7.
(b) Let=(b
1b )where=dim(V). Then theth column of the transition matrix from
tois[b
].Now,[b ]·[b]=b·b(by Exercise 19, sinceis an orthonormal basis). This
equals0if6 =and equals1if=sinceis orthonormal. Hence, the columns of the transition
matrix form an orthonormal set ofvectors inR
and so this matrix is orthogonal by part (2)
of Theorem 6.7.
(c) Let=(c
1c )where=dim(V). Thenc ·c=[c]·[c](by Exercise 19 sinceis
orthonormal)=(P[c
])·(P[c ])(by Theorem 6.9 sincePis orthogonal) =[c ]·[c](since
Pis the transition matrix fromto)=e
·ewhich equals0if6 =and equals1if=
Henceis orthonormal.
(21) First note thatA
Ais an×matrix. If6 =,the( )entry ofA
Aequals(th row ofA
)·(th
column ofA)=(th column ofA)·(th column ofA)=0, because the columns ofAare orthogonal
to each other. Thus,A
Ais a diagonal matrix. Finally, the( )entry ofA
Aequals(th row of
A
)·(th column ofA)=(th column ofA)·(th column ofA)=1,becausetheth column ofAis
aunitvector.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 119
(15) Use part (2) of Theorem 6.7. To be a unit vector, thefirst column must be[±100]. To be a unit
vector orthogonal to thefirst column, the second column must be[0±10],etc.
(16) (a) By Theorem 6.5, we can expand the orthonormal set{u}to an orthonormal basis forR
.Form
the matrix using these basis vectors as rows. Then use part (1) of Theorem 6.7.
(b)
⎡
⎢
⎢
⎣
√
6
6
√
6
3
√
6
6
√
30
6
−
√
30
15
−
√
30
30
0
√
5
5
−
2
√
5
5
⎤
⎥
⎥
⎦
is one possible answer.
(17) Proof of the other half of part (1) of Theorem 6.7:( )entry ofAA
=(th row ofA)·(th row ofA)
=
½
1=
06 =
(since the rows ofAform an orthonormal set). HenceAA
=I,andAis orthogonal.
Proof of part (2):Ais orthogonal iffA
is orthogonal (by part (2) of Theorem 6.6) iffthe rows of
A
form an orthonormal basis forR
(by part (1) of Theorem 6.7) iffthe columns ofAform an
orthonormal basis forR
.
(18) (a)kvk
2
=v·v=Av·Av(by Theorem 6.9)=kAvk
2
. Then take square roots.
(b) Using part (a) and Theorem 6.9, we have
v·w
(kvk)(kwk)
=
(Av)·(Aw)
(kAvk)(kAwk)
and so the cosines of the appropriate angles are equal.
(19) A detailed proof of this can be found in the beginning of the proof of Theorem 6.19 in Appendix A.
(That portion of the proof uses no results beyond Section 6.1.)
(20) (a) LetQ
1andQ 2be the matrices whose columns are the vectors inand, respectively. Then
Q
1is orthogonal by part (2) of Theorem 6.7. Also,Q 1(respectively,Q 2) is the transition matrix
from(respectively,) to standard coordinates. By Theorem 4.19,Q
−1
2
is the transition matrix
from standard coordinates to. Theorem 4.18 then impliesP=Q
−1
2
Q1Hence,Q 2=Q 1P
−1
By parts (2) and (3) of Theorem 6.6,Q
2is orthogonal, since bothPandQ 1are orthogonal. Thus
is an orthonormal basis by part (2) of Theorem 6.7.
(b) Let=(b
1b )where=dim(V). Then theth column of the transition matrix from
tois[b
].Now,[b ]·[b]=b·b(by Exercise 19, sinceis an orthonormal basis). This
equals0if6 =and equals1if=sinceis orthonormal. Hence, the columns of the transition
matrix form an orthonormal set ofvectors inR
and so this matrix is orthogonal by part (2)
of Theorem 6.7.
(c) Let=(c
1c )where=dim(V). Thenc ·c=[c]·[c](by Exercise 19 sinceis
orthonormal)=(P[c
])·(P[c ])(by Theorem 6.9 sincePis orthogonal) =[c ]·[c](since
Pis the transition matrix fromto)=e
·ewhich equals0if6 =and equals1if=
Henceis orthonormal.
(21) First note thatA
Ais an×matrix. If6 =,the( )entry ofA
Aequals(th row ofA
)·(th
column ofA)=(th column ofA)·(th column ofA)=0, because the columns ofAare orthogonal
to each other. Thus,A
Ais a diagonal matrix. Finally, the( )entry ofA
Aequals(th row of
A
)·(th column ofA)=(th column ofA)·(th column ofA)=1,becausetheth column ofAis
aunitvector.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 119
Answers to Exercises Section 6.2
(22) (a)F (b)T (c)T (d)F (e)T (f)T (g)T (h)F (i)T (j)T
Section 6.2
(1) (a)W
⊥
=span({[23]})
(b)W
⊥
=span({[52−1][012]})
(c)W
⊥
=span({[237]})
(d)W
⊥
=span({[3−14]})
(e)W
⊥
=span({[−25−1]})
(f)W
⊥
=span({[71−2−3][04−12]})
(g)W
⊥
=span({[3−241]})
(2) (a)w
1=proj
Wv=[−
33
35
111
35
12
7
];w2=[−
2
35
−
6
35
2
7
]
(b)w
1=proj
Wv=[−
17
9
−
10
9
14
9
];w2=[
26
9
−
26
9
13
9
]
(c)w
1=proj
Wv=[
1
7
−
3
7
−
2
7
];w2=[
13
7
17
7
−
19
7
]
(d)w
1=proj
Wv=[−
54
35
4
35
42
35
92
35
];w2=[
19
35
101
35
63
35
−
22
35
]
(3) Use the orthonormal basis{ij}forW.
(4) (a)
3
√
129
43
(b)
√
3806
22
(c)
2
√
247
13
(d)
8
√
17
17
(5) (a) Orthogonal projection onto3++=0
(b) Orthogonal reflection through−2−2=0
(c) Neither.
()=(−1)(+1)
2
. 1=span{[−795]}, −1=span{[−720][1102]}.
Eigenspaces have wrong dimensions.
(d) Neither.
()=(−1)
2
(+1). 1=span{[−140][−704]}, −1=span{[213]}.
Eigenspaces are not orthogonal.
(6)
1
9
⎡
⎣
52−4
28 2
−42 5
⎤
⎦
(7)
1
7
⎡
⎣
−23−6
36 2
−62 3
⎤
⎦
(8) (a)
⎡
⎢
⎣
1
3
−
1
3
−
1
3
−
1
3
5
6
−
1
6
−
1
3
−
1
6
5
6
⎤
⎥
⎦
(b) This is straightforward.
(9) (a) Characteristic polynomial=
3
−2
2
+
(b) Characteristic polynomial=
3
−
2
(c) Characteristic polynomial=
3
−
2
−+1
Copyrightc°2016 Elsevier Ltd. All rights reserved. 120
(22) (a)F (b)T (c)T (d)F (e)T (f)T (g)T (h)F (i)T (j)T
Section 6.2
(1) (a)W
⊥
=span({[23]})
(b)W
⊥
=span({[52−1][012]})
(c)W
⊥
=span({[237]})
(d)W
⊥
=span({[3−14]})
(e)W
⊥
=span({[−25−1]})
(f)W
⊥
=span({[71−2−3][04−12]})
(g)W
⊥
=span({[3−241]})
(2) (a)w
1=proj
Wv=[−
33
35
111
35
12
7
];w2=[−
2
35
−
6
35
2
7
]
(b)w
1=proj
Wv=[−
17
9
−
10
9
14
9
];w2=[
26
9
−
26
9
13
9
]
(c)w
1=proj
Wv=[
1
7
−
3
7
−
2
7
];w2=[
13
7
17
7
−
19
7
]
(d)w
1=proj
Wv=[−
54
35
4
35
42
35
92
35
];w2=[
19
35
101
35
63
35
−
22
35
]
(3) Use the orthonormal basis{ij}forW.
(4) (a)
3
√
129
43
(b)
√
3806
22
(c)
2
√
247
13
(d)
8
√
17
17
(5) (a) Orthogonal projection onto3++=0
(b) Orthogonal reflection through−2−2=0
(c) Neither.
()=(−1)(+1)
2
. 1=span{[−795]}, −1=span{[−720][1102]}.
Eigenspaces have wrong dimensions.
(d) Neither.
()=(−1)
2
(+1). 1=span{[−140][−704]}, −1=span{[213]}.
Eigenspaces are not orthogonal.
(6)
1
9
⎡
⎣
52−4
28 2
−42 5
⎤
⎦
(7)
1
7
⎡
⎣
−23−6
36 2
−62 3
⎤
⎦
(8) (a)
⎡
⎢
⎣
1
3
−
1
3
−
1
3
−
1
3
5
6
−
1
6
−
1
3
−
1
6
5
6
⎤
⎥
⎦
(b) This is straightforward.
(9) (a) Characteristic polynomial=
3
−2
2
+
(b) Characteristic polynomial=
3
−
2
(c) Characteristic polynomial=
3
−
2
−+1
Copyrightc°2016 Elsevier Ltd. All rights reserved. 120
Answers to Exercises Section 6.2
(10) (a)
1
59
⎡
⎣
50−21−3
−21 10 −7
−3−758
⎤
⎦
(b)
1
17
⎡
⎣
86 −6
613 4
−6413
⎤
⎦
(c)
1
9
⎡
⎢
⎢
⎣
22 3 −1
2802
30 6 −3
−12−32
⎤
⎥
⎥
⎦
(d)
1
15
⎡
⎢
⎢
⎣
93 −3−6
31 −1−2
−3−112
−6−224
⎤
⎥
⎥
⎦
(11)W
⊥
1
=W
⊥
2
=⇒(W
⊥
1
)
⊥
=(W
⊥
2
)
⊥
=⇒W 1=W 2(by Corollary 6.14).
(12) Letv∈W
⊥
2
and letw∈W 1.Thenw∈W 2,andsov·w=0, which impliesv∈W
⊥
1
.
(13) Both parts rely on the uniqueness assertion in Corollary 6.16, and the equationv=v+0.
(14)v∈W
⊥
=⇒proj
Wv=0(see Exercise 13)=⇒minimum distance betweenandW=kvk(by
Theorem 6.18).
Conversely, supposekvkis the minimum distance fromtoW.Letw
1=proj
Wvand
w
2=v−proj
Wv.Thenkvk=kw 2k, by Theorem 6.18. Hence
kvk
2
=kw 1+w2k
2
=(w 1+w2)·(w 1+w2)
=w
1·w1+2w 1·w2+w2·w2
=kw 1k
2
+kw 2k
2
(sincew 1⊥w2)
Subtractingkvk
2
=kw 2k
2
from both sides yields0=kw 1k
2
, implyingw 1=0. Hencev=w 2∈W
⊥
.
(15) LetA∈V,B∈W.Then
“A·B”=
X
=1
X
=1
=
X
=2
−1
X
=1
+
X
=1
+
−1
X
=1
X
=+1
=
X
=2
−1
X
=1
+0+
−1
X
=1
X
=+1
(−)(since =0)
=
X
=2
−1
X
=1
−
X
=2
−1
X
=1
=0
Now use Exercise 14(b) in Section 4.6 and follow the hint in the text.
(16) Letu=
a
kak
.Then{u}is an orthonormal basis forW. Henceproj
Wb=(b·u)u=(b·
a
kak
)
a
kak
=
(
b·a
kak
2)a=proj
ab, according to Section 1.2.
(17) Letbe a spanning set forW. Clearly, ifv∈W
⊥
andu∈thenv·u=0.
Conversely, supposev·u=0for allu∈.Letw∈W.Thenw=
1u1+···+ ufor some
u
1u ∈. Hence,v·w= 1(v·u 1)+···+ (v·u )=0+···+0=0.Thusv∈W
⊥
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 121
(10) (a)
1
59
⎡
⎣
50−21−3
−21 10 −7
−3−758
⎤
⎦
(b)
1
17
⎡
⎣
86 −6
613 4
−6413
⎤
⎦
(c)
1
9
⎡
⎢
⎢
⎣
22 3 −1
2802
30 6 −3
−12−32
⎤
⎥
⎥
⎦
(d)
1
15
⎡
⎢
⎢
⎣
93 −3−6
31 −1−2
−3−112
−6−224
⎤
⎥
⎥
⎦
(11)W
⊥
1
=W
⊥
2
=⇒(W
⊥
1
)
⊥
=(W
⊥
2
)
⊥
=⇒W 1=W 2(by Corollary 6.14).
(12) Letv∈W
⊥
2
and letw∈W 1.Thenw∈W 2,andsov·w=0, which impliesv∈W
⊥
1
.
(13) Both parts rely on the uniqueness assertion in Corollary 6.16, and the equationv=v+0.
(14)v∈W
⊥
=⇒proj
Wv=0(see Exercise 13)=⇒minimum distance betweenandW=kvk(by
Theorem 6.18).
Conversely, supposekvkis the minimum distance fromtoW.Letw
1=proj
Wvand
w
2=v−proj
Wv.Thenkvk=kw 2k, by Theorem 6.18. Hence
kvk
2
=kw 1+w2k
2
=(w 1+w2)·(w 1+w2)
=w
1·w1+2w 1·w2+w2·w2
=kw 1k
2
+kw 2k
2
(sincew 1⊥w2)
Subtractingkvk
2
=kw 2k
2
from both sides yields0=kw 1k
2
, implyingw 1=0. Hencev=w 2∈W
⊥
.
(15) LetA∈V,B∈W.Then
“A·B”=
X
=1
X
=1
=
X
=2
−1
X
=1
+
X
=1
+
−1
X
=1
X
=+1
=
X
=2
−1
X
=1
+0+
−1
X
=1
X
=+1
(−)(since =0)
=
X
=2
−1
X
=1
−
X
=2
−1
X
=1
=0
Now use Exercise 14(b) in Section 4.6 and follow the hint in the text.
(16) Letu=
a
kak
.Then{u}is an orthonormal basis forW. Henceproj
Wb=(b·u)u=(b·
a
kak
)
a
kak
=
(
b·a
kak
2)a=proj
ab, according to Section 1.2.
(17) Letbe a spanning set forW. Clearly, ifv∈W
⊥
andu∈thenv·u=0.
Conversely, supposev·u=0for allu∈.Letw∈W.Thenw=
1u1+···+ ufor some
u
1u ∈. Hence,v·w= 1(v·u 1)+···+ (v·u )=0+···+0=0.Thusv∈W
⊥
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 121
Answers to Exercises Section 6.2
(18) First, showW⊆(W
⊥
)
⊥
.Letw∈W. We need to show thatw∈(W
⊥
)
⊥
. That is, we must prove
thatw⊥vfor allv∈W
⊥
. But, by the definition ofW
⊥
,w·v=0,sincew∈W,whichcompletes
this half of the proof.
Next, proveW=(W
⊥
)
⊥
. By Corollary 6.13,dim(W)=−dim(W
⊥
)=−(−dim((W
⊥
)
⊥
)) =
dim((W
⊥
)
⊥
). Thus, by Theorem 4.13,W=(W
⊥
)
⊥
.
(19) The proof thatis a linear operator is straightforward.
Suppose{u
1u }is an orthonormal basis forW.
First, we prove thatrange()=W.Notethatifw∈W,thenw=w+0,wherew∈Wand
0∈W
⊥
. Hence, by the uniqueness statement in Corollary 6.16,(w)=proj
Ww=w. Therefore,
everywinWis in the range of. Similarly, Corollary 6.16 implies thatproj
Wv∈Wfor every
v∈R
. Hence,range()=W.
To showW
⊥
=ker(),wefirst prove thatW
⊥
⊆ker().Ifv∈W
⊥
,thensincev·w=0for every
w∈W,(v)=proj
W(v)=(v·u 1)u1+···+(v·u )u=0u 1+···+0u =0. Hence,v∈ker().
Finally, by the Dimension Theorem, we have
dim(ker()) =−dim(range()) =−dim(W)=dim(W
⊥
)
SinceW
⊥
⊆ker(), Theorem 4.13 implies thatW
⊥
=ker().
(20) Since for eachv∈ker()wehaveAv=0, each row ofAis orthogonal tov. Hence,
ker()⊆(row space ofA)
⊥
.Also,dim(ker())=−rank(A) (by part (2) of Theorem 5.9)
=−dim(row space ofA)=dim((row space ofA)
⊥
) (by Corollary 6.13). Then apply Theorem 4.13.
(21) Supposev∈(ker())
⊥
and(v)=0.Then(v)=0,sov∈ker(). Apply Theorem 6.11 to show
ker()={0}.
(22) First,a∈W
⊥
by Corollary 6.16. Also, sinceproj
Wv∈Wandw∈W(sinceis inW), we have
b=(proj
Wv)−w∈W. Finally,
kv−wk
2
=kv−proj
Wv+(proj
Wv)−wk
2
=ka+bk
2
=(a+b)·(a+b)
=kak
2
+kbk
2
(sincea·b=0)
≥kak
2
=kv−proj
Wvk
2
(23) (a) Mimic the proof of Theorem 6.11.
(b) Mimic the proofs of Theorem 6.12 and Corollary 6.13.
(24) Note that, for anyv∈R
,ifweconsidervto be an×1matrix, thenv
vis the1×1matrix whose
single entry isv·v.Thus,wecanequatev·vwithv
v.
(a)v·(Aw)=v
(Aw)=(v
A)w=(A
v)
w=(A
v)·w
(b) Letv∈ker(
2). We must show thatvis orthogonal to every vector inrange( 1). An arbitrary
element ofrange(
1)is of the form 1(w),forsomew∈R
.Now,
v·
1(w)= 2(v)·w (by part (a))
=0·w (becausev∈ker(
2))
=0
Hence,v∈(range(
1))
⊥
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 122
(18) First, showW⊆(W
⊥
)
⊥
.Letw∈W. We need to show thatw∈(W
⊥
)
⊥
. That is, we must prove
thatw⊥vfor allv∈W
⊥
. But, by the definition ofW
⊥
,w·v=0,sincew∈W,whichcompletes
this half of the proof.
Next, proveW=(W
⊥
)
⊥
. By Corollary 6.13,dim(W)=−dim(W
⊥
)=−(−dim((W
⊥
)
⊥
)) =
dim((W
⊥
)
⊥
). Thus, by Theorem 4.13,W=(W
⊥
)
⊥
.
(19) The proof thatis a linear operator is straightforward.
Suppose{u
1u }is an orthonormal basis forW.
First, we prove thatrange()=W.Notethatifw∈W,thenw=w+0,wherew∈Wand
0∈W
⊥
. Hence, by the uniqueness statement in Corollary 6.16,(w)=proj
Ww=w. Therefore,
everywinWis in the range of. Similarly, Corollary 6.16 implies thatproj
Wv∈Wfor every
v∈R
. Hence,range()=W.
To showW
⊥
=ker(),wefirst prove thatW
⊥
⊆ker().Ifv∈W
⊥
,thensincev·w=0for every
w∈W,(v)=proj
W(v)=(v·u 1)u1+···+(v·u )u=0u 1+···+0u =0. Hence,v∈ker().
Finally, by the Dimension Theorem, we have
dim(ker()) =−dim(range()) =−dim(W)=dim(W
⊥
)
SinceW
⊥
⊆ker(), Theorem 4.13 implies thatW
⊥
=ker().
(20) Since for eachv∈ker()wehaveAv=0, each row ofAis orthogonal tov. Hence,
ker()⊆(row space ofA)
⊥
.Also,dim(ker())=−rank(A) (by part (2) of Theorem 5.9)
=−dim(row space ofA)=dim((row space ofA)
⊥
) (by Corollary 6.13). Then apply Theorem 4.13.
(21) Supposev∈(ker())
⊥
and(v)=0.Then(v)=0,sov∈ker(). Apply Theorem 6.11 to show
ker()={0}.
(22) First,a∈W
⊥
by Corollary 6.16. Also, sinceproj
Wv∈Wandw∈W(sinceis inW), we have
b=(proj
Wv)−w∈W. Finally,
kv−wk
2
=kv−proj
Wv+(proj
Wv)−wk
2
=ka+bk
2
=(a+b)·(a+b)
=kak
2
+kbk
2
(sincea·b=0)
≥kak
2
=kv−proj
Wvk
2
(23) (a) Mimic the proof of Theorem 6.11.
(b) Mimic the proofs of Theorem 6.12 and Corollary 6.13.
(24) Note that, for anyv∈R
,ifweconsidervto be an×1matrix, thenv
vis the1×1matrix whose
single entry isv·v.Thus,wecanequatev·vwithv
v.
(a)v·(Aw)=v
(Aw)=(v
A)w=(A
v)
w=(A
v)·w
(b) Letv∈ker(
2). We must show thatvis orthogonal to every vector inrange( 1). An arbitrary
element ofrange(
1)is of the form 1(w),forsomew∈R
.Now,
v·
1(w)= 2(v)·w (by part (a))
=0·w (becausev∈ker(
2))
=0
Hence,v∈(range(
1))
⊥
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 122
Answers to Exercises Section 6.3
(c) By Theorem 5.9,dim(ker( 2)) =−rank(A
)=−rank(A)(by Corollary 5.11). Also,
dim(range(
1)
⊥
)=−dim(range( 1))(by Corollary 6.13)=−rank(A)(by Theorem 5.9).
Hence,dim(ker(
2)) = dim(range( 1)
⊥
). This, together with part (b) and Theorem 4.13 com-
pletes the proof.
(d) The row space ofA=the column space ofA
= range( 2). Switching the roles ofAandA
in
parts (a), (b) and (c), we see that(range(
2))
⊥
=ker( 1). Taking the orthogonal complement
of both sides of this equality and using Corollary 6.14 yieldsrange(
2)=(ker( 1))
⊥
,which
completes the proof.
(25) (a) T
(b) T
(c) F
(d) F
(e) T
(f) F
(g) T
(h) T
(i) T
(j) F
(k) T
(l) T
Section 6.3
(1) Parts (a) and (g) are symmetric, because the matrix foris symmetric.
Part (b) is not symmetric, because the matrix forwith respect to the standard basis is not symmetric.
Parts (c), (d), and (f) are symmetric, sinceis orthogonally diagonalizable.
Part (e) is not symmetric, sinceis not diagonalizable, and hence not orthogonally diagonalizable.
(2) (a)
1
25
∙
−724
24 7
¸
(b)
1
11
⎡
⎣
11−66
−6180
604
⎤
⎦
(c)
1
49
⎡
⎣
58−6−18
−653 12
−18 12 85
⎤
⎦
(d)
1
169
⎡
⎢
⎢
⎣
−119−72−96 0
−72 119 0 96
−96 0 119 −72
096 −72−119
⎤
⎥
⎥
⎦
(3) (a)=(
1
13
[512]
1
13
[−125]),P=
1
13
∙
5−12
12 5
¸
,D=
∙
00
0 169
¸
(b)=(
1
5
[43]
1
5
[3−4]),P=
1
5
∙
43
3−4
¸
,D=
∙
30
0−1
¸
(c)=
³
1
√
2
[−110]
1
3
√
2
[114]
1
3
[−2−21]
´
(other bases are possible, since
1is two-dimensional),
P=
⎡
⎢
⎢
⎣
−
1
√
2
1
3
√
2
−
2
3
1
√
2
1
3
√
2
−
2
3
0
4
3
√
2
1
3
⎤
⎥
⎥
⎦
,D=
⎡
⎣
100
010
003
⎤
⎦.
Another possibility is
=(
1
3
[−122]
1
3
[2−12]
1
3
[22−1]),P=
1
3
⎡
⎣
−12 −2
2−1−2
221
⎤
⎦,D=
⎡
⎣
100
010
003
⎤
⎦
(d)=(
1
9
[−814]
1
9
[447]
1
9
[1−84]),
P=
1
9
⎡
⎣
−84 1
14−8
47 4
⎤
⎦,D=
⎡
⎣
000
0−30
009
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 123
(c) By Theorem 5.9,dim(ker( 2)) =−rank(A
)=−rank(A)(by Corollary 5.11). Also,
dim(range(
1)
⊥
)=−dim(range( 1))(by Corollary 6.13)=−rank(A)(by Theorem 5.9).
Hence,dim(ker(
2)) = dim(range( 1)
⊥
). This, together with part (b) and Theorem 4.13 com-
pletes the proof.
(d) The row space ofA=the column space ofA
= range( 2). Switching the roles ofAandA
in
parts (a), (b) and (c), we see that(range(
2))
⊥
=ker( 1). Taking the orthogonal complement
of both sides of this equality and using Corollary 6.14 yieldsrange(
2)=(ker( 1))
⊥
,which
completes the proof.
(25) (a) T
(b) T
(c) F
(d) F
(e) T
(f) F
(g) T
(h) T
(i) T
(j) F
(k) T
(l) T
Section 6.3
(1) Parts (a) and (g) are symmetric, because the matrix foris symmetric.
Part (b) is not symmetric, because the matrix forwith respect to the standard basis is not symmetric.
Parts (c), (d), and (f) are symmetric, sinceis orthogonally diagonalizable.
Part (e) is not symmetric, sinceis not diagonalizable, and hence not orthogonally diagonalizable.
(2) (a)
1
25
∙
−724
24 7
¸
(b)
1
11
⎡
⎣
11−66
−6180
604
⎤
⎦
(c)
1
49
⎡
⎣
58−6−18
−653 12
−18 12 85
⎤
⎦
(d)
1
169
⎡
⎢
⎢
⎣
−119−72−96 0
−72 119 0 96
−96 0 119 −72
096 −72−119
⎤
⎥
⎥
⎦
(3) (a)=(
1
13
[512]
1
13
[−125]),P=
1
13
∙
5−12
12 5
¸
,D=
∙
00
0 169
¸
(b)=(
1
5
[43]
1
5
[3−4]),P=
1
5
∙
43
3−4
¸
,D=
∙
30
0−1
¸
(c)=
³
1
√
2
[−110]
1
3
√
2
[114]
1
3
[−2−21]
´
(other bases are possible, since
1is two-dimensional),
P=
⎡
⎢
⎢
⎣
−
1
√
2
1
3
√
2
−
2
3
1
√
2
1
3
√
2
−
2
3
0
4
3
√
2
1
3
⎤
⎥
⎥
⎦
,D=
⎡
⎣
100
010
003
⎤
⎦.
Another possibility is
=(
1
3
[−122]
1
3
[2−12]
1
3
[22−1]),P=
1
3
⎡
⎣
−12 −2
2−1−2
221
⎤
⎦,D=
⎡
⎣
100
010
003
⎤
⎦
(d)=(
1
9
[−814]
1
9
[447]
1
9
[1−84]),
P=
1
9
⎡
⎣
−84 1
14−8
47 4
⎤
⎦,D=
⎡
⎣
000
0−30
009
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 123
Answers to Exercises Section 6.3
(e)=
³
1
√
14
[3210]
1
√
14
[−2301]
1
√
14
[10−32]
1
√
14
[0−123]
´
,
P=
1
√
14
⎡
⎢
⎢
⎣
3−210
230 −1
10 −32
0123
⎤
⎥
⎥
⎦
,D=
⎡
⎢
⎢
⎣
20 00
02 00
00−30
00 05
⎤
⎥
⎥
⎦
(f)=(
1
√
26
[413]
1
5
[30−4]
1
5
√
26
[4−253])
(other bases are possible, since
−13is two-dimensional),
P=
⎡
⎢
⎢
⎢
⎣
4
√
26
3
5
4
5
√
26
1
√
26
0−
25
5
√
26
3
√
26
−
4
5
3
5
√
26
⎤
⎥
⎥
⎥
⎦
,D=
⎡
⎣
1300
0−13 0
00 −13
⎤
⎦
(g)=
³
1
√
5
[120]
1
√
6
[−211]
1
√
30
[2−15]
´
,
P=
⎡
⎢
⎢
⎢
⎣
1
√
5
−
2
√
6
2
√
30
2
√
5
1
√
6
−
1
√
30
0
1
√
6
5
√
30
⎤
⎥
⎥
⎥
⎦
,D=
⎡
⎣
15 0 0
015 0
00 −15
⎤
⎦
(4) (a)=(
1
19
[−10156]
1
19
[15610]),=(
1
19
√
5
[202726]
1
19
√
5
[35−24−2]),
A=
∙
−22
21
¸
,P=
1
√
5
∙
1−2
21
¸
,D=
∙
20
0−3
¸
(b)=(
1
2
[1−111]
1
2
[−1111]
1
2
[111−1]),
=(
1
2
√
5
[3−131]
1
√
30
[−2431]
1
6
[−1−102])
A=
⎡
⎣
2−12
−122
22 −1
⎤
⎦,P=
⎡
⎢
⎢
⎢
⎣
2
√
5
−
1
√
30
1
√
6
0
5
√
30
1
√
6
1
√
5
2
√
30
−
2
√
6
⎤
⎥
⎥
⎥
⎦
,D=
⎡
⎣
30 0
03 0
00−3
⎤
⎦
(5) (a)
1
25
∙
23−36
−36 2
¸
(b)
1
17
∙
353−120
−120 514
¸
(c)
1
3
⎡
⎣
11 4 −4
417 −8
−4−817
⎤
⎦
(6) For example, the matrixAin Example 7 of Section 5.6 is diagonalizable but not symmetric and hence
not orthogonally diagonalizable.
(7)
1
2
∙
++
p
(−)
2
+4
2
0
0 +−
p
(−)
2
+4
2
¸
(Note: You only need to compute the eigen-
values. You do not need the corresponding eigenvectors.)
(8) (a) Sinceis diagonalizable and=1is the only eigenvalue,
1=R
. Hence, for everyv∈R
,
(v)=1v=v. Therefore,is the identity operator.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 124
(e)=
³
1
√
14
[3210]
1
√
14
[−2301]
1
√
14
[10−32]
1
√
14
[0−123]
´
,
P=
1
√
14
⎡
⎢
⎢
⎣
3−210
230 −1
10 −32
0123
⎤
⎥
⎥
⎦
,D=
⎡
⎢
⎢
⎣
20 00
02 00
00−30
00 05
⎤
⎥
⎥
⎦
(f)=(
1
√
26
[413]
1
5
[30−4]
1
5
√
26
[4−253])
(other bases are possible, since
−13is two-dimensional),
P=
⎡
⎢
⎢
⎢
⎣
4
√
26
3
5
4
5
√
26
1
√
26
0−
25
5
√
26
3
√
26
−
4
5
3
5
√
26
⎤
⎥
⎥
⎥
⎦
,D=
⎡
⎣
1300
0−13 0
00 −13
⎤
⎦
(g)=
³
1
√
5
[120]
1
√
6
[−211]
1
√
30
[2−15]
´
,
P=
⎡
⎢
⎢
⎢
⎣
1
√
5
−
2
√
6
2
√
30
2
√
5
1
√
6
−
1
√
30
0
1
√
6
5
√
30
⎤
⎥
⎥
⎥
⎦
,D=
⎡
⎣
15 0 0
015 0
00 −15
⎤
⎦
(4) (a)=(
1
19
[−10156]
1
19
[15610]),=(
1
19
√
5
[202726]
1
19
√
5
[35−24−2]),
A=
∙
−22
21
¸
,P=
1
√
5
∙
1−2
21
¸
,D=
∙
20
0−3
¸
(b)=(
1
2
[1−111]
1
2
[−1111]
1
2
[111−1]),
=(
1
2
√
5
[3−131]
1
√
30
[−2431]
1
6
[−1−102])
A=
⎡
⎣
2−12
−122
22 −1
⎤
⎦,P=
⎡
⎢
⎢
⎢
⎣
2
√
5
−
1
√
30
1
√
6
0
5
√
30
1
√
6
1
√
5
2
√
30
−
2
√
6
⎤
⎥
⎥
⎥
⎦
,D=
⎡
⎣
30 0
03 0
00−3
⎤
⎦
(5) (a)
1
25
∙
23−36
−36 2
¸
(b)
1
17
∙
353−120
−120 514
¸
(c)
1
3
⎡
⎣
11 4 −4
417 −8
−4−817
⎤
⎦
(6) For example, the matrixAin Example 7 of Section 5.6 is diagonalizable but not symmetric and hence
not orthogonally diagonalizable.
(7)
1
2
∙
++
p
(−)
2
+4
2
0
0 +−
p
(−)
2
+4
2
¸
(Note: You only need to compute the eigen-
values. You do not need the corresponding eigenvectors.)
(8) (a) Sinceis diagonalizable and=1is the only eigenvalue,
1=R
. Hence, for everyv∈R
,
(v)=1v=v. Therefore,is the identity operator.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 124
Answers to Exercises Section 6.3
(b)must be the zero linear operator. Sinceis diagonalizable, the eigenspace for0is all ofV.
(9) Let
1and 2be symmetric operators onR
. Then, note that for allxy∈R
,
(
2◦1)(x)·y= 2(1(x))·y= 1(x)· 2(y)=x·( 1◦2)(y)
Now
2◦1is symmetric ifffor allxy∈R
,x·( 1◦2)(y)=( 2◦1)(x)·y=x·( 2◦1)(y)iff
for ally∈R
,(1◦2)(y)=( 2◦1)(y)iff( 1◦2)=( 2◦1).
(This argument uses the fact that if for allx∈R
,x·y=x·z,theny=z.Proof:Letx=y−z.
Then,
(y−z)·y=(y−z)·z=⇒((y−z)·y)−((y−z)·z)=0
=⇒(y−z)·(y−z)=0
=⇒ky−zk
2
=0
=⇒y−z=0=⇒y=z)
(10) (i)=⇒(ii): This is Exercise 6 in Section 3.4.
(ii)=⇒(iii): SinceAandBare symmetric, both are orthogonally similar to diagonal matricesD
1
andD 2, respectively, by Theorems 6.19 and 6.22. Now, sinceAandBhave the same characteristic
polynomial, and hence the same eigenvalues with the same algebraic multiplicities (and geometric
multiplicities, sinceAandBare diagonalizable), we can assumeD
1=D 2(by listing the eigenvectors
in an appropriate order when diagonalizing). Thus, ifD
1=P
−1
1
AP1andD 2=P
−1
2
BP2,forsome
orthogonal matricesP
1andP 2,then(P 1P
−1
2
)
−1
A(P 1P
−1
2
)=B.NotethatP 1P
−1
2
is orthogonal by
parts (2) and (3) of Theorem 6.6, soAandBare orthogonally similar.
(iii)=⇒(i): Trivial.
(11)(v
1)·v2=v1·(v 2)=⇒( 1v1)·v2=v1·(2v2)=⇒( 2−1)(v1·v2)=0 =⇒v 1·v2=0,since
2−16 =0.
(12) SupposeAis orthogonal, andis an eigenvalue forAwith corresponding unit eigenvectoru.Then
1=u·u=Au·Au(by Theorem 6.9)=(u)·(u)=
2
(u·u)=
2
. Hence=±1.
Conversely, suppose thatAis symmetric with all eigenvalues equal to±1.LetPbe an orthogonal
matrix withP
−1
AP=D, a diagonal matrix with all entries equal to±1on the main diagonal. Note
thatD
2
=I. Then, sinceA=PDP
−1
,AA
=AA=PD
2
P
−1
=PI P
−1
=I. HenceAis
orthogonal.
(13) (a)−A(A−I
)
=−A(A
−I
)=−AA
+A=−I +A=A−I . Therefore,
|A−I
|=
¯
¯−A(A−I )
¯
¯=|−A|
¯
¯(A−I
)
¯
¯=(−1)
|A||(A−I )|=−|A−I |
sinceis odd and|A|=1.Thisimpliesthat|A−I
|=0,andsoA−I is singular.
(b) By part (a),A−I
is singular. Hence, there is a nonzero vectorvsuch that(A−I )v=0.
Thus,Av=I
v=v,ansovis an eigenvector for the eigenvalue=1.
(c) Supposevis a unit eigenvector forAcorresponding to=1.Expandtheset {v}to an
orthonormal basis={wxv}forR
3
. Notice that we listed the vectorvlast. LetQbe the
orthogonal matrix whose columns arewxandv, in that order. LetP=Q
AQ.Now,thelast
column ofPequalsQ
Av=Q
v(sinceAv=v)=e 3,sincevis a unit vector orthogonal to
thefirst two rows ofQ
.Next,sincePis the product of orthogonal matrices,Pis orthogonal,
and since|Q
|=|Q|=±1we must have|P|=|A|=1.Also,thefirst two columns ofPare
Copyrightc°2016 Elsevier Ltd. All rights reserved. 125
(b)must be the zero linear operator. Sinceis diagonalizable, the eigenspace for0is all ofV.
(9) Let
1and 2be symmetric operators onR
. Then, note that for allxy∈R
,
(
2◦1)(x)·y= 2(1(x))·y= 1(x)· 2(y)=x·( 1◦2)(y)
Now
2◦1is symmetric ifffor allxy∈R
,x·( 1◦2)(y)=( 2◦1)(x)·y=x·( 2◦1)(y)iff
for ally∈R
,(1◦2)(y)=( 2◦1)(y)iff( 1◦2)=( 2◦1).
(This argument uses the fact that if for allx∈R
,x·y=x·z,theny=z.Proof:Letx=y−z.
Then,
(y−z)·y=(y−z)·z=⇒((y−z)·y)−((y−z)·z)=0
=⇒(y−z)·(y−z)=0
=⇒ky−zk
2
=0
=⇒y−z=0=⇒y=z)
(10) (i)=⇒(ii): This is Exercise 6 in Section 3.4.
(ii)=⇒(iii): SinceAandBare symmetric, both are orthogonally similar to diagonal matricesD
1
andD 2, respectively, by Theorems 6.19 and 6.22. Now, sinceAandBhave the same characteristic
polynomial, and hence the same eigenvalues with the same algebraic multiplicities (and geometric
multiplicities, sinceAandBare diagonalizable), we can assumeD
1=D 2(by listing the eigenvectors
in an appropriate order when diagonalizing). Thus, ifD
1=P
−1
1
AP1andD 2=P
−1
2
BP2,forsome
orthogonal matricesP
1andP 2,then(P 1P
−1
2
)
−1
A(P 1P
−1
2
)=B.NotethatP 1P
−1
2
is orthogonal by
parts (2) and (3) of Theorem 6.6, soAandBare orthogonally similar.
(iii)=⇒(i): Trivial.
(11)(v
1)·v2=v1·(v 2)=⇒( 1v1)·v2=v1·(2v2)=⇒( 2−1)(v1·v2)=0 =⇒v 1·v2=0,since
2−16 =0.
(12) SupposeAis orthogonal, andis an eigenvalue forAwith corresponding unit eigenvectoru.Then
1=u·u=Au·Au(by Theorem 6.9)=(u)·(u)=
2
(u·u)=
2
. Hence=±1.
Conversely, suppose thatAis symmetric with all eigenvalues equal to±1.LetPbe an orthogonal
matrix withP
−1
AP=D, a diagonal matrix with all entries equal to±1on the main diagonal. Note
thatD
2
=I. Then, sinceA=PDP
−1
,AA
=AA=PD
2
P
−1
=PI P
−1
=I. HenceAis
orthogonal.
(13) (a)−A(A−I
)
=−A(A
−I
)=−AA
+A=−I +A=A−I . Therefore,
|A−I
|=
¯
¯−A(A−I )
¯
¯=|−A|
¯
¯(A−I
)
¯
¯=(−1)
|A||(A−I )|=−|A−I |
sinceis odd and|A|=1.Thisimpliesthat|A−I
|=0,andsoA−I is singular.
(b) By part (a),A−I
is singular. Hence, there is a nonzero vectorvsuch that(A−I )v=0.
Thus,Av=I
v=v,ansovis an eigenvector for the eigenvalue=1.
(c) Supposevis a unit eigenvector forAcorresponding to=1.Expandtheset {v}to an
orthonormal basis={wxv}forR
3
. Notice that we listed the vectorvlast. LetQbe the
orthogonal matrix whose columns arewxandv, in that order. LetP=Q
AQ.Now,thelast
column ofPequalsQ
Av=Q
v(sinceAv=v)=e 3,sincevis a unit vector orthogonal to
thefirst two rows ofQ
.Next,sincePis the product of orthogonal matrices,Pis orthogonal,
and since|Q
|=|Q|=±1we must have|P|=|A|=1.Also,thefirst two columns ofPare
Copyrightc°2016 Elsevier Ltd. All rights reserved. 125
Answers to Exercises Chapter 6 Review
orthogonal to its last column,e 3,makingthe(31)and(32)entries ofPequal to0.Sincethe
first column ofPis a unit vector with third coordinate0,itcanbeexpressedas[cossin0],
for some value of,with0≤2. The second column ofPmust be a unit vector with
third coordinate0, orthogonal to thefirst column. The only possibilities are±[−sincos0].
Choosing the minus sign makes|P|=−1, so the second column must be[−sincos0].Thus,
P=Q
AQhas the desired form.
(d) The matrix for the linear operatoronR
3
given by(v)=Avwith respect to the standard
basis has a matrix that is orthogonal with determinant1. But, by part (c), and the Orthogonal
Diagonalization Method, the matrix forwith respect to the ordered orthonormal basis
=(wxv)is
⎡
⎣
cos−sin0
sincos0
001
⎤
⎦. According to Table 5.1 in Section 5.2, this is a coun-
terclockwise rotation around the-axis through the angle. But since we are working in-
coordinates, the-axis corresponds to the vectorv,thethirdvectorin. Because the basisis
orthonormal, the plane of the rotation is perpendicular to the axis of rotation.
(e) The axis of rotation is in the direction of the vector[−133]The angle of rotation is≈278
◦
.
This is a counterclockwise rotation about the vector[−133]asyoulookdownfromthepoint
(−133)toward the plane through the origin spanned by[611]and[01−1].
(f) LetGbe the matrix with respect to the standard basis for any chosen orthogonal reflection
through a plane inR
3
. Then,Gis orthogonal,|G|=−1(since it has eigenvalues−111), and
G
2
=I3LetC=AG.ThenCis orthogonal since it is the product of orthogonal matrices. Since
|A|=−1, it follows that|C|=1.Thus,A=AG
2
=CG,whereCrepresents a rotation about
some axis inR
3
by part (d) of this exercise.
(14) (a) T (b) F (c) T (d) T (e) T (f) T
Chapter 6 Review Exercises
(1) (a)[v] =[−142] (b)[v] =[323]
(2) (a){[1−1−11][1111]}
(b){[13431][−1−1011][−13−43−1]}
(3){[63−6][366][2−21]}
(4) (a){[4704][201
−2][29−28−1820][04−14−7]}
(b){
1
9
[4704]
1
3
[201−2]
1
9
√
29
[29−28−1820]
1
3
√
29
[04−14−7]}
(c)
⎡
⎢
⎢
⎢
⎢
⎢
⎣
4
9
7
9
0
4
9
2
3
0
1
3
−
2
3
√
29
9
−
28
9
√
29
−
2
√
29
20
9
√
29
0
4
3
√
29
−
14
3
√
29
−
7
3
√
29
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(5) Using the hint, note thatv·w=v·(A
Aw).Now,A
Aeis theth column ofA
A,soe ·(A
Ae)
is the( )entry ofA
A. Since this equalse ·e,weseethatA
A=I . Hence,A
=A
−1
,andA
is orthogonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 126
orthogonal to its last column,e 3,makingthe(31)and(32)entries ofPequal to0.Sincethe
first column ofPis a unit vector with third coordinate0,itcanbeexpressedas[cossin0],
for some value of,with0≤2. The second column ofPmust be a unit vector with
third coordinate0, orthogonal to thefirst column. The only possibilities are±[−sincos0].
Choosing the minus sign makes|P|=−1, so the second column must be[−sincos0].Thus,
P=Q
AQhas the desired form.
(d) The matrix for the linear operatoronR
3
given by(v)=Avwith respect to the standard
basis has a matrix that is orthogonal with determinant1. But, by part (c), and the Orthogonal
Diagonalization Method, the matrix forwith respect to the ordered orthonormal basis
=(wxv)is
⎡
⎣
cos−sin0
sincos0
001
⎤
⎦. According to Table 5.1 in Section 5.2, this is a coun-
terclockwise rotation around the-axis through the angle. But since we are working in-
coordinates, the-axis corresponds to the vectorv,thethirdvectorin. Because the basisis
orthonormal, the plane of the rotation is perpendicular to the axis of rotation.
(e) The axis of rotation is in the direction of the vector[−133]The angle of rotation is≈278
◦
.
This is a counterclockwise rotation about the vector[−133]asyoulookdownfromthepoint
(−133)toward the plane through the origin spanned by[611]and[01−1].
(f) LetGbe the matrix with respect to the standard basis for any chosen orthogonal reflection
through a plane inR
3
. Then,Gis orthogonal,|G|=−1(since it has eigenvalues−111), and
G
2
=I3LetC=AG.ThenCis orthogonal since it is the product of orthogonal matrices. Since
|A|=−1, it follows that|C|=1.Thus,A=AG
2
=CG,whereCrepresents a rotation about
some axis inR
3
by part (d) of this exercise.
(14) (a) T (b) F (c) T (d) T (e) T (f) T
Chapter 6 Review Exercises
(1) (a)[v] =[−142] (b)[v] =[323]
(2) (a){[1−1−11][1111]}
(b){[13431][−1−1011][−13−43−1]}
(3){[63−6][366][2−21]}
(4) (a){[4704][201
−2][29−28−1820][04−14−7]}
(b){
1
9
[4704]
1
3
[201−2]
1
9
√
29
[29−28−1820]
1
3
√
29
[04−14−7]}
(c)
⎡
⎢
⎢
⎢
⎢
⎢
⎣
4
9
7
9
0
4
9
2
3
0
1
3
−
2
3
√
29
9
−
28
9
√
29
−
2
√
29
20
9
√
29
0
4
3
√
29
−
14
3
√
29
−
7
3
√
29
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(5) Using the hint, note thatv·w=v·(A
Aw).Now,A
Aeis theth column ofA
A,soe ·(A
Ae)
is the( )entry ofA
A. Since this equalse ·e,weseethatA
A=I . Hence,A
=A
−1
,andA
is orthogonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 126
Answers to Exercises Chapter 6 Review
(6) (a)w 1=proj
Wv=[0−918];w 2=proj
W
⊥v=[2168]
(b)w
1=proj
Wv=
£
7
2
23
2
5
2
−
3
2
¤
;w
2=proj
W
⊥v=
£
−
32
−
3
2
9
2
−
15
2
¤
(7) (a) Distance≈10141294 (b) Distance≈14248050
(8){[10−2−2][01−22]}
(9)
1
3
⎡
⎣
2−11
−121
112
⎤
⎦
(10)
1
7
⎡
⎣
36 −2
6−23
−236
⎤
⎦
(11) (a)
()=(−1)
2
=
3
−2
2
+ (b) ()=
2
(−1) =
3
−
2
(12) (a) Not a symmetric operator, since the matrix forwith respect to the standard basis is
⎡
⎢
⎢
⎣
0000
3000
0200
0010
⎤
⎥
⎥
⎦
which is not symmetric.
(b) Symmetric operator, since the matrix foris orthogonally diagonalizable.
(c) Symmetric operator, since the matrix forwith respect to the standard basis is symmetric.
(13) (a)=
³
1
√
6
[−1−21]
1
√
30
[125]
1
√
5
[−210]
´
;P=
⎡
⎢
⎢
⎣
−
1
√
6
1
√
30
−
2
√
5
−
2
√
6
2
√
30
1
√
5
1
√
6
5
√
30
0
⎤
⎥
⎥
⎦
;D=
⎡
⎣
10 0
02 0
00−1
⎤
⎦
(b)=
³
1
√
3
[1−11]
1
√
2
[110]
1
√
6
[−112]
´
;P=
⎡
⎢
⎢
⎢
⎣
1
√
3
1
√
2
−
1
√
6
−
1
√
3
1
√
2
1
√
6
1
√
3
0
2
√
6
⎤
⎥
⎥
⎥
⎦
;D=
⎡
⎣
000
030
003
⎤
⎦
(14) Any diagonalizable4×4matrix that is not symmetric works. For example, chose any non-diagonal
upper triangular4×4matrix with4distinct entries on the main diagonal.
(15) Because
¡
A
A
¢
=A
¡
A
¢
=A
A,weseethatA
Ais symmetric. ThereforeA
Ais orthogo-
nally diagonalizable by Corollary 6.23.
(16) (a) T
(b) T
(c) T
(d) T
(e) F
(f) F
(g) T
(h) T
(i) T
(j) T
(k) T
(l) T
(m) F
(n) T
(o) T
(p) T
(q) F
(r) F
(s) T
(t) T
(u) F
(v) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 127
(6) (a)w 1=proj
Wv=[0−918];w 2=proj
W
⊥v=[2168]
(b)w
1=proj
Wv=
£
7
2
23
2
5
2
−
3
2
¤
;w
2=proj
W
⊥v=
£
−
32
−
3
2
9
2
−
15
2
¤
(7) (a) Distance≈10141294 (b) Distance≈14248050
(8){[10−2−2][01−22]}
(9)
1
3
⎡
⎣
2−11
−121
112
⎤
⎦
(10)
1
7
⎡
⎣
36 −2
6−23
−236
⎤
⎦
(11) (a)
()=(−1)
2
=
3
−2
2
+ (b) ()=
2
(−1) =
3
−
2
(12) (a) Not a symmetric operator, since the matrix forwith respect to the standard basis is
⎡
⎢
⎢
⎣
0000
3000
0200
0010
⎤
⎥
⎥
⎦
which is not symmetric.
(b) Symmetric operator, since the matrix foris orthogonally diagonalizable.
(c) Symmetric operator, since the matrix forwith respect to the standard basis is symmetric.
(13) (a)=
³
1
√
6
[−1−21]
1
√
30
[125]
1
√
5
[−210]
´
;P=
⎡
⎢
⎢
⎣
−
1
√
6
1
√
30
−
2
√
5
−
2
√
6
2
√
30
1
√
5
1
√
6
5
√
30
0
⎤
⎥
⎥
⎦
;D=
⎡
⎣
10 0
02 0
00−1
⎤
⎦
(b)=
³
1
√
3
[1−11]
1
√
2
[110]
1
√
6
[−112]
´
;P=
⎡
⎢
⎢
⎢
⎣
1
√
3
1
√
2
−
1
√
6
−
1
√
3
1
√
2
1
√
6
1
√
3
0
2
√
6
⎤
⎥
⎥
⎥
⎦
;D=
⎡
⎣
000
030
003
⎤
⎦
(14) Any diagonalizable4×4matrix that is not symmetric works. For example, chose any non-diagonal
upper triangular4×4matrix with4distinct entries on the main diagonal.
(15) Because
¡
A
A
¢
=A
¡
A
¢
=A
A,weseethatA
Ais symmetric. ThereforeA
Ais orthogo-
nally diagonalizable by Corollary 6.23.
(16) (a) T
(b) T
(c) T
(d) T
(e) F
(f) F
(g) T
(h) T
(i) T
(j) T
(k) T
(l) T
(m) F
(n) T
(o) T
(p) T
(q) F
(r) F
(s) T
(t) T
(u) F
(v) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 127
Answers to Exercises Section 7.1
Chapter 7
Section 7.1
(1) (a)[1 + 41+6−]
(b)[−12−32−7+3053−29]
(c)[5 +2−3]
(d)[−24−12−28−8−32]
(e)1+28
(f)3+77
(2) Letz
1=[1],andletz 2=[1].
(a) Part (1):z
1·z2=
P
=1
=
P
=1
(()) =
P
=1
(()) =
P
=1
(
)=
P
=1
(
)=z2·z1
Part (2):z 1·z1=
P
=1
=
P
=1
||
2
≥0
Also,z
1·z1=
P
=1
||
2
=
³
p
P
=1
||
2
´
2
=kz 1k
2
(b)
(z1·z2)=
P
=1
=
P
=1
=
P
=1
=z1·(z 2)
(3) (a)
∙
11 + 4−4−2
2−4 12
¸
(b)
⎡
⎣
1−26−4
03−5
1007+3
⎤
⎦
(c)
⎡
⎣
1−010
23−0
6−457+3
⎤
⎦
(d)
∙
−3−15−39
9−603+12
¸
(e)
∙
−7+6−9−
−3−34−6
¸
(f)
∙
1+40−
4−14
13−5023 + 21
¸
(g)
∙
−12 + 6−16−3−11 + 41
−15 + 23−20 + 5−61 + 15
¸
(h)
∙
86−336−39
61 + 3613 + 9
¸
(i)
⎡
⎣
4+36−5+39
1−7−6−4
5+40−7−5
⎤
⎦
(j)
⎡
⎣
40 + 5850−20 + 80
4+88−6 −20
56−1050 + 1080 + 102
⎤
⎦
(4) (a) ( )entryof(Z)
∗
=( )entryof
(Z)
=( )entryof(Z)==
=(( )entryofZ)=(( )entryofZ
)=(( )entryofZ
∗
)
(b) LetZbe an×complexmatrixandletWbe an×matrix. Note that(ZW)
∗
andW
∗
Z
∗
are both×matrices. We present two methods of proof that(ZW)
∗
=W
∗
Z
∗
First method: Notice that the rule(AB)
=B
A
holds for complex matrices, since matrix
multiplication for complex matrices is defined in exactly the same manner as for real matrices,
and hence the proof of Theorem 1.18 is valid for complex matrices as well. Thus we have:
(ZW)
∗
=
(ZW)
=W
Z
=(W
)(Z
)(by part (4) of Theorem 7.2)=W
∗
Z
∗
Second method: We begin by computing the( )entry of(ZW)
∗
Now, the( )entry of
ZW=
11+···+ . Hence,
( )entry of(ZW)
∗
=
(( )entry ofZW)
=(11+···+ )=11+···+
Copyrightc°2016 Elsevier Ltd. All rights reserved. 128
Chapter 7
Section 7.1
(1) (a)[1 + 41+6−]
(b)[−12−32−7+3053−29]
(c)[5 +2−3]
(d)[−24−12−28−8−32]
(e)1+28
(f)3+77
(2) Letz
1=[1],andletz 2=[1].
(a) Part (1):z
1·z2=
P
=1
=
P
=1
(()) =
P
=1
(()) =
P
=1
(
)=
P
=1
(
)=z2·z1
Part (2):z 1·z1=
P
=1
=
P
=1
||
2
≥0
Also,z
1·z1=
P
=1
||
2
=
³
p
P
=1
||
2
´
2
=kz 1k
2
(b)
(z1·z2)=
P
=1
=
P
=1
=
P
=1
=z1·(z 2)
(3) (a)
∙
11 + 4−4−2
2−4 12
¸
(b)
⎡
⎣
1−26−4
03−5
1007+3
⎤
⎦
(c)
⎡
⎣
1−010
23−0
6−457+3
⎤
⎦
(d)
∙
−3−15−39
9−603+12
¸
(e)
∙
−7+6−9−
−3−34−6
¸
(f)
∙
1+40−
4−14
13−5023 + 21
¸
(g)
∙
−12 + 6−16−3−11 + 41
−15 + 23−20 + 5−61 + 15
¸
(h)
∙
86−336−39
61 + 3613 + 9
¸
(i)
⎡
⎣
4+36−5+39
1−7−6−4
5+40−7−5
⎤
⎦
(j)
⎡
⎣
40 + 5850−20 + 80
4+88−6 −20
56−1050 + 1080 + 102
⎤
⎦
(4) (a) ( )entryof(Z)
∗
=( )entryof
(Z)
=( )entryof(Z)==
=(( )entryofZ)=(( )entryofZ
)=(( )entryofZ
∗
)
(b) LetZbe an×complexmatrixandletWbe an×matrix. Note that(ZW)
∗
andW
∗
Z
∗
are both×matrices. We present two methods of proof that(ZW)
∗
=W
∗
Z
∗
First method: Notice that the rule(AB)
=B
A
holds for complex matrices, since matrix
multiplication for complex matrices is defined in exactly the same manner as for real matrices,
and hence the proof of Theorem 1.18 is valid for complex matrices as well. Thus we have:
(ZW)
∗
=
(ZW)
=W
Z
=(W
)(Z
)(by part (4) of Theorem 7.2)=W
∗
Z
∗
Second method: We begin by computing the( )entry of(ZW)
∗
Now, the( )entry of
ZW=
11+···+ . Hence,
( )entry of(ZW)
∗
=
(( )entry ofZW)
=(11+···+ )=11+···+
Copyrightc°2016 Elsevier Ltd. All rights reserved. 128
Answers to Exercises Section 7.1
Next, we compute the( )entry ofW
∗
Z
∗
to show that it equals the( )entry of(ZW)
∗
.Let
A=Z
∗
,an×matrix, andB=W
∗
,a×matrix. Then =
and =. Hence,
( )entry ofW
∗
Z
∗
=( )entry ofBA= 11+···+
=
11+···+=11+···+
Therefore,(ZW)
∗
=W
∗
Z
∗
.
(5) (a) Skew-Hermitian
(b) Neither
(c) Hermitian
(d) Skew-Hermitian
(e) Hermitian
(6) (a)H
∗
=(
1
2
(Z+Z
∗
))
∗
=
1
2
(Z+Z
∗
)
∗
(by part (3) of Theorem 7.2)=
1
2
(Z
∗
+Z)(by part (2) of
Theorem 7.2)=H. A similar calculation shows thatKis skew-Hermitian.
(b) Part (a) proves existence for the decomposition by providing specific matricesHandK.For
uniqueness, supposeZ=H+K,whereHis Hermitian andKis skew-Hermitian. Our goal is to
show thatHandKmust satisfy the formulas from part (a). NowZ
∗
=(H+K)
∗
=H
∗
+K
∗
=
H−K. HenceZ+Z
∗
=(H+K)+(H−K)=2H,whichgivesH=
1
2
(Z+Z
∗
). Similarly,
Z−Z
∗
=2K,soK=
1
2
(Z−Z
∗
).
(7) (a)HJis Hermitian iff(HJ)
∗
=HJiffJ
∗
H
∗
=HJiffJH=HJ(sinceHandJare Hermitian).
(b) Use induction on.
Base Step(=1):H
1
=His given to be Hermitian.
Inductive Step: AssumeH
is Hermitian and prove thatH
+1
is Hermitian. ButH
H=H
+1
=H
1+
=HH
(by part (1) of Theorem 1.17), and so by part (a),H
+1
=H
His Hermitian.
(c)(P
∗
HP)
∗
=P
∗
H
∗
(P
∗
)
∗
=P
∗
HP(sinceHis Hermitian).
(8)(AA
∗
)
∗
=(A
∗
)
∗
A
∗
=AA
∗
(andA
∗
Ais handled similarly).
(9) LetZbe Hermitian. ThenZZ
∗
=ZZ=Z
∗
Z. Similarly, ifZis skew-Hermitian,
ZZ
∗
=Z(−Z)=−(ZZ)=(−Z)Z=Z
∗
Z
(10) LetZbe normal. ConsiderH
1=
1
2
(Z+Z
∗
)andH 2=
1
2
(Z−Z
∗
).NotethatH 1is Hermitian andH 2
is skew-Hermitian (by Exercise 6(a)). ClearlyZ=H 1+H2.Now,
H
1H2=(
1
2
)(
1
2
)(Z+Z
∗
)(Z−Z
∗
)
=
1
4
(Z
2
+Z
∗
Z−ZZ
∗
−(Z
∗
)
2
)
=
1
4
(Z
2
−(Z
∗
)
2
)(sinceZis normal)
A similar computation shows thatH
2H1is also equal to
1
4
(Z
2
−(Z
∗
)
2
). HenceH 1H2=H 2H1.
Conversely, letH
1be Hermitian andH 2be skew-Hermitian withH 1H2=H 2H1.IfZ=H 1+H 2,
then
ZZ
∗
=(H 1+H2)(H1+H2)
∗
=(H 1+H2)(H
∗
1
+H
∗
2
)(by part (2) of Theorem 7.2)
=(H
1+H2)(H1−H2)(sinceH 1is Hermitian andH 2is skew-Hermitian)
=H
2
1
+H2H1−H1H2−H
2
2
=H
2
1
−H
2
2
(sinceH 1H2=H 2H1)
A similar argument showsZ
∗
Z=H
2
1
−H
2
2
also, and soZZ
∗
=Z
∗
Z,andZis normal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 129
Next, we compute the( )entry ofW
∗
Z
∗
to show that it equals the( )entry of(ZW)
∗
.Let
A=Z
∗
,an×matrix, andB=W
∗
,a×matrix. Then =
and =. Hence,
( )entry ofW
∗
Z
∗
=( )entry ofBA= 11+···+
=
11+···+=11+···+
Therefore,(ZW)
∗
=W
∗
Z
∗
.
(5) (a) Skew-Hermitian
(b) Neither
(c) Hermitian
(d) Skew-Hermitian
(e) Hermitian
(6) (a)H
∗
=(
1
2
(Z+Z
∗
))
∗
=
1
2
(Z+Z
∗
)
∗
(by part (3) of Theorem 7.2)=
1
2
(Z
∗
+Z)(by part (2) of
Theorem 7.2)=H. A similar calculation shows thatKis skew-Hermitian.
(b) Part (a) proves existence for the decomposition by providing specific matricesHandK.For
uniqueness, supposeZ=H+K,whereHis Hermitian andKis skew-Hermitian. Our goal is to
show thatHandKmust satisfy the formulas from part (a). NowZ
∗
=(H+K)
∗
=H
∗
+K
∗
=
H−K. HenceZ+Z
∗
=(H+K)+(H−K)=2H,whichgivesH=
1
2
(Z+Z
∗
). Similarly,
Z−Z
∗
=2K,soK=
1
2
(Z−Z
∗
).
(7) (a)HJis Hermitian iff(HJ)
∗
=HJiffJ
∗
H
∗
=HJiffJH=HJ(sinceHandJare Hermitian).
(b) Use induction on.
Base Step(=1):H
1
=His given to be Hermitian.
Inductive Step: AssumeH
is Hermitian and prove thatH
+1
is Hermitian. ButH
H=H
+1
=H
1+
=HH
(by part (1) of Theorem 1.17), and so by part (a),H
+1
=H
His Hermitian.
(c)(P
∗
HP)
∗
=P
∗
H
∗
(P
∗
)
∗
=P
∗
HP(sinceHis Hermitian).
(8)(AA
∗
)
∗
=(A
∗
)
∗
A
∗
=AA
∗
(andA
∗
Ais handled similarly).
(9) LetZbe Hermitian. ThenZZ
∗
=ZZ=Z
∗
Z. Similarly, ifZis skew-Hermitian,
ZZ
∗
=Z(−Z)=−(ZZ)=(−Z)Z=Z
∗
Z
(10) LetZbe normal. ConsiderH
1=
1
2
(Z+Z
∗
)andH 2=
1
2
(Z−Z
∗
).NotethatH 1is Hermitian andH 2
is skew-Hermitian (by Exercise 6(a)). ClearlyZ=H 1+H2.Now,
H
1H2=(
1
2
)(
1
2
)(Z+Z
∗
)(Z−Z
∗
)
=
1
4
(Z
2
+Z
∗
Z−ZZ
∗
−(Z
∗
)
2
)
=
1
4
(Z
2
−(Z
∗
)
2
)(sinceZis normal)
A similar computation shows thatH
2H1is also equal to
1
4
(Z
2
−(Z
∗
)
2
). HenceH 1H2=H 2H1.
Conversely, letH
1be Hermitian andH 2be skew-Hermitian withH 1H2=H 2H1.IfZ=H 1+H 2,
then
ZZ
∗
=(H 1+H2)(H1+H2)
∗
=(H 1+H2)(H
∗
1
+H
∗
2
)(by part (2) of Theorem 7.2)
=(H
1+H2)(H1−H2)(sinceH 1is Hermitian andH 2is skew-Hermitian)
=H
2
1
+H2H1−H1H2−H
2
2
=H
2
1
−H
2
2
(sinceH 1H2=H 2H1)
A similar argument showsZ
∗
Z=H
2
1
−H
2
2
also, and soZZ
∗
=Z
∗
Z,andZis normal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 129
Answers to Exercises Section 7.2
(11) (a) F (b) F (c) T (d) F (e) F (f) T
Section 7.2
(1) (a)=
1
5
+
13
5
;=
28
5
−
3
5
(b) No solutions: After applying the row reduction method to thefirst3columns, the matrix changes
to:
⎡
⎣
11+0
001
000
1−
3+4
4−2
⎤
⎦
(c) Matrix row reduces to
⎡
⎣
104 −3
01 −
00 0
2+5
5+2
0
⎤
⎦;
Solution set={((2+5)−(4−3)(5 + 2)+ )|∈C}
(d)=7+;=−6+5
(e) No solutions: After applying the row reduction method to thefirst column, the matrix changes
to:
∙
12+3
00
−1+3
3+4
¸
(f) Matrix row reduces to
∙
10−3+2
01 4+7
−
2−5
¸
;
Solution set={((−)+(3−2)(2−5)−(4 + 7) )|∈C}
(2) (a)|A|=0;|A
∗
|=0=
|A|
(b)|A|=−15−23;|A
∗
|=−15 + 23=
|A|
(c)|A|=3−5;|A
∗
|=3+5=
|A|
(3) Your computations may produce complex scalar multiples of the vectors given here, although they
might not be immediately recognized as such.
(a)
A()=
2
+(1−)−=(−)(+1); eigenvalues 1=, 2=−1;
={[1 +2]|∈C}; −1={[7 + 617]|∈C}.
(b)
A()=
3
−11
2
+44−34; eigenvalues: 1=5+3, 2=5−3, 3=1;
1
={[1−351−3]|∈C}; 2
={[1 + 351+3]|∈C}; 3
={[122]|∈C}.
(c)
A()=
3
+(2−2)
2
−(1 + 4)−2=(−)
2
(+2); 1=, 2−2;
={[(−3−2)02] +[110]| ∈C}; −2={[−11]|∈C}.
(d)
A()=
3
−2
2
−+2+(−2
2
+4)=(−)
2
(−2);eigenvalues: 1=, 2=2;
={[110]|∈C}; 2={[01]|∈C}. (Note thatAis not diagonalizable.)
(4) (a) LetAbe the matrix from Exercise 3(a).Ais diagonalizable sinceAhas2distinct eigenvalues.
P=
∙
1+7+6
217
¸
;D=
∙
0
0−1
¸
(b) LetAbe the matrix from Exercise 3(d). The solution to Exercise 3(d) shows that the Diago-
nalization Method of Section 3.4 only produces two fundamental eigenvectors. Hence,Ais not
diagonalizable by Step 4 of that method.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 130
(11) (a) F (b) F (c) T (d) F (e) F (f) T
Section 7.2
(1) (a)=
1
5
+
13
5
;=
28
5
−
3
5
(b) No solutions: After applying the row reduction method to thefirst3columns, the matrix changes
to:
⎡
⎣
11+0
001
000
1−
3+4
4−2
⎤
⎦
(c) Matrix row reduces to
⎡
⎣
104 −3
01 −
00 0
2+5
5+2
0
⎤
⎦;
Solution set={((2+5)−(4−3)(5 + 2)+ )|∈C}
(d)=7+;=−6+5
(e) No solutions: After applying the row reduction method to thefirst column, the matrix changes
to:
∙
12+3
00
−1+3
3+4
¸
(f) Matrix row reduces to
∙
10−3+2
01 4+7
−
2−5
¸
;
Solution set={((−)+(3−2)(2−5)−(4 + 7) )|∈C}
(2) (a)|A|=0;|A
∗
|=0=
|A|
(b)|A|=−15−23;|A
∗
|=−15 + 23=
|A|
(c)|A|=3−5;|A
∗
|=3+5=
|A|
(3) Your computations may produce complex scalar multiples of the vectors given here, although they
might not be immediately recognized as such.
(a)
A()=
2
+(1−)−=(−)(+1); eigenvalues 1=, 2=−1;
={[1 +2]|∈C}; −1={[7 + 617]|∈C}.
(b)
A()=
3
−11
2
+44−34; eigenvalues: 1=5+3, 2=5−3, 3=1;
1
={[1−351−3]|∈C}; 2
={[1 + 351+3]|∈C}; 3
={[122]|∈C}.
(c)
A()=
3
+(2−2)
2
−(1 + 4)−2=(−)
2
(+2); 1=, 2−2;
={[(−3−2)02] +[110]| ∈C}; −2={[−11]|∈C}.
(d)
A()=
3
−2
2
−+2+(−2
2
+4)=(−)
2
(−2);eigenvalues: 1=, 2=2;
={[110]|∈C}; 2={[01]|∈C}. (Note thatAis not diagonalizable.)
(4) (a) LetAbe the matrix from Exercise 3(a).Ais diagonalizable sinceAhas2distinct eigenvalues.
P=
∙
1+7+6
217
¸
;D=
∙
0
0−1
¸
(b) LetAbe the matrix from Exercise 3(d). The solution to Exercise 3(d) shows that the Diago-
nalization Method of Section 3.4 only produces two fundamental eigenvectors. Hence,Ais not
diagonalizable by Step 4 of that method.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 130
Answers to Exercises Section 7.3
(c) LetAbe the matrix in Exercise 3(b). A()=
3
−11
2
+44−34. There are 3 eigenvalues,
5+35−3and1, each producing one complex fundamental eigenvector. Thus, there are three
fundamental complex eigenvectors, and soAis diagonalizable as a complex matrix.
On the other hand,Ahas only onerealeigenvalue=1, and sinceproduces only one real
fundamental eigenvector,Ais not diagonalizable as a real matrix.
(5) By the Fundamental Theorem of Algebra, the characteristic polynomial
A()of a complex×
matrixAmust factor intolinear factors. Since each root of
A()has multiplicity1,theremustbe
distinct roots. Each of these roots is an eigenvalue, and each eigenvalue yields at least one fundamental
eigenvector. Thus, we have a total offundamental eigenvectors, showing thatAis diagonalizable.
(6) (a) T (b) F (c) F (d) F
Section 7.3
(1) (a) Mimic the proof in Example 6 in Section 4.1 of the textbook, restricting the discussion to poly-
nomial functions of degree≤, and using complex-valued functions and complex scalars instead
of their real counterparts.
(b) This is the complex analog of Theorem 1.12. Property (1) is proven for the real case just after
the statement of Theorem 1.12 in Section 1.4 of thetextbook. Generalizethisprooftocomplex
matrices. The other properties are similarly proved.
(2) (a) Linearly independent; dim=2
(b) Not linearly independent; dim=1.(Note:[−3+639]=3[2 +−3])
(c) Not linearly independent; dim=2. (Note: Using the given vectors as columns, the matrix row
reduces to
⎡
⎣
10 1
01−2
00 0
⎤
⎦)
(d) Not linearly independent, dim=2. (Note: Using the given vectors as columns, the matrix row
reduces to
⎡
⎣
10
01−2
00 0
⎤
⎦)
(3) (a) Linearly independent; dim=2.(Note:[−3+−1]is not a scalar multiple of[2 +−3])
(b) Linearly independent; dim=2.(Note:[−3+639]is not arealscalar multiple of[2 +−
3])
(c) Not linearly independent; dim=2.
(Note:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
311
−11 −3
1−25
202
04 −8
−11 −3
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
10 1
01−2
00 0
00 0
00 0
00 0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 131
(c) LetAbe the matrix in Exercise 3(b). A()=
3
−11
2
+44−34. There are 3 eigenvalues,
5+35−3and1, each producing one complex fundamental eigenvector. Thus, there are three
fundamental complex eigenvectors, and soAis diagonalizable as a complex matrix.
On the other hand,Ahas only onerealeigenvalue=1, and sinceproduces only one real
fundamental eigenvector,Ais not diagonalizable as a real matrix.
(5) By the Fundamental Theorem of Algebra, the characteristic polynomial
A()of a complex×
matrixAmust factor intolinear factors. Since each root of
A()has multiplicity1,theremustbe
distinct roots. Each of these roots is an eigenvalue, and each eigenvalue yields at least one fundamental
eigenvector. Thus, we have a total offundamental eigenvectors, showing thatAis diagonalizable.
(6) (a) T (b) F (c) F (d) F
Section 7.3
(1) (a) Mimic the proof in Example 6 in Section 4.1 of the textbook, restricting the discussion to poly-
nomial functions of degree≤, and using complex-valued functions and complex scalars instead
of their real counterparts.
(b) This is the complex analog of Theorem 1.12. Property (1) is proven for the real case just after
the statement of Theorem 1.12 in Section 1.4 of thetextbook. Generalizethisprooftocomplex
matrices. The other properties are similarly proved.
(2) (a) Linearly independent; dim=2
(b) Not linearly independent; dim=1.(Note:[−3+639]=3[2 +−3])
(c) Not linearly independent; dim=2. (Note: Using the given vectors as columns, the matrix row
reduces to
⎡
⎣
10 1
01−2
00 0
⎤
⎦)
(d) Not linearly independent, dim=2. (Note: Using the given vectors as columns, the matrix row
reduces to
⎡
⎣
10
01−2
00 0
⎤
⎦)
(3) (a) Linearly independent; dim=2.(Note:[−3+−1]is not a scalar multiple of[2 +−3])
(b) Linearly independent; dim=2.(Note:[−3+639]is not arealscalar multiple of[2 +−
3])
(c) Not linearly independent; dim=2.
(Note:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
311
−11 −3
1−25
202
04 −8
−11 −3
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
10 1
01−2
00 0
00 0
00 0
00 0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 131
Answers to Exercises Section 7.4
(d) Linearly independent; dim=3.
(Note:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
313
−111
1−2−2
205
043
−11 −8
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
100
010
001
000
000
000
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
)
(4) (a) The3×3complex matrix whose columns (or rows) are the given vectors row reduces toI
3.
(b)[1+−1]
(5) Ordered basis=([10][0][01][0]);matrix=
⎡
⎢
⎢
⎣
0010
000 −1
1000
0−10 0
⎤
⎥
⎥
⎦
(6) Span: Letv∈V. Then there exists
1∈Csuch thatv= 1v1+···+ v. Suppose, for each
,
=+,forsome ∈R.Then
v=(
1+1)v1+···+( +)v=1v1+1(v1)+···+ v+(v)
which shows thatv∈span({v
1v1v v}).
Linear Independence: Suppose
1v1+1(v1)+···+ v+(v)=0.Then
(
1+1)v1+···+( +)v=0
implying
(
1+1)=···=( +)=0
(since{v
1v }is linearly independent). Hence, 1=1=2=2=···= ==0.
(7) Exercise 6 clearly implies that ifVis an-dimensional complex vector space, then it is2-dimensional
when considered as a real vector space. Hence,the real dimension must be even. Therefore,R
3
cannot
be considered as a complex vector space since its real dimension is odd.
(8)
⎡
⎢
⎢
⎣
−3+−
2
5
−
11
5
1
2
−
3
2
−
−
1
2
+
7
2
−
8
5
−
4
5
⎤
⎥
⎥
⎦
(9) (a) T (b) F (c) T (d) F
Section 7.4
(1) (a) Not orthogonal;[1 + 2−3−]·[4−23+]=−10 + 10
(b) Not orthogonal;[1−−1+1−]·[−22]=−5−5
(c) Orthogonal (d) Orthogonal
Copyrightc°2016 Elsevier Ltd. All rights reserved. 132
(d) Linearly independent; dim=3.
(Note:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
313
−111
1−2−2
205
043
−11 −8
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
row reduces to
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
100
010
001
000
000
000
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
)
(4) (a) The3×3complex matrix whose columns (or rows) are the given vectors row reduces toI
3.
(b)[1+−1]
(5) Ordered basis=([10][0][01][0]);matrix=
⎡
⎢
⎢
⎣
0010
000 −1
1000
0−10 0
⎤
⎥
⎥
⎦
(6) Span: Letv∈V. Then there exists
1∈Csuch thatv= 1v1+···+ v. Suppose, for each
,
=+,forsome ∈R.Then
v=(
1+1)v1+···+( +)v=1v1+1(v1)+···+ v+(v)
which shows thatv∈span({v
1v1v v}).
Linear Independence: Suppose
1v1+1(v1)+···+ v+(v)=0.Then
(
1+1)v1+···+( +)v=0
implying
(
1+1)=···=( +)=0
(since{v
1v }is linearly independent). Hence, 1=1=2=2=···= ==0.
(7) Exercise 6 clearly implies that ifVis an-dimensional complex vector space, then it is2-dimensional
when considered as a real vector space. Hence,the real dimension must be even. Therefore,R
3
cannot
be considered as a complex vector space since its real dimension is odd.
(8)
⎡
⎢
⎢
⎣
−3+−
2
5
−
11
5
1
2
−
3
2
−
−
1
2
+
7
2
−
8
5
−
4
5
⎤
⎥
⎥
⎦
(9) (a) T (b) F (c) T (d) F
Section 7.4
(1) (a) Not orthogonal;[1 + 2−3−]·[4−23+]=−10 + 10
(b) Not orthogonal;[1−−1+1−]·[−22]=−5−5
(c) Orthogonal (d) Orthogonal
Copyrightc°2016 Elsevier Ltd. All rights reserved. 132
Answers to Exercises Section 7.4
(2)( z)·( z)= (z·z)(by parts (4) and (5) of Theorem 7.1)= (0)(sincez is orthogonal to
z
)=0so zis orthogonal to zAlso,
||
z||
2
=(z)·( z)=
(z·z)=| |
2
||z||
2
=1
since|
|=1andz is a unit vector. Hence, each zis also a unit vector.
(3) (a){[1 + 1][2−1−−1+][01]}
(b)
⎡
⎢
⎢
⎢
⎣
1+
2
2
1
2
2
√
8
−1−
√
8
−1+
√
8
0
1
√
2
√
2
⎤
⎥
⎥
⎥
⎦
(4) For any×matrixB,|B|=|B|by part (3) of Theorem 7.5. Therefore,
¯
¯
¯
¯
|A|
¯
¯
¯
¯
2
=|A|
|A|=|A||A|=|A||(A)
|=|A||A
∗
|=|AA
∗
|=|I |=1
(5) (a)Aunitary⇒A
∗
=A
−1
⇒(A
∗
)
−1
=A⇒(A
∗
)
−1
=(A
∗
)
∗
⇒A
∗
is unitary.
(b)(AB)
∗
=B
∗
A
∗
(by part (5) of Theorem 7.2)=B
−1
A
−1
=(AB)
−1
.
(6) (a)A
∗
=A
−1
iff(
A)
=A
−1
iff(A)
=A
−1
iff(A)
∗
=(A)
−1
(using the fact thatA
−1
=(A)
−1
,
sinceAA
−1
=I=⇒
AA
−1
=I=I=⇒(A)(A
−1
)=I =⇒A
−1
=(A)
−1
).
(b)(A
)
∗
=(
(A
))
=((A)
)
(by part (4) of Theorem 7.2)=((A)
)
=(A
∗
)
=(A
−1
)
=
(A
)
−1
.
(c)A
2
=IiffA=A
−1
iffA=A
∗
(sinceAis unitary) iffAis Hermitian.
(7) (a)Ais unitary iffA
∗
=A
−1
iff
A
=A
−1
iff(A
)
=(A
−1
)
iff(A
)
∗
=(A
)
−1
iffA
is
unitary.
(b) Follow the hint in the textbook.
(8) Modify the proof of Theorem 6.8.
(9)Z
∗
=
1
3
⎡
⎣
−1−32−22
2−2−2−2
−22 1−4
⎤
⎦andZZ
∗
=Z
∗
Z=
1
9
⎡
⎣
22 8 10
8162
−10−225
⎤
⎦.
(10) (a) LetAbe the given matrix for.Ais normal since
A
∗
=
∙
1+62+10
−10 + 2 5
¸
andAA
∗
=A
∗
A=
∙
141 12−12
12 + 12129
¸
HenceAis unitarily diagonalizable by Theorem 7.9.
(b)P=
"
−1+
√
6
1−
√
3
2
√
6
1
√
3
#
;P
−1
AP=P
∗
AP=
∙
9+6 0
0 −3−12
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 133
(2)( z)·( z)= (z·z)(by parts (4) and (5) of Theorem 7.1)= (0)(sincez is orthogonal to
z
)=0so zis orthogonal to zAlso,
||
z||
2
=(z)·( z)=
(z·z)=| |
2
||z||
2
=1
since|
|=1andz is a unit vector. Hence, each zis also a unit vector.
(3) (a){[1 + 1][2−1−−1+][01]}
(b)
⎡
⎢
⎢
⎢
⎣
1+
2
2
1
2
2
√
8
−1−
√
8
−1+
√
8
0
1
√
2
√
2
⎤
⎥
⎥
⎥
⎦
(4) For any×matrixB,|B|=|B|by part (3) of Theorem 7.5. Therefore,
¯
¯
¯
¯
|A|
¯
¯
¯
¯
2
=|A|
|A|=|A||A|=|A||(A)
|=|A||A
∗
|=|AA
∗
|=|I |=1
(5) (a)Aunitary⇒A
∗
=A
−1
⇒(A
∗
)
−1
=A⇒(A
∗
)
−1
=(A
∗
)
∗
⇒A
∗
is unitary.
(b)(AB)
∗
=B
∗
A
∗
(by part (5) of Theorem 7.2)=B
−1
A
−1
=(AB)
−1
.
(6) (a)A
∗
=A
−1
iff(
A)
=A
−1
iff(A)
=A
−1
iff(A)
∗
=(A)
−1
(using the fact thatA
−1
=(A)
−1
,
sinceAA
−1
=I=⇒
AA
−1
=I=I=⇒(A)(A
−1
)=I =⇒A
−1
=(A)
−1
).
(b)(A
)
∗
=(
(A
))
=((A)
)
(by part (4) of Theorem 7.2)=((A)
)
=(A
∗
)
=(A
−1
)
=
(A
)
−1
.
(c)A
2
=IiffA=A
−1
iffA=A
∗
(sinceAis unitary) iffAis Hermitian.
(7) (a)Ais unitary iffA
∗
=A
−1
iff
A
=A
−1
iff(A
)
=(A
−1
)
iff(A
)
∗
=(A
)
−1
iffA
is
unitary.
(b) Follow the hint in the textbook.
(8) Modify the proof of Theorem 6.8.
(9)Z
∗
=
1
3
⎡
⎣
−1−32−22
2−2−2−2
−22 1−4
⎤
⎦andZZ
∗
=Z
∗
Z=
1
9
⎡
⎣
22 8 10
8162
−10−225
⎤
⎦.
(10) (a) LetAbe the given matrix for.Ais normal since
A
∗
=
∙
1+62+10
−10 + 2 5
¸
andAA
∗
=A
∗
A=
∙
141 12−12
12 + 12129
¸
HenceAis unitarily diagonalizable by Theorem 7.9.
(b)P=
"
−1+
√
6
1−
√
3
2
√
6
1
√
3
#
;P
−1
AP=P
∗
AP=
∙
9+6 0
0 −3−12
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 133
Answers to Exercises Section 7.4
(11) (a)Ais normal sinceA
∗
=
⎡
⎣
−4−52−24−4
2−2−1−8−2+2
4−4−2+2−4−5
⎤
⎦andAA
∗
=A
∗
A=81I 3.
HenceAis unitarily diagonalizable by Theorem 7.9.
(b)P=
1
3
⎡
⎣
−212
1−22
22
⎤
⎦andP
−1
AP=P
∗
AP=
⎡
⎣
−90 0
090
009
⎤
⎦.
(12) (a) Note that
Az·Az=z·z=
(z·z)(by part (5) of Theorem 7.1)
=(z·z) (by part (4) of Theorem 7.1),
while
Az·Az=z·(A
∗
(Az))(by Theorem 7.3)
=z·((A
∗
A)z)
=z·((A
−1
A)z)
=z·z
Thus
(z·z)=z·z,andso=1,whichgives||
2
=1, and hence||=1.
(b) LetAbe unitary, and letbe an eigenvalue ofA. By part (a),||=1.
Now, supposeAis Hermitian. Thenis real by Theorem 7.11. Hence=±1.
Conversely, suppose every eigenvalue ofAis±1.SinceAis unitary,A
∗
=A
−1
,andsoAA
∗
=
A
∗
A=I, which implies thatAis normal. ThusAis unitarily diagonalizable (by Theorem 7.9).
LetA=PDP
∗
,wherePis unitary andDis diagonal. But since the eigenvalues ofAare±1
(real),D
∗
=D.ThusA
∗
=(PDP
∗
)
∗
=PDP
∗
=A,andAis Hermitian.
(13) The eigenvalues are−4,2+
√
6,and2−
√
6.
(14) (a) SinceAis normal,Ais unitarily diagonalizable (by Theorem 7.9). HenceA=PDP
∗
for some
diagonal matrixDand some unitary matrixP. Since all eigenvalues ofAare real, the main
diagonal elements ofDare real, and soD
∗
=D.Thus,
A
∗
=(PDP
∗
)
∗
=PD
∗
P
∗
=PDP
∗
=A
and soAis Hermitian.
(b) SinceAis normal,Ais unitarily diagonalizable (by Theorem 7.9). Hence,A=PDP
∗
for some
diagonal matrixDand some unitary matrixP.ThusPP
∗
=I.Alsonotethatif
D=
⎡
⎢
⎣
1···0
.
.
.
.
.
.
.
.
.
0···
⎤
⎥
⎦thenD
∗
=
⎡
⎢
⎣
1···0
.
.
.
.
.
.
.
.
.
0···
⎤
⎥
⎦
and soDD
∗
=
⎡
⎢
⎣
1
1···0
.
.
.
.
.
.
.
.
.
0···
⎤
⎥
⎦=
⎡
⎢
⎣
|
1|
2
···0
.
.
.
.
.
.
.
.
.
0··· |
|
2
⎤
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 134
(11) (a)Ais normal sinceA
∗
=
⎡
⎣
−4−52−24−4
2−2−1−8−2+2
4−4−2+2−4−5
⎤
⎦andAA
∗
=A
∗
A=81I 3.
HenceAis unitarily diagonalizable by Theorem 7.9.
(b)P=
1
3
⎡
⎣
−212
1−22
22
⎤
⎦andP
−1
AP=P
∗
AP=
⎡
⎣
−90 0
090
009
⎤
⎦.
(12) (a) Note that
Az·Az=z·z=
(z·z)(by part (5) of Theorem 7.1)
=(z·z) (by part (4) of Theorem 7.1),
while
Az·Az=z·(A
∗
(Az))(by Theorem 7.3)
=z·((A
∗
A)z)
=z·((A
−1
A)z)
=z·z
Thus
(z·z)=z·z,andso=1,whichgives||
2
=1, and hence||=1.
(b) LetAbe unitary, and letbe an eigenvalue ofA. By part (a),||=1.
Now, supposeAis Hermitian. Thenis real by Theorem 7.11. Hence=±1.
Conversely, suppose every eigenvalue ofAis±1.SinceAis unitary,A
∗
=A
−1
,andsoAA
∗
=
A
∗
A=I, which implies thatAis normal. ThusAis unitarily diagonalizable (by Theorem 7.9).
LetA=PDP
∗
,wherePis unitary andDis diagonal. But since the eigenvalues ofAare±1
(real),D
∗
=D.ThusA
∗
=(PDP
∗
)
∗
=PDP
∗
=A,andAis Hermitian.
(13) The eigenvalues are−4,2+
√
6,and2−
√
6.
(14) (a) SinceAis normal,Ais unitarily diagonalizable (by Theorem 7.9). HenceA=PDP
∗
for some
diagonal matrixDand some unitary matrixP. Since all eigenvalues ofAare real, the main
diagonal elements ofDare real, and soD
∗
=D.Thus,
A
∗
=(PDP
∗
)
∗
=PD
∗
P
∗
=PDP
∗
=A
and soAis Hermitian.
(b) SinceAis normal,Ais unitarily diagonalizable (by Theorem 7.9). Hence,A=PDP
∗
for some
diagonal matrixDand some unitary matrixP.ThusPP
∗
=I.Alsonotethatif
D=
⎡
⎢
⎣
1···0
.
.
.
.
.
.
.
.
.
0···
⎤
⎥
⎦thenD
∗
=
⎡
⎢
⎣
1···0
.
.
.
.
.
.
.
.
.
0···
⎤
⎥
⎦
and soDD
∗
=
⎡
⎢
⎣
1
1···0
.
.
.
.
.
.
.
.
.
0···
⎤
⎥
⎦=
⎡
⎢
⎣
|
1|
2
···0
.
.
.
.
.
.
.
.
.
0··· |
|
2
⎤
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 134
Answers to Exercises Section 7.5
Since the main diagonal elements ofDare the (not necessarily distinct) eigenvalues ofA,and
these eigenvalues all have absolute value1, it follows thatDD
∗
=I. Then,
AA
∗
=(PDP
∗
)(PDP
∗
)
∗
=(PDP
∗
)(PD
∗
P
∗
)=P(D(P
∗
P)D
∗
)P
∗
=P(DD
∗
)P
∗
=PP
∗
=I
Hence,A
∗
=A
−1
,andAis unitary.
(c) BecauseAis unitary,A
−1
=A
∗
. Hence,AA
∗
=I=A
∗
A.Therefore,Ais normal.
(15) (a) F (b) T (c) T (d) T (e) F
Section 7.5
(1) (a) Note thathxxi=(Ax)·(Ax)=kAxk
2
≥0.
However,kAxk
2
=0⇐⇒Ax=0⇐⇒A
−1
Ax=A
−1
0⇐⇒x=0.
Also,hxyi=(Ax)·(Ay)=(Ay)·(Ax)=hyxiand
hx+yzi=(A(x+y))·(Az)=(Ax+Ay)·Az
=((Ax)·(Az)) + ((Ay)·(Az)) =hxzi+hyzi.
Finally,hxyi=(A(x))·(Ay)=((Ax
))·(Ay)=((Ax)·(Ay)) =hxyi.
(b)hxyi=−183,kxk=
√
314
(2) Note thathp
1p1i=
2
+···+
2
1
+
2
0
≥0.
Clearly,
2
+···+
2
1
+
2
0
=0if and only if each =0.
Also,hp
1p2i= +···+ 11+00=+···+ 11+00=hp 2p1iand
hp
1+p2p3i=h( +)
+···+( 0+0)
+···+ 0i
=(
+)+···+( 0+0)0=( +)+···+( 00+00)
=(
+···+ 00)+( +···+ 00)=hp 1p3i+hp 2p3i.
Finally,hp
1p2i=( )+···+( 1)1+( 0)0=( +···+ 11+00)=hp 1p2i.
(3) (a) Note thathffi=
R
(f())
2
≥0. Also, we know from calculus that a nonnegative continuous
function on an interval has integral zero if and only if the function is the zero function.
Also,hfgi=
R
f()g()=
R
g()f()=hgfiand
hf+ghi=
R
(f()+g())h()=
R
(f()h()+g()h())
=
R
f()h()+
R
g()h()=hfhi+hghi.
Finally,hfgi=
R
(f())g()=
R
f()g()=hfgi.
(b)hfgi=
1
2
(
+1);kfk=
q
1
2
(
2
−1)
(4) Note thathAAi=trace(A
A), which by Exercise 28 in Section 1.5 is the sum of the squares of all
the entries ofA, and hence is nonnegative. Also, this is clearly equal to zero if and only ifA=O
.
Next,hABi=trace(A
B)=trace((A
B)
) (by Exercise 13 in Section 1.4)
=trace(B
A)=hBAi.
Also,hA+BCi=trace((A+B)
C)=trace((A
+B
)C)=trace(A
C+B
C)
=trace(A
C)+trace(B
C)=hACi+hBCi.
Finally,hABi=trace((A)
B)=trace((A
)B)=trace((A
B))
=(trace(A
B))=hABi.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 135
Since the main diagonal elements ofDare the (not necessarily distinct) eigenvalues ofA,and
these eigenvalues all have absolute value1, it follows thatDD
∗
=I. Then,
AA
∗
=(PDP
∗
)(PDP
∗
)
∗
=(PDP
∗
)(PD
∗
P
∗
)=P(D(P
∗
P)D
∗
)P
∗
=P(DD
∗
)P
∗
=PP
∗
=I
Hence,A
∗
=A
−1
,andAis unitary.
(c) BecauseAis unitary,A
−1
=A
∗
. Hence,AA
∗
=I=A
∗
A.Therefore,Ais normal.
(15) (a) F (b) T (c) T (d) T (e) F
Section 7.5
(1) (a) Note thathxxi=(Ax)·(Ax)=kAxk
2
≥0.
However,kAxk
2
=0⇐⇒Ax=0⇐⇒A
−1
Ax=A
−1
0⇐⇒x=0.
Also,hxyi=(Ax)·(Ay)=(Ay)·(Ax)=hyxiand
hx+yzi=(A(x+y))·(Az)=(Ax+Ay)·Az
=((Ax)·(Az)) + ((Ay)·(Az)) =hxzi+hyzi.
Finally,hxyi=(A(x))·(Ay)=((Ax
))·(Ay)=((Ax)·(Ay)) =hxyi.
(b)hxyi=−183,kxk=
√
314
(2) Note thathp
1p1i=
2
+···+
2
1
+
2
0
≥0.
Clearly,
2
+···+
2
1
+
2
0
=0if and only if each =0.
Also,hp
1p2i= +···+ 11+00=+···+ 11+00=hp 2p1iand
hp
1+p2p3i=h( +)
+···+( 0+0)
+···+ 0i
=(
+)+···+( 0+0)0=( +)+···+( 00+00)
=(
+···+ 00)+( +···+ 00)=hp 1p3i+hp 2p3i.
Finally,hp
1p2i=( )+···+( 1)1+( 0)0=( +···+ 11+00)=hp 1p2i.
(3) (a) Note thathffi=
R
(f())
2
≥0. Also, we know from calculus that a nonnegative continuous
function on an interval has integral zero if and only if the function is the zero function.
Also,hfgi=
R
f()g()=
R
g()f()=hgfiand
hf+ghi=
R
(f()+g())h()=
R
(f()h()+g()h())
=
R
f()h()+
R
g()h()=hfhi+hghi.
Finally,hfgi=
R
(f())g()=
R
f()g()=hfgi.
(b)hfgi=
1
2
(
+1);kfk=
q
1
2
(
2
−1)
(4) Note thathAAi=trace(A
A), which by Exercise 28 in Section 1.5 is the sum of the squares of all
the entries ofA, and hence is nonnegative. Also, this is clearly equal to zero if and only ifA=O
.
Next,hABi=trace(A
B)=trace((A
B)
) (by Exercise 13 in Section 1.4)
=trace(B
A)=hBAi.
Also,hA+BCi=trace((A+B)
C)=trace((A
+B
)C)=trace(A
C+B
C)
=trace(A
C)+trace(B
C)=hACi+hBCi.
Finally,hABi=trace((A)
B)=trace((A
)B)=trace((A
B))
=(trace(A
B))=hABi.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 135
Answers to Exercises Section 7.5
(5) (a)h0xi=h0+0xi=h0xi+h0xi.Sinceh0xi∈C,weknowthat“−h0xi”exists. Adding
“−h0xi” to both sides gives0=h0xi. Then,hx0i=0, by property (3) in the definition of an
inner product.
(b) For complex inner product spaces,hxyi=hyxi(by property (3) of an inner product space)
=hyxi(by property (5) of an inner product space)=(hyxi)=hxyi(byproperty(3)of
an inner product space).
A similar proof works in real inner product spaces.
(6)kxk=
p
hxxi=
p
hxxi(by property (5) of an inner product space)=
q
hxxi(by part
(3) of Theorem 7.12)=
p
||
2
hxxi=||
p
hxxi=||kxk.
(7) (a) We have
kx+yk
2
=hx+yx+yi
=hxxi+hxyi+hyxi+hyyi
=kxk
2
+2hxyi+kyk
2
(by property (3) of an inner product space).
(b) Use part (a) and the fact thatxandyare orthogonal iffhxyi=0.
(c) A proof similar to that in part (a) yieldskx−yk
2
=kxk
2
−2hxyi+kyk
2
.
Add this to the equation forkx+yk
2
in part (a).
(8) (a) Subtract the equation forkx−yk
2
intheanswertoExercise7(c)fromtheequationforkx+yk
2
in Exercise 7(a). Then multiply by
1
4
.
(b) In the complex case,
kx+yk
2
=hx+yx+yi
=hxxi+hxyi+hyxi+hyyi
=hxxi+hxyi+
hxyi+hyyi
Similarly,kx−yk
2
=hxxi−hxyi−
hxyi+hyyi
kx+yk
2
=hxxi−hxyi+
hxyi+hyyi
andkx−yk
2
=hxxi+hxyi−
hxyi+hyyi
Then,kx+yk
2
−kx−yk
2
=2hxyi+2
hxyi
Also,(kx+yk
2
−kx−yk
2
)= (−2hxyi+2
hxyi)
=2hxyi−2hxyi
Adding these and multiplying by
1
4
produces the desired result.
(9) (a)
q
3
3
−
3
2
(b)0941radians, or,539
◦
(10) (a)
√
174 (b)0586radians, or,336
◦
(11) (a) If eitherxoryis0, then the theorem is obviously true. We only need to examine the case when
bothkxkandkykare nonzero. We need to prove−kxkkyk≤|hxyi|≤kxkkyk. By property
Copyrightc°2016 Elsevier Ltd. All rights reserved. 136
(5) (a)h0xi=h0+0xi=h0xi+h0xi.Sinceh0xi∈C,weknowthat“−h0xi”exists. Adding
“−h0xi” to both sides gives0=h0xi. Then,hx0i=0, by property (3) in the definition of an
inner product.
(b) For complex inner product spaces,hxyi=hyxi(by property (3) of an inner product space)
=hyxi(by property (5) of an inner product space)=(hyxi)=hxyi(byproperty(3)of
an inner product space).
A similar proof works in real inner product spaces.
(6)kxk=
p
hxxi=
p
hxxi(by property (5) of an inner product space)=
q
hxxi(by part
(3) of Theorem 7.12)=
p
||
2
hxxi=||
p
hxxi=||kxk.
(7) (a) We have
kx+yk
2
=hx+yx+yi
=hxxi+hxyi+hyxi+hyyi
=kxk
2
+2hxyi+kyk
2
(by property (3) of an inner product space).
(b) Use part (a) and the fact thatxandyare orthogonal iffhxyi=0.
(c) A proof similar to that in part (a) yieldskx−yk
2
=kxk
2
−2hxyi+kyk
2
.
Add this to the equation forkx+yk
2
in part (a).
(8) (a) Subtract the equation forkx−yk
2
intheanswertoExercise7(c)fromtheequationforkx+yk
2
in Exercise 7(a). Then multiply by
1
4
.
(b) In the complex case,
kx+yk
2
=hx+yx+yi
=hxxi+hxyi+hyxi+hyyi
=hxxi+hxyi+
hxyi+hyyi
Similarly,kx−yk
2
=hxxi−hxyi−
hxyi+hyyi
kx+yk
2
=hxxi−hxyi+
hxyi+hyyi
andkx−yk
2
=hxxi+hxyi−
hxyi+hyyi
Then,kx+yk
2
−kx−yk
2
=2hxyi+2
hxyi
Also,(kx+yk
2
−kx−yk
2
)= (−2hxyi+2
hxyi)
=2hxyi−2hxyi
Adding these and multiplying by
1
4
produces the desired result.
(9) (a)
q
3
3
−
3
2
(b)0941radians, or,539
◦
(10) (a)
√
174 (b)0586radians, or,336
◦
(11) (a) If eitherxoryis0, then the theorem is obviously true. We only need to examine the case when
bothkxkandkykare nonzero. We need to prove−kxkkyk≤|hxyi|≤kxkkyk. By property
Copyrightc°2016 Elsevier Ltd. All rights reserved. 136
Answers to Exercises Section 7.5
(5) of an inner product space, and part (3) of Theorem 7.12, this is equivalent to proving
¯
¯
¯
¯
¿
x
kxk
y
kyk
À¯
¯
¯
¯
≤1
Hence, it is enough to show|habi|≤1for anyunitvectorsaandb.
Lethabi=∈C. We need to show||≤1.If=0,wearedone. If6 =0,thenletv=
||
a.
Then
hvbi=
¿
||
ab
À
=
||
habi=
||
=||∈R
Also, since
¯
¯
¯
||
¯
¯
¯=1,v=
||
ais a unit vector. Now,
0≤kv−bk
2
=hv−bv−bi
=kvk
2
−hvbi−
hvbi+kbk
2
=1−||−||+1
=2−2||
Hence−2≤−2||,or1≥||, completing the proof.
(b) Note that
kx+yk
2
=hx+yx+yi
=hxxi+hxyi+hyxi+hyyi
=kxk
2
+hxyi+
hxyi+kyk
2
(by property (3) of an inner product space)
=kxk
2
+2(real part ofhxyi)+kyk
2
≤kxk
2
+2|hxyi|+kyk
2
≤kxk
2
+2kxkkyk+kyk
2
(by the Cauchy-Schwarz Inequality)
=(kxk+kyk)
2
(12) The given inequality follows directly from the Cauchy-Schwarz Inequality.
(13) (i)(xy)=kx−yk=k(−1)(x−y)k(by Theorem 7.13)=ky−xk=(yx).
(ii)(xy)=kx−yk=
p
hx−yx−yi≥0.
Also,(xy)=0iffkx−yk=0iffhx−yx−yi=0iffx−y=0(by property (2) of an inner
product space) iffx=y.
(iii)(xy)=kx−yk=k(x−z)+(z−y)k≤kx−zk+kz−yk
(by the Triangle Inequality, part (2) of Theorem 7.14)=(xz)+(z
y).
(14) (a) Orthogonal (b) Orthogonal (c) Not orthogonal (d) Orthogonal
(15) Supposex∈,wherexis a linear combination of{x
1x }⊆.Thenx= 1x1+···+ x.
Consider
hxxi=hx
1x1i+···+hx xi(by part (2) of Theorem 7.12)
=
1hxx 1i+···+hxx i(by part (3) of Theorem 7.12)
=1(0) +···+(0) (since vectors inare orthogonal)
=0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 137
(5) of an inner product space, and part (3) of Theorem 7.12, this is equivalent to proving
¯
¯
¯
¯
¿
x
kxk
y
kyk
À¯
¯
¯
¯
≤1
Hence, it is enough to show|habi|≤1for anyunitvectorsaandb.
Lethabi=∈C. We need to show||≤1.If=0,wearedone. If6 =0,thenletv=
||
a.
Then
hvbi=
¿
||
ab
À
=
||
habi=
||
=||∈R
Also, since
¯
¯
¯
||
¯
¯
¯=1,v=
||
ais a unit vector. Now,
0≤kv−bk
2
=hv−bv−bi
=kvk
2
−hvbi−
hvbi+kbk
2
=1−||−||+1
=2−2||
Hence−2≤−2||,or1≥||, completing the proof.
(b) Note that
kx+yk
2
=hx+yx+yi
=hxxi+hxyi+hyxi+hyyi
=kxk
2
+hxyi+
hxyi+kyk
2
(by property (3) of an inner product space)
=kxk
2
+2(real part ofhxyi)+kyk
2
≤kxk
2
+2|hxyi|+kyk
2
≤kxk
2
+2kxkkyk+kyk
2
(by the Cauchy-Schwarz Inequality)
=(kxk+kyk)
2
(12) The given inequality follows directly from the Cauchy-Schwarz Inequality.
(13) (i)(xy)=kx−yk=k(−1)(x−y)k(by Theorem 7.13)=ky−xk=(yx).
(ii)(xy)=kx−yk=
p
hx−yx−yi≥0.
Also,(xy)=0iffkx−yk=0iffhx−yx−yi=0iffx−y=0(by property (2) of an inner
product space) iffx=y.
(iii)(xy)=kx−yk=k(x−z)+(z−y)k≤kx−zk+kz−yk
(by the Triangle Inequality, part (2) of Theorem 7.14)=(xz)+(z
y).
(14) (a) Orthogonal (b) Orthogonal (c) Not orthogonal (d) Orthogonal
(15) Supposex∈,wherexis a linear combination of{x
1x }⊆.Thenx= 1x1+···+ x.
Consider
hxxi=hx
1x1i+···+hx xi(by part (2) of Theorem 7.12)
=
1hxx 1i+···+hxx i(by part (3) of Theorem 7.12)
=1(0) +···+(0) (since vectors inare orthogonal)
=0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 137
Answers to Exercises Section 7.5
a contradiction, sincexis assumed to be nonzero.
(16) (a) We have
Z
−
cos =
1
(sin)|
−
=
1
(sin−sin(−))
=
1
(0) = 0(sincethesineofanintegralmultipleofis0)
Similarly,
Z
−
sin =−
1
(cos)|
−
=−
1
(cos−cos(−))
=−
1
(cos−cos)(sincecos(−)=cos)
=−
1
(0) = 0
(b) Use the trigonometric identities
coscos=
1
2
(cos(−)+cos(+))
andsinsin=
1
2
(cos(−)−cos(+))
Then,
Z
−
coscos =
1
2
Z
−
cos(−)+
1
2
Z
−
cos(+)=
1
2
(0) +
1
2
(0) = 0
by part (a), since±is an integer. Also,
Z
−
sinsin =
1
2
Z
−
cos(−)−
1
2
Z
−
cos(+)
=
1
2
(0)−
1
2
(0) = 0by part (a).
(c) Use the trigonometric identitysincos=
1
2
(sin(+)+sin(−)). Then,
Z
−
cossin =
1
2
Z
−
sin(+)+
1
2
Z
−
sin(−)
=
1
2
(0) +
1
2
(0) = 0
by part (a), since±is an integer.
(d) Obvious from parts (a), (b), and (c) with the real inner product of Example 5 (or Example 8)
with=−,=:hfgi=
R
−
()().
Copyrightc°2016 Elsevier Ltd. All rights reserved. 138
a contradiction, sincexis assumed to be nonzero.
(16) (a) We have
Z
−
cos =
1
(sin)|
−
=
1
(sin−sin(−))
=
1
(0) = 0(sincethesineofanintegralmultipleofis0)
Similarly,
Z
−
sin =−
1
(cos)|
−
=−
1
(cos−cos(−))
=−
1
(cos−cos)(sincecos(−)=cos)
=−
1
(0) = 0
(b) Use the trigonometric identities
coscos=
1
2
(cos(−)+cos(+))
andsinsin=
1
2
(cos(−)−cos(+))
Then,
Z
−
coscos =
1
2
Z
−
cos(−)+
1
2
Z
−
cos(+)=
1
2
(0) +
1
2
(0) = 0
by part (a), since±is an integer. Also,
Z
−
sinsin =
1
2
Z
−
cos(−)−
1
2
Z
−
cos(+)
=
1
2
(0)−
1
2
(0) = 0by part (a).
(c) Use the trigonometric identitysincos=
1
2
(sin(+)+sin(−)). Then,
Z
−
cossin =
1
2
Z
−
sin(+)+
1
2
Z
−
sin(−)
=
1
2
(0) +
1
2
(0) = 0
by part (a), since±is an integer.
(d) Obvious from parts (a), (b), and (c) with the real inner product of Example 5 (or Example 8)
with=−,=:hfgi=
R
−
()().
Copyrightc°2016 Elsevier Ltd. All rights reserved. 138
Answers to Exercises Section 7.5
(17) Let=(v 1v )be an orthogonal ordered basis for a subspaceWofV,andletv∈W.Let
[v]
=[1]. The goal is to show that =hvv ikv k
2
.Now,v= 1v1+···+ v. Hence,
hvv
i=h 1v1+···+ vvi
=
1hv1vi+···+ hvvi+···+ hvvi
(by properties (4) and (5) of an inner product)
=
1(0) +···+ −1(0) + kvk
2
++1(0) +···+ (0)
(sinceis orthogonal)
=
kvk
2
Hence,
=hvv ikv k
2
. Also, ifis orthonormal, then eachkv k=1,so =hvv i.
(18) Letv=
1v1+···+ vandw= 1v1+···+ v,where =hvv iand =hwv i(by
Theorem 7.16). Then
hvwi=h
1v1+···+ v1v1+···+ vi
=
1
1hv1v1i+ 22hv2v2i+···+ hvvi
(by property (5) of an inner product and part (3) of Theorem 7.12,
and sincehv
viequals0whenandare distinct)
=
1
1+22+···+ (sincekv k=1for1≤≤)
=hvv
1i
hwv 1i+hvv 2ihwv 2i+···+hvv ihwv i
(19) Usingw
1=
2
−+1,w 2=1,andw 3=yields the orthogonal basis{v 1v2v3},withv 1=
2
−+1,
v
2=−20
2
+20+13,andv 3=15
2
+4−5.
(20){[−9−48][2711−22][53−4]}
(21) The proof is totally analogous to the (long) proof of Theorem 6.4 in Section 6.1 of the textbook. Use
Theorem 7.15 in the proof in place of Theorem 6.1.
(22) (a) Follow the proof of Theorem 6.11, using properties (4) and (5) of an inner product. In the last
step, use the fact thathwwi=0 =⇒w=0(by property (2) of an inner product).
(b) Prove part (4) of Theorem 7.19first, using a proof similar to the proof of Theorem 6.12. (In
that proof, use Theorem 7.16 in place of Theorem 6.3.) Then prove part (5) of Theorem 7.19 as
follows:
LetWbe a subspace ofVof dimension. By Theorem 7.17,Whas an orthogonal basis
{v
1v }. Expand this basis to an orthogonal basis for all ofV.(Thatis,first expand to
any basis forVby Theorem 4.15, then use the Gram-Schmidt Process. Since thefirstvectors
are already orthogonal, this expands{v
1v }to an orthogonal basis{v 1v }forV.)
Then, by part (4) of Theorem 7.19,{v
+1v }is a basis forW
⊥
, and so dim(W
⊥
)=−.
Hence dim(W)+dim(W
⊥
)=.
(c) Since any vector inWis orthogonal to all vectors inW
⊥
,W⊆(W
⊥
)
⊥
(d) By part (3) of Theorem 7.19,W⊆(W
⊥
)
⊥
. Next, by part (5) of Theorem 7.19,
dim(W)=−dim(W
⊥
)=−(−dim((W
⊥
)
⊥
)) = dim((W
⊥
)
⊥
)
Thus, by Theorem 4.13, or its complex analog,W=(W
⊥
)
⊥
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 139
(17) Let=(v 1v )be an orthogonal ordered basis for a subspaceWofV,andletv∈W.Let
[v]
=[1]. The goal is to show that =hvv ikv k
2
.Now,v= 1v1+···+ v. Hence,
hvv
i=h 1v1+···+ vvi
=
1hv1vi+···+ hvvi+···+ hvvi
(by properties (4) and (5) of an inner product)
=
1(0) +···+ −1(0) + kvk
2
++1(0) +···+ (0)
(sinceis orthogonal)
=
kvk
2
Hence,
=hvv ikv k
2
. Also, ifis orthonormal, then eachkv k=1,so =hvv i.
(18) Letv=
1v1+···+ vandw= 1v1+···+ v,where =hvv iand =hwv i(by
Theorem 7.16). Then
hvwi=h
1v1+···+ v1v1+···+ vi
=
1
1hv1v1i+ 22hv2v2i+···+ hvvi
(by property (5) of an inner product and part (3) of Theorem 7.12,
and sincehv
viequals0whenandare distinct)
=
1
1+22+···+ (sincekv k=1for1≤≤)
=hvv
1i
hwv 1i+hvv 2ihwv 2i+···+hvv ihwv i
(19) Usingw
1=
2
−+1,w 2=1,andw 3=yields the orthogonal basis{v 1v2v3},withv 1=
2
−+1,
v
2=−20
2
+20+13,andv 3=15
2
+4−5.
(20){[−9−48][2711−22][53−4]}
(21) The proof is totally analogous to the (long) proof of Theorem 6.4 in Section 6.1 of the textbook. Use
Theorem 7.15 in the proof in place of Theorem 6.1.
(22) (a) Follow the proof of Theorem 6.11, using properties (4) and (5) of an inner product. In the last
step, use the fact thathwwi=0 =⇒w=0(by property (2) of an inner product).
(b) Prove part (4) of Theorem 7.19first, using a proof similar to the proof of Theorem 6.12. (In
that proof, use Theorem 7.16 in place of Theorem 6.3.) Then prove part (5) of Theorem 7.19 as
follows:
LetWbe a subspace ofVof dimension. By Theorem 7.17,Whas an orthogonal basis
{v
1v }. Expand this basis to an orthogonal basis for all ofV.(Thatis,first expand to
any basis forVby Theorem 4.15, then use the Gram-Schmidt Process. Since thefirstvectors
are already orthogonal, this expands{v
1v }to an orthogonal basis{v 1v }forV.)
Then, by part (4) of Theorem 7.19,{v
+1v }is a basis forW
⊥
, and so dim(W
⊥
)=−.
Hence dim(W)+dim(W
⊥
)=.
(c) Since any vector inWis orthogonal to all vectors inW
⊥
,W⊆(W
⊥
)
⊥
(d) By part (3) of Theorem 7.19,W⊆(W
⊥
)
⊥
. Next, by part (5) of Theorem 7.19,
dim(W)=−dim(W
⊥
)=−(−dim((W
⊥
)
⊥
)) = dim((W
⊥
)
⊥
)
Thus, by Theorem 4.13, or its complex analog,W=(W
⊥
)
⊥
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 139
Answers to Exercises Section 7.5
(23)W
⊥
=span({
3
−
2
+1})
(24) Using1andas the vectors in the Gram-Schmidt Process, the basis obtained forW
⊥
is
{−5
2
+10−215
2
−14+2}.
(25) As in the hint, let{v
1v
}be an orthonormal basis forW.Nowifv∈V,let
w
1=hvv 1iv1+···+hvv iv
andw 2=v−w 1. Then,w 1∈Wbecausew 1is a linear combination of basis vectors forW.Weclaim
thatw
2∈W
⊥
.Toseethis,letu∈W.Thenu= 1v1+···+ vfor some 1.Then
huw
2i=huv−w 1i=h 1v1+···+ vv−(hvv 1iv1+···+hvv iv)i
=h
1v1+···+ vvi−h 1v1+···+ vhvv 1iv1+···+hvv ivi
=
X
=1
hvvi−
X
=1
X
=1
hvv ihvvi
Buthv
vi=0when6 =andhv vi=1when=,since{v 1v }is an orthonormal set.
Hence,
huw
2i=
X
=1
hvvi−
X
=1
hvv i
=
X
=1
hvvi−
X
=1
hvvi=0
Since this is true for everyu∈W,weconcludethatw
2∈W
⊥
.
Finally, we want to show uniqueness of decomposition. Suppose thatv=w
1+w2andv=w
0
1
+w
0
2
wherew
1w
0
1
∈Wandw 2,w
0
2
∈W
⊥
. We want to show thatw 1=w
0
1
andw 2=w
0
2
Now,
w
1−w
0
1
=w
0
2
−w2.Also,w 1−w
0
1
∈W,butw
0
2
−w2∈W
⊥
Thus,w 1−w
0
1
=w
0
2
−w2∈W∩W
⊥
By part (2) of Theorem 7.19,w
1−w
0
1
=w
0
2
−w2=0Hence,w 1=w
0
1
andw 2=w
0
2
.
(26)w
1=
1
2
(sin−cos),w 2=
1
−
1
2
sin+
1
2
cos
(27) Orthonormal basis forW=
nq
15
14
(2
2
−1)
q
3
322
(10
2
+7+2)
o
.Thenv=4
2
−+3=w 1+w2,
wherew
1=
1
23
(32
2
+73+ 57)∈Wandw 2=
12
23
(5
2
−8+1)∈W
⊥
.
(28) LetW=span({v
1v }).Noticethatw 1=
P
=1
hvv iv(by Theorem 7.16, since{v 1v }
is an orthonormal basis forWby Theorem 7.15). Then, using the hint in the text yields
kvk
2
=hvvi=hw 1+w2w1+w2i
=hw
1w1i+hw 2w1i+hw 1w2i+hw 2w2i
=kw
1k
2
+kw 2k
2
(sincew 1∈Wandw 2∈W
⊥
)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 140
(23)W
⊥
=span({
3
−
2
+1})
(24) Using1andas the vectors in the Gram-Schmidt Process, the basis obtained forW
⊥
is
{−5
2
+10−215
2
−14+2}.
(25) As in the hint, let{v
1v
}be an orthonormal basis forW.Nowifv∈V,let
w
1=hvv 1iv1+···+hvv iv
andw 2=v−w 1. Then,w 1∈Wbecausew 1is a linear combination of basis vectors forW.Weclaim
thatw
2∈W
⊥
.Toseethis,letu∈W.Thenu= 1v1+···+ vfor some 1.Then
huw
2i=huv−w 1i=h 1v1+···+ vv−(hvv 1iv1+···+hvv iv)i
=h
1v1+···+ vvi−h 1v1+···+ vhvv 1iv1+···+hvv ivi
=
X
=1
hvvi−
X
=1
X
=1
hvv ihvvi
Buthv
vi=0when6 =andhv vi=1when=,since{v 1v }is an orthonormal set.
Hence,
huw
2i=
X
=1
hvvi−
X
=1
hvv i
=
X
=1
hvvi−
X
=1
hvvi=0
Since this is true for everyu∈W,weconcludethatw
2∈W
⊥
.
Finally, we want to show uniqueness of decomposition. Suppose thatv=w
1+w2andv=w
0
1
+w
0
2
wherew
1w
0
1
∈Wandw 2,w
0
2
∈W
⊥
. We want to show thatw 1=w
0
1
andw 2=w
0
2
Now,
w
1−w
0
1
=w
0
2
−w2.Also,w 1−w
0
1
∈W,butw
0
2
−w2∈W
⊥
Thus,w 1−w
0
1
=w
0
2
−w2∈W∩W
⊥
By part (2) of Theorem 7.19,w
1−w
0
1
=w
0
2
−w2=0Hence,w 1=w
0
1
andw 2=w
0
2
.
(26)w
1=
1
2
(sin−cos),w 2=
1
−
1
2
sin+
1
2
cos
(27) Orthonormal basis forW=
nq
15
14
(2
2
−1)
q
3
322
(10
2
+7+2)
o
.Thenv=4
2
−+3=w 1+w2,
wherew
1=
1
23
(32
2
+73+ 57)∈Wandw 2=
12
23
(5
2
−8+1)∈W
⊥
.
(28) LetW=span({v
1v }).Noticethatw 1=
P
=1
hvv iv(by Theorem 7.16, since{v 1v }
is an orthonormal basis forWby Theorem 7.15). Then, using the hint in the text yields
kvk
2
=hvvi=hw 1+w2w1+w2i
=hw
1w1i+hw 2w1i+hw 1w2i+hw 2w2i
=kw
1k
2
+kw 2k
2
(sincew 1∈Wandw 2∈W
⊥
)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 140
Answers to Exercises Chapter 7 Review
Hence,
kvk
2
≥kw 1k
2
=
*Ã
X
=1
hvv iv
!
⎛
⎝
X
=1
hvv iv
⎞
⎠
+
=
X
=1
*
(hvv
iv)
⎛
⎝
X
=1
hvv iv
⎞
⎠
+
=
X
=1
X
=1
h(hvv iv)(hvv iv)i
=
X
=1
X
=1
((hvv ihvv i)hv vi)
=
X
=1
(hvv ihvv i)(because{v 1v }is an orthonormal set)
=
X
=1
hvv i
2
(29) (a) Let{v
1v }be an orthonormal basis forW.Letxy∈V.Then
proj
Wx=hxv 1iv1+···+hxv iv
proj
Wy=hyv 1iv1+···+hyv iv
proj
W(x+y)= hx+yv 1iv1+···+hx+yv ivand
proj
W(x)= hxv 1iv1+···+hxv iv
Clearly, by properties (4) and (5) of an inner product,proj
W(x+y)=proj
Wx+proj
Wy,and
proj
W(x)=(proj
Wx). Henceis a linear transformation.
(b) ker()=W
⊥
, range()=W
(c) First, we establish that ifv∈W,then(v)=v. Note that forv∈W,v=v+0,wherev∈W
and0∈W
⊥
. But by Theorem 7.20, any decompositionv=w 1+w2withw 1∈Wandw 2∈W
⊥
is unique. Hence,w 1=vandw 2=0. Then, sincew 1=proj
Wv,weget(v)=v.
Finally, letv∈Vand letx=(v)=proj
Wv∈W.Then
(◦)(v)=((v)) =(x)=proj
Wx=x(sincex∈W)=(v)
Hence,◦=.
(30) (a) F (b) F (c) F (d) T (e) F
Chapter 7 Review Exercises
(1) (a)0
(b)(1 + 2)(v·z) = ((1 + 2)v)·z=−21 + 43,v·((1 + 2)z)=47−9
Copyrightc°2016 Elsevier Ltd. All rights reserved. 141
Hence,
kvk
2
≥kw 1k
2
=
*Ã
X
=1
hvv iv
!
⎛
⎝
X
=1
hvv iv
⎞
⎠
+
=
X
=1
*
(hvv
iv)
⎛
⎝
X
=1
hvv iv
⎞
⎠
+
=
X
=1
X
=1
h(hvv iv)(hvv iv)i
=
X
=1
X
=1
((hvv ihvv i)hv vi)
=
X
=1
(hvv ihvv i)(because{v 1v }is an orthonormal set)
=
X
=1
hvv i
2
(29) (a) Let{v
1v }be an orthonormal basis forW.Letxy∈V.Then
proj
Wx=hxv 1iv1+···+hxv iv
proj
Wy=hyv 1iv1+···+hyv iv
proj
W(x+y)= hx+yv 1iv1+···+hx+yv ivand
proj
W(x)= hxv 1iv1+···+hxv iv
Clearly, by properties (4) and (5) of an inner product,proj
W(x+y)=proj
Wx+proj
Wy,and
proj
W(x)=(proj
Wx). Henceis a linear transformation.
(b) ker()=W
⊥
, range()=W
(c) First, we establish that ifv∈W,then(v)=v. Note that forv∈W,v=v+0,wherev∈W
and0∈W
⊥
. But by Theorem 7.20, any decompositionv=w 1+w2withw 1∈Wandw 2∈W
⊥
is unique. Hence,w 1=vandw 2=0. Then, sincew 1=proj
Wv,weget(v)=v.
Finally, letv∈Vand letx=(v)=proj
Wv∈W.Then
(◦)(v)=((v)) =(x)=proj
Wx=x(sincex∈W)=(v)
Hence,◦=.
(30) (a) F (b) F (c) F (d) T (e) F
Chapter 7 Review Exercises
(1) (a)0
(b)(1 + 2)(v·z) = ((1 + 2)v)·z=−21 + 43,v·((1 + 2)z)=47−9
Copyrightc°2016 Elsevier Ltd. All rights reserved. 141
Answers to Exercises Chapter 7 Review
(c) Since(v·z)=(13+17),
(1 + 2)(v·z) = ((1 + 2)v)·z=(1 + 2)(13 + 17)=−21 + 43
while
v·((1 + 2)z)=(1 + 2)(13 + 17)=(1−2)(13 + 17)=47−9
(d)w·z=−35 + 24;w·(v+z)=−35 + 24,sincew·v=0
(2) (a)H=
⎡
⎣
21+3 7−
1−3 34 −10−10
7+−10 + 10 30
⎤
⎦
(b)AA
∗
=
∙
32 −2+14
−2−14 34
¸
, which is clearly Hermitian.
(3)(Az)·w=z·(A
∗
w)(by Theorem 7.3)=z·(−Aw)(sinceAis skew-Hermitian)=−z·(Aw)(by
Theorem 7.1, part (5))
(4) (a)=4+3,=−2
(b)=2+7,=0,=3−2
(c) No solutions
(d){[(2 +)−(3−)(7 +)− ]|∈C}={[(2−3)+(1+)7+(1−) ]|∈C}
(5) By part (3) of Theorem 7.5,|A
∗
|=
|A|. Hence,|A
∗
A|=|A
∗
|·|A|=|A|·|A|=
¯
¯
¯
¯
|A|
¯
¯
¯
¯
2
,whichisreal
and nonnegative. This equals zero if and only if|A|=0, which occurs if and only ifAis singular (by
part (4) of Theorem 7.5).
(6) (a)
A()=
3
−
2
+−1=(
2
+1)(−1) = (−)(+)(−1);
D=
⎡
⎣
00
0−0
001
⎤
⎦;P=
⎡
⎣
−2−−2+0
1−1+2
111
⎤
⎦
(b)
A()=
3
−2
2
−=(−)
2
; not diagonalizable. Eigenspace for=is one-dimensional,
with fundamental eigenvector[1−3−11]. Fundamental eigenvector for=0is[−11].
(7) (a) One possibility: Consider:C→Cgiven by(z)=
z.Notethat(v+w)=(v+w)=v+w=
(v)+(w).Butis not a linear operator onCbecause()=−, but(1) =(1) =,sothe
rule “(v)=(v)”isnotsatisfied.
(b) The example given in part (a)isalinearoperatoronC, thought of as arealvector space. In
that case we may use only real scalars, and so, ifv=+,then(v)=(+)=−=
(−)=(v).
Note: for any functionfrom any vector space to itself (real or complex), the rule “(v+w)=
(v)+(w)” implies that(
v)=(v)for anyrationalscalar.Thus,anyexampleforreal
vector spaces for which(v+w)=(v)+(w)is satisfied, but(v)=(v)is not satisfied
for some, must involve anirrationalvalue of.
One possibility: ConsiderRas a vector space over the rationals (as the scalarfield). Choose
an uncountably infinite basisforRoverQwith1∈andπ∈.Define:R→Rby letting
(1)=1,but(v)=0for everyv∈withv6 =1.(Defining
on a basis uniquely determines
Copyrightc°2016 Elsevier Ltd. All rights reserved. 142
(c) Since(v·z)=(13+17),
(1 + 2)(v·z) = ((1 + 2)v)·z=(1 + 2)(13 + 17)=−21 + 43
while
v·((1 + 2)z)=(1 + 2)(13 + 17)=(1−2)(13 + 17)=47−9
(d)w·z=−35 + 24;w·(v+z)=−35 + 24,sincew·v=0
(2) (a)H=
⎡
⎣
21+3 7−
1−3 34 −10−10
7+−10 + 10 30
⎤
⎦
(b)AA
∗
=
∙
32 −2+14
−2−14 34
¸
, which is clearly Hermitian.
(3)(Az)·w=z·(A
∗
w)(by Theorem 7.3)=z·(−Aw)(sinceAis skew-Hermitian)=−z·(Aw)(by
Theorem 7.1, part (5))
(4) (a)=4+3,=−2
(b)=2+7,=0,=3−2
(c) No solutions
(d){[(2 +)−(3−)(7 +)− ]|∈C}={[(2−3)+(1+)7+(1−) ]|∈C}
(5) By part (3) of Theorem 7.5,|A
∗
|=
|A|. Hence,|A
∗
A|=|A
∗
|·|A|=|A|·|A|=
¯
¯
¯
¯
|A|
¯
¯
¯
¯
2
,whichisreal
and nonnegative. This equals zero if and only if|A|=0, which occurs if and only ifAis singular (by
part (4) of Theorem 7.5).
(6) (a)
A()=
3
−
2
+−1=(
2
+1)(−1) = (−)(+)(−1);
D=
⎡
⎣
00
0−0
001
⎤
⎦;P=
⎡
⎣
−2−−2+0
1−1+2
111
⎤
⎦
(b)
A()=
3
−2
2
−=(−)
2
; not diagonalizable. Eigenspace for=is one-dimensional,
with fundamental eigenvector[1−3−11]. Fundamental eigenvector for=0is[−11].
(7) (a) One possibility: Consider:C→Cgiven by(z)=
z.Notethat(v+w)=(v+w)=v+w=
(v)+(w).Butis not a linear operator onCbecause()=−, but(1) =(1) =,sothe
rule “(v)=(v)”isnotsatisfied.
(b) The example given in part (a)isalinearoperatoronC, thought of as arealvector space. In
that case we may use only real scalars, and so, ifv=+,then(v)=(+)=−=
(−)=(v).
Note: for any functionfrom any vector space to itself (real or complex), the rule “(v+w)=
(v)+(w)” implies that(
v)=(v)for anyrationalscalar.Thus,anyexampleforreal
vector spaces for which(v+w)=(v)+(w)is satisfied, but(v)=(v)is not satisfied
for some, must involve anirrationalvalue of.
One possibility: ConsiderRas a vector space over the rationals (as the scalarfield). Choose
an uncountably infinite basisforRoverQwith1∈andπ∈.Define:R→Rby letting
(1)=1,but(v)=0for everyv∈withv6 =1.(Defining
on a basis uniquely determines
Copyrightc°2016 Elsevier Ltd. All rights reserved. 142
Answers to Exercises Chapter 7 Review
.) Thisis a linear operator onRoverQ,andso(v+w)=(v)+(w)is satisfied. But, by
definition()=0, but(1)=(1)=.Thus,is not a linear operator on therealvector
spaceR.
(8) (a)={[11−][4 + 57−4 1][10−2+6−8−−1+8][001]}
(b)={
1
2
[11−]
1
6
√
3
[4 + 57−4 1]
1
3
√
30
[10−2+6−8−−1+8]
1
√
2
[001]}
(c)
⎡
⎢
⎢
⎢
⎢
⎣
1
2
−
1
2
1
2
1
2
1
6
√
3
(4−5)
1
6
√
3
(7 + 4) −
1
6
√
3
1
6
√
3
10
3
√
30
1
3
√
30
(−2−6)
1
3
√
30
(−8+)
1
3
√
30
(−1−8)
00
1
√
2
−
1
√
2
⎤
⎥
⎥
⎥
⎥
⎦
(9) (a)
A()=
2
−5+4=(−1)(−4);D=
∙
10
04
¸
;P=
⎡
⎣
1
√
6
(−1−)
1
√
3
(1 +)
2
√
6
1
√
3
⎤
⎦
(b)
A()=
3
+(−98 + 98)
2
−4802=(−49 + 49)
2
;D=
⎡
⎣
00 0
049−49 0
0049 −49
⎤
⎦;
P=
⎡
⎢
⎢
⎢
⎣
6
7
1
√
5
−4
7
√
5
3
7
2
√
5
−2
7
√
5
2
7
0
15
7
√
5
⎤
⎥
⎥
⎥
⎦
is a probable answer. Another possibility:P=
1
7
⎡
⎣
3−2−6
263
632
⎤
⎦
(10) BothAA
∗
andA
∗
Aequal
⎡
⎣
80 105 + 45 60
105−45 185 60−60
60 60 + 60 95
⎤
⎦,andsoAis normal. Hence, by
Theorem 7.9,Ais unitarily diagonalizable.
(11) Now,A
∗
A=
⎡
⎢
⎢
⎢
⎢
⎣
459 −46 + 459−459 + 46−11−925−83 + 2227
−46−459 473 92 + 445 −918 + 1032248−139
−459−4692−445 473 −81 + 932305−2206
−11 + 925−918−103−81−932 1871 −4482−218
−83−22272248 + 139305 + 2206−4482 + 218 10843
⎤
⎥
⎥
⎥
⎥
⎦
butAA
∗
=
⎡
⎢
⎢
⎢
⎢
⎣
7759 −120 + 23953850−1881−1230−3862−95−2905
−120−2395 769 −648−1165−1188 + 445−906 + 75
3850 + 1881−648 + 1165 2371 329 −2220660−1467
−1230 + 3862−1188−445329 + 2220 2127 1467 + 415
−95 + 2905−906−75660 + 14671467−415 1093
⎤
⎥
⎥
⎥
⎥
⎦
soAis not normal. Hence,Ais not unitarily diagonalizable, by Theorem 7.9. (Note:Aisdiagonal-
izable (with eigenvalues0,±1,and
±), but the eigenspaces are not orthogonal to each other.)
(12) IfUis a unitary matrix, thenU
∗
=U
−1
. Hence,U
∗
U=UU
∗
=I.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 143
.) Thisis a linear operator onRoverQ,andso(v+w)=(v)+(w)is satisfied. But, by
definition()=0, but(1)=(1)=.Thus,is not a linear operator on therealvector
spaceR.
(8) (a)={[11−][4 + 57−4 1][10−2+6−8−−1+8][001]}
(b)={
1
2
[11−]
1
6
√
3
[4 + 57−4 1]
1
3
√
30
[10−2+6−8−−1+8]
1
√
2
[001]}
(c)
⎡
⎢
⎢
⎢
⎢
⎣
1
2
−
1
2
1
2
1
2
1
6
√
3
(4−5)
1
6
√
3
(7 + 4) −
1
6
√
3
1
6
√
3
10
3
√
30
1
3
√
30
(−2−6)
1
3
√
30
(−8+)
1
3
√
30
(−1−8)
00
1
√
2
−
1
√
2
⎤
⎥
⎥
⎥
⎥
⎦
(9) (a)
A()=
2
−5+4=(−1)(−4);D=
∙
10
04
¸
;P=
⎡
⎣
1
√
6
(−1−)
1
√
3
(1 +)
2
√
6
1
√
3
⎤
⎦
(b)
A()=
3
+(−98 + 98)
2
−4802=(−49 + 49)
2
;D=
⎡
⎣
00 0
049−49 0
0049 −49
⎤
⎦;
P=
⎡
⎢
⎢
⎢
⎣
6
7
1
√
5
−4
7
√
5
3
7
2
√
5
−2
7
√
5
2
7
0
15
7
√
5
⎤
⎥
⎥
⎥
⎦
is a probable answer. Another possibility:P=
1
7
⎡
⎣
3−2−6
263
632
⎤
⎦
(10) BothAA
∗
andA
∗
Aequal
⎡
⎣
80 105 + 45 60
105−45 185 60−60
60 60 + 60 95
⎤
⎦,andsoAis normal. Hence, by
Theorem 7.9,Ais unitarily diagonalizable.
(11) Now,A
∗
A=
⎡
⎢
⎢
⎢
⎢
⎣
459 −46 + 459−459 + 46−11−925−83 + 2227
−46−459 473 92 + 445 −918 + 1032248−139
−459−4692−445 473 −81 + 932305−2206
−11 + 925−918−103−81−932 1871 −4482−218
−83−22272248 + 139305 + 2206−4482 + 218 10843
⎤
⎥
⎥
⎥
⎥
⎦
butAA
∗
=
⎡
⎢
⎢
⎢
⎢
⎣
7759 −120 + 23953850−1881−1230−3862−95−2905
−120−2395 769 −648−1165−1188 + 445−906 + 75
3850 + 1881−648 + 1165 2371 329 −2220660−1467
−1230 + 3862−1188−445329 + 2220 2127 1467 + 415
−95 + 2905−906−75660 + 14671467−415 1093
⎤
⎥
⎥
⎥
⎥
⎦
soAis not normal. Hence,Ais not unitarily diagonalizable, by Theorem 7.9. (Note:Aisdiagonal-
izable (with eigenvalues0,±1,and
±), but the eigenspaces are not orthogonal to each other.)
(12) IfUis a unitary matrix, thenU
∗
=U
−1
. Hence,U
∗
U=UU
∗
=I.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 143
Answers to Exercises Chapter 7 Review
(13) Distance=
q
8
105
≈0276
(14){[100][430][542]}
(15) Orthogonal basis forW:{sin cos+
1
2
sin};w 1=2sin−
36
4
2
+3
(cos+
1
2
sin);w 2=−w 1
(16) (a) T
(b) F
(c) T
(d) F
(e) F
(f) T
(g) F
(h) T
(i) F
(j) T
(k) T
(l) T
(m) F
(n) T
(o) T
(p) T
(q) T
(r) T
(s) T
(t) T
(u) T
(v) T
(w) F
Copyrightc°2016 Elsevier Ltd. All rights reserved. 144
(13) Distance=
q
8
105
≈0276
(14){[100][430][542]}
(15) Orthogonal basis forW:{sin cos+
1
2
sin};w 1=2sin−
36
4
2
+3
(cos+
1
2
sin);w 2=−w 1
(16) (a) T
(b) F
(c) T
(d) F
(e) F
(f) T
(g) F
(h) T
(i) F
(j) T
(k) T
(l) T
(m) F
(n) T
(o) T
(p) T
(q) T
(r) T
(s) T
(t) T
(u) T
(v) T
(w) F
Copyrightc°2016 Elsevier Ltd. All rights reserved. 144
Answers to Exercises Section 8.1
Chapter 8
Section 8.1
(1) Symmetric:(a),(b),(c),(d)
(a)
1:
⎡
⎢
⎢
⎣
0111
1011
1101
1110
⎤
⎥
⎥
⎦ (b)
2:
⎡
⎢
⎢
⎢
⎢
⎣
12001
20010
00101
01001
10111
⎤
⎥
⎥
⎥
⎥
⎦
(c)
3:
⎡
⎣
000
000
000
⎤
⎦
(d)
4:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
010100
102100
020011
110020
001201
001010
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(e)
1:
⎡
⎢
⎢
⎣
0100
0010
1001
1100
⎤
⎥
⎥
⎦
(f) 2:
⎡
⎢
⎢
⎣
0210
0111
0101
0001
⎤
⎥
⎥
⎦
(g)
3:
⎡
⎢
⎢
⎢
⎢
⎣
02200
10001
10020
00101
02020
⎤
⎥
⎥
⎥
⎥
⎦
(h)
4:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
02000000
00100000
00010000
00001000
00000200
00000010
00000001
10000000
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(2) Allfigures for this exercise appear on the next page.
Ccan be the adjacency matrix for a digraph (only) (see Figure 12).
Fcan be the adjacency matrix for either a graph or digraph (see Figure 13).
Gcan be the adjacency matrix for a digraph (only) (see Figure 14).
Hcan be the adjacency matrix for a digraph (only) (see Figure 15).
Ican be the adjacency matrix for a graph or digraph (see Figure 16).
Jcan be the adjacency matrix for a graph or digraph (see Figure 17).
Kcan be the adjacency matrix for a graph or digraph (see Figure 18).
(3) The digraph is shown in Figure 19 (see the next page), and the adjacency matrix is
A
B
C
D
E
F
ABCDEF
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
001000
000110
010010
100001
110100
000100
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
The transpose gives no new information. But it does suggest a different interpretation of the
results: namely, the( )entry of the transpose equals1if authorinfluences author.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 145
Chapter 8
Section 8.1
(1) Symmetric:(a),(b),(c),(d)
(a)
1:
⎡
⎢
⎢
⎣
0111
1011
1101
1110
⎤
⎥
⎥
⎦ (b)
2:
⎡
⎢
⎢
⎢
⎢
⎣
12001
20010
00101
01001
10111
⎤
⎥
⎥
⎥
⎥
⎦
(c)
3:
⎡
⎣
000
000
000
⎤
⎦
(d)
4:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
010100
102100
020011
110020
001201
001010
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(e)
1:
⎡
⎢
⎢
⎣
0100
0010
1001
1100
⎤
⎥
⎥
⎦
(f) 2:
⎡
⎢
⎢
⎣
0210
0111
0101
0001
⎤
⎥
⎥
⎦
(g)
3:
⎡
⎢
⎢
⎢
⎢
⎣
02200
10001
10020
00101
02020
⎤
⎥
⎥
⎥
⎥
⎦
(h)
4:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
02000000
00100000
00010000
00001000
00000200
00000010
00000001
10000000
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(2) Allfigures for this exercise appear on the next page.
Ccan be the adjacency matrix for a digraph (only) (see Figure 12).
Fcan be the adjacency matrix for either a graph or digraph (see Figure 13).
Gcan be the adjacency matrix for a digraph (only) (see Figure 14).
Hcan be the adjacency matrix for a digraph (only) (see Figure 15).
Ican be the adjacency matrix for a graph or digraph (see Figure 16).
Jcan be the adjacency matrix for a graph or digraph (see Figure 17).
Kcan be the adjacency matrix for a graph or digraph (see Figure 18).
(3) The digraph is shown in Figure 19 (see the next page), and the adjacency matrix is
A
B
C
D
E
F
ABCDEF
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
001000
000110
010010
100001
110100
000100
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
The transpose gives no new information. But it does suggest a different interpretation of the
results: namely, the( )entry of the transpose equals1if authorinfluences author.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 145
Answers to Exercises Section 8.1
(4) (a)15
(b)56
(c)74 = 1 + 2 + 15 + 56
(d)89 = 0 + 4 + 10 + 75
(e) Length2
(f) Length2
(5) (a)4
(b)0
(c)7=1+2+4
(d)18=2+0+8+8
(e) Nosuchpathexists
(f) Length2
(6) (a)8 (b)114 (c)92=0+6+12+74
(7) (a)3 (b)10 (c)19=1+2+4+12
(8) (a) If the vertex is theth vertex, then theth row andth column entries of the adjacency matrix
all equal zero, except possibly for the( )entry.
(b) If the vertex is theth vertex, then theth row entries of the adjacency matrix all equal zero,
except possibly for the( )entry. (Note: Theth column entries may be nonzero.)
(9) (a) The trace equals the total number of loops in the graph or digraph.
(b) The trace equals the total number of cycles of lengthin the graph or digraph. (See Exercise
6 in the textbook for the definition of a cycle.)
(10) Connected: (b), (c); Disconnected: (a), (d)
(11) (a) Since
2hasthesameedgesas 1, along with one new loop at each vertex, 2has the same
number of edges connecting any two distinct vertices as
1.Thus,theentriesoffthe main
diagonal of the adjacency matrices for the two graphs are the same. But
2has one more
loop at each vertex than
1has. Hence, the entries on the main diagonal of the adjacency
matrix for
2are all1larger than the entries ofA, so the adjacency matrix for 2is found
by addingI
toA. This does not change any entries offthe main diagonal, but adds1to
every entry on the main diagonal.
(b) Connectivity only involves the existence of a path betweendistinctvertices. If there is a path
from
toin 1,for6 =, the same path connects these two vertices in 2.Ifthereisa
path from
toin 2, then a similar path can be found from toin 1merely by
deleting any loops that might appear in the path from
toin 2.Thus, is connected
by a path to
in 1if and only if is connected by a path to in 2. Hence, 1is
connected if and only if
2is connected.
(c) Suppose there is a path of lengthfrom
toin 1,where≤. Then, we canfind a
path of lengthfrom
toin 2byfirst following the known path of lengthfrom
tothat exists in 1(and hence in 2), and then going(−)times around the newly
addedloopat
.
(d) In the discussion in the text for Theorem 8.3, we saw that a graph is connected if and only if
every pair of distinct vertices are connected by a path of length≤(−1). However, using an
argument similar to that in part (c), if there is a path of length less than(−1)connecting
two vertices in
2, then there is a path of length equal to(−1)connecting those same two
vertices. Hence,
2is connected if and only if each pair of distinct vertices is connected by
a path of length(−1). (Note: Because of the newly added loops, there is a path of every
length≥1connecting each vertex in
2to itself. Thus, all of the main diagonal entries of
any positive power of(A+I
)are nonzero.)
(e) By parts (a) and (d), including the note at the end of the answer for (d) above, and Theorem
8.1,
2is connected if and only if(A+I )
−1
has no zero entries. But, by part (b), 1is
connected if and only if
2is connected. Hence, 1is connected if and only if(A+I )
−1
has no zero entries.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 147
(4) (a)15
(b)56
(c)74 = 1 + 2 + 15 + 56
(d)89 = 0 + 4 + 10 + 75
(e) Length2
(f) Length2
(5) (a)4
(b)0
(c)7=1+2+4
(d)18=2+0+8+8
(e) Nosuchpathexists
(f) Length2
(6) (a)8 (b)114 (c)92=0+6+12+74
(7) (a)3 (b)10 (c)19=1+2+4+12
(8) (a) If the vertex is theth vertex, then theth row andth column entries of the adjacency matrix
all equal zero, except possibly for the( )entry.
(b) If the vertex is theth vertex, then theth row entries of the adjacency matrix all equal zero,
except possibly for the( )entry. (Note: Theth column entries may be nonzero.)
(9) (a) The trace equals the total number of loops in the graph or digraph.
(b) The trace equals the total number of cycles of lengthin the graph or digraph. (See Exercise
6 in the textbook for the definition of a cycle.)
(10) Connected: (b), (c); Disconnected: (a), (d)
(11) (a) Since
2hasthesameedgesas 1, along with one new loop at each vertex, 2has the same
number of edges connecting any two distinct vertices as
1.Thus,theentriesoffthe main
diagonal of the adjacency matrices for the two graphs are the same. But
2has one more
loop at each vertex than
1has. Hence, the entries on the main diagonal of the adjacency
matrix for
2are all1larger than the entries ofA, so the adjacency matrix for 2is found
by addingI
toA. This does not change any entries offthe main diagonal, but adds1to
every entry on the main diagonal.
(b) Connectivity only involves the existence of a path betweendistinctvertices. If there is a path
from
toin 1,for6 =, the same path connects these two vertices in 2.Ifthereisa
path from
toin 2, then a similar path can be found from toin 1merely by
deleting any loops that might appear in the path from
toin 2.Thus, is connected
by a path to
in 1if and only if is connected by a path to in 2. Hence, 1is
connected if and only if
2is connected.
(c) Suppose there is a path of lengthfrom
toin 1,where≤. Then, we canfind a
path of lengthfrom
toin 2byfirst following the known path of lengthfrom
tothat exists in 1(and hence in 2), and then going(−)times around the newly
addedloopat
.
(d) In the discussion in the text for Theorem 8.3, we saw that a graph is connected if and only if
every pair of distinct vertices are connected by a path of length≤(−1). However, using an
argument similar to that in part (c), if there is a path of length less than(−1)connecting
two vertices in
2, then there is a path of length equal to(−1)connecting those same two
vertices. Hence,
2is connected if and only if each pair of distinct vertices is connected by
a path of length(−1). (Note: Because of the newly added loops, there is a path of every
length≥1connecting each vertex in
2to itself. Thus, all of the main diagonal entries of
any positive power of(A+I
)are nonzero.)
(e) By parts (a) and (d), including the note at the end of the answer for (d) above, and Theorem
8.1,
2is connected if and only if(A+I )
−1
has no zero entries. But, by part (b), 1is
connected if and only if
2is connected. Hence, 1is connected if and only if(A+I )
−1
has no zero entries.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 147
Answers to Exercises Section 8.1
(f) Part (a):(A+I 4)
3
=
⎡
⎢
⎢
⎣
4103240
0800
3202532
4003237
⎤
⎥
⎥
⎦
; disconnected;
Part (b):(A+I
4)
3
=
⎡
⎢
⎢
⎣
13 4 9 1
4765
9683
1534
⎤
⎥
⎥
⎦
; connected;
Part (c):(A+I
5)
4
=
⎡
⎢
⎢
⎢
⎢
⎣
912 812 6
12 93 120 28 120
8 120 252 42 168
12 28 42 21 24
6 120 168 24 172
⎤
⎥
⎥
⎥
⎥
⎦
; connected;
Part (d):(A+I
5)
4
=
⎡
⎢
⎢
⎢
⎢
⎣
86 0 0 85 85
0 100 78 0 0
0786100
85 0 0 86 85
85 0 0 85 86
⎤
⎥
⎥
⎥
⎥
⎦
; disconnected.
(12) If the graph is connected, then vertex
must have at least one edge connecting it to some other
distinct vertex
. (Otherwise there could be no paths at all connecting to any other vertex.)
But then, the path
→ → is a path of length2connecting to itself. Hence, the( )
entry ofA
2
is at least1. Since this is true for each, all entries on the main diagonal ofA
2
are
nonzero.
(13) (a) The digraph
2in Figure 8.2 is strongly connected since there is a path from any vertex to any
other vertex. (This can be verified by inspection, or by using part (b) of this exercise — that
is, lettingArepresent the adjacency matrix for this digraph and noting thatA+A
2
+A
3
has all nonzero entries offthe main diagonal.) The digraph in Figure 8.7 is not strongly
connected, since there is no path directed to
5from any other vertex.
(b) Use the fact that if a path exists between a given pair
,of vertices, then there must be
a path of length at most−1between
and . (This is because any longer path would
involve returning to a vertex
that was already visited earlier in the path since there are
onlyvertices. Hence, any portion of the longer path that involves travelling from
to
itself could be removed from that path to yield a shorter path between
and .) Then use
Corollary 8.2.
(c) Create a new digraph by considering each directed edge of the given digraph and adding
a corresponding directed edge going in the opposite direction. If we include loops in this
process, the adjacency matrix for the new digraph will be(A+A
). Next, convert the new
digraph into a graph by replacing each pair of directed edges (that is, each directed edge
together with its opposite) by a single (undirected) edge. The entries of the adjacency matrix
of this graph that are not on the main diagonal would agree with those of(A+A
), while
the main diagonal entries of the adjacency matrix of this graph would agree with those ofA
(These actually represent the number of loops at each vertex in the original digraph.) But
loops are irrelevant in determining connectivity since they do not connect distinct vertices.
Hence, by Theorem 8.3, this new graph is connected if and only if
¡
A+A
¢
+
¡
A+A
¢
2
+
¡
A+A
¢
3
+···+
¡
A+A
¢
−1
has all entries nonzero offthe main diagonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 148
(f) Part (a):(A+I 4)
3
=
⎡
⎢
⎢
⎣
4103240
0800
3202532
4003237
⎤
⎥
⎥
⎦
; disconnected;
Part (b):(A+I
4)
3
=
⎡
⎢
⎢
⎣
13 4 9 1
4765
9683
1534
⎤
⎥
⎥
⎦
; connected;
Part (c):(A+I
5)
4
=
⎡
⎢
⎢
⎢
⎢
⎣
912 812 6
12 93 120 28 120
8 120 252 42 168
12 28 42 21 24
6 120 168 24 172
⎤
⎥
⎥
⎥
⎥
⎦
; connected;
Part (d):(A+I
5)
4
=
⎡
⎢
⎢
⎢
⎢
⎣
86 0 0 85 85
0 100 78 0 0
0786100
85 0 0 86 85
85 0 0 85 86
⎤
⎥
⎥
⎥
⎥
⎦
; disconnected.
(12) If the graph is connected, then vertex
must have at least one edge connecting it to some other
distinct vertex
. (Otherwise there could be no paths at all connecting to any other vertex.)
But then, the path
→ → is a path of length2connecting to itself. Hence, the( )
entry ofA
2
is at least1. Since this is true for each, all entries on the main diagonal ofA
2
are
nonzero.
(13) (a) The digraph
2in Figure 8.2 is strongly connected since there is a path from any vertex to any
other vertex. (This can be verified by inspection, or by using part (b) of this exercise — that
is, lettingArepresent the adjacency matrix for this digraph and noting thatA+A
2
+A
3
has all nonzero entries offthe main diagonal.) The digraph in Figure 8.7 is not strongly
connected, since there is no path directed to
5from any other vertex.
(b) Use the fact that if a path exists between a given pair
,of vertices, then there must be
a path of length at most−1between
and . (This is because any longer path would
involve returning to a vertex
that was already visited earlier in the path since there are
onlyvertices. Hence, any portion of the longer path that involves travelling from
to
itself could be removed from that path to yield a shorter path between
and .) Then use
Corollary 8.2.
(c) Create a new digraph by considering each directed edge of the given digraph and adding
a corresponding directed edge going in the opposite direction. If we include loops in this
process, the adjacency matrix for the new digraph will be(A+A
). Next, convert the new
digraph into a graph by replacing each pair of directed edges (that is, each directed edge
together with its opposite) by a single (undirected) edge. The entries of the adjacency matrix
of this graph that are not on the main diagonal would agree with those of(A+A
), while
the main diagonal entries of the adjacency matrix of this graph would agree with those ofA
(These actually represent the number of loops at each vertex in the original digraph.) But
loops are irrelevant in determining connectivity since they do not connect distinct vertices.
Hence, by Theorem 8.3, this new graph is connected if and only if
¡
A+A
¢
+
¡
A+A
¢
2
+
¡
A+A
¢
3
+···+
¡
A+A
¢
−1
has all entries nonzero offthe main diagonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 148
Answers to Exercises Section 8.2
(d) For thefirst adjacency matrix, letC=(A+A
)Then
C=
⎡
⎢
⎢
⎢
⎢
⎣
00020
02001
00030
20300
01002
⎤
⎥
⎥
⎥
⎥
⎦
,and,C+C
2
+C
3
+C
4
=
⎡
⎢
⎢
⎢
⎢
⎣
56084280
0620058
84 0 126 42 0
28 0 42 182 0
0580062
⎤
⎥
⎥
⎥
⎥
⎦
However, some of the entries offthe main diagonal of this sum are zero, so the digraph forA
is not weakly connected.
For the second adjacency matrix, letD=(B+B
)Then
D=
⎡
⎢
⎢
⎢
⎢
⎣
00220
00010
20001
21000
01102
⎤
⎥
⎥
⎥
⎥
⎦
,and,D+D
2
+D
3
+D
4
=
⎡
⎢
⎢
⎢
⎢
⎣
80 20 24 20 32
206464
24 4524432
20 6444612
32 4321252
⎤
⎥
⎥
⎥
⎥
⎦
Since this sum has no zero entries offthe main diagonal, the digraph forBis weakly connected.
(14) (a) Use part (c).
(b) Yes, it is a dominance digraph, because no tie games are possible and because each team
plays every other team. Thus if
and are two given teams, either defeats or vice
versa.
(c) Since the entries of bothAandA
are all zeroes and ones, then, with6 =,the( )entry
ofA+A
=+=1iff =1or =1, but not both. Also, the( )entry of
A+A
=2 =0iff =0. Hence,A+A
has the desired properties (all main diagonal
entries equal to0, and all other entries equal to1)iffthere is always exactly one directed edge
between two distinct vertices (
=1or =1, but not both, for6 =), and the digraph
has no loops (
=0). (This is the definition of a dominance digraph.)
(15) We prove Theorem 8.1 by induction on. For the base step,=1. Thiscaseistruebythe
definition of the adjacency matrix. For the inductive step, letB=A
.ThenA
+1
=BA.Note
that the( )entry ofA
+1
=(th row ofB)·(th column ofA)=
X
=1
=
X
=1
(number of paths of lengthfrom to)·(number of paths of length1from to)
=(number of paths of length+1from
to)
(16) (a) T
(b) F
(c) T
(d) F
(e) T
(f) T
(g) T
(h) F
(i) F
(j) F
(k) F
(l) F
(m) T
Section 8.2
(1) (a) 1=8, 2=5, 3=3
(b)
1=10, 2=4, 3=5, 4=1, 5=6
(c)
1=12, 2=5, 3=3, 4=2, 5=2, 6=7
(d)
1=64, 2=48, 3=16, 4=36, 5=12, 6=28
Copyrightc°2016 Elsevier Ltd. All rights reserved. 149
(d) For thefirst adjacency matrix, letC=(A+A
)Then
C=
⎡
⎢
⎢
⎢
⎢
⎣
00020
02001
00030
20300
01002
⎤
⎥
⎥
⎥
⎥
⎦
,and,C+C
2
+C
3
+C
4
=
⎡
⎢
⎢
⎢
⎢
⎣
56084280
0620058
84 0 126 42 0
28 0 42 182 0
0580062
⎤
⎥
⎥
⎥
⎥
⎦
However, some of the entries offthe main diagonal of this sum are zero, so the digraph forA
is not weakly connected.
For the second adjacency matrix, letD=(B+B
)Then
D=
⎡
⎢
⎢
⎢
⎢
⎣
00220
00010
20001
21000
01102
⎤
⎥
⎥
⎥
⎥
⎦
,and,D+D
2
+D
3
+D
4
=
⎡
⎢
⎢
⎢
⎢
⎣
80 20 24 20 32
206464
24 4524432
20 6444612
32 4321252
⎤
⎥
⎥
⎥
⎥
⎦
Since this sum has no zero entries offthe main diagonal, the digraph forBis weakly connected.
(14) (a) Use part (c).
(b) Yes, it is a dominance digraph, because no tie games are possible and because each team
plays every other team. Thus if
and are two given teams, either defeats or vice
versa.
(c) Since the entries of bothAandA
are all zeroes and ones, then, with6 =,the( )entry
ofA+A
=+=1iff =1or =1, but not both. Also, the( )entry of
A+A
=2 =0iff =0. Hence,A+A
has the desired properties (all main diagonal
entries equal to0, and all other entries equal to1)iffthere is always exactly one directed edge
between two distinct vertices (
=1or =1, but not both, for6 =), and the digraph
has no loops (
=0). (This is the definition of a dominance digraph.)
(15) We prove Theorem 8.1 by induction on. For the base step,=1. Thiscaseistruebythe
definition of the adjacency matrix. For the inductive step, letB=A
.ThenA
+1
=BA.Note
that the( )entry ofA
+1
=(th row ofB)·(th column ofA)=
X
=1
=
X
=1
(number of paths of lengthfrom to)·(number of paths of length1from to)
=(number of paths of length+1from
to)
(16) (a) T
(b) F
(c) T
(d) F
(e) T
(f) T
(g) T
(h) F
(i) F
(j) F
(k) F
(l) F
(m) T
Section 8.2
(1) (a) 1=8, 2=5, 3=3
(b)
1=10, 2=4, 3=5, 4=1, 5=6
(c)
1=12, 2=5, 3=3, 4=2, 5=2, 6=7
(d)
1=64, 2=48, 3=16, 4=36, 5=12, 6=28
Copyrightc°2016 Elsevier Ltd. All rights reserved. 149
Answers to Exercises Section 8.3
(2) (a) T (b) T
Section 8.3
(1) (a)=−08−33;≈−73when=5
(b)=112+105;≈665when=5
(c)=−15+38;≈−37when=5
(2) (a)=0375
2
+035+360
(b)=−083
2
+147−18
(c)=−0042
2
+0633+0266
(3) (a)=
1
4
3
+
25
28
2
+
25
14
+
37
35
(b)=−
1
6
3
−
1
14
2
−
1
3
+
117
35
(4) (a)=44286
2
−20571
(b)=07116
2
+16858+08989
(c)=−01014
2
+09633−08534
(d)=−01425
3
+09882
(e)=0
3
−03954
2
+09706
(5) (a)=02+274; the angle reaches434
◦
in the8th month.
(b) The angle reaches514
◦
in the12th month.
(c)=0007143
2
+01286+28614; the tower will be leaning at543
◦
in the12th month.
(d) The quadratic approximation will probably be more accurate because the amount of change in
the angle from the vertical is generally increasing with each passing month.
(e) The angle reaches382
◦
after54months.
(f) The angle reaches20
◦
after408months.
(6) The answers to (a) and (b) assume that the years are renumbered as suggested in the textbook.
(a)=2581+ 15097; predicted population in2005is29293million.
(b) Predicted population in2030is35745million.
(c)=097
2
+19+ 16005; predicted population in2030is37413million.
(7) The least-squares polynomial is=
4
5
2
−
2
5
+2,whichistheexactquadratic through the given three
points.
(8) When=1, the system in Theorem 8.4 is
∙
11 ···1
12···
¸
⎡
⎢
⎢
⎢
⎣
1
1
1 2
.
.
.
.
.
.
1
⎤
⎥
⎥
⎥
⎦
∙
0
1
¸
=
∙
11 ···1
12···
¸
⎡
⎢
⎢
⎢
⎣
1
2
.
.
.
⎤
⎥
⎥
⎥
⎦
which becomes
"
P
=1
P
=1
P
=1
2
#
∙
0
1
¸
=
"P
=1
P
=1
#
the desired system.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 150
(2) (a) T (b) T
Section 8.3
(1) (a)=−08−33;≈−73when=5
(b)=112+105;≈665when=5
(c)=−15+38;≈−37when=5
(2) (a)=0375
2
+035+360
(b)=−083
2
+147−18
(c)=−0042
2
+0633+0266
(3) (a)=
1
4
3
+
25
28
2
+
25
14
+
37
35
(b)=−
1
6
3
−
1
14
2
−
1
3
+
117
35
(4) (a)=44286
2
−20571
(b)=07116
2
+16858+08989
(c)=−01014
2
+09633−08534
(d)=−01425
3
+09882
(e)=0
3
−03954
2
+09706
(5) (a)=02+274; the angle reaches434
◦
in the8th month.
(b) The angle reaches514
◦
in the12th month.
(c)=0007143
2
+01286+28614; the tower will be leaning at543
◦
in the12th month.
(d) The quadratic approximation will probably be more accurate because the amount of change in
the angle from the vertical is generally increasing with each passing month.
(e) The angle reaches382
◦
after54months.
(f) The angle reaches20
◦
after408months.
(6) The answers to (a) and (b) assume that the years are renumbered as suggested in the textbook.
(a)=2581+ 15097; predicted population in2005is29293million.
(b) Predicted population in2030is35745million.
(c)=097
2
+19+ 16005; predicted population in2030is37413million.
(7) The least-squares polynomial is=
4
5
2
−
2
5
+2,whichistheexactquadratic through the given three
points.
(8) When=1, the system in Theorem 8.4 is
∙
11 ···1
12···
¸
⎡
⎢
⎢
⎢
⎣
1
1
1 2
.
.
.
.
.
.
1
⎤
⎥
⎥
⎥
⎦
∙
0
1
¸
=
∙
11 ···1
12···
¸
⎡
⎢
⎢
⎢
⎣
1
2
.
.
.
⎤
⎥
⎥
⎥
⎦
which becomes
"
P
=1
P
=1
P
=1
2
#
∙
0
1
¸
=
"P
=1
P
=1
#
the desired system.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 150
Answers to Exercises Section 8.4
(9) (a) 1=
230
39
,2=
155
39
;
⎧
⎪
⎨
⎪
⎩
4
1−3 2=11
2
3
which is almost12
2
1+5 2=31
2
3
which is almost32
3
1+ 2=21
2
3
which is close to21
(b)
1=
46
11
,2=−
12
11
,3=
62
33
;
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
2
1− 2+ 3=11
1
3
which is almost11
−
1+3 2− 3=−9
1
3
which is almost−9
1−2 2+3 3=12
3
1−4 2+2 3=20
2
3
which is almost21
(10) (a) T (b) F (c) F (d) F
Section 8.4
(1)Ais not stochastic, sinceAis not square;Ais not regular, sinceAis not stochastic.
Bis not stochastic, since the entries of column2do not sum to1;Bis not regular, sinceBis not
stochastic.
Cis stochastic;Cis regular, sinceCis stochastic and has all nonzero entries.
Dis stochastic;Dis not regular, since every positive power ofDis a matrix whose rows are the rows
ofDpermuted in some order, and hence every such power contains zero entries.
Eis not stochastic, since the entries of column1do not sum to1;Eis not regular, sinceEis not
stochastic.
Fis stochastic;Fis not regular, since every positive power ofFhas all second row entries zero.
Gis not stochastic, sinceGis not square;Gis not regular, sinceGis not stochastic.
His stochastic;His regular, since
H
2
=
⎡
⎢
⎢
⎣
1
2
1
4
1
4
1
4
1
2
1
4
1
4
1
4
1
2
⎤
⎥
⎥
⎦
which has no zero entries.
(2) (a)p
1=[
5
18
13
18
],p2=[
67
216
149
216
]
(b)p
1=[
2
9
13
36
5
12
],p2=[
25
108
97
216
23
72
]
(c)p
1=[
17
48
1
3
5
16
],p2=[
205
576
49
144
175
576
]
(3) (a)[
2
5
3
5
] (b)[
18
59
20
59
21
59
] (c)[
1
4
3
10
3
10
3
20
]
(4) (a)[0406] (b)[030503390356]
(5) (a)[03401750340145]in the next election;
[03555018750287501695]in the election after that
(b) The steady-state vector is[036020024020]; in a century, the votes would be36%for Party
Aand24%for Party C.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 151
(9) (a) 1=
230
39
,2=
155
39
;
⎧
⎪
⎨
⎪
⎩
4
1−3 2=11
2
3
which is almost12
2
1+5 2=31
2
3
which is almost32
3
1+ 2=21
2
3
which is close to21
(b)
1=
46
11
,2=−
12
11
,3=
62
33
;
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
2
1− 2+ 3=11
1
3
which is almost11
−
1+3 2− 3=−9
1
3
which is almost−9
1−2 2+3 3=12
3
1−4 2+2 3=20
2
3
which is almost21
(10) (a) T (b) F (c) F (d) F
Section 8.4
(1)Ais not stochastic, sinceAis not square;Ais not regular, sinceAis not stochastic.
Bis not stochastic, since the entries of column2do not sum to1;Bis not regular, sinceBis not
stochastic.
Cis stochastic;Cis regular, sinceCis stochastic and has all nonzero entries.
Dis stochastic;Dis not regular, since every positive power ofDis a matrix whose rows are the rows
ofDpermuted in some order, and hence every such power contains zero entries.
Eis not stochastic, since the entries of column1do not sum to1;Eis not regular, sinceEis not
stochastic.
Fis stochastic;Fis not regular, since every positive power ofFhas all second row entries zero.
Gis not stochastic, sinceGis not square;Gis not regular, sinceGis not stochastic.
His stochastic;His regular, since
H
2
=
⎡
⎢
⎢
⎣
1
2
1
4
1
4
1
4
1
2
1
4
1
4
1
4
1
2
⎤
⎥
⎥
⎦
which has no zero entries.
(2) (a)p
1=[
5
18
13
18
],p2=[
67
216
149
216
]
(b)p
1=[
2
9
13
36
5
12
],p2=[
25
108
97
216
23
72
]
(c)p
1=[
17
48
1
3
5
16
],p2=[
205
576
49
144
175
576
]
(3) (a)[
2
5
3
5
] (b)[
18
59
20
59
21
59
] (c)[
1
4
3
10
3
10
3
20
]
(4) (a)[0406] (b)[030503390356]
(5) (a)[03401750340145]in the next election;
[03555018750287501695]in the election after that
(b) The steady-state vector is[036020024020]; in a century, the votes would be36%for Party
Aand24%for Party C.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 151
Answers to Exercises Section 8.4
(6) (a)
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
2
1
6
1
6
1
5
0
1
8
1
2
00
1
5
1
8
0
1
2
1
10
1
10
1
4
0
1
6
1
2
1
5
0
1
3
1
6
1
5
1
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(b) IfMis the stochastic matrix in part (a), thenM
2
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
41
120
1
6
1
5
13
60
9
100
1
8
27
80
13
240
13
200
1
5
3
20
13
240
73
240
29
200
3
25
13
48
13
120
29
120
107
300
13
60
9
80
1
3
1
5
13
60
28
75
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
which has
all entries nonzero. Thus,Mis regular.
(c)
29
120
, since the probability vector after two time intervals is[
1
5
13
240
73
240
29
120
1
5
]
(d)[
1
5
3
20
3
20
1
4
1
4
]; over time, the rat frequents rooms B and C the least and rooms D and E the
most.
(7) The limit vector for any initial input is[100].However,Mis not regular, since every positive power
ofMis upper triangular, and hence is zero below the main diagonal. The uniquefixed point is[100].
(8) (a) Any steady-state vector is a solution of the system
µ∙
1−
1−
¸
−
∙
10
01
¸¶∙
1
2
¸
=
∙
0
0
¸
which simplifies to
∙
−
−
¸∙
1
2
¸
=
∙
0
0
¸
. Using the fact that
1+2=1yields a
larger system whose augmented matrix is
⎡
⎣
−
−
11
¯
¯
¯
¯
¯
¯
0
0
1
⎤
⎦. This reduces to
⎡
⎢
⎢
⎣
10
01
00
¯
¯
¯
¯
¯
¯
¯
¯
+
+
0
⎤
⎥
⎥
⎦
,
assumingandare not both zero. (Begin the row reduction with the row operationh1i↔h3i
in case=0.) Hence,
1=
+
,2=
+
istheuniquesteady-statevector.
(b) Here,=
1
2
,=
1
3
, and so the steady-state vector is
6
5
[
1
3
1
2
]=[
2
5
3
5
], which agrees with the result
from Exercise 3(a).
(9) It is enough to show that the product of any two stochastic matrices is stochastic. LetAandBbe
×stochastic matrices. Then the entries ofABare clearly nonnegative, since the entries of bothA
Copyrightc°2016 Elsevier Ltd. All rights reserved. 152
(6) (a)
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
2
1
6
1
6
1
5
0
1
8
1
2
00
1
5
1
8
0
1
2
1
10
1
10
1
4
0
1
6
1
2
1
5
0
1
3
1
6
1
5
1
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(b) IfMis the stochastic matrix in part (a), thenM
2
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
41
120
1
6
1
5
13
60
9
100
1
8
27
80
13
240
13
200
1
5
3
20
13
240
73
240
29
200
3
25
13
48
13
120
29
120
107
300
13
60
9
80
1
3
1
5
13
60
28
75
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
which has
all entries nonzero. Thus,Mis regular.
(c)
29
120
, since the probability vector after two time intervals is[
1
5
13
240
73
240
29
120
1
5
]
(d)[
1
5
3
20
3
20
1
4
1
4
]; over time, the rat frequents rooms B and C the least and rooms D and E the
most.
(7) The limit vector for any initial input is[100].However,Mis not regular, since every positive power
ofMis upper triangular, and hence is zero below the main diagonal. The uniquefixed point is[100].
(8) (a) Any steady-state vector is a solution of the system
µ∙
1−
1−
¸
−
∙
10
01
¸¶∙
1
2
¸
=
∙
0
0
¸
which simplifies to
∙
−
−
¸∙
1
2
¸
=
∙
0
0
¸
. Using the fact that
1+2=1yields a
larger system whose augmented matrix is
⎡
⎣
−
−
11
¯
¯
¯
¯
¯
¯
0
0
1
⎤
⎦. This reduces to
⎡
⎢
⎢
⎣
10
01
00
¯
¯
¯
¯
¯
¯
¯
¯
+
+
0
⎤
⎥
⎥
⎦
,
assumingandare not both zero. (Begin the row reduction with the row operationh1i↔h3i
in case=0.) Hence,
1=
+
,2=
+
istheuniquesteady-statevector.
(b) Here,=
1
2
,=
1
3
, and so the steady-state vector is
6
5
[
1
3
1
2
]=[
2
5
3
5
], which agrees with the result
from Exercise 3(a).
(9) It is enough to show that the product of any two stochastic matrices is stochastic. LetAandBbe
×stochastic matrices. Then the entries ofABare clearly nonnegative, since the entries of bothA
Copyrightc°2016 Elsevier Ltd. All rights reserved. 152
Answers to Exercises Section 8.5
andBare nonnegative. Furthermore, the sum of the entries in theth column ofAB=
X
=1
(AB) =
X
=1
(th row ofA)·(th column ofB)
=
X
=1
(11+22+···+ )
=
Ã
X
=1
1
!
1+
Ã
X
=1
2
!
2+···+
Ã
X
=1
!
=(1) 1+(1) 2+···+(1)
(sinceAis stochastic, and each of the summations
is a sum of the entries of a column ofA)
=1
sinceBis stochastic. HenceABis stochastic.
(10) SupposeMis a×matrix.
Base Step (=1):th entry inMp=(th row ofM)·p=
P
=1
=
X
=1
(probability of moving from state to)(probability of being in state )
=probability of being in state
after1step of the process
=th entry ofp
1
Inductive Step: Assumep
=M
pis the probability vector aftersteps of the process. Then, after
an additional step of the process, the probability vectorp
+1=Mp
=M(M
p)=M
+1
p.
(11) (a) F (b) T (c) T (d) T (e) F
Section 8.5
(1) (a)−24−46−15−30 10 16 39 62
26 42 51 84 24 37 −11−23
(b)97 177 146 96 169 143 113
201 171 93 168 133 175 311
254 175 312 256 238 (430) (357)
(where “27” was used twice to pad the last vector)
(2) (a) HOMEWORK IS GOOD FOR THE SOUL
(b) WHO IS BURIED IN GRANT
(
’
)
STOMB—
(c) DOCTOR, I HAVE A CODE
(d) TOMAKEASLOWHORSEFASTDON
(
’
)
T FEED IT— —
Copyrightc°2016 Elsevier Ltd. All rights reserved. 153
andBare nonnegative. Furthermore, the sum of the entries in theth column ofAB=
X
=1
(AB) =
X
=1
(th row ofA)·(th column ofB)
=
X
=1
(11+22+···+ )
=
Ã
X
=1
1
!
1+
Ã
X
=1
2
!
2+···+
Ã
X
=1
!
=(1) 1+(1) 2+···+(1)
(sinceAis stochastic, and each of the summations
is a sum of the entries of a column ofA)
=1
sinceBis stochastic. HenceABis stochastic.
(10) SupposeMis a×matrix.
Base Step (=1):th entry inMp=(th row ofM)·p=
P
=1
=
X
=1
(probability of moving from state to)(probability of being in state )
=probability of being in state
after1step of the process
=th entry ofp
1
Inductive Step: Assumep
=M
pis the probability vector aftersteps of the process. Then, after
an additional step of the process, the probability vectorp
+1=Mp
=M(M
p)=M
+1
p.
(11) (a) F (b) T (c) T (d) T (e) F
Section 8.5
(1) (a)−24−46−15−30 10 16 39 62
26 42 51 84 24 37 −11−23
(b)97 177 146 96 169 143 113
201 171 93 168 133 175 311
254 175 312 256 238 (430) (357)
(where “27” was used twice to pad the last vector)
(2) (a) HOMEWORK IS GOOD FOR THE SOUL
(b) WHO IS BURIED IN GRANT
(
’
)
STOMB—
(c) DOCTOR, I HAVE A CODE
(d) TOMAKEASLOWHORSEFASTDON
(
’
)
T FEED IT— —
Copyrightc°2016 Elsevier Ltd. All rights reserved. 153
Answers to Exercises Section 8.6
(3) (a) T (b) T (c) F
Section 8.6
(1) (a)=
1
2
arctan(−
√
3
3
)=−
12
;P=
"
√
6+
√
2
4
√
6−
√
2
4
√
2−
√
6
4
√
6+
√
2
4
#
;
equation in-coordinates:
2
−
2
=2or,
2
2
−
2
2
=1;
center in-coordinates:(00);centerin-coordinates:(00);
see Figures 20 and 21.
Figure 20: Rotated Hyperbola
Figure 21: Original Hyperbola
Copyrightc°2016 Elsevier Ltd. All rights reserved. 154
(3) (a) T (b) T (c) F
Section 8.6
(1) (a)=
1
2
arctan(−
√
3
3
)=−
12
;P=
"
√
6+
√
2
4
√
6−
√
2
4
√
2−
√
6
4
√
6+
√
2
4
#
;
equation in-coordinates:
2
−
2
=2or,
2
2
−
2
2
=1;
center in-coordinates:(00);centerin-coordinates:(00);
see Figures 20 and 21.
Figure 20: Rotated Hyperbola
Figure 21: Original Hyperbola
Copyrightc°2016 Elsevier Ltd. All rights reserved. 154
Answers to Exercises Section 8.6
(b)=
4
,P=
"
√
2
2
−
√
2
2
√
2
2
√
2
2
#
;
equation in-coordinates:4
2
+9
2
−8=32,or,
(−1)
2
9
+
2
4
=1;
center in-coordinates:(10);centerin-coordinates:(
√
2
2
√
2
2
);
see Figures 22 and 23.
Figure 22: Rotated Ellipse
Figure 23: Original Ellipse
Copyrightc°2016 Elsevier Ltd. All rights reserved. 155
(b)=
4
,P=
"
√
2
2
−
√
2
2
√
2
2
√
2
2
#
;
equation in-coordinates:4
2
+9
2
−8=32,or,
(−1)
2
9
+
2
4
=1;
center in-coordinates:(10);centerin-coordinates:(
√
2
2
√
2
2
);
see Figures 22 and 23.
Figure 22: Rotated Ellipse
Figure 23: Original Ellipse
Copyrightc°2016 Elsevier Ltd. All rights reserved. 155
Answers to Exercises Section 8.6
(c)=
1
2
arctan(−
√
3) =−
6
;P=
"
√
3
2
1
2
−
1
2
√
3
2
#
;
equation in-coordinates:=2
2
−12+13,or(+5)=2(−3)
2
;
vertex in-coordinates:(3−5);vertexin-coordinates:(00981−5830);
see Figures 24 and 25.
Figure 24: Rotated Parabola
Figure 25: Original Parabola
Copyrightc°2016 Elsevier Ltd. All rights reserved. 156
(c)=
1
2
arctan(−
√
3) =−
6
;P=
"
√
3
2
1
2
−
1
2
√
3
2
#
;
equation in-coordinates:=2
2
−12+13,or(+5)=2(−3)
2
;
vertex in-coordinates:(3−5);vertexin-coordinates:(00981−5830);
see Figures 24 and 25.
Figure 24: Rotated Parabola
Figure 25: Original Parabola
Copyrightc°2016 Elsevier Ltd. All rights reserved. 156
Answers to Exercises Section 8.6
(d)≈06435radians (about36
◦
52
0
);P=
"
4
5
−
3
5
3
5
4
5
#
;
equation in-coordinates:4
2
−16+9
2
+18=11or,
(−2)
2
9
+
(+1)
2
4
=1;
center in-coordinates:(2−1); center in-coordinates=(
11
5
2
5
);
see Figures 26 and 27.
Figure 26: Rotated Ellipse
Figure 27: Original Ellipse
Copyrightc°2016 Elsevier Ltd. All rights reserved. 157
(d)≈06435radians (about36
◦
52
0
);P=
"
4
5
−
3
5
3
5
4
5
#
;
equation in-coordinates:4
2
−16+9
2
+18=11or,
(−2)
2
9
+
(+1)
2
4
=1;
center in-coordinates:(2−1); center in-coordinates=(
11
5
2
5
);
see Figures 26 and 27.
Figure 26: Rotated Ellipse
Figure 27: Original Ellipse
Copyrightc°2016 Elsevier Ltd. All rights reserved. 157
Answers to Exercises Section 8.6
(e)≈−06435radians (about−36
◦
52
0
);P=
"
4
5
3
5
−
3
5
4
5
#
;
equation in-coordinates:12=
2
+12or,(+3)=
1
12
(+6)
2
;
vertex in-coordinates:(−6−3);vertexin-coordinates=(−
33
5
6
5
);
see Figures 28 and 29.
Figure 28: Rotated Parabola
Figure 29: Original Parabola
Copyrightc°2016 Elsevier Ltd. All rights reserved. 158
(e)≈−06435radians (about−36
◦
52
0
);P=
"
4
5
3
5
−
3
5
4
5
#
;
equation in-coordinates:12=
2
+12or,(+3)=
1
12
(+6)
2
;
vertex in-coordinates:(−6−3);vertexin-coordinates=(−
33
5
6
5
);
see Figures 28 and 29.
Figure 28: Rotated Parabola
Figure 29: Original Parabola
Copyrightc°2016 Elsevier Ltd. All rights reserved. 158
Answers to Exercises Section 8.7
(f) (All answers are rounded to four significant digits.)
≈04442radians (about25
◦
27
0
);P=
∙
09029−04298
04298 09029
¸
;
equation in-coordinates:
2
(1770)
2−
2
(2050)
2=1;
center in-coordinates:(00);centerin-coordinates=(00);
see Figures 30 and 31.
Figure 30: Rotated Hyperbola
Figure 31: Original Hyperbola
(2) (a) T (b) F (c) T (d) F
Section 8.7
(1) (a)(91)(95)(121)(125)(143)
(b)(35)(19)(57)(310)(69)
(c)(−25)(09)(−57)(−210)(−510)
(d)(206)(2014)(326)(3214)(4010)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 159
(f) (All answers are rounded to four significant digits.)
≈04442radians (about25
◦
27
0
);P=
∙
09029−04298
04298 09029
¸
;
equation in-coordinates:
2
(1770)
2−
2
(2050)
2=1;
center in-coordinates:(00);centerin-coordinates=(00);
see Figures 30 and 31.
Figure 30: Rotated Hyperbola
Figure 31: Original Hyperbola
(2) (a) T (b) F (c) T (d) F
Section 8.7
(1) (a)(91)(95)(121)(125)(143)
(b)(35)(19)(57)(310)(69)
(c)(−25)(09)(−57)(−210)(−510)
(d)(206)(2014)(326)(3214)(4010)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 159
Answers to Exercises Section 8.7
(2) (a)(311)(59)(711)(117)(1511);seeFigure32
(b)(−82)(−75)(−106)(−911)(−1413);seeFigure33
(c)(81)(84)(114)(1010)(1611);seeFigure34
(d)(318)(412)(518)(76)(918);seeFigure35
Figure 32
Figure 33
Copyrightc°2016 Elsevier Ltd. All rights reserved. 160
(2) (a)(311)(59)(711)(117)(1511);seeFigure32
(b)(−82)(−75)(−106)(−911)(−1413);seeFigure33
(c)(81)(84)(114)(1010)(1611);seeFigure34
(d)(318)(412)(518)(76)(918);seeFigure35
Figure 32
Figure 33
Copyrightc°2016 Elsevier Ltd. All rights reserved. 160
Answers to Exercises Section 8.7
Figure 34
Figure 35
(3) (a)(3−4)(3−10)(7−6)(9−9)(10−3)
(b)(43)(103)(67)(99)(310)
(c)(−26)(08)(−817)(−1022)(−1625)
(d)(64)(117)(112)(141)(10−3)
(4) (a)(149)(10
6)(1111)(89)(68)(1114)
(b)(−2−3)(−6−3)(−3−6)(−6−6)(−8−6)(−2−9)
(c)(24)(26)(85)(86)(twice),(144)
(5) (a)(47)(69)(108)(92)(114)
(b)
(05)(17)(011)(−58)(−410)
(c)(73)(712)(918)(103)(1012)
(6) (a)(220)(317)(514)(619)(616)(914);seeFigure36
(b)(1717)(1714)(1810)(2015)(1912)(2210);seeFigure37
Copyrightc°2016 Elsevier Ltd. All rights reserved. 161
Figure 34
Figure 35
(3) (a)(3−4)(3−10)(7−6)(9−9)(10−3)
(b)(43)(103)(67)(99)(310)
(c)(−26)(08)(−817)(−1022)(−1625)
(d)(64)(117)(112)(141)(10−3)
(4) (a)(149)(10
6)(1111)(89)(68)(1114)
(b)(−2−3)(−6−3)(−3−6)(−6−6)(−8−6)(−2−9)
(c)(24)(26)(85)(86)(twice),(144)
(5) (a)(47)(69)(108)(92)(114)
(b)
(05)(17)(011)(−58)(−410)
(c)(73)(712)(918)(103)(1012)
(6) (a)(220)(317)(514)(619)(616)(914);seeFigure36
(b)(1717)(1714)(1810)(2015)(1912)(2210);seeFigure37
Copyrightc°2016 Elsevier Ltd. All rights reserved. 161
Answers to Exercises Section 8.7
(c)(118)(−313)(−68)(−217)(−512)(−610);seeFigure38
(d)(−196)(−167)(−159)(−277)(−218)(−2710);seeFigure39
Figure 36
Figure 37
Figure 38
Copyrightc°2016 Elsevier Ltd. All rights reserved. 162
(c)(118)(−313)(−68)(−217)(−512)(−610);seeFigure38
(d)(−196)(−167)(−159)(−277)(−218)(−2710);seeFigure39
Figure 36
Figure 37
Figure 38
Copyrightc°2016 Elsevier Ltd. All rights reserved. 162
Answers to Exercises Section 8.7
Figure 39
(7) (a) Show that the given matrix is equal to the product
⎡
⎣
10
01
001
⎤
⎦
⎡
⎣
cos−sin0
sincos0
001
⎤
⎦
⎡
⎣
10−
01−
00 1
⎤
⎦
(b) Show that the given matrix is equal to the product
⎡
⎣
100
01
001
⎤
⎦
µ
1
1+
2
¶
⎡
⎣
1− 2
2 0
2
2
−10
001
⎤
⎦
⎡
⎣
10 0
01−
00 1
⎤
⎦
(c) Show that the given matrix is equal to the product
⎡
⎣
10
01
001
⎤
⎦
⎡
⎣
00
00
001
⎤
⎦
⎡
⎣
10−
01−
00 1
⎤
⎦
(8) Show that the given matrix is equal to the product
⎡
⎣
10
010
001
⎤
⎦
⎡
⎣
−100
010
001
⎤
⎦
⎡
⎣
10−
01 0
00 1
⎤
⎦
(9) (a)(47)(69)(108)(92)(114)
(b)(05)(1
7)(011)(−58)(−410)
(c)(73)(712)(918)(103)(1012)
(10) (a) Multiplying the two matrices yieldsI
3.
(b) Thefirst matrix represents a translation along the vector[ ], while the second is the inverse
translation along the vector[−−].
(c) See theanswerfor Exercise 7(a), and substitutefor,andfor, and then use part (a) of this
Exercise.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 163
Figure 39
(7) (a) Show that the given matrix is equal to the product
⎡
⎣
10
01
001
⎤
⎦
⎡
⎣
cos−sin0
sincos0
001
⎤
⎦
⎡
⎣
10−
01−
00 1
⎤
⎦
(b) Show that the given matrix is equal to the product
⎡
⎣
100
01
001
⎤
⎦
µ
1
1+
2
¶
⎡
⎣
1− 2
2 0
2
2
−10
001
⎤
⎦
⎡
⎣
10 0
01−
00 1
⎤
⎦
(c) Show that the given matrix is equal to the product
⎡
⎣
10
01
001
⎤
⎦
⎡
⎣
00
00
001
⎤
⎦
⎡
⎣
10−
01−
00 1
⎤
⎦
(8) Show that the given matrix is equal to the product
⎡
⎣
10
010
001
⎤
⎦
⎡
⎣
−100
010
001
⎤
⎦
⎡
⎣
10−
01 0
00 1
⎤
⎦
(9) (a)(47)(69)(108)(92)(114)
(b)(05)(1
7)(011)(−58)(−410)
(c)(73)(712)(918)(103)(1012)
(10) (a) Multiplying the two matrices yieldsI
3.
(b) Thefirst matrix represents a translation along the vector[ ], while the second is the inverse
translation along the vector[−−].
(c) See theanswerfor Exercise 7(a), and substitutefor,andfor, and then use part (a) of this
Exercise.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 163
Answers to Exercises Section 8.7
(11) (a) Show that the matrices in parts (a) and (c) of Exercise 7 commute with each other when=.
Both products equal
⎡
⎣
(cos)−(sin)(1−(cos)) +(sin)
(sin)(cos)(1−(cos))−(sin)
00 1
⎤
⎦
(b) Consider the reflection about the-axis, whose matrix is given in Section 8.7 of the textbook as
A=
⎡
⎣
−100
010
001
⎤
⎦, and a counterclockwise rotation of90
◦
about the origin, whose matrix is
B=
⎡
⎣
0−10
100
001
⎤
⎦.ThenAB=
⎡
⎣
010
100
001
⎤
⎦, butBA=
⎡
⎣
0−10
−100
001
⎤
⎦,andsoAandB
do not commute. In particular, starting from the point(10), performing the rotation and then
the reflection yields(01). However, performing the reflection followed by the rotation produces
(0−1).
(c) Consider a reflection about=−and scaling by a factor ofin the-direction andin the
-direction, with6 =Now,
(reflection)◦(scaling)([10]) = (reflection)([0]) = [0−]
But,
(scaling
)◦(reflection)([10]) = (scaling)([0−1]) = [0−]
Hence,
(reflection)◦(scaling)6 =(scaling)◦(reflection)
(12) (a) The matrices are
∙
cos−sin
sincos
¸
and
⎡
⎣
cos−sin0
sincos0
001
⎤
⎦.IfArepresents either matrix, we
can verifyAis orthogonal by directly computingAA
.
(b) IfA=
⎡
⎣
0−118
10 −6
001
⎤
⎦,thenAA
=
⎡
⎣
325−108 18
−108 37 −6
18−61
⎤
⎦6=I 3.
(c) Letbe the slope of a nonvertical line. Then, the matrices are, respectively,
⎡
⎣
1−
2
1+
2
2
1+
2
2
1+
2
2
−1
1+
2
⎤
⎦and
⎡
⎢
⎢
⎢
⎣
1−
2
1+
2
2
1+
20
2
1+
2
2
−1
1+
20
001
⎤
⎥
⎥
⎥
⎦
For a vertical line, the matrices are
∙
−10
01
¸
and
⎡
⎣
−100
010
001
⎤
⎦. All of these matrices are
obviously symmetric, and since a reflection is its own inverse, all of these matrices are their own
inverses. Hence, ifAis any one of these matrices, thenAA
=AA=I,soAis orthogonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 164
(11) (a) Show that the matrices in parts (a) and (c) of Exercise 7 commute with each other when=.
Both products equal
⎡
⎣
(cos)−(sin)(1−(cos)) +(sin)
(sin)(cos)(1−(cos))−(sin)
00 1
⎤
⎦
(b) Consider the reflection about the-axis, whose matrix is given in Section 8.7 of the textbook as
A=
⎡
⎣
−100
010
001
⎤
⎦, and a counterclockwise rotation of90
◦
about the origin, whose matrix is
B=
⎡
⎣
0−10
100
001
⎤
⎦.ThenAB=
⎡
⎣
010
100
001
⎤
⎦, butBA=
⎡
⎣
0−10
−100
001
⎤
⎦,andsoAandB
do not commute. In particular, starting from the point(10), performing the rotation and then
the reflection yields(01). However, performing the reflection followed by the rotation produces
(0−1).
(c) Consider a reflection about=−and scaling by a factor ofin the-direction andin the
-direction, with6 =Now,
(reflection)◦(scaling)([10]) = (reflection)([0]) = [0−]
But,
(scaling
)◦(reflection)([10]) = (scaling)([0−1]) = [0−]
Hence,
(reflection)◦(scaling)6 =(scaling)◦(reflection)
(12) (a) The matrices are
∙
cos−sin
sincos
¸
and
⎡
⎣
cos−sin0
sincos0
001
⎤
⎦.IfArepresents either matrix, we
can verifyAis orthogonal by directly computingAA
.
(b) IfA=
⎡
⎣
0−118
10 −6
001
⎤
⎦,thenAA
=
⎡
⎣
325−108 18
−108 37 −6
18−61
⎤
⎦6=I 3.
(c) Letbe the slope of a nonvertical line. Then, the matrices are, respectively,
⎡
⎣
1−
2
1+
2
2
1+
2
2
1+
2
2
−1
1+
2
⎤
⎦and
⎡
⎢
⎢
⎢
⎣
1−
2
1+
2
2
1+
20
2
1+
2
2
−1
1+
20
001
⎤
⎥
⎥
⎥
⎦
For a vertical line, the matrices are
∙
−10
01
¸
and
⎡
⎣
−100
010
001
⎤
⎦. All of these matrices are
obviously symmetric, and since a reflection is its own inverse, all of these matrices are their own
inverses. Hence, ifAis any one of these matrices, thenAA
=AA=I,soAis orthogonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 164
Answers to Exercises Section 8.8
(13) (a) F (b) F (c) F (d) T (e) T (f) F
Section 8.8
(1) (a) 1
∙
7
3
¸
+
2
−
∙
2
1
¸
(b)
1
3
∙
1
1
¸
+
2
−2
∙
3
4
¸
(c)
1
⎡
⎣
0
−1
1
⎤
⎦+
2
⎡
⎣
1
−1
1
⎤
⎦+
3
3
⎡
⎣
2
0
1
⎤
⎦
(d)
1
⎡
⎣
−1
1
0
⎤
⎦+
2
⎡
⎣
5
0
2
⎤
⎦+
3
4
⎡
⎣
1
1
1
⎤
⎦
(There are other possible answers. For example, thefirst two vectors in the sum could be any
basis for the two-dimensional eigenspace corresponding to the eigenvalue1.)
(e)
1
⎡
⎢
⎢
⎣
−1
−1
1
0
⎤
⎥
⎥
⎦
+
2
⎡
⎢
⎢
⎣
1
3
0
1
⎤
⎥
⎥
⎦
+
3
⎡
⎢
⎢
⎣
2
3
0
1
⎤
⎥
⎥
⎦
+
4
−3
⎡
⎢
⎢
⎣
0
1
1
1
⎤
⎥
⎥
⎦
(There are other possible answers. For example, thefirst two vectors in the sum could be any
basis for the two-dimensional eigenspace corresponding to the eigenvalue1.)
(2) (a)=
1
2
+2
−3
(b)= 1
+2
−
+3
5
(c)= 1
2
+2
−2
+3
√
2
+4
−
√
2
(3) Using the fact thatAv =v, it can be easily seen that, withF()= 1
1
v1+···+
v,both
sides ofF
0
()=AF()yield 11
1
v1+···+
v.
(4) (a) Suppose(v
1v )is an ordered basis forR
consisting of eigenvectors ofAcorresponding
to the eigenvalues
1. Then, by Theorem 8.9, all solutions toF
0
()=AF()are of the
formF()=
1
1
v1+···+
v. Substituting=0yieldsv=F(0) = 1v1+···+
v.
But since(v
1v
)formsabasisforR
,1are uniquely determined (by Theorem 4.9).
Thus,F()is uniquely determined.
(b)F()=2
5
⎡
⎣
1
0
1
⎤
⎦−
⎡
⎣
−1
2
0
⎤
⎦−2
−
⎡
⎣
1
1
1
⎤
⎦.
(5) (a) The equations
1()=, 2()=
0
()=
(−1)
translate into these(−1)equations:
0
1
()= 2(),
0
2
()= 3()
0
−1
()= ().Also,
()
+−1
(−1)
+···+ 1
0
+0=0
translates into
0
()=− 01()− 12()−···− −1().Theseequations taken together
are easily seen to be represented byF
0
()=AF(),whereAandFare as given.
(b) Base Step(=1):A=[−
0],I1−A=[+ 0],andso A()=+ 0.
Inductive Step: Assume true for.Provetruefor+1. For the case+1,
A=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
0100 ···0
0010 ···0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0000 ···1
−
0−1−2−3···−
⎤
⎥
⎥
⎥
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 165
(13) (a) F (b) F (c) F (d) T (e) T (f) F
Section 8.8
(1) (a) 1
∙
7
3
¸
+
2
−
∙
2
1
¸
(b)
1
3
∙
1
1
¸
+
2
−2
∙
3
4
¸
(c)
1
⎡
⎣
0
−1
1
⎤
⎦+
2
⎡
⎣
1
−1
1
⎤
⎦+
3
3
⎡
⎣
2
0
1
⎤
⎦
(d)
1
⎡
⎣
−1
1
0
⎤
⎦+
2
⎡
⎣
5
0
2
⎤
⎦+
3
4
⎡
⎣
1
1
1
⎤
⎦
(There are other possible answers. For example, thefirst two vectors in the sum could be any
basis for the two-dimensional eigenspace corresponding to the eigenvalue1.)
(e)
1
⎡
⎢
⎢
⎣
−1
−1
1
0
⎤
⎥
⎥
⎦
+
2
⎡
⎢
⎢
⎣
1
3
0
1
⎤
⎥
⎥
⎦
+
3
⎡
⎢
⎢
⎣
2
3
0
1
⎤
⎥
⎥
⎦
+
4
−3
⎡
⎢
⎢
⎣
0
1
1
1
⎤
⎥
⎥
⎦
(There are other possible answers. For example, thefirst two vectors in the sum could be any
basis for the two-dimensional eigenspace corresponding to the eigenvalue1.)
(2) (a)=
1
2
+2
−3
(b)= 1
+2
−
+3
5
(c)= 1
2
+2
−2
+3
√
2
+4
−
√
2
(3) Using the fact thatAv =v, it can be easily seen that, withF()= 1
1
v1+···+
v,both
sides ofF
0
()=AF()yield 11
1
v1+···+
v.
(4) (a) Suppose(v
1v )is an ordered basis forR
consisting of eigenvectors ofAcorresponding
to the eigenvalues
1. Then, by Theorem 8.9, all solutions toF
0
()=AF()are of the
formF()=
1
1
v1+···+
v. Substituting=0yieldsv=F(0) = 1v1+···+
v.
But since(v
1v
)formsabasisforR
,1are uniquely determined (by Theorem 4.9).
Thus,F()is uniquely determined.
(b)F()=2
5
⎡
⎣
1
0
1
⎤
⎦−
⎡
⎣
−1
2
0
⎤
⎦−2
−
⎡
⎣
1
1
1
⎤
⎦.
(5) (a) The equations
1()=, 2()=
0
()=
(−1)
translate into these(−1)equations:
0
1
()= 2(),
0
2
()= 3()
0
−1
()= ().Also,
()
+−1
(−1)
+···+ 1
0
+0=0
translates into
0
()=− 01()− 12()−···− −1().Theseequations taken together
are easily seen to be represented byF
0
()=AF(),whereAandFare as given.
(b) Base Step(=1):A=[−
0],I1−A=[+ 0],andso A()=+ 0.
Inductive Step: Assume true for.Provetruefor+1. For the case+1,
A=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
0100 ···0
0010 ···0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0000 ···1
−
0−1−2−3···−
⎤
⎥
⎥
⎥
⎥
⎥
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 165
Answers to Exercises Section 8.8
Then
(I
+1−A)=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
−10 0 ···00
0−10 ···00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
00 00 ··· −1
0123··· −1 +
⎤
⎥
⎥
⎥
⎥
⎥
⎦
Using a cofactor expansion on thefirst column yields
|I
+1−A|=
¯
¯
¯
¯
¯
¯
¯
¯
¯
−10 ···00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
000 ··· −1
123··· −1 +
¯
¯
¯
¯
¯
¯
¯
¯
¯
+(−1)
(+1)+1
0
¯
¯
¯
¯
¯
¯
¯
¯
¯
−10 ···00
−1···00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
00 ···−1
¯
¯
¯
¯
¯
¯
¯
¯
¯
Thefirst determinant equals|I
−B|,where
B=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
010 ···00
001 ···00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
000 ···01
−
1−2−3···− −1 −
⎤
⎥
⎥
⎥
⎥
⎥
⎦
Now, using the inductive hypothesis on thefirst determinant, and Theorems 3.2 and 3.10 on the
second determinant (since its corresponding matrix is lower triangular) yields
A()=(
+
−1
+···+ 2+ 1)+(−1)
+2
0(−1)
=
+1
+
+···+ 1+ 0
(6) (a)[
23−01−12−···− −1]
(b) Using Theorem 8.10,
0=
A()=
+−1
−1
+···+ 1+ 0
Thus,
=− 0−1−···− −1
−1
.Lettingx=[1
2
−1
]and substitutingxfor
[
1]in part (a) gives
Ax=[
2
−1
−0−1−···− −1
−1
]
=[
2
−1
]=x
(c) Letv=[
23].Thenv=[ 2]. By part (a),
Av=[
23−0−···− −1]
Equating thefirst−1coordinates ofAvandvyields
2= 3=2 =−1
Copyrightc°2016 Elsevier Ltd. All rights reserved. 166
Then
(I
+1−A)=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
−10 0 ···00
0−10 ···00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
00 00 ··· −1
0123··· −1 +
⎤
⎥
⎥
⎥
⎥
⎥
⎦
Using a cofactor expansion on thefirst column yields
|I
+1−A|=
¯
¯
¯
¯
¯
¯
¯
¯
¯
−10 ···00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
000 ··· −1
123··· −1 +
¯
¯
¯
¯
¯
¯
¯
¯
¯
+(−1)
(+1)+1
0
¯
¯
¯
¯
¯
¯
¯
¯
¯
−10 ···00
−1···00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
00 ···−1
¯
¯
¯
¯
¯
¯
¯
¯
¯
Thefirst determinant equals|I
−B|,where
B=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
010 ···00
001 ···00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
000 ···01
−
1−2−3···− −1 −
⎤
⎥
⎥
⎥
⎥
⎥
⎦
Now, using the inductive hypothesis on thefirst determinant, and Theorems 3.2 and 3.10 on the
second determinant (since its corresponding matrix is lower triangular) yields
A()=(
+
−1
+···+ 2+ 1)+(−1)
+2
0(−1)
=
+1
+
+···+ 1+ 0
(6) (a)[
23−01−12−···− −1]
(b) Using Theorem 8.10,
0=
A()=
+−1
−1
+···+ 1+ 0
Thus,
=− 0−1−···− −1
−1
.Lettingx=[1
2
−1
]and substitutingxfor
[
1]in part (a) gives
Ax=[
2
−1
−0−1−···− −1
−1
]
=[
2
−1
]=x
(c) Letv=[
23].Thenv=[ 2]. By part (a),
Av=[
23−0−···− −1]
Equating thefirst−1coordinates ofAvandvyields
2= 3=2 =−1
Copyrightc°2016 Elsevier Ltd. All rights reserved. 166
Answers to Exercises Section 8.9
Further recursive substitution gives
2= 3=
2
=
−1
Hencev=[1
−1
].
(d) By parts (b) and (c),{[1
−1
]}is a basis for .
(7) (a) T (b) T (c) T (d) F
Section 8.9
(1) (a) Unique least-squares solution:v=
£
23
30
11
10
¤
;||Av−b||= √
6
6
≈0408;||Az−b||=1
(b) Unique least-squares solution:v=
£
229
75
1
3
¤
;||Av−b||=
59
√
3
15
≈6813;||Az−b||=7
(c) Infinite number of least-squares solutions, all of the formv=
£
7+
17
3
−13−
23
3
¤
;two par-
ticular least-squares solutions are
£
17
3
−
23
3
0
¤
and
£
8−12
1
3
¤
;withvas either of these vectors,
||Av−b||=
√
6
3
≈0816;||Az−b||=3
(d) Infinite number of least-squares solutions, all of the formv=
£
12
+
5
4
−1−
3
2
−
11
4
+
19
3
¤
;
two particular least-squares solutions are
£
−
1
2
29
6
10
¤
and
£
1
4
43
12
01
¤
;withvas either of these
vectors,||Av−b||=
√
6
3
≈0816;||Az−b||=
√
5≈2236
(2) (a) Infinite number of least-squares solutions, all of the formv=
£
−
4
7
+
19
42
8
7
−
5
21
¤
with
5
24
≤≤
19
24
(b) Infinite number of least-squares solutions, all of the formv=
£
−
6
5
+
19
10
−
1
5
+
19
35
¤
with
0≤≤
19
12
(3) (a)v≈[158−058]; (
0
I−C)v≈[0025−0015](Actual nearest eigenvalue is2+
√
3≈3732)
(b)v≈[046−036090]; (
0
I−C)v≈[003−004007](Actual nearest eigenvalue is
√
2≈1414)
(c)v≈[145−005−040]; (
0
I−C)v≈[−009−006−015](Actual nearest eigenvalue is
3
√
12≈
2289)
(4) IfAX=bis consistent, thenbis in the subspaceWof Theorem 8.13. Thus,proj
Wb=bFinally,
the fact that statement (3) of Theorem 8.13 is equivalent to statement (1) of Theorem 8.13 shows that
the actual solutions are the same as the least-squares solutions in this case.
(5) Letb=Av
1.ThenA
Av1=A
b.Also,A
Av2=A
Av1=A
b. Hencev 1andv 2both satisfy
part (3) of Theorem 8.13. Therefore, by part (1) of Theorem 8.13, ifW={Ax|x∈R
},then
Av
1=proj
Wb=Av 2.
(6) Letb=[
1 ],letAbe as in Theorem 8.4, and letW={Ax|x∈R
}Sinceproj
Wb∈W
exists, there must be somev∈R
such thatAv=proj
Wb. Hence,vsatisfies part (1) of Theorem
8.13, so(A
A)v=A
bby part (3) of Theorem 8.13. This shows that the system(A
A)X=A
b
is consistent, which proves part (2) of Theorem 8.4.
Next, let()=
0+1+···+
andz=[ 01]A short computation shows that
||Az−b||
2
=the sum of the squares of the vertical distances illustrated in Figure 8.10, just before
the definition of a least-squares polynomial in Section 8.3 of the textbook. Hence, minimizing||Az−b||
over all possible(+1)-vectorszgives the coefficients of a degreeleast-squares polynomial for the
given points(
11)( )However, parts (2) and (3) of Theorem 8.13 show that such a minimal
Copyrightc°2016 Elsevier Ltd. All rights reserved. 167
Further recursive substitution gives
2= 3=
2
=
−1
Hencev=[1
−1
].
(d) By parts (b) and (c),{[1
−1
]}is a basis for .
(7) (a) T (b) T (c) T (d) F
Section 8.9
(1) (a) Unique least-squares solution:v=
£
23
30
11
10
¤
;||Av−b||= √
6
6
≈0408;||Az−b||=1
(b) Unique least-squares solution:v=
£
229
75
1
3
¤
;||Av−b||=
59
√
3
15
≈6813;||Az−b||=7
(c) Infinite number of least-squares solutions, all of the formv=
£
7+
17
3
−13−
23
3
¤
;two par-
ticular least-squares solutions are
£
17
3
−
23
3
0
¤
and
£
8−12
1
3
¤
;withvas either of these vectors,
||Av−b||=
√
6
3
≈0816;||Az−b||=3
(d) Infinite number of least-squares solutions, all of the formv=
£
12
+
5
4
−1−
3
2
−
11
4
+
19
3
¤
;
two particular least-squares solutions are
£
−
1
2
29
6
10
¤
and
£
1
4
43
12
01
¤
;withvas either of these
vectors,||Av−b||=
√
6
3
≈0816;||Az−b||=
√
5≈2236
(2) (a) Infinite number of least-squares solutions, all of the formv=
£
−
4
7
+
19
42
8
7
−
5
21
¤
with
5
24
≤≤
19
24
(b) Infinite number of least-squares solutions, all of the formv=
£
−
6
5
+
19
10
−
1
5
+
19
35
¤
with
0≤≤
19
12
(3) (a)v≈[158−058]; (
0
I−C)v≈[0025−0015](Actual nearest eigenvalue is2+
√
3≈3732)
(b)v≈[046−036090]; (
0
I−C)v≈[003−004007](Actual nearest eigenvalue is
√
2≈1414)
(c)v≈[145−005−040]; (
0
I−C)v≈[−009−006−015](Actual nearest eigenvalue is
3
√
12≈
2289)
(4) IfAX=bis consistent, thenbis in the subspaceWof Theorem 8.13. Thus,proj
Wb=bFinally,
the fact that statement (3) of Theorem 8.13 is equivalent to statement (1) of Theorem 8.13 shows that
the actual solutions are the same as the least-squares solutions in this case.
(5) Letb=Av
1.ThenA
Av1=A
b.Also,A
Av2=A
Av1=A
b. Hencev 1andv 2both satisfy
part (3) of Theorem 8.13. Therefore, by part (1) of Theorem 8.13, ifW={Ax|x∈R
},then
Av
1=proj
Wb=Av 2.
(6) Letb=[
1 ],letAbe as in Theorem 8.4, and letW={Ax|x∈R
}Sinceproj
Wb∈W
exists, there must be somev∈R
such thatAv=proj
Wb. Hence,vsatisfies part (1) of Theorem
8.13, so(A
A)v=A
bby part (3) of Theorem 8.13. This shows that the system(A
A)X=A
b
is consistent, which proves part (2) of Theorem 8.4.
Next, let()=
0+1+···+
andz=[ 01]A short computation shows that
||Az−b||
2
=the sum of the squares of the vertical distances illustrated in Figure 8.10, just before
the definition of a least-squares polynomial in Section 8.3 of the textbook. Hence, minimizing||Az−b||
over all possible(+1)-vectorszgives the coefficients of a degreeleast-squares polynomial for the
given points(
11)( )However, parts (2) and (3) of Theorem 8.13 show that such a minimal
Copyrightc°2016 Elsevier Ltd. All rights reserved. 167
Answers to Exercises Section 8.10
solution is found by solving(A
A)v=A
bthus proving part (1) of Theorem 8.4.
Finally, whenA
Ais row equivalent toI +1, the uniqueness condition holds by Theorems 2.15
and 2.16, which proves part (3) of Theorem 8.4.
(7) (a) F (b) T (c) T (d) T (e) T
Section 8.10
(1) (a)C=
∙
812
0−9
¸
,A=
∙
86
6−9
¸
(b)C=
∙
7−17
011
¸
,A=
"
7−
17
2
−
17
2
11
#
(c)C=
⎡
⎣
54 −3
0−25
000
⎤
⎦,A=
⎡
⎢
⎢
⎣
52 −
3
2
2−2
5
2
−
3
2
5
2
0
⎤
⎥
⎥
⎦
(2) (a)A=
∙
43−24
−24 57
¸
,P=
1
5
∙
−34
43
¸
,D=
∙
75 0
025
¸
,
=
¡
1
5
[−34]
1
5
[43]
¢
,[x] =[−7−4],(x) = 4075
(b)A=
⎡
⎣
−51640
16 37 16
40 16 49
⎤
⎦,P=
1
9
⎡
⎣
1−84
−814
447
⎤
⎦,D=
⎡
⎣
27 0 0
0−27 0
0081
⎤
⎦,
=
¡
1
9
[1−84]
1
9
[−814]
1
9
[447]
¢
,[x] =[3−63],(x)=0
(c)A=
⎡
⎣
18 48 −30
48−68 18
−30 18 1
⎤
⎦,P=
1
7
⎡
⎣
2−63
3−2−6
632
⎤
⎦,D=
⎡
⎣
00 0
049 0
00 −98
⎤
⎦,
=
¡
1
7
[236]
1
7
[−6−23]
1
7
[3−62]
¢
,[x] =[506],(x)=−3528
(d)A=
⎡
⎢
⎢
⎣
10 −12 12
0 5 60 60
−12 60 864 576
12 60 576 864
⎤
⎥
⎥
⎦
,P=
1
17
⎡
⎢
⎢
⎣
10 12 −12
01 −12−12
−12 12 1 0
12 12 0 1
⎤
⎥
⎥
⎦
,D=
⎡
⎢
⎢
⎣
289 0 0 0
0 1445 0 0
0000
0000
⎤
⎥
⎥
⎦
,
=
¡
1
17
[10−1212]
1
17
[011212]
1
17
[12−1210]
1
17
[−12−1201]
¢
,
[x]
=[1−3−3−10],(x) = 13294
(3) First,
=e
Ae=e
Be=.Also,for6 =,letx=e +e.Then
+++=x
Ax=x
Bx= +++
Using
=and =,weget +=+. Hence, sinceAandBare symmetric,
2
=2 ,andso =.
(4) Yes; if(x)=Σ
,1≤≤≤,thenx
C1xandC 1upper triangular imply that the( )
entry forC
1is zero ifand if≤. A similar argument describesC 2.Thus,C 1=C2.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 168
solution is found by solving(A
A)v=A
bthus proving part (1) of Theorem 8.4.
Finally, whenA
Ais row equivalent toI +1, the uniqueness condition holds by Theorems 2.15
and 2.16, which proves part (3) of Theorem 8.4.
(7) (a) F (b) T (c) T (d) T (e) T
Section 8.10
(1) (a)C=
∙
812
0−9
¸
,A=
∙
86
6−9
¸
(b)C=
∙
7−17
011
¸
,A=
"
7−
17
2
−
17
2
11
#
(c)C=
⎡
⎣
54 −3
0−25
000
⎤
⎦,A=
⎡
⎢
⎢
⎣
52 −
3
2
2−2
5
2
−
3
2
5
2
0
⎤
⎥
⎥
⎦
(2) (a)A=
∙
43−24
−24 57
¸
,P=
1
5
∙
−34
43
¸
,D=
∙
75 0
025
¸
,
=
¡
1
5
[−34]
1
5
[43]
¢
,[x] =[−7−4],(x) = 4075
(b)A=
⎡
⎣
−51640
16 37 16
40 16 49
⎤
⎦,P=
1
9
⎡
⎣
1−84
−814
447
⎤
⎦,D=
⎡
⎣
27 0 0
0−27 0
0081
⎤
⎦,
=
¡
1
9
[1−84]
1
9
[−814]
1
9
[447]
¢
,[x] =[3−63],(x)=0
(c)A=
⎡
⎣
18 48 −30
48−68 18
−30 18 1
⎤
⎦,P=
1
7
⎡
⎣
2−63
3−2−6
632
⎤
⎦,D=
⎡
⎣
00 0
049 0
00 −98
⎤
⎦,
=
¡
1
7
[236]
1
7
[−6−23]
1
7
[3−62]
¢
,[x] =[506],(x)=−3528
(d)A=
⎡
⎢
⎢
⎣
10 −12 12
0 5 60 60
−12 60 864 576
12 60 576 864
⎤
⎥
⎥
⎦
,P=
1
17
⎡
⎢
⎢
⎣
10 12 −12
01 −12−12
−12 12 1 0
12 12 0 1
⎤
⎥
⎥
⎦
,D=
⎡
⎢
⎢
⎣
289 0 0 0
0 1445 0 0
0000
0000
⎤
⎥
⎥
⎦
,
=
¡
1
17
[10−1212]
1
17
[011212]
1
17
[12−1210]
1
17
[−12−1201]
¢
,
[x]
=[1−3−3−10],(x) = 13294
(3) First,
=e
Ae=e
Be=.Also,for6 =,letx=e +e.Then
+++=x
Ax=x
Bx= +++
Using
=and =,weget +=+. Hence, sinceAandBare symmetric,
2
=2 ,andso =.
(4) Yes; if(x)=Σ
,1≤≤≤,thenx
C1xandC 1upper triangular imply that the( )
entry forC
1is zero ifand if≤. A similar argument describesC 2.Thus,C 1=C2.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 168
Answers to Exercises Section 8.10
(5) (a) Consider the expression(x)= 11
2
1
+···+
2
in Step 3 of the Quadratic Form Method,
where
11are the eigenvalues ofA,and[x] =[1]. Clearly, if all of 11
are positive andx6 =0,then(x)must be positive. (Obviously,(0)=0.)
Conversely, if(x)is positive for allx6 =0, then choosexso that[x]
=eto yield(x)= ,
thus proving that the eigenvalue
is positive.
(b) Replace “positive” with “nonnegative” throughout the solution to part (a).
(6) (a) T (b) F (c) F (d) T (e) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 169
(5) (a) Consider the expression(x)= 11
2
1
+···+
2
in Step 3 of the Quadratic Form Method,
where
11are the eigenvalues ofA,and[x] =[1]. Clearly, if all of 11
are positive andx6 =0,then(x)must be positive. (Obviously,(0)=0.)
Conversely, if(x)is positive for allx6 =0, then choosexso that[x]
=eto yield(x)= ,
thus proving that the eigenvalue
is positive.
(b) Replace “positive” with “nonnegative” throughout the solution to part (a).
(6) (a) T (b) F (c) F (d) T (e) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 169
Answers to Exercises Section 9.1
Chapter 9
Section 9.1
(1) (a) Solution tofirst system:(6021500); solution to second system:(302750). The systems are
ill-conditioned, because a very small change in the coefficient ofleads to a very large change in
the solution.
(b) Solution tofirst system:(4008002000); solution to second system:(336666
2
3
1600). Systems
are ill-conditioned.
(2) Answers to this problem may differ significantly from the following, depending on where the rounding
is performed in the algorithm:
(a) Without partial pivoting:(32100765); with partial pivoting:(32300767).Actualsolutionis
(32140765).
(b) Without partial pivoting:(3040107−521); with partial pivoting:(3010101−503).(Actual
solution is(300010−5).)
(c) Without partial pivoting:(226101−211); with partial pivoting:(277−327595).Actual
solution is(267−315573).
(3) Answers to this problem may differ significantly from the following, depending on where the rounding
is performed in the algorithm:
(a) Without partial pivoting:(321407651)
; with partial pivoting:(321307648). Actual solution is
(32140765).
(b) Without partial pivoting:(300110−5); with partial pivoting:(30009995−4999).(Actual
solution is(300010−5).)
(c) Without partial pivoting:(−23808801−1630); with partial pivoting:(2678−31595746).
Actual solution is(267−315573).
(4) (a)
1 2
Initial Values00000000
After1Step5200−6000
After2Steps6400−8229
After3Steps6846−8743
After4Steps6949−8934
After5Steps6987−8978
After6Steps6996−8994
After7Steps6999−8998
After8Steps7000−9000
After9Steps7000−9000
Copyrightc°2016 Elsevier Ltd. All rights reserved. 170
Chapter 9
Section 9.1
(1) (a) Solution tofirst system:(6021500); solution to second system:(302750). The systems are
ill-conditioned, because a very small change in the coefficient ofleads to a very large change in
the solution.
(b) Solution tofirst system:(4008002000); solution to second system:(336666
2
3
1600). Systems
are ill-conditioned.
(2) Answers to this problem may differ significantly from the following, depending on where the rounding
is performed in the algorithm:
(a) Without partial pivoting:(32100765); with partial pivoting:(32300767).Actualsolutionis
(32140765).
(b) Without partial pivoting:(3040107−521); with partial pivoting:(3010101−503).(Actual
solution is(300010−5).)
(c) Without partial pivoting:(226101−211); with partial pivoting:(277−327595).Actual
solution is(267−315573).
(3) Answers to this problem may differ significantly from the following, depending on where the rounding
is performed in the algorithm:
(a) Without partial pivoting:(321407651)
; with partial pivoting:(321307648). Actual solution is
(32140765).
(b) Without partial pivoting:(300110−5); with partial pivoting:(30009995−4999).(Actual
solution is(300010−5).)
(c) Without partial pivoting:(−23808801−1630); with partial pivoting:(2678−31595746).
Actual solution is(267−315573).
(4) (a)
1 2
Initial Values00000000
After1Step5200−6000
After2Steps6400−8229
After3Steps6846−8743
After4Steps6949−8934
After5Steps6987−8978
After6Steps6996−8994
After7Steps6999−8998
After8Steps7000−9000
After9Steps7000−9000
Copyrightc°2016 Elsevier Ltd. All rights reserved. 170
Answers to Exercises Section 9.1
(b)
1 2 3
Initial Values000000000000
After1Step−0778−43752000
After2Steps−1042−48192866
After3Steps−0995−49942971
After4Steps−1003−49952998
After5Steps−1000−50002999
After6Steps−1000−50003000
After7Steps−1000−50003000
(c)
1 2 3
Initial Values000000000000
After1Step −88574500−4333
After2Steps−107383746−8036
After3Steps−116884050−8537
After4Steps−118753975−8904
After5Steps−119694005−8954
After6Steps−119883998−8991
After7Steps−119974001−8996
After8Steps−119994000−8999
After9Steps−120004000−9000
After10Steps−120004000−9000
(d)
1 2 3 4
Initial Values0000000000000000
After1Step0900−16673000−2077
After2Steps1874−09723792−2044
After3Steps1960−09993966−2004
After4Steps1993−09983989−1999
After5Steps1997−10013998−2000
After6Steps2000−10003999−2000
After7Steps2000−10004000−2000
After8Steps2000−10004000−2000
(5) (a)
1 2
Initial Values00000000
After1Step5200−8229
After2Steps6846−8934
After3Steps6987−8994
After4Steps6999−9000
After5Steps7000−9000
After6Steps7000−9000
Copyrightc°2016 Elsevier Ltd. All rights reserved. 171
(b)
1 2 3
Initial Values000000000000
After1Step−0778−43752000
After2Steps−1042−48192866
After3Steps−0995−49942971
After4Steps−1003−49952998
After5Steps−1000−50002999
After6Steps−1000−50003000
After7Steps−1000−50003000
(c)
1 2 3
Initial Values000000000000
After1Step −88574500−4333
After2Steps−107383746−8036
After3Steps−116884050−8537
After4Steps−118753975−8904
After5Steps−119694005−8954
After6Steps−119883998−8991
After7Steps−119974001−8996
After8Steps−119994000−8999
After9Steps−120004000−9000
After10Steps−120004000−9000
(d)
1 2 3 4
Initial Values0000000000000000
After1Step0900−16673000−2077
After2Steps1874−09723792−2044
After3Steps1960−09993966−2004
After4Steps1993−09983989−1999
After5Steps1997−10013998−2000
After6Steps2000−10003999−2000
After7Steps2000−10004000−2000
After8Steps2000−10004000−2000
(5) (a)
1 2
Initial Values00000000
After1Step5200−8229
After2Steps6846−8934
After3Steps6987−8994
After4Steps6999−9000
After5Steps7000−9000
After6Steps7000−9000
Copyrightc°2016 Elsevier Ltd. All rights reserved. 171
Answers to Exercises Section 9.1
(b)
1 2 3
Initial Values000000000000
After1Step−0778−45692901
After2Steps−0963−49782993
After3Steps−0998−49993000
After4Steps−1000−50003000
After5Steps−1000−50003000
(c)
1 2 3
Initial Values000000000000
After1Step −88573024−7790
After2Steps−115153879−8818
After3Steps−119313981−8974
After4Steps−119903997−8996
After5Steps−119984000−8999
After6Steps−120004000−9000
After7Steps−120004000−9000
(d)
1 2 3 4
Initial Values0000000000000000
After1Step0900−17673510−2012
After2Steps1980−10504003−2002
After3Steps2006−10004002−2000
After4Steps2000−10004000−2000
After5Steps2000−10004000−2000
(6) Strictly diagonally dominant: (a), (c)
(7) (a) Put the third equationfirst, and move the other two down to get the following:
1 2 3
Initial Values000000000000
After1Step3125−04811461
After2Steps2517−05001499
After3Steps2500−05001500
After4Steps2500−05001500
(b) Put thefirst equation last (and leave the other two alone) to get:
1 2 3
Initial Values000000000000
After1Step3700−68564965
After2Steps3889−79805045
After3Steps3993−80095004
After4Steps4000−80015000
After5Steps4000−80005000
After6Steps4000−80005000
(c) Put the second equationfirst, the fourth equation second, thefirst equation third, and the third
equation fourth to get the following:
Copyrightc°2016 Elsevier Ltd. All rights reserved. 172
(b)
1 2 3
Initial Values000000000000
After1Step−0778−45692901
After2Steps−0963−49782993
After3Steps−0998−49993000
After4Steps−1000−50003000
After5Steps−1000−50003000
(c)
1 2 3
Initial Values000000000000
After1Step −88573024−7790
After2Steps−115153879−8818
After3Steps−119313981−8974
After4Steps−119903997−8996
After5Steps−119984000−8999
After6Steps−120004000−9000
After7Steps−120004000−9000
(d)
1 2 3 4
Initial Values0000000000000000
After1Step0900−17673510−2012
After2Steps1980−10504003−2002
After3Steps2006−10004002−2000
After4Steps2000−10004000−2000
After5Steps2000−10004000−2000
(6) Strictly diagonally dominant: (a), (c)
(7) (a) Put the third equationfirst, and move the other two down to get the following:
1 2 3
Initial Values000000000000
After1Step3125−04811461
After2Steps2517−05001499
After3Steps2500−05001500
After4Steps2500−05001500
(b) Put thefirst equation last (and leave the other two alone) to get:
1 2 3
Initial Values000000000000
After1Step3700−68564965
After2Steps3889−79805045
After3Steps3993−80095004
After4Steps4000−80015000
After5Steps4000−80005000
After6Steps4000−80005000
(c) Put the second equationfirst, the fourth equation second, thefirst equation third, and the third
equation fourth to get the following:
Copyrightc°2016 Elsevier Ltd. All rights reserved. 172
Answers to Exercises Section 9.1
1 2 3 4
Initial Values000000000000 0000
After1Step 5444−53799226−10447
After2Steps8826−843510808−11698
After3Steps9820−892010961−11954
After4Steps9973−898610994−11993
After5Steps9995−899810999−11999
After6Steps9999−900011000−12000
After7Steps10000−900011000−12000
After8Steps10000−900011000−12000
(8) The Jacobi Method yields the following:
1 2 3
Initial Values00 00 00
After1Step 160−130 120
After2Steps−370 590 −870
After3Steps 2240−610 2120
After4Steps−770 9070−14950
After5Steps3056025150 −3560
After6Steps122350190350−238950
The Gauss-Seidel Method yields the following:
1 2 3
Initial Values 00 00 00
After1Step 160 830 −1830
After2Steps 2480 18410 −35650
After3Steps 56560410530 −806330
After4Steps12464809091410−17816650
The actual solution is(2−31).
(9) (a)
1 2 3
Initial Values000000000000
After1Step 350022501625
After2Steps156324062516
After3Steps103922232869
After4Steps095420892979
After5Steps096620283003
After6Steps098520063005
After7Steps099520003003
After8Steps099919993001
After9Steps100020003000
After10Steps100020003000
Copyrightc°2016 Elsevier Ltd. All rights reserved. 173
1 2 3 4
Initial Values000000000000 0000
After1Step 5444−53799226−10447
After2Steps8826−843510808−11698
After3Steps9820−892010961−11954
After4Steps9973−898610994−11993
After5Steps9995−899810999−11999
After6Steps9999−900011000−12000
After7Steps10000−900011000−12000
After8Steps10000−900011000−12000
(8) The Jacobi Method yields the following:
1 2 3
Initial Values00 00 00
After1Step 160−130 120
After2Steps−370 590 −870
After3Steps 2240−610 2120
After4Steps−770 9070−14950
After5Steps3056025150 −3560
After6Steps122350190350−238950
The Gauss-Seidel Method yields the following:
1 2 3
Initial Values 00 00 00
After1Step 160 830 −1830
After2Steps 2480 18410 −35650
After3Steps 56560410530 −806330
After4Steps12464809091410−17816650
The actual solution is(2−31).
(9) (a)
1 2 3
Initial Values000000000000
After1Step 350022501625
After2Steps156324062516
After3Steps103922232869
After4Steps095420892979
After5Steps096620283003
After6Steps098520063005
After7Steps099520003003
After8Steps099919993001
After9Steps100020003000
After10Steps100020003000
Copyrightc°2016 Elsevier Ltd. All rights reserved. 173
Answers to Exercises Section 9.2
(b)
1 2 3
Initial Values000000000000
After1Step 350040004500
After2Steps−075000000750
After3Steps312540004875
After4Steps−093800000938
After5Steps303140004969
After6Steps−098400000985
After7Steps300840004992
After8Steps−099600000996
(10) (a) T (b) F (c) F (d) T (e) F (f) F
Section 9.2
(1) (a)LDU=
∙
10
−31
¸∙
20
05
¸∙
1−2
01
¸
(b)LDU=
∙
10
1
2
1
¸∙
30
0−2
¸∙
1
1
3
01
¸
(c)LDU=
⎡
⎣
100
−210
−241
⎤
⎦
⎡
⎣
−100
020
003
⎤
⎦
⎡
⎣
1−42
01 −4
001
⎤
⎦
(d)LDU=
⎡
⎢
⎣
100
5
2
10
1
2
−
3
2
1
⎤
⎥
⎦
⎡
⎣
200
0−40
003
⎤
⎦
⎡
⎣
13−2
01−5
00 1
⎤
⎦
(e)LDU=
⎡
⎢
⎢
⎢
⎣
1 000
−
4
3
100
−2−
3
2
10
2
3
−201
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
−3000
0−
2
3
00
00
1
2
0
0001
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
1−
1
3
−
1
3
1
3
01
5
2
−
11
2
001 3
000 1
⎤
⎥
⎥
⎥
⎦
(f)LDU=
⎡
⎢
⎢
⎣
1000
2100
−3−310
−12 −21
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
−3000
0−20 0
0010
000 −3
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
14−2−3
01 0 −1
0015
0001
⎤
⎥
⎥
⎦
(2) (a) With the given values ofL,D,andU,LDU=
∙
+
¸
.Thefirst row ofLDUcannot
equal[01], since this would give=0,forcing=06 =1.
(b) The matrix
∙
01
10
¸
cannot be reduced to row echelon form using only Type (I) and lower Type
(II) row operations.
(3) (a)
KU=
∙
−10
2−3
¸∙
1−5
01
¸
;solution={(4−1)}.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 174
(b)
1 2 3
Initial Values000000000000
After1Step 350040004500
After2Steps−075000000750
After3Steps312540004875
After4Steps−093800000938
After5Steps303140004969
After6Steps−098400000985
After7Steps300840004992
After8Steps−099600000996
(10) (a) T (b) F (c) F (d) T (e) F (f) F
Section 9.2
(1) (a)LDU=
∙
10
−31
¸∙
20
05
¸∙
1−2
01
¸
(b)LDU=
∙
10
1
2
1
¸∙
30
0−2
¸∙
1
1
3
01
¸
(c)LDU=
⎡
⎣
100
−210
−241
⎤
⎦
⎡
⎣
−100
020
003
⎤
⎦
⎡
⎣
1−42
01 −4
001
⎤
⎦
(d)LDU=
⎡
⎢
⎣
100
5
2
10
1
2
−
3
2
1
⎤
⎥
⎦
⎡
⎣
200
0−40
003
⎤
⎦
⎡
⎣
13−2
01−5
00 1
⎤
⎦
(e)LDU=
⎡
⎢
⎢
⎢
⎣
1 000
−
4
3
100
−2−
3
2
10
2
3
−201
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
−3000
0−
2
3
00
00
1
2
0
0001
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
1−
1
3
−
1
3
1
3
01
5
2
−
11
2
001 3
000 1
⎤
⎥
⎥
⎥
⎦
(f)LDU=
⎡
⎢
⎢
⎣
1000
2100
−3−310
−12 −21
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
−3000
0−20 0
0010
000 −3
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
14−2−3
01 0 −1
0015
0001
⎤
⎥
⎥
⎦
(2) (a) With the given values ofL,D,andU,LDU=
∙
+
¸
.Thefirst row ofLDUcannot
equal[01], since this would give=0,forcing=06 =1.
(b) The matrix
∙
01
10
¸
cannot be reduced to row echelon form using only Type (I) and lower Type
(II) row operations.
(3) (a)
KU=
∙
−10
2−3
¸∙
1−5
01
¸
;solution={(4−1)}.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 174
Answers to Exercises Section 9.3
(b)KU=
⎡
⎣
200
2−10
1−33
⎤
⎦
⎡
⎣
1−25
013
001
⎤
⎦;solution={(5−12)}.
(c)KU=
⎡
⎣
−10 0
43 0
−25−2
⎤
⎦
⎡
⎣
1−32
01 −5
001
⎤
⎦;solution={(2−31)}.
(d)KU=
⎡
⎢
⎢
⎣
3000
1−20 0
−5−14 0
130 −3
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
1−522
01 −
30
001 −2
0001
⎤
⎥
⎥
⎦
;solution={(2−213)}.
(4) (a) F (b) T (c) F (d) F
Section 9.3
(1) (a) After nine iterations, eigenvector=[060080]and eigenvalue=50.
(b) Afterfive iterations, eigenvector=[091042]and eigenvalue=531.
(c) After seven iterations, eigenvector=[041041082]and eigenvalue=30.
(d) Afterfive iterations, eigenvector=[058058000058]and eigenvalue=600.
(e) Afterfifteen iterations, eigenvector=[0346085201850346]and eigenvalue=5405
(f) After six iterations, eigenvector=[04455−05649−0
6842−01193]and eigenvalue=57323.
(2) For the matrices in both parts, the Power Method fails to converge, even after many iterations. The
matrix in part (a) is not diagonalizable, with1as its only eigenvalue and dim(
1)=1. The matrix in
part (b) has eigenvalues1,3,and−3, none of which are strictly dominant.
(3) (a)u
is derived fromu −1asu=Au−1,where is a normalizing constant. A proof by induction
shows thatu
=A
u0for some nonzero constant .Ifu 0=0v1+0v2,then
u
=0A
v1+0A
v2=0
1
v1+0
2
v2
Hence
=0
1
and =0
2
.Thus,
|
|
||
=
¯
¯
¯
¯
¯
0
1
0
2
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
1
2
¯
¯
¯
¯
|0|
|0|
(b) Let
1be the eigenvalues ofAwith| 1|| |,for2≤≤.Let{v 1v }be as given
in the exercise. Suppose the initial vector in the Power Method isu
0=01v1+···+ 0vand
theth iteration yieldsu
=1v1+···+ v. As in part (a), a proof by induction shows that
u
=A
u0for some nonzero constant . Therefore,
u
= 01A
v1+02A
v2+···+ 0A
v
= 01
1
v1+02
2
v2+···+ 0
v
Hence,
=0
.Thus,for2≤≤, 6 =0,and 06 =0,wehave
|
1|
||
=
¯
¯
01
1
¯
¯
¯
¯
0
¯
¯
=
¯
¯
¯
¯
1
¯
¯
¯
¯
|01|
|0|
Copyrightc°2016 Elsevier Ltd. All rights reserved. 175
(b)KU=
⎡
⎣
200
2−10
1−33
⎤
⎦
⎡
⎣
1−25
013
001
⎤
⎦;solution={(5−12)}.
(c)KU=
⎡
⎣
−10 0
43 0
−25−2
⎤
⎦
⎡
⎣
1−32
01 −5
001
⎤
⎦;solution={(2−31)}.
(d)KU=
⎡
⎢
⎢
⎣
3000
1−20 0
−5−14 0
130 −3
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
1−522
01 −
30
001 −2
0001
⎤
⎥
⎥
⎦
;solution={(2−213)}.
(4) (a) F (b) T (c) F (d) F
Section 9.3
(1) (a) After nine iterations, eigenvector=[060080]and eigenvalue=50.
(b) Afterfive iterations, eigenvector=[091042]and eigenvalue=531.
(c) After seven iterations, eigenvector=[041041082]and eigenvalue=30.
(d) Afterfive iterations, eigenvector=[058058000058]and eigenvalue=600.
(e) Afterfifteen iterations, eigenvector=[0346085201850346]and eigenvalue=5405
(f) After six iterations, eigenvector=[04455−05649−0
6842−01193]and eigenvalue=57323.
(2) For the matrices in both parts, the Power Method fails to converge, even after many iterations. The
matrix in part (a) is not diagonalizable, with1as its only eigenvalue and dim(
1)=1. The matrix in
part (b) has eigenvalues1,3,and−3, none of which are strictly dominant.
(3) (a)u
is derived fromu −1asu=Au−1,where is a normalizing constant. A proof by induction
shows thatu
=A
u0for some nonzero constant .Ifu 0=0v1+0v2,then
u
=0A
v1+0A
v2=0
1
v1+0
2
v2
Hence
=0
1
and =0
2
.Thus,
|
|
||
=
¯
¯
¯
¯
¯
0
1
0
2
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
1
2
¯
¯
¯
¯
|0|
|0|
(b) Let
1be the eigenvalues ofAwith| 1|| |,for2≤≤.Let{v 1v }be as given
in the exercise. Suppose the initial vector in the Power Method isu
0=01v1+···+ 0vand
theth iteration yieldsu
=1v1+···+ v. As in part (a), a proof by induction shows that
u
=A
u0for some nonzero constant . Therefore,
u
= 01A
v1+02A
v2+···+ 0A
v
= 01
1
v1+02
2
v2+···+ 0
v
Hence,
=0
.Thus,for2≤≤, 6 =0,and 06 =0,wehave
|
1|
||
=
¯
¯
01
1
¯
¯
¯
¯
0
¯
¯
=
¯
¯
¯
¯
1
¯
¯
¯
¯
|01|
|0|
Copyrightc°2016 Elsevier Ltd. All rights reserved. 175
Answers to Exercises Section 9.4
(4) (a) F (b) T (c) T (d) F
Section 9.4
(1) (a)Q=
1
3
⎡
⎣
21−2
−22−1
12 2
⎤
⎦;R=
⎡
⎣
36 3
06−9
00 3
⎤
⎦
(b)Q=
1
11
⎡
⎣
6−2−9
7−66
692
⎤
⎦;R=
⎡
⎣
11 22 11
01122
0011
⎤
⎦
(c)Q=
⎡
⎢
⎢
⎣
√
6
6
√
3
3
√
2
2
−
√
6
3
√
3
3
0
√
6
6
√
3
3
−
√
2
2
⎤
⎥
⎥
⎦
;R=
⎡
⎢
⎢
⎣
√
63
√
6−
2
√
6
3
02
√
3−
10
√
3
3
00
√
2
⎤
⎥
⎥
⎦
(d)Q=
1
3
⎡
⎢
⎢
⎣
220
2−21
0−1−2
10 −2
⎤
⎥
⎥
⎦
;R=
⎡
⎣
6−39
0912
0015
⎤
⎦
(e)Q=
1
105
⎡
⎢
⎢
⎣
14 99 −232
70 0 −70−35
77−18 64 26
03045 −90
⎤
⎥
⎥
⎦
;R=
⎡
⎢
⎢
⎣
105 105−105 210
0 210 105 315
0 0 105 420
0 0 0 210
⎤
⎥
⎥
⎦
(2) (a)Q=
1
13
⎡
⎣
34
4−12
12 3
⎤
⎦;R=
∙
13 26
013
¸
;
∙
¸
=
1
169
∙
940
−362
¸
≈
∙
5562
−2142
¸
(b)Q=
1
3
⎡
⎢
⎢
⎣
2−21
10 −2
0−1−2
220
⎤
⎥
⎥
⎦
;R=
⎡
⎣
39 6
06 9
0012
⎤
⎦;
⎡
⎣
⎤
⎦=
5
72
⎡
⎣
43
3
62
⎤
⎦≈
⎡
⎣
2986
0208
4306
⎤
⎦
(c)Q=
1
9
⎡
⎢
⎢
⎢
⎢
⎢
⎣
18 2
√
2
40
7
2
√
2
04−
9
2
√
2
−81 2
√
2
⎤
⎥
⎥
⎥
⎥
⎥
⎦
;R=
⎡
⎣
9−99
018 −9
0018
√
2
⎤
⎦;
⎡
⎣
⎤
⎦=
1
108
⎡
⎣
−61
65
66
⎤
⎦≈
⎡
⎣
−0565
0602
0611
⎤
⎦
(d)Q=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
3
2
3
2
19
√
19
2
9
4
9
−
1
19
√
19
4
9
2
9
−
3
19
√
19
4
9
−
4
9
−
1
19
√
19
2
3
−
1
3
2
19
√
19
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
;R=
⎡
⎣
9−18 18
09 9
002
√
19
⎤
⎦;
⎡
⎣
⎤
⎦=
1
1026
⎡
⎣
2968
6651
3267
⎤
⎦≈
⎡
⎣
2893
6482
3184
⎤
⎦
(3) We will show that the entries ofQandRare uniquely determined by the given requirements. We will
proceed column by column, using a proof by induction on the column number.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 176
(4) (a) F (b) T (c) T (d) F
Section 9.4
(1) (a)Q=
1
3
⎡
⎣
21−2
−22−1
12 2
⎤
⎦;R=
⎡
⎣
36 3
06−9
00 3
⎤
⎦
(b)Q=
1
11
⎡
⎣
6−2−9
7−66
692
⎤
⎦;R=
⎡
⎣
11 22 11
01122
0011
⎤
⎦
(c)Q=
⎡
⎢
⎢
⎣
√
6
6
√
3
3
√
2
2
−
√
6
3
√
3
3
0
√
6
6
√
3
3
−
√
2
2
⎤
⎥
⎥
⎦
;R=
⎡
⎢
⎢
⎣
√
63
√
6−
2
√
6
3
02
√
3−
10
√
3
3
00
√
2
⎤
⎥
⎥
⎦
(d)Q=
1
3
⎡
⎢
⎢
⎣
220
2−21
0−1−2
10 −2
⎤
⎥
⎥
⎦
;R=
⎡
⎣
6−39
0912
0015
⎤
⎦
(e)Q=
1
105
⎡
⎢
⎢
⎣
14 99 −232
70 0 −70−35
77−18 64 26
03045 −90
⎤
⎥
⎥
⎦
;R=
⎡
⎢
⎢
⎣
105 105−105 210
0 210 105 315
0 0 105 420
0 0 0 210
⎤
⎥
⎥
⎦
(2) (a)Q=
1
13
⎡
⎣
34
4−12
12 3
⎤
⎦;R=
∙
13 26
013
¸
;
∙
¸
=
1
169
∙
940
−362
¸
≈
∙
5562
−2142
¸
(b)Q=
1
3
⎡
⎢
⎢
⎣
2−21
10 −2
0−1−2
220
⎤
⎥
⎥
⎦
;R=
⎡
⎣
39 6
06 9
0012
⎤
⎦;
⎡
⎣
⎤
⎦=
5
72
⎡
⎣
43
3
62
⎤
⎦≈
⎡
⎣
2986
0208
4306
⎤
⎦
(c)Q=
1
9
⎡
⎢
⎢
⎢
⎢
⎢
⎣
18 2
√
2
40
7
2
√
2
04−
9
2
√
2
−81 2
√
2
⎤
⎥
⎥
⎥
⎥
⎥
⎦
;R=
⎡
⎣
9−99
018 −9
0018
√
2
⎤
⎦;
⎡
⎣
⎤
⎦=
1
108
⎡
⎣
−61
65
66
⎤
⎦≈
⎡
⎣
−0565
0602
0611
⎤
⎦
(d)Q=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
3
2
3
2
19
√
19
2
9
4
9
−
1
19
√
19
4
9
2
9
−
3
19
√
19
4
9
−
4
9
−
1
19
√
19
2
3
−
1
3
2
19
√
19
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
;R=
⎡
⎣
9−18 18
09 9
002
√
19
⎤
⎦;
⎡
⎣
⎤
⎦=
1
1026
⎡
⎣
2968
6651
3267
⎤
⎦≈
⎡
⎣
2893
6482
3184
⎤
⎦
(3) We will show that the entries ofQandRare uniquely determined by the given requirements. We will
proceed column by column, using a proof by induction on the column number.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 176
Answers to Exercises Section 9.4
Base Step:=1.Here,
[1st column ofA]=Q·[1st column ofR]=[1st column ofQ](
11)
becauseRis upper triangular. But, since the1st column ofQis a unit vector and
11is positive,
11=k[1st column ofA]kand[1st column ofQ]=
1
11
[1st column ofA]
Hence, thefirst column in each ofQandRis uniquely determined.
Inductive Step: Assume1and that columns1(−1)of bothQandRare uniquely
determined. We will show that theth column in each ofQandRis uniquely determined. Now,
[th column ofA]=Q·[th column ofR]=
X
=1
[th column ofQ]( )
Letv=
P
−1
=1
[th column ofQ]( ). By the Inductive Hypothesis,vis uniquely determined. Now,
[th column ofQ](
)=[th column ofA]−v
But since theth column ofQis a unit vector and
is positive,
=k[th column ofA]−vkand[th column ofQ]=
1
([th column ofA]−v)
Hence, theth column in each ofQandRis uniquely determined.
(4) (a) IfAis square, thenA=QR,whereQis an orthogonal matrix andRhas nonnegative entries
along its main diagonal. SettingU=R,weseethat
A
A=U
Q
QU=U
(I)U=U
U
(b) SupposeA=QR(where this is aQRfactorization ofA)andA
A=U
U.ThenRandR
must be nonsingular. Also,P=Q(R
)
−1
U
is an orthogonal matrix, because
PP
=Q(R
)
−1
U
¡
Q(R
)
−1
U
¢
=Q(R
)
−1
U
UR
−1
Q
=Q(R
)
−1
A
AR
−1
Q
=Q(R
)
−1
(R
Q
)(QR)R
−1
Q
=Q((R
)
−1
R
)(Q
Q)(RR
−1
)Q
=QI
IIQ
=QQ
=I
Finally,
PU=Q(R
)
−1
U
U=(Q
)
−1
(R
)
−1
A
A
=(R
Q
)
−1
A
A=(A
)
−1
A
A=I A=A
and soPUis aQRfactorization ofA, which is unique by Exercise 3. Hence,Uis uniquely
determined.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 177
Base Step:=1.Here,
[1st column ofA]=Q·[1st column ofR]=[1st column ofQ](
11)
becauseRis upper triangular. But, since the1st column ofQis a unit vector and
11is positive,
11=k[1st column ofA]kand[1st column ofQ]=
1
11
[1st column ofA]
Hence, thefirst column in each ofQandRis uniquely determined.
Inductive Step: Assume1and that columns1(−1)of bothQandRare uniquely
determined. We will show that theth column in each ofQandRis uniquely determined. Now,
[th column ofA]=Q·[th column ofR]=
X
=1
[th column ofQ]( )
Letv=
P
−1
=1
[th column ofQ]( ). By the Inductive Hypothesis,vis uniquely determined. Now,
[th column ofQ](
)=[th column ofA]−v
But since theth column ofQis a unit vector and
is positive,
=k[th column ofA]−vkand[th column ofQ]=
1
([th column ofA]−v)
Hence, theth column in each ofQandRis uniquely determined.
(4) (a) IfAis square, thenA=QR,whereQis an orthogonal matrix andRhas nonnegative entries
along its main diagonal. SettingU=R,weseethat
A
A=U
Q
QU=U
(I)U=U
U
(b) SupposeA=QR(where this is aQRfactorization ofA)andA
A=U
U.ThenRandR
must be nonsingular. Also,P=Q(R
)
−1
U
is an orthogonal matrix, because
PP
=Q(R
)
−1
U
¡
Q(R
)
−1
U
¢
=Q(R
)
−1
U
UR
−1
Q
=Q(R
)
−1
A
AR
−1
Q
=Q(R
)
−1
(R
Q
)(QR)R
−1
Q
=Q((R
)
−1
R
)(Q
Q)(RR
−1
)Q
=QI
IIQ
=I
Finally,
PU=Q(R
)
−1
U
U=(Q
)
−1
(R
)
−1
A
A
=(R
Q
)
−1
A
A=(A
)
−1
A
A=I A=A
and soPUis aQRfactorization ofA, which is unique by Exercise 3. Hence,Uis uniquely
determined.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 177
Answers to Exercises Section 9.5
(5)(a)T(b)T(c)T(d)F (e)T
Section 9.5
(1) For each part, one possibility is given.
(a)U=
1
√
2
∙
11
1−1
¸
,Σ=
∙
2
√
10 0
0
√
10
¸
,V=
1
√
5
∙
2−1
12
¸
(b)U=
1
5
∙
−3−4
−43
¸
,Σ=
∙
15
√
20
010
√
2
¸
,V=
1
√
2
∙
−11
11
¸
(c)U=
1
√
10
∙
−31
13
¸
,Σ=
∙
9
√
10 0 0
03
√
10 0
¸
,V=
1
3
⎡
⎣
−1−22
−221
212
⎤
⎦
(d)U=
1
√
5
∙
−1−2
−21
¸
,Σ=
∙
5
√
500
05
√
50
¸
,V=
1
5
⎡
⎣
−304
403
050
⎤
⎦
(e)U=
⎡
⎢
⎢
⎢
⎣
2
√
13
18
7
√
13
−
3
7
3
√
13
−12
7
√
13
2
7
0
13
7
√
13
6
7
⎤
⎥
⎥
⎥
⎦
,Σ=
⎡
⎣
100
010
000
⎤
⎦,V=U
(Note:Arepresents the orthogonal projection onto the plane−3+2+6=0.)
(f)U=
1
7
⎡
⎣
623
3−6−2
23 −6
⎤
⎦,Σ=
⎡
⎣
2
√
20
0
√
2
00
⎤
⎦,V=
1
√
2
∙
1−1
11
¸
(g)U=
1
11
⎡
⎣
26 9
−96−2
67−6
⎤
⎦,Σ=
⎡
⎣
30
02
00
⎤
⎦,V=
∙
01
10
¸
(h)U=
⎡
⎢
⎢
⎣
−
2
5
√
2−
8
15
√
2
1
3
1
2
√
2−
1
6
√
2
2
3
3
10
√
2−
13
30
√
2−
2
3
⎤
⎥
⎥
⎦
,Σ=
⎡
⎣
2000
0200
0000
⎤
⎦,V=
1
√
2
⎡
⎢
⎢
⎣
0−101
−1 010
1 010
0 101
⎤
⎥
⎥
⎦
(2) (a)A
+
=
1
2250
∙
104 70 122
−158 110 31
¸
,v=
1
2250
∙
5618
3364
¸
,A
Av=A
b=
1
15
∙
6823
3874
¸
(b)A
+
=
1
450
∙
843−11
−624−48
¸
,v=
1
90
∙
173
264
¸
,A
Av=A
b=
∙
65
170
¸
(c)A
+
=
1
84
⎡
⎣
36 24 12 0
12 36 −24 0
−31−23 41 49
⎤
⎦,v=
1
14
⎡
⎣
44
−18
71
⎤
⎦,A
Av=A
b=
1
7
⎡
⎣
127
−30
60
⎤
⎦
(d)A
+
=
1
54
⎡
⎣
20 4 4
86 4 1
46−4−7
⎤
⎦,v=
1
54
⎡
⎣
26
59
7
⎤
⎦,A
Av=A
b=
⎡
⎣
13
24
−2
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 178
(5)(a)T(b)T(c)T(d)F (e)T
Section 9.5
(1) For each part, one possibility is given.
(a)U=
1
√
2
∙
11
1−1
¸
,Σ=
∙
2
√
10 0
0
√
10
¸
,V=
1
√
5
∙
2−1
12
¸
(b)U=
1
5
∙
−3−4
−43
¸
,Σ=
∙
15
√
20
010
√
2
¸
,V=
1
√
2
∙
−11
11
¸
(c)U=
1
√
10
∙
−31
13
¸
,Σ=
∙
9
√
10 0 0
03
√
10 0
¸
,V=
1
3
⎡
⎣
−1−22
−221
212
⎤
⎦
(d)U=
1
√
5
∙
−1−2
−21
¸
,Σ=
∙
5
√
500
05
√
50
¸
,V=
1
5
⎡
⎣
−304
403
050
⎤
⎦
(e)U=
⎡
⎢
⎢
⎢
⎣
2
√
13
18
7
√
13
−
3
7
3
√
13
−12
7
√
13
2
7
0
13
7
√
13
6
7
⎤
⎥
⎥
⎥
⎦
,Σ=
⎡
⎣
100
010
000
⎤
⎦,V=U
(Note:Arepresents the orthogonal projection onto the plane−3+2+6=0.)
(f)U=
1
7
⎡
⎣
623
3−6−2
23 −6
⎤
⎦,Σ=
⎡
⎣
2
√
20
0
√
2
00
⎤
⎦,V=
1
√
2
∙
1−1
11
¸
(g)U=
1
11
⎡
⎣
26 9
−96−2
67−6
⎤
⎦,Σ=
⎡
⎣
30
02
00
⎤
⎦,V=
∙
01
10
¸
(h)U=
⎡
⎢
⎢
⎣
−
2
5
√
2−
8
15
√
2
1
3
1
2
√
2−
1
6
√
2
2
3
3
10
√
2−
13
30
√
2−
2
3
⎤
⎥
⎥
⎦
,Σ=
⎡
⎣
2000
0200
0000
⎤
⎦,V=
1
√
2
⎡
⎢
⎢
⎣
0−101
−1 010
1 010
0 101
⎤
⎥
⎥
⎦
(2) (a)A
+
=
1
2250
∙
104 70 122
−158 110 31
¸
,v=
1
2250
∙
5618
3364
¸
,A
Av=A
b=
1
15
∙
6823
3874
¸
(b)A
+
=
1
450
∙
843−11
−624−48
¸
,v=
1
90
∙
173
264
¸
,A
Av=A
b=
∙
65
170
¸
(c)A
+
=
1
84
⎡
⎣
36 24 12 0
12 36 −24 0
−31−23 41 49
⎤
⎦,v=
1
14
⎡
⎣
44
−18
71
⎤
⎦,A
Av=A
b=
1
7
⎡
⎣
127
−30
60
⎤
⎦
(d)A
+
=
1
54
⎡
⎣
20 4 4
86 4 1
46−4−7
⎤
⎦,v=
1
54
⎡
⎣
26
59
7
⎤
⎦,A
Av=A
b=
⎡
⎣
13
24
−2
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 178
Answers to Exercises Section 9.5
(3) (a)A=2
√
10
µ
1
√
2
∙
1
1
¸¶
³
1√
5
£
21
¤
´
+
√
10
µ
1
√
2
∙
1
−1
¸¶
³
1√
5
£
−12
¤
´
(b)A=9
√
10
µ
1
√
10
∙
−3
1
¸¶
¡
13
£
−1−22
¤¢
+3
√
10
µ
1
√
10
∙
1
3
¸¶
¡
13
£
−221
¤¢
(c)A=2
√
2
⎛
⎝
1
7
⎡ ⎣
6
3
2
⎤
⎦
⎞
⎠
³
1
√
2
£
11
¤
´
+
√
2
⎛
⎝
1
7
⎡ ⎣
2
−6
3
⎤
⎦
⎞
⎠
³
1
√
2
£
−11
¤
´
(d)A=3
⎛
⎝
1
11
⎡ ⎣
2
−9
6
⎤
⎦
⎞
⎠
£
01
¤
+2
⎛
⎝
1
11
⎡ ⎣
6
6
7
⎤
⎦
⎞
⎠
£
10
¤
(e)A=2
⎡
⎢
⎣
−
2
5
√
2
1
2
√
2
3
10
√
2
⎤
⎥
⎦
³
1
√
2
£
0−110
¤
´
+2
⎡
⎢
⎣
−
8
15
√
2
−
1
6
√
2
−
13
30
√
2
⎤
⎥
⎦
³
1
√
2
£
−1001
¤
´
(4) IfAis orthogonal, thenA
A=I . Therefore,=1is the only eigenvalue forA
A,andthe
eigenspace is all ofR
. Therefore,{v 1v }can be any orthonormal basis forR
.Ifwetake
{v
1v }to be the standard basis forR
, then the corresponding Singular Value Decomposition for
AisAI
I. If, instead, we use therowsofA, which form an orthonormal basis forR
,torepresent
{v
1v }, then the corresponding Singular Value Decomposition forAisI IA.
(5) IfA=UΣV
,thenA
A=VΣ
U
UΣV
=VΣ
ΣV
.For≤,let be the( )entry ofΣ.
Thus, for≤,A
Av=VΣ
ΣV
v=VΣ
Σe=VΣ
(e)=V(
2
e)=
2
v. (Note that in
this proof, “e
” has been used to represent both a vector inR
and a vector inR
.) If,then
A
Av=VΣ
ΣV
v=VΣ
Σe=VΣ
(0)=0. This proves the claims made in the exercise.
(6) BecauseAis symmetric, it is orthogonally diagonalizable. LetD=P
APbe an orthogonal diago-
nalization forA, with the eigenvalues ofAalong the main diagonal ofD.Thus,A=PDP
.By
Exercise 5, the columns ofPform an orthonormal basis of eigenvectors forA
A, and the correspond-
ing eigenvalues are the squares of the diagonal entries inD. Hence, the singular values ofAare the
square roots of these eigenvalues, which are the absolute values of the diagonal entries ofD.Sincethe
diagonal entries ofDare the eigenvalues ofA, this completes the proof.
(7) Expressv∈R
asv= 1v1+···+ v,where{v 1v }are right singular vectors forA.Since
{v
1v }is an orthonormal basis forR
,kvk=
p
2
1
+···+
2
.Thus,
kAvk
2
=(Av)·(Av)
=(
1Av1+···+ Av)·( 1Av1+···+ Av)
=
2
1
2
1
+···+
2
2
(by parts (2) and (3) of Lemma 9.4)
≤
2
1
2
1
+
2
2
2
1
+···+
2
2
1
=
2
1
¡
2
1
+
2
2
+···+
2
¢
=
2
1
kvk
2
Therefore,kAvk≤
1kvk.
(8) In all parts, assumerepresents the number of nonzero singular values.
(a) Theth column ofVis the right singular vectorv
, which is a unit eigenvector corresponding to
the eigenvalue
ofA
A.But−v is also a unit eigenvector corresponding to the eigenvalue
Copyrightc°2016 Elsevier Ltd. All rights reserved. 179
(3) (a)A=2
√
10
µ
1
√
2
∙
1
1
¸¶
³
1√
5
£
21
¤
´
+
√
10
µ
1
√
2
∙
1
−1
¸¶
³
1√
5
£
−12
¤
´
(b)A=9
√
10
µ
1
√
10
∙
−3
1
¸¶
¡
13
£
−1−22
¤¢
+3
√
10
µ
1
√
10
∙
1
3
¸¶
¡
13
£
−221
¤¢
(c)A=2
√
2
⎛
⎝
1
7
⎡ ⎣
6
3
2
⎤
⎦
⎞
⎠
³
1
√
2
£
11
¤
´
+
√
2
⎛
⎝
1
7
⎡ ⎣
2
−6
3
⎤
⎦
⎞
⎠
³
1
√
2
£
−11
¤
´
(d)A=3
⎛
⎝
1
11
⎡ ⎣
2
−9
6
⎤
⎦
⎞
⎠
£
01
¤
+2
⎛
⎝
1
11
⎡ ⎣
6
6
7
⎤
⎦
⎞
⎠
£
10
¤
(e)A=2
⎡
⎢
⎣
−
2
5
√
2
1
2
√
2
3
10
√
2
⎤
⎥
⎦
³
1
√
2
£
0−110
¤
´
+2
⎡
⎢
⎣
−
8
15
√
2
−
1
6
√
2
−
13
30
√
2
⎤
⎥
⎦
³
1
√
2
£
−1001
¤
´
(4) IfAis orthogonal, thenA
A=I . Therefore,=1is the only eigenvalue forA
A,andthe
eigenspace is all ofR
. Therefore,{v 1v }can be any orthonormal basis forR
.Ifwetake
{v
1v }to be the standard basis forR
, then the corresponding Singular Value Decomposition for
AisAI
I. If, instead, we use therowsofA, which form an orthonormal basis forR
,torepresent
{v
1v }, then the corresponding Singular Value Decomposition forAisI IA.
(5) IfA=UΣV
,thenA
A=VΣ
U
UΣV
=VΣ
ΣV
.For≤,let be the( )entry ofΣ.
Thus, for≤,A
Av=VΣ
ΣV
v=VΣ
Σe=VΣ
(e)=V(
2
e)=
2
v. (Note that in
this proof, “e
” has been used to represent both a vector inR
and a vector inR
.) If,then
A
Av=VΣ
ΣV
v=VΣ
Σe=VΣ
(0)=0. This proves the claims made in the exercise.
(6) BecauseAis symmetric, it is orthogonally diagonalizable. LetD=P
APbe an orthogonal diago-
nalization forA, with the eigenvalues ofAalong the main diagonal ofD.Thus,A=PDP
.By
Exercise 5, the columns ofPform an orthonormal basis of eigenvectors forA
A, and the correspond-
ing eigenvalues are the squares of the diagonal entries inD. Hence, the singular values ofAare the
square roots of these eigenvalues, which are the absolute values of the diagonal entries ofD.Sincethe
diagonal entries ofDare the eigenvalues ofA, this completes the proof.
(7) Expressv∈R
asv= 1v1+···+ v,where{v 1v }are right singular vectors forA.Since
{v
1v }is an orthonormal basis forR
,kvk=
p
2
1
+···+
2
.Thus,
kAvk
2
=(Av)·(Av)
=(
1Av1+···+ Av)·( 1Av1+···+ Av)
=
2
1
2
1
+···+
2
2
(by parts (2) and (3) of Lemma 9.4)
≤
2
1
2
1
+
2
2
2
1
+···+
2
2
1
=
2
1
¡
2
1
+
2
2
+···+
2
¢
=
2
1
kvk
2
Therefore,kAvk≤
1kvk.
(8) In all parts, assumerepresents the number of nonzero singular values.
(a) Theth column ofVis the right singular vectorv
, which is a unit eigenvector corresponding to
the eigenvalue
ofA
A.But−v is also a unit eigenvector corresponding to the eigenvalue
Copyrightc°2016 Elsevier Ltd. All rights reserved. 179
Answers to Exercises Section 9.5
ofA
A.Thus,ifanyvectorv is replaced with its opposite vector, the set{v 1v }is still
an orthonormal basis forR
consisting of eigenvectors forA
A. Since the vectors are kept in the
same order, the
do not increase, and thus{v 1v }fulfills all the necessary conditions to be
a set of right singular vectors forA.For≤, the left singular vectoru
=
1
Av,sowhenwe
change the sign ofv
, we must adjustUby changing the sign ofu as well. For,changing
the sign ofv
has no effect onU, but still produces a valid Singular Value Decomposition.
(b) If the eigenspace
forA
Ahas dimension higher than1, then the corresponding right singular
vectors can be replaced by any orthonormal basis for
, for which there is an infinite number of
choices. Then the associated left singular vectorsu
1u must be adjusted accordingly.
(c) Let{v
1v }be a set of right singular vectors forA.ByExercise5,theth diagonal entry of
ΣmustbethesquarerootofaneigenvalueofA
Acorresponding to the eigenvectorv .Since
{v
1v }is an orthonormal basis of eigenvectors, it must correspond to a complete set of
eigenvalues forA
A. Also by Exercise 5, all of the vectorsv ,for, correspond to the
eigenvalue0. Hence, all of the square roots of the nonzero eigenvalues ofA
Amust lie on the
diagonal ofΣ. Thus, the values that appear on the diagonal ofΣare uniquely determined. Finally,
the order in which these numbers appear is determined by the requirement for the Singular Value
Decomposition that they appear in non-increasing order.
(d) By definition, for1≤≤,u
=
1
Av, and so these left singular vectors are completely
determined by the choices made forv
1v ,thefirstcolumns ofV.
(e) Columns+1throughofUare the left singular vectorsu
+1u ,whichcanbeany
orthonormal basis for the orthogonal complement of the column space ofA(by parts (2) and (3)
of Theorem 9.5). If=+1, this orthogonal complement to the column space is one-dimensional,
and so there are only two choices foru
+1, which are opposites of each other, becauseu +1must
be a unit vector. If+1, the orthogonal complement to the column space has dimension
greater than1,andthereisaninfinite number of choices for its orthonormal basis.
(9) (a) Each right singular vectorv
,for1≤≤, must be an eigenvector forA
A.Performingthe
Gram-Schmidt Process on the rows ofA, eliminating zero vectors, and normalizing will produce
an orthonormal basis for the row space ofA, but there is no guarantee that it will consist of
eigenvectors forA
A. For example, ifA=
∙
11
01
¸
, performing the Gram-Schmidt Process on
the rows ofAproduces the two vectors[11]and
£
−
1
2
1
2
¤
, neither of which is an eigenvector for
A
A=
∙
11
12
¸
.
(b) The right singular vectorsv
+1v form an orthonormal basis for the eigenspace 0ofA
A.
Any orthonormal basis for
0will do. By part (5) of Theorem 9.5, 0equals the kernel of the
linear transformationwhose matrix with respect to the standard bases isA.Abasisforker()
can be found by using the Kernel Method. That basis can be turned into an orthonormal basis
forker()by applying the Gram-Schmidt Process and normalizing.
(10) IfA=UΣV
, as given in the exercise, thenA
+
=VΣ
+
U
,andso
A
+
A=VΣ
+
U
UΣV
=VΣ
+
ΣV
Note thatΣ
+
Σis an×diagonal matrix whosefirstdiagonal entries equal1, with the remaining
diagonal entries equal to0. Note also that since the columnsv
1v ofVare orthonormal,
V
v=e,for1≤≤.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 180
ofA
A.Thus,ifanyvectorv is replaced with its opposite vector, the set{v 1v }is still
an orthonormal basis forR
consisting of eigenvectors forA
A. Since the vectors are kept in the
same order, the
do not increase, and thus{v 1v }fulfills all the necessary conditions to be
a set of right singular vectors forA.For≤, the left singular vectoru
=
1
Av,sowhenwe
change the sign ofv
, we must adjustUby changing the sign ofu as well. For,changing
the sign ofv
has no effect onU, but still produces a valid Singular Value Decomposition.
(b) If the eigenspace
forA
Ahas dimension higher than1, then the corresponding right singular
vectors can be replaced by any orthonormal basis for
, for which there is an infinite number of
choices. Then the associated left singular vectorsu
1u must be adjusted accordingly.
(c) Let{v
1v }be a set of right singular vectors forA.ByExercise5,theth diagonal entry of
ΣmustbethesquarerootofaneigenvalueofA
Acorresponding to the eigenvectorv .Since
{v
1v }is an orthonormal basis of eigenvectors, it must correspond to a complete set of
eigenvalues forA
A. Also by Exercise 5, all of the vectorsv ,for, correspond to the
eigenvalue0. Hence, all of the square roots of the nonzero eigenvalues ofA
Amust lie on the
diagonal ofΣ. Thus, the values that appear on the diagonal ofΣare uniquely determined. Finally,
the order in which these numbers appear is determined by the requirement for the Singular Value
Decomposition that they appear in non-increasing order.
(d) By definition, for1≤≤,u
=
1
Av, and so these left singular vectors are completely
determined by the choices made forv
1v ,thefirstcolumns ofV.
(e) Columns+1throughofUare the left singular vectorsu
+1u ,whichcanbeany
orthonormal basis for the orthogonal complement of the column space ofA(by parts (2) and (3)
of Theorem 9.5). If=+1, this orthogonal complement to the column space is one-dimensional,
and so there are only two choices foru
+1, which are opposites of each other, becauseu +1must
be a unit vector. If+1, the orthogonal complement to the column space has dimension
greater than1,andthereisaninfinite number of choices for its orthonormal basis.
(9) (a) Each right singular vectorv
,for1≤≤, must be an eigenvector forA
A.Performingthe
Gram-Schmidt Process on the rows ofA, eliminating zero vectors, and normalizing will produce
an orthonormal basis for the row space ofA, but there is no guarantee that it will consist of
eigenvectors forA
A. For example, ifA=
∙
11
01
¸
, performing the Gram-Schmidt Process on
the rows ofAproduces the two vectors[11]and
£
−
1
2
1
2
¤
, neither of which is an eigenvector for
A
A=
∙
11
12
¸
.
(b) The right singular vectorsv
+1v form an orthonormal basis for the eigenspace 0ofA
A.
Any orthonormal basis for
0will do. By part (5) of Theorem 9.5, 0equals the kernel of the
linear transformationwhose matrix with respect to the standard bases isA.Abasisforker()
can be found by using the Kernel Method. That basis can be turned into an orthonormal basis
forker()by applying the Gram-Schmidt Process and normalizing.
(10) IfA=UΣV
, as given in the exercise, thenA
+
=VΣ
+
U
,andso
A
+
A=VΣ
+
U
UΣV
=VΣ
+
ΣV
Note thatΣ
+
Σis an×diagonal matrix whosefirstdiagonal entries equal1, with the remaining
diagonal entries equal to0. Note also that since the columnsv
1v ofVare orthonormal,
V
v=e,for1≤≤.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 180
Answers to Exercises Section 9.5
(a) If1≤≤,thenA
+
Av=VΣ
+
ΣV
v=VΣ
+
Σe=Ve =v.
(b) If,thenA
+
Av=VΣ
+
ΣV
v=VΣ
+
Σe=V(0)=0.
(c) By parts (4) and (5) of Theorem 9.5,{v
1v }is an orthonormal basis for(ker())
⊥
,and
{v
+1v }is an orthonormal basis forker(). The orthogonal projection onto(ker())
⊥
sends every vector in(ker())
⊥
to itself, and every vector orthogonal to(ker())
⊥
(that is, every
vector inker())to0. Parts (a) and (b) prove that this is precisely the result obtained by
multiplyingA
+
Aby the bases for(ker())
⊥
andker().
(d) Using part (a), if1≤≤,then
AA
+
Av=A
¡
A
+
Av
¢
=Av
By part (b), if,
AA
+
Av=A
¡
A
+
Av
¢
=A(0)=0
But, if,v
∈ker(),andsoAv =0as well. Hence,AA
+
AandArepresent linear
transformations fromR
toR
that agree on a basis forR
. Therefore, they are equal as
matrices.
(e) By part (d),AA
+
A=A.IfAis nonsingular, we can multiply both sides byA
−1
(on the left)
to obtainA
+
A=I . Multiplying byA
−1
again (on the right) yieldsA
+
=A
−1
.
(11) IfA=UΣV
, as given in the exercise, thenA
+
=VΣ
+
U
.Notethatsincethecolumnsu 1u
ofUare orthonormal,U
u=e,for1≤≤.
(a) If1≤≤,then
A
+
u=VΣ
+
U
u=VΣ
+
e=V
µ
1
e
¶
=
1
v
Thus,
AA
+
u=A
¡
A
+
u
¢
=A
µ
1
v
¶
=
1
Av=u
(b) If,then
A
+
u=VΣ
+
U
u=VΣ
+
e=V(0)=0
Thus,
AA
+
u=A
¡
A
+
u
¢
=A(0)=0
(c) By Theorem 9.5,{u
1u }is an orthonormal basis forrange(),and{u +1u }is an
orthonormal basis for(range())
⊥
. The orthogonal projection ontorange()sends every vector
inrange()to itself, and every vector in(range())
⊥
to0. Parts (a) and (b) prove that this is
precisely the action of multiplying byAA
+
by showing how it acts on bases forrange()and
(range())
⊥
.
(d) Using part (a), if1≤≤,then
A
+
AA
+
u=A
+
¡
AA
+
u
¢
=A
+
u
By part (b), if,
A
+
AA
+
u=A
+
¡
AA
+
u
¢
=A
+
(0)=0
But, if,A
+
u=0as well. Hence,A
+
AA
+
andA
+
represent linear transformations from
R
toR
that agree on a basis forR
. Therefore, they are equal as matrices.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 181
(a) If1≤≤,thenA
+
Av=VΣ
+
ΣV
v=VΣ
+
Σe=Ve =v.
(b) If,thenA
+
Av=VΣ
+
ΣV
v=VΣ
+
Σe=V(0)=0.
(c) By parts (4) and (5) of Theorem 9.5,{v
1v }is an orthonormal basis for(ker())
⊥
,and
{v
+1v }is an orthonormal basis forker(). The orthogonal projection onto(ker())
⊥
sends every vector in(ker())
⊥
to itself, and every vector orthogonal to(ker())
⊥
(that is, every
vector inker())to0. Parts (a) and (b) prove that this is precisely the result obtained by
multiplyingA
+
Aby the bases for(ker())
⊥
andker().
(d) Using part (a), if1≤≤,then
AA
+
Av=A
¡
A
+
Av
¢
=Av
By part (b), if,
AA
+
Av=A
¡
A
+
Av
¢
=A(0)=0
But, if,v
∈ker(),andsoAv =0as well. Hence,AA
+
AandArepresent linear
transformations fromR
toR
that agree on a basis forR
. Therefore, they are equal as
matrices.
(e) By part (d),AA
+
A=A.IfAis nonsingular, we can multiply both sides byA
−1
(on the left)
to obtainA
+
A=I . Multiplying byA
−1
again (on the right) yieldsA
+
=A
−1
.
(11) IfA=UΣV
, as given in the exercise, thenA
+
=VΣ
+
U
.Notethatsincethecolumnsu 1u
ofUare orthonormal,U
u=e,for1≤≤.
(a) If1≤≤,then
A
+
u=VΣ
+
U
u=VΣ
+
e=V
µ
1
e
¶
=
1
v
Thus,
AA
+
u=A
¡
A
+
u
¢
=A
µ
1
v
¶
=
1
Av=u
(b) If,then
A
+
u=VΣ
+
U
u=VΣ
+
e=V(0)=0
Thus,
AA
+
u=A
¡
A
+
u
¢
=A(0)=0
(c) By Theorem 9.5,{u
1u }is an orthonormal basis forrange(),and{u +1u }is an
orthonormal basis for(range())
⊥
. The orthogonal projection ontorange()sends every vector
inrange()to itself, and every vector in(range())
⊥
to0. Parts (a) and (b) prove that this is
precisely the action of multiplying byAA
+
by showing how it acts on bases forrange()and
(range())
⊥
.
(d) Using part (a), if1≤≤,then
A
+
AA
+
u=A
+
¡
AA
+
u
¢
=A
+
u
By part (b), if,
A
+
AA
+
u=A
+
¡
AA
+
u
¢
=A
+
(0)=0
But, if,A
+
u=0as well. Hence,A
+
AA
+
andA
+
represent linear transformations from
R
toR
that agree on a basis forR
. Therefore, they are equal as matrices.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 181
Answers to Exercises Section 9.5
(e) Let 1:(ker())
⊥
→range()be given by 1(v)=Av. Parts (2) and (3) of Theorem 9.5 shows
thatdim((ker())
⊥
)== dim(range()). Part (a) of this exercise and part (2) of Theorem 9.5
shows that
:range()→range()given by(u)=
1(A
+
u)
is the identity linear transformation. Hence,
1must be onto, and so by Corollary 5.13, 1is an
isomorphism. Becauseis the identity transformation, the inverse of
1hasA
+
as its matrix
with respect to the standard basis. This shows thatA
+
uis uniquely determined foru∈range().
Thus, since the subspacerange()depends only onA, not on the Singular Value Decomposition
ofA,A
+
uis uniquely determined on the subspacerange(), independently from which Singular
Value Decomposition is used forAto computeA
+
. Similarly, part (3) of Theorem 9.5 and part
(b) of this exercise shows thatA
+
u=0on a basis for(range())
⊥
,andsoA
+
u=0forall
vectors in(range())
⊥
– again, independent of which Singular Value Decomposition ofAis
used. Thus,A
+
uis uniquely determined foru∈range()andu∈(range())
⊥
, and thus, on a
basis forR
which can be chosen from these two subspaces. Thus,A
+
is uniquely determined.
(12) By part (a) of Exercise 28 in Section 1.5,trace
³
AA
´
equals the sum of the squares of the entries of
A.IfA=UΣV
is a Singular Value Decomposition ofA,then
AA
=UΣV
VΣ
U
=UΣΣ
U
Using part (c) of Exercise 28 in Section 1.5, we see that
trace
³
AA
´
=trace
³
UΣΣ
U
´
=trace
³
UU
ΣΣ
´
=trace
³
ΣΣ
´
=
2
1
+···+
2
(13) LetV
1be the×matrix whose columns are
v
v v1v −1v+1v
in that order. LetU
1be the×matrix whose columns are
u
u u1u −1u+1u
in that order. (The notation used here assumes that1,and, but the construction
ofV
1andU 1can be adjusted in an obvious way if=1,=,or=.) LetΣ 1be the diagonal
×matrix with
in thefirst−+1diagonal entries and zero on the remaining diagonal
entries. We claim thatU
1Σ1V
1
is a Singular Value Decomposition forA . To show this, because
U
1,Σ1,andV 1are of the proper form, it is enough to show thatA =U 1Σ1V
1
.
If≤≤,then
A
v=
¡
uv
+···+ uv
¢
v
=u
and
¡
U
1Σ1V
1
¢
v
=U 1Σ1e−+1=U 1e−+1=u
Ifor,then
A
v=
¡
uv
+···+ uv
¢
v
=0
If,
¡
U
1Σ1V
1
¢
v
=U 1Σ1e+−+1 =U 10=0
and if,
¡
U
1Σ1V
1
¢
v
=U 1Σ1e=U 10=0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 182
(e) Let 1:(ker())
⊥
→range()be given by 1(v)=Av. Parts (2) and (3) of Theorem 9.5 shows
thatdim((ker())
⊥
)== dim(range()). Part (a) of this exercise and part (2) of Theorem 9.5
shows that
:range()→range()given by(u)=
1(A
+
u)
is the identity linear transformation. Hence,
1must be onto, and so by Corollary 5.13, 1is an
isomorphism. Becauseis the identity transformation, the inverse of
1hasA
+
as its matrix
with respect to the standard basis. This shows thatA
+
uis uniquely determined foru∈range().
Thus, since the subspacerange()depends only onA, not on the Singular Value Decomposition
ofA,A
+
uis uniquely determined on the subspacerange(), independently from which Singular
Value Decomposition is used forAto computeA
+
. Similarly, part (3) of Theorem 9.5 and part
(b) of this exercise shows thatA
+
u=0on a basis for(range())
⊥
,andsoA
+
u=0forall
vectors in(range())
⊥
– again, independent of which Singular Value Decomposition ofAis
used. Thus,A
+
uis uniquely determined foru∈range()andu∈(range())
⊥
, and thus, on a
basis forR
which can be chosen from these two subspaces. Thus,A
+
is uniquely determined.
(12) By part (a) of Exercise 28 in Section 1.5,trace
³
AA
´
equals the sum of the squares of the entries of
A.IfA=UΣV
is a Singular Value Decomposition ofA,then
AA
=UΣV
VΣ
U
=UΣΣ
U
Using part (c) of Exercise 28 in Section 1.5, we see that
trace
³
AA
´
=trace
³
UΣΣ
U
´
=trace
³
UU
ΣΣ
´
=trace
³
ΣΣ
´
=
2
1
+···+
2
(13) LetV
1be the×matrix whose columns are
v
v v1v −1v+1v
in that order. LetU
1be the×matrix whose columns are
u
u u1u −1u+1u
in that order. (The notation used here assumes that1,and, but the construction
ofV
1andU 1can be adjusted in an obvious way if=1,=,or=.) LetΣ 1be the diagonal
×matrix with
in thefirst−+1diagonal entries and zero on the remaining diagonal
entries. We claim thatU
1Σ1V
1
is a Singular Value Decomposition forA . To show this, because
U
1,Σ1,andV 1are of the proper form, it is enough to show thatA =U 1Σ1V
1
.
If≤≤,then
A
v=
¡
uv
+···+ uv
¢
v
=u
and
¡
U
1Σ1V
1
¢
v
=U 1Σ1e−+1=U 1e−+1=u
Ifor,then
A
v=
¡
uv
+···+ uv
¢
v
=0
If,
¡
U
1Σ1V
1
¢
v
=U 1Σ1e+−+1 =U 10=0
and if,
¡
U
1Σ1V
1
¢
v
=U 1Σ1e=U 10=0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 182
Answers to Exercises Section 9.5
Hence,A v=
¡
U 1Σ1V
1
¢
v
for every,andsoA =U 1Σ1V
1
.
Finally, since we know a Singular Value Decomposition forA
,weknowthat are the
singular values forA
, and by part (1) of Theorem 9.5,rank(A )=−+1.
(14) (a)A=
⎡
⎢
⎢
⎢
⎢
⎣
40−515 −15 5 −30
1831 212−061 8
50 5 45 −45−5−60
−24−15090933−24
425−2560 −60 25−375
⎤
⎥
⎥
⎥
⎥
⎦
(b)A
1=
⎡
⎢
⎢
⎢
⎢
⎣
25 0 25−25 0−25
000000
50 0 50−50 0−50
000000
50 0 50−50 0−50
⎤
⎥
⎥
⎥
⎥
⎦
A 2=
⎡
⎢
⎢
⎢
⎢
⎣
35015 −15 0−35
00 0 00 0
55045 −45 0−55
00 0 00 0
40060 −60 0−40
⎤
⎥
⎥
⎥
⎥
⎦
A
3=
⎡
⎢
⎢
⎢
⎢
⎣
40−515−15 5 −30
00000 0
50 5 45 −45−5−60
00000 0
425−2560−60 25−375
⎤
⎥
⎥
⎥
⎥
⎦
A 4=
⎡
⎢
⎢
⎢
⎢
⎣
40−515−15 5 −30
18180 0 −181 8
50 5 45 −45−5−60
−24−240 0 2 4−24
425−2560−60 25−375
⎤
⎥
⎥
⎥
⎥
⎦
(c)(A)≈15385;(A−A
1)(A)≈02223;(A−A 2)(A)≈01068;(A−A 3)(A)≈
00436;(A−A
4)(A)≈00195
(d) The method described in the text for the compression of digital images takes the matrix describing
the image and alters it by zeroing out some of the lower singular values. This exercise illustrates
how the matricesA
that use only thefirstsingular values for a matrixAget closer to approx-
imatingAasincreases. The matrix for a digital image is, of course, much larger than the5×6
matrix considered in this exercise. Also, you can frequently get a very good approximation of the
image using a small enough number of singular values so that less data needs to be saved. Using
the outer product form of the Singular Value Decomposition, only the singular values and the
relevant singular vectors are needed to construct eachA
, so not allentries ofA need to be
kept in storage.
(15) These are the steps to process a digital image in MATLAB, as described in the textbook:
—Enter the command:edit
—Use the text editor to enter the following MATLAB program:
function totalmat = RevisedPic(U, S, V, k)
T=V’
totalmat = 0;
for i = 1:k
totalmat = totalmat + U(:,i)*T(i,:)*S(i,i);
end
—Save this program under the nameRevisedPic.m
—Enter the command:A=imread(’picturefilename’)
where “picturefilename” is the name of thefile containing the picture, including itsfile
extension (preferably .tif or .jpg).
—Enter the command:ndims(A)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 183
Hence,A v=
¡
U 1Σ1V
1
¢
v
for every,andsoA =U 1Σ1V
1
.
Finally, since we know a Singular Value Decomposition forA
,weknowthat are the
singular values forA
, and by part (1) of Theorem 9.5,rank(A )=−+1.
(14) (a)A=
⎡
⎢
⎢
⎢
⎢
⎣
40−515 −15 5 −30
1831 212−061 8
50 5 45 −45−5−60
−24−15090933−24
425−2560 −60 25−375
⎤
⎥
⎥
⎥
⎥
⎦
(b)A
1=
⎡
⎢
⎢
⎢
⎢
⎣
25 0 25−25 0−25
000000
50 0 50−50 0−50
000000
50 0 50−50 0−50
⎤
⎥
⎥
⎥
⎥
⎦
A 2=
⎡
⎢
⎢
⎢
⎢
⎣
35015 −15 0−35
00 0 00 0
55045 −45 0−55
00 0 00 0
40060 −60 0−40
⎤
⎥
⎥
⎥
⎥
⎦
A
3=
⎡
⎢
⎢
⎢
⎢
⎣
40−515−15 5 −30
00000 0
50 5 45 −45−5−60
00000 0
425−2560−60 25−375
⎤
⎥
⎥
⎥
⎥
⎦
A 4=
⎡
⎢
⎢
⎢
⎢
⎣
40−515−15 5 −30
18180 0 −181 8
50 5 45 −45−5−60
−24−240 0 2 4−24
425−2560−60 25−375
⎤
⎥
⎥
⎥
⎥
⎦
(c)(A)≈15385;(A−A
1)(A)≈02223;(A−A 2)(A)≈01068;(A−A 3)(A)≈
00436;(A−A
4)(A)≈00195
(d) The method described in the text for the compression of digital images takes the matrix describing
the image and alters it by zeroing out some of the lower singular values. This exercise illustrates
how the matricesA
that use only thefirstsingular values for a matrixAget closer to approx-
imatingAasincreases. The matrix for a digital image is, of course, much larger than the5×6
matrix considered in this exercise. Also, you can frequently get a very good approximation of the
image using a small enough number of singular values so that less data needs to be saved. Using
the outer product form of the Singular Value Decomposition, only the singular values and the
relevant singular vectors are needed to construct eachA
, so not allentries ofA need to be
kept in storage.
(15) These are the steps to process a digital image in MATLAB, as described in the textbook:
—Enter the command:edit
—Use the text editor to enter the following MATLAB program:
function totalmat = RevisedPic(U, S, V, k)
T=V’
totalmat = 0;
for i = 1:k
totalmat = totalmat + U(:,i)*T(i,:)*S(i,i);
end
—Save this program under the nameRevisedPic.m
—Enter the command:A=imread(’picturefilename’)
where “picturefilename” is the name of thefile containing the picture, including itsfile
extension (preferably .tif or .jpg).
—Enter the command:ndims(A)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 183
Answers to Exercises Section 9.5
—If the response is “3”, then MATLAB is treating your picture as if it is in color, even if it appears
to be black-and-white.
If this is the case, enter the command:B = A(:,:,1);
(to use only thefirst color of the three used for color pictures)
followed by the command:C = double(B);
(to convert the integer format to decimal)
Otherwise, just enter:C = double(A);
—Enter the command:[U,S,V] = svd(C);
(This computes the Singular Value Decomposition of C.)
—Enter the command:W = RevisedPic(U,S,V,100);
(Here, the number “100” represents the number of singular values you are using.
You may change this to any value you like.)
—Enter the command:R = uint8(round(W));
(This converts the decimal output to the correct integer format.)
—Enter the command:imwrite(R,’Revised100.tif’,’tif’)
(This will write the revised picture out to afile named “Revised100.tif”.
Of course, you can use anyfile name you would like, or use the .jpg extension instead of .tif.)
—You may repeat the steps:
W=RevisedPic(U,S,V,k);
R = uint8(round(W));
imwrite(R,’Revisedk.tif’,’tif’)
with different values for k to see how the picture gets more refined as more singular values
are used.
—Output can be viewed using any appropriate software you may have on your computer for viewing
suchfiles.
(16) (a) F
(b) T
(c) F
(d) F
(e) F
(f) T
(g) F
(h) T
(i) F
(j) T
(k) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 184
—If the response is “3”, then MATLAB is treating your picture as if it is in color, even if it appears
to be black-and-white.
If this is the case, enter the command:B = A(:,:,1);
(to use only thefirst color of the three used for color pictures)
followed by the command:C = double(B);
(to convert the integer format to decimal)
Otherwise, just enter:C = double(A);
—Enter the command:[U,S,V] = svd(C);
(This computes the Singular Value Decomposition of C.)
—Enter the command:W = RevisedPic(U,S,V,100);
(Here, the number “100” represents the number of singular values you are using.
You may change this to any value you like.)
—Enter the command:R = uint8(round(W));
(This converts the decimal output to the correct integer format.)
—Enter the command:imwrite(R,’Revised100.tif’,’tif’)
(This will write the revised picture out to afile named “Revised100.tif”.
Of course, you can use anyfile name you would like, or use the .jpg extension instead of .tif.)
—You may repeat the steps:
W=RevisedPic(U,S,V,k);
R = uint8(round(W));
imwrite(R,’Revisedk.tif’,’tif’)
with different values for k to see how the picture gets more refined as more singular values
are used.
—Output can be viewed using any appropriate software you may have on your computer for viewing
suchfiles.
(16) (a) F
(b) T
(c) F
(d) F
(e) F
(f) T
(g) F
(h) T
(i) F
(j) T
(k) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 184
Answers to Exercises Appendix B
Appendices
Appendix B
(1) (a) Not a function; undefined for1
(b) Function; range={∈R|≥0}; image of2=1; pre-images of2={−35}
(c) Not a function; two values are assigned to each6 =1
(d) Function; range={−4−2024}; image of2=−2; pre-images of2={67}
(e) Not a function (undefined at=
2
)
(f) Function; range=all prime numbers; image of2=2; pre-image of2={012}
(g) Not a function;2is assigned to two different values (−1and6)
(2) (a){−15−10−551015}
(b){−9−8−7−6−556789}
(c){−8−6−4−202468}
(3)(◦)(
)=
1
4
√
75
2
−30+35;(◦)()=
1
4
(5
√
3
2
+2−1)
(4)(◦)
µ∙
¸¶
=
∙
−824
28
¸∙
¸
;(◦)
µ∙
¸¶
=
∙
−12 8
−412
¸∙
¸
(5) (a) Let:→be given by(1) = 4,(2) = 5,(3) = 6and:→be given by(4) =
(7) = 8,(5) = 9,(6) = 10.Then◦is onto, but
−1
({7})is empty, sois not onto.
(b) Use the example from part (a). Note that◦is one-to-one, but(4) =(7),sois not
one-to-one.
(6) The functionis onto because, given any real number,the×diagonal matrixAwith
11=
and
=1for2≤≤has determinant.Also,is not one-to-one because for every6 =0,any
other matrix obtained by performing a Type (II) operation onA(for example,
1←2+1) also has determinant.
(7) The functionis not onto because only symmetric3×3matrices are in the range of((A)=
A+A
=(A
+A)
=((A))
). Also,is not one-to-one because ifAis any nonsymmetric3×3
matrix, then(A)=A+A
=A
+(A
)
=(A
), butA6 =A
.
(8)is not one-to-one, because(+1)=(+3)=1;is not onto, because there is no pre-image for
.For≥3the pre-image ofP 2isP3.
(9) If(
1)=( 2),then3
3
1
−5=3
3
2
−5=⇒3
3
1
=3
3
2
=⇒
3
1
=
3
2
=⇒ 1=2.Sois one-to-one.
Also, if∈R,then
³
¡
+5
3
¢1
3
´
=,sois onto. Inverse of=
−1
()=
¡
+5
3
¢1
3
.
(10) The functionis one-to-one, because
(A
1)=(A 2)
=⇒B
−1
A1B=B
−1
A2B
=⇒B(B
−1
A1B)B
−1
=B(B
−1
A2B)B
−1
=⇒(BB
−1
)A1(BB
−1
)=(BB
−1
)A2(BB
−1
)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 185
Appendices
Appendix B
(1) (a) Not a function; undefined for1
(b) Function; range={∈R|≥0}; image of2=1; pre-images of2={−35}
(c) Not a function; two values are assigned to each6 =1
(d) Function; range={−4−2024}; image of2=−2; pre-images of2={67}
(e) Not a function (undefined at=
2
)
(f) Function; range=all prime numbers; image of2=2; pre-image of2={012}
(g) Not a function;2is assigned to two different values (−1and6)
(2) (a){−15−10−551015}
(b){−9−8−7−6−556789}
(c){−8−6−4−202468}
(3)(◦)(
)=
1
4
√
75
2
−30+35;(◦)()=
1
4
(5
√
3
2
+2−1)
(4)(◦)
µ∙
¸¶
=
∙
−824
28
¸∙
¸
;(◦)
µ∙
¸¶
=
∙
−12 8
−412
¸∙
¸
(5) (a) Let:→be given by(1) = 4,(2) = 5,(3) = 6and:→be given by(4) =
(7) = 8,(5) = 9,(6) = 10.Then◦is onto, but
−1
({7})is empty, sois not onto.
(b) Use the example from part (a). Note that◦is one-to-one, but(4) =(7),sois not
one-to-one.
(6) The functionis onto because, given any real number,the×diagonal matrixAwith
11=
and
=1for2≤≤has determinant.Also,is not one-to-one because for every6 =0,any
other matrix obtained by performing a Type (II) operation onA(for example,
1←2+1) also has determinant.
(7) The functionis not onto because only symmetric3×3matrices are in the range of((A)=
A+A
=(A
+A)
=((A))
). Also,is not one-to-one because ifAis any nonsymmetric3×3
matrix, then(A)=A+A
=A
+(A
)
=(A
), butA6 =A
.
(8)is not one-to-one, because(+1)=(+3)=1;is not onto, because there is no pre-image for
.For≥3the pre-image ofP 2isP3.
(9) If(
1)=( 2),then3
3
1
−5=3
3
2
−5=⇒3
3
1
=3
3
2
=⇒
3
1
=
3
2
=⇒ 1=2.Sois one-to-one.
Also, if∈R,then
³
¡
+5
3
¢1
3
´
=,sois onto. Inverse of=
−1
()=
¡
+5
3
¢1
3
.
(10) The functionis one-to-one, because
(A
1)=(A 2)
=⇒B
−1
A1B=B
−1
A2B
=⇒B(B
−1
A1B)B
−1
=B(B
−1
A2B)B
−1
=⇒(BB
−1
)A1(BB
−1
)=(BB
−1
)A2(BB
−1
)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 185
Answers to Exercises Appendix C
=⇒I A1I=IA2I
=⇒A 1=A 2.
The functionis onto, because for anyC∈M
,(BCB
−1
)=B
−1
(BCB
−1
)B=C.Also,
−1
(A)
=BAB
−1
.
(11) (a) Let∈.Since◦is onto, there is some∈such that(◦)()=.Then(()) =.
Let()=.Then()=,sois onto. (Exercise 5(a) shows thatis not necessarily onto.)
(b) Let
12∈such that( 1)=( 2).Then(( 1)) =(( 2)), implying(◦)( 1)=
(◦)(
2). But since◦is one-to-one, 1=2. Hence,is one-to-one. (Exercise 5(b) shows
thatis not necessarily one-to-one.)
(12) (a) F (b) T (c) F (d) F (e) F (f) F (g) F (h) F
Appendix C
(1) (a)11−
(b)24−32
(c)20−12
(d)18−9
(e)9+19
(f)2+42
(g)−17−19
(h)5−4
(i)9+2
(j)−6
(k)16 + 22
(l)
√
73
(m)
√
53
(n)5
(2) (a)
3
20
+
1
20
(b)
3
25
−
4
25
(c)−
4
17
−
1
17
(d)−
5
34
+
3
34
(3) In all parts, let
1=1+1and 2=2+2.
(a) Part (1):
1+2=(1+1)+( 2+2)
=(1+2)+( 1+2)
=(
1+2)−( 1+2)
=(
1−1)+( 2−2)
=
1+2
Part (2):
(12)=(1+1)(2+2)
=(12−12)+( 12+21)
=(
12−12)−( 12+21)
=(
12−(− 1)(− 2)) + ( 1(−2)+ 2(−1))
=(
1−1)(2−2)
=
12
(b) If
16 =0,then
1
1
exists. Hence,
1
1
(12)=
1
1
(0), implying 2=0.
(c) Part (4):
1=
1⇐⇒ 1+1= 1−1⇐⇒ 1=− 1⇐⇒2 1=0⇐⇒ 1=0⇐⇒ 1is
real.
Part (5):
1=−1⇐⇒ 1+1=−( 1−1)⇐⇒ 1+1=− 1+1⇐⇒ 1=− 1⇐⇒
2
1=0⇐⇒ 1=0⇐⇒ 1is pure imaginary.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 186
=⇒I A1I=IA2I
=⇒A 1=A 2.
The functionis onto, because for anyC∈M
,(BCB
−1
)=B
−1
(BCB
−1
)B=C.Also,
−1
(A)
=BAB
−1
.
(11) (a) Let∈.Since◦is onto, there is some∈such that(◦)()=.Then(()) =.
Let()=.Then()=,sois onto. (Exercise 5(a) shows thatis not necessarily onto.)
(b) Let
12∈such that( 1)=( 2).Then(( 1)) =(( 2)), implying(◦)( 1)=
(◦)(
2). But since◦is one-to-one, 1=2. Hence,is one-to-one. (Exercise 5(b) shows
thatis not necessarily one-to-one.)
(12) (a) F (b) T (c) F (d) F (e) F (f) F (g) F (h) F
Appendix C
(1) (a)11−
(b)24−32
(c)20−12
(d)18−9
(e)9+19
(f)2+42
(g)−17−19
(h)5−4
(i)9+2
(j)−6
(k)16 + 22
(l)
√
73
(m)
√
53
(n)5
(2) (a)
3
20
+
1
20
(b)
3
25
−
4
25
(c)−
4
17
−
1
17
(d)−
5
34
+
3
34
(3) In all parts, let
1=1+1and 2=2+2.
(a) Part (1):
1+2=(1+1)+( 2+2)
=(1+2)+( 1+2)
=(
1+2)−( 1+2)
=(
1−1)+( 2−2)
=
1+2
Part (2):
(12)=(1+1)(2+2)
=(12−12)+( 12+21)
=(
12−12)−( 12+21)
=(
12−(− 1)(− 2)) + ( 1(−2)+ 2(−1))
=(
1−1)(2−2)
=
12
(b) If
16 =0,then
1
1
exists. Hence,
1
1
(12)=
1
1
(0), implying 2=0.
(c) Part (4):
1=
1⇐⇒ 1+1= 1−1⇐⇒ 1=− 1⇐⇒2 1=0⇐⇒ 1=0⇐⇒ 1is
real.
Part (5):
1=−1⇐⇒ 1+1=−( 1−1)⇐⇒ 1+1=− 1+1⇐⇒ 1=− 1⇐⇒
2
1=0⇐⇒ 1=0⇐⇒ 1is pure imaginary.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 186
Answers to Exercises Appendix D
(4) In all parts, let 1=1+1and 2=2+2.Alsonotethat 11=|1|
2
because
1
1=( 1+1)(1−1)=
2
1
+
2
1
.
(a)|
12|
2
=(12)
(12)(by the above)=( 12)(12)(by part (2) of Theorem C.1)
=(
1
1)(22)=| 1|
2
|2|
2
=(| 1||2|)
2
. Now take square roots.
(b) We have
¯
¯
¯
¯
1
1
¯
¯
¯
¯
=
¯
¯
¯
¯
1
|1|
2
¯
¯
¯
¯
(by the boxed equation just before Theorem C.1 in Appendix C)
=
¯
¯
¯
¯
1
2
1
+
2
1
−
1
2
1
+
2
1
¯
¯
¯
¯
=
s
µ
1
2
1
+
2
1
¶
2
+
µ
1
2
1
+
2
1
¶
2
=
s
µ
1
2
1
+
2
1
¶
2
(
2
1
+
2
1
)
=
1
p
2
1
+
2
1
=
1|1|
(c) We have
µ
1
2
¶
=
µ
1
2
22
¶
=
µ
1
2
|2|
2
¶
=
µ
(
1+1)(2−2)
2
2
+
2
2
¶
=
(12−12)
2
2
+
2
2
+
(
12−12)
2
2
+
2
2
=
(
12−12)
2
2
+
2
2
−
(
12−12)
2
2
+
2
2
=
(
12−12)
2
2
+
2
2
+
(−
12+12)
2
2
+
2
2
=
(
1−1)(2+2)
2
2
+
2
2
=
12
22
=
1
2
(5) (a) F (b) F (c) T (d) T (e) F
Appendix D
(1) (a) (III):h2i↔h3i; inverse operation is (III):h2i↔h3i.
The matrix is its own inverse.
(b) (I):h2i←−2h2i; inverse operation is (I):h2i←−
1
2
h2i.
The inverse matrix is
⎡
⎣
100
0−
1
2
0
001
⎤
⎦.
(c) (II):h3i←−4h1i+h3i; inverse operation is (II):h3i←4h1i+h3i.
The inverse matrix is
⎡
⎣
100
010
401
⎤
⎦.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 187
(4) In all parts, let 1=1+1and 2=2+2.Alsonotethat 11=|1|
2
because
1
1=( 1+1)(1−1)=
2
1
+
2
1
.
(a)|
12|
2
=(12)
(12)(by the above)=( 12)(12)(by part (2) of Theorem C.1)
=(
1
1)(22)=| 1|
2
|2|
2
=(| 1||2|)
2
. Now take square roots.
(b) We have
¯
¯
¯
¯
1
1
¯
¯
¯
¯
=
¯
¯
¯
¯
1
|1|
2
¯
¯
¯
¯
(by the boxed equation just before Theorem C.1 in Appendix C)
=
¯
¯
¯
¯
1
2
1
+
2
1
−
1
2
1
+
2
1
¯
¯
¯
¯
=
s
µ
1
2
1
+
2
1
¶
2
+
µ
1
2
1
+
2
1
¶
2
=
s
µ
1
2
1
+
2
1
¶
2
(
2
1
+
2
1
)
=
1
p
2
1
+
2
1
=
1|1|
(c) We have
µ
1
2
¶
=
µ
1
2
22
¶
=
µ
1
2
|2|
2
¶
=
µ
(
1+1)(2−2)
2
2
+
2
2
¶
=
(12−12)
2
2
+
2
2
+
(
12−12)
2
2
+
2
2
=
(
12−12)
2
2
+
2
2
−
(
12−12)
2
2
+
2
2
=
(
12−12)
2
2
+
2
2
+
(−
12+12)
2
2
+
2
2
=
(
1−1)(2+2)
2
2
+
2
2
=
12
22
=
1
2
(5) (a) F (b) F (c) T (d) T (e) F
Appendix D
(1) (a) (III):h2i↔h3i; inverse operation is (III):h2i↔h3i.
The matrix is its own inverse.
(b) (I):h2i←−2h2i; inverse operation is (I):h2i←−
1
2
h2i.
The inverse matrix is
⎡
⎣
100
0−
1
2
0
001
⎤
⎦.
(c) (II):h3i←−4h1i+h3i; inverse operation is (II):h3i←4h1i+h3i.
The inverse matrix is
⎡
⎣
100
010
401
⎤
⎦.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 187
Answers to Exercises Appendix D
(d) (I):h2i←6h2i; inverse operation is (I):h2i←
1
6
h2i.
The inverse matrix is
⎡
⎢
⎢
⎢
⎢
⎣
1000
0
1
6
00
0010
0001
⎤
⎥
⎥
⎥
⎥
⎦
.
(e) (II):h3i←−2h4i+h3i; inverse operation is (II):h3i←2h4i+h3i.
The inverse matrix is
⎡
⎢
⎢
⎣
1000
0100
0012
0001
⎤
⎥
⎥
⎦
.
(f) (III):h1i↔h4i; inverse operation is (III):h1i↔h4i.
The matrix is its own inverse.
(2) (a)
∙
49
37
¸
=
∙
40
01
¸∙
10
31
¸∙
10
0
1
4
¸∙
1
94
01
¸∙
10
01
¸
(b) Not possible, since the matrix is singular.
(c) The product of the following matrices in the order listed:
⎡
⎢
⎢
⎣
0100
1000
0010
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
−3000
0100
0010
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
1000
0100
0010
3001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
1000
0010
0100
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
1000
0600
0010
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
1000
0100
0050
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎢
⎢
⎣
10 00
01−
5
3
0
00 10
00 01
⎤
⎥
⎥
⎥
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎢
⎢
⎣
100
2
3
010 0
001 0
000 1
⎤
⎥
⎥
⎥
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎢
⎢
⎣
100 0
010 −
1
6
001 0
000 1
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(3) IfAandBare row equivalent, thenB=E
···E 2E1Afor some elementary matricesE 1E ,by
Theorem D.3. HenceB=PA,withP=E
···E 1. By Corollary D.4,Pis nonsingular.
Conversely, ifB=PAfor a nonsingular matrixP, then by Corollary D.4,P=E
···E 1for some
elementary matricesE
1E . HenceAandBare row equivalent by Theorem D.3.
(4) Follow the hint in the textbook. BecauseUis upper triangular with all nonzero diagonal entries, when
row reducingUtoI
, no Type (III) row operations are needed, and all Type (II) row operations used
will be of the formhi←hi+hi,where(the pivots are all below the targets). Thus, none of the
Type (II) row operations change a diagonal entry, since
=0when. Hence, only Type (I) row
operations make changes on the main diagonal, and no main diagonal entry will be made zero. Also,
the elementary matrices for the Type (II) row operations mentioned are upper triangular: nonzero
on the main diagonal, and perhaps in the( )entry, with. Since the elementary matrices for
Type (I) row operations are also upper triangular, we see thatU
−1
, which is the product of all these
elementary matrices, is the product of upper triangular matrices. Therefore, it is also upper triangular.
(5) For Type (I) and (III) operations,E=E
.IfEcorresponds to a Type (II) operationhi←hi+hi,
thenE
corresponds tohi←hi+hi,soE
is also an elementary matrix.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 188
(d) (I):h2i←6h2i; inverse operation is (I):h2i←
1
6
h2i.
The inverse matrix is
⎡
⎢
⎢
⎢
⎢
⎣
1000
0
1
6
00
0010
0001
⎤
⎥
⎥
⎥
⎥
⎦
.
(e) (II):h3i←−2h4i+h3i; inverse operation is (II):h3i←2h4i+h3i.
The inverse matrix is
⎡
⎢
⎢
⎣
1000
0100
0012
0001
⎤
⎥
⎥
⎦
.
(f) (III):h1i↔h4i; inverse operation is (III):h1i↔h4i.
The matrix is its own inverse.
(2) (a)
∙
49
37
¸
=
∙
40
01
¸∙
10
31
¸∙
10
0
1
4
¸∙
1
94
01
¸∙
10
01
¸
(b) Not possible, since the matrix is singular.
(c) The product of the following matrices in the order listed:
⎡
⎢
⎢
⎣
0100
1000
0010
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
−3000
0100
0010
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
1000
0100
0010
3001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
1000
0010
0100
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
1000
0600
0010
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎣
1000
0100
0050
0001
⎤
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎢
⎢
⎣
10 00
01−
5
3
0
00 10
00 01
⎤
⎥
⎥
⎥
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎢
⎢
⎣
100
2
3
010 0
001 0
000 1
⎤
⎥
⎥
⎥
⎥
⎥
⎦
,
⎡
⎢
⎢
⎢
⎢
⎢
⎣
100 0
010 −
1
6
001 0
000 1
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(3) IfAandBare row equivalent, thenB=E
···E 2E1Afor some elementary matricesE 1E ,by
Theorem D.3. HenceB=PA,withP=E
···E 1. By Corollary D.4,Pis nonsingular.
Conversely, ifB=PAfor a nonsingular matrixP, then by Corollary D.4,P=E
···E 1for some
elementary matricesE
1E . HenceAandBare row equivalent by Theorem D.3.
(4) Follow the hint in the textbook. BecauseUis upper triangular with all nonzero diagonal entries, when
row reducingUtoI
, no Type (III) row operations are needed, and all Type (II) row operations used
will be of the formhi←hi+hi,where(the pivots are all below the targets). Thus, none of the
Type (II) row operations change a diagonal entry, since
=0when. Hence, only Type (I) row
operations make changes on the main diagonal, and no main diagonal entry will be made zero. Also,
the elementary matrices for the Type (II) row operations mentioned are upper triangular: nonzero
on the main diagonal, and perhaps in the( )entry, with. Since the elementary matrices for
Type (I) row operations are also upper triangular, we see thatU
−1
, which is the product of all these
elementary matrices, is the product of upper triangular matrices. Therefore, it is also upper triangular.
(5) For Type (I) and (III) operations,E=E
.IfEcorresponds to a Type (II) operationhi←hi+hi,
thenE
corresponds tohi←hi+hi,soE
is also an elementary matrix.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 188
Answers to Exercises Appendix D
(6) Note that(AF
)
=FA
. Now, multiplying byFon the left performs a row operation onA
.Taking
the transpose again shows that this is a corresponding column operation onA.Thefirst equation show
that this is the same as multiplying byF
on the right side ofA.
(7)Ais nonsingular iffrank(A)=(by Theorem 2.15) iffAis row equivalent toI
(by the definition of
rank) iffA=E
···E 1I=E···E 1for some elementary matricesE 1E (by Theorem D.3).
(8)AX=Ohas a nontrivial solution iffrank(A)(by Theorem 2.7) iffAis singular (by Theorem
2.15) iffAcannot be expressed as a product of elementary matrices (by Corollary D.4).
(9) (a) By Theorem D.3,E
···E 2E1Ais row equivalent toA, so both have the same reduced row echelon
form, and thus have the same rank (by definition of rank).
(b) The reduced row echelon form ofAcannot have more nonzero rows thanA.
(c) IfAhasrows of zeroes, thenrank(A)=−.ButABhas at leastrows of zeroes, so
rank(AB)≤−.
(d) LetA=E
···E 1D,whereDis in reduced row echelon form. Then
rank(AB)=rank( E
···E 1DB)
=rank(DB)( by repeated use of part (a))
≤rank(D)( by part (c))
=rank(A)( by definition of rank)
(e) Exercise 18 in Section 2.3 proves the same results as those in parts (a) through (d), except that it
is phrased in terms of the underlying row operations rather than in terms of elementary matrices.
(10) (a) T (b) F (c) F (d) T (e) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 189
(6) Note that(AF
)
=FA
. Now, multiplying byFon the left performs a row operation onA
.Taking
the transpose again shows that this is a corresponding column operation onA.Thefirst equation show
that this is the same as multiplying byF
on the right side ofA.
(7)Ais nonsingular iffrank(A)=(by Theorem 2.15) iffAis row equivalent toI
(by the definition of
rank) iffA=E
···E 1I=E···E 1for some elementary matricesE 1E (by Theorem D.3).
(8)AX=Ohas a nontrivial solution iffrank(A)(by Theorem 2.7) iffAis singular (by Theorem
2.15) iffAcannot be expressed as a product of elementary matrices (by Corollary D.4).
(9) (a) By Theorem D.3,E
···E 2E1Ais row equivalent toA, so both have the same reduced row echelon
form, and thus have the same rank (by definition of rank).
(b) The reduced row echelon form ofAcannot have more nonzero rows thanA.
(c) IfAhasrows of zeroes, thenrank(A)=−.ButABhas at leastrows of zeroes, so
rank(AB)≤−.
(d) LetA=E
···E 1D,whereDis in reduced row echelon form. Then
rank(AB)=rank( E
···E 1DB)
=rank(DB)( by repeated use of part (a))
≤rank(D)( by part (c))
=rank(A)( by definition of rank)
(e) Exercise 18 in Section 2.3 proves the same results as those in parts (a) through (d), except that it
is phrased in terms of the underlying row operations rather than in terms of elementary matrices.
(10) (a) T (b) F (c) F (d) T (e) T
Copyrightc°2016 Elsevier Ltd. All rights reserved. 189
190
Chapter Tests
We include here three sample tests for each of
Chapters 1 through 7. Answer keys for all tests follow
the test collection itself. The answer keys also include
optional hints that an instructor might want to divulge for
some of the more complicated or time -consuming
problems.
Instructors who are supervising independent study
students can use these tests to measure their students’
progress at regular intervals. However, in a typical
classroom situation, we do not expect instructors to give
a test immediately after every chapter is covered, since
that would amount to six or seven tests throughout the
semester. Rather, we envision these tests as being used
as a test bank; that is, a supply of questions, or ideas for
questions, to assist instructors in composing their own
tests.
Note that most of the tests, as printed here, would
take a student much more than an hour to complete, even
with the use of software on a computer and/or calculator.
Hence, we expect instructors to choose appropriate
subsets of these tests to fulfill their classroom needs.
Chapter Tests
We include here three sample tests for each of
Chapters 1 through 7. Answer keys for all tests follow
the test collection itself. The answer keys also include
optional hints that an instructor might want to divulge for
some of the more complicated or time -consuming
problems.
Instructors who are supervising independent study
students can use these tests to measure their students’
progress at regular intervals. However, in a typical
classroom situation, we do not expect instructors to give
a test immediately after every chapter is covered, since
that would amount to six or seven tests throughout the
semester. Rather, we envision these tests as being used
as a test bank; that is, a supply of questions, or ideas for
questions, to assist instructors in composing their own
tests.
Note that most of the tests, as printed here, would
take a student much more than an hour to complete, even
with the use of software on a computer and/or calculator.
Hence, we expect instructors to choose appropriate
subsets of these tests to fulfill their classroom needs.
Andrilli/Hecker - Chapter Tests Chapter 1 - Version A
Test for Chapter 1 – Version A
(1) A rower can propel a boat5km/hr on a calm river. If the rower rows southeastward against a current
of2km/hr northward, what is the net velocity of the boat? Also, what is the net speed of the boat?
(2) Use a calculator tofind the angle(to the nearest degree) between the vectorsx=[3−25]and
y=[−41−1].
(3) Prove that, for any vectorsxy∈R
,
kx+yk
2
+kx−yk
2
=2(kxk
2
+kyk
2
)
(4) Letx=[−342]represent the force on an object in a three-dimensional coordinate system, and let
a=[5−13]be a given vector. Useproj
axto decomposexinto two component forces in directions
parallel and orthogonal toa. Verify that your answer is correct.
(5) State the contrapositive, converse, and inverse of the following statement:
Ifkx+yk6 =kxk+kyk,thenxis not parallel toy.
Which one of these is logically equivalent to the original statement?
(6) Give the negation of the following statement:
kxk=kykor(x−y)·(x+y)6 =0.
(7) Use a proof by induction to show that if the×matricesA
1A are diagonal, then
P
=1
Ais
diagonal.
(8) DecomposeA=
⎡
⎣
−43 −2
6−17
24 −3
⎤
⎦into the sum ofSandV,whereSis a symmetric matrix andV
is a skew-symmetric matrix.
(9) Given the following information about the employees of a certain TV network, calculate the total
amount of salaries and perks paid out by the network for each TV show:
TV Show 1
TV Show 2
TV Show 3
TV Show 4
Actors Writers Directors
⎡
⎢
⎢
⎣
12 4 2
10 2 3
635
941
⎤
⎥
⎥
⎦
Actor
Writer
Director
Salary Perks
⎡
⎣
$50000 $40000
$60000 $30000
$80000 $25000
⎤
⎦
(10) LetAandBbe symmetric×matrices. Prove thatABis symmetric if and only ifAandB
commute.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 191
Test for Chapter 1 – Version A
(1) A rower can propel a boat5km/hr on a calm river. If the rower rows southeastward against a current
of2km/hr northward, what is the net velocity of the boat? Also, what is the net speed of the boat?
(2) Use a calculator tofind the angle(to the nearest degree) between the vectorsx=[3−25]and
y=[−41−1].
(3) Prove that, for any vectorsxy∈R
,
kx+yk
2
+kx−yk
2
=2(kxk
2
+kyk
2
)
(4) Letx=[−342]represent the force on an object in a three-dimensional coordinate system, and let
a=[5−13]be a given vector. Useproj
axto decomposexinto two component forces in directions
parallel and orthogonal toa. Verify that your answer is correct.
(5) State the contrapositive, converse, and inverse of the following statement:
Ifkx+yk6 =kxk+kyk,thenxis not parallel toy.
Which one of these is logically equivalent to the original statement?
(6) Give the negation of the following statement:
kxk=kykor(x−y)·(x+y)6 =0.
(7) Use a proof by induction to show that if the×matricesA
1A are diagonal, then
P
=1
Ais
diagonal.
(8) DecomposeA=
⎡
⎣
−43 −2
6−17
24 −3
⎤
⎦into the sum ofSandV,whereSis a symmetric matrix andV
is a skew-symmetric matrix.
(9) Given the following information about the employees of a certain TV network, calculate the total
amount of salaries and perks paid out by the network for each TV show:
TV Show 1
TV Show 2
TV Show 3
TV Show 4
Actors Writers Directors
⎡
⎢
⎢
⎣
12 4 2
10 2 3
635
941
⎤
⎥
⎥
⎦
Actor
Writer
Director
Salary Perks
⎡
⎣
$50000 $40000
$60000 $30000
$80000 $25000
⎤
⎦
(10) LetAandBbe symmetric×matrices. Prove thatABis symmetric if and only ifAandB
commute.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 191
Andrilli/Hecker - Chapter Tests Chapter 1 - Version B
Test for Chapter 1 – Version B
(1) Three people are competing in a three-way tug-of-war, using three ropes all attached to a brass ring.
Player A is pulling due north with a force of100lbs. Player B is pulling due east, and Player C is
pulling at an angle of30degrees offdue south, toward the west side. The ring is not moving. At what
magnitude of force are Players B and C pulling?
(2) Use a calculator tofind the angle(to the nearest degree) between the vectorsx=[2−5]and
y=[−65].
(3) Prove that, for any vectorsaandbinR
witha6 =0,thevectorb−proj abis orthogonal toa.
(4) Letx=[5−2−4]represent the force on an object in a three-dimensional coordinate system, and let
a=[−23−3]be a given vector. Useproj
axto decomposexinto two component forces in directions
parallel and orthogonal toa. Verify that your answer is correct.
(5) State the contrapositive, converse, and inverse of the following statement:
Ifkxk
2
+2(x·y)0,thenkx+ykkyk.
Which one of these is logically equivalent to the original statement?
(6) Give the negation of the following statement:
There is a unit vectorxinR
3
such thatxis parallel to[−231].
(7) Use a proof by induction to show that if the×matricesA
1A are lower triangular, then
P
=1
Ais lower triangular.
(8) DecomposeA=
⎡
⎣
58−2
−63−3
94 1
⎤
⎦into the sum ofSandV,whereSis a symmetric matrix andVis
a skew-symmetric matrix.
(9) Given the following information about the amount of foods (in ounces) eaten by three cats each week,
and the percentage of certain nutrients in each food type,find the total intake of each type of nutrient
for each cat each week:
Cat 1
Cat 2
Cat 3
Food A Food B Food C Food D
⎡
⎣
94 78
63 104
46 97
⎤
⎦
Food A
Food B
Food C
Food D
Nutrient 1 Nutrient 2 Nutrient 3
⎡
⎢
⎢
⎣
5% 4% 8%
2% 3% 2%
9% 2% 6%
6% 0% 8%
⎤
⎥
⎥
⎦
(10) Prove that ifAandBare both×skew-symmetric matrices, then(AB)
=BA.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 192
Test for Chapter 1 – Version B
(1) Three people are competing in a three-way tug-of-war, using three ropes all attached to a brass ring.
Player A is pulling due north with a force of100lbs. Player B is pulling due east, and Player C is
pulling at an angle of30degrees offdue south, toward the west side. The ring is not moving. At what
magnitude of force are Players B and C pulling?
(2) Use a calculator tofind the angle(to the nearest degree) between the vectorsx=[2−5]and
y=[−65].
(3) Prove that, for any vectorsaandbinR
witha6 =0,thevectorb−proj abis orthogonal toa.
(4) Letx=[5−2−4]represent the force on an object in a three-dimensional coordinate system, and let
a=[−23−3]be a given vector. Useproj
axto decomposexinto two component forces in directions
parallel and orthogonal toa. Verify that your answer is correct.
(5) State the contrapositive, converse, and inverse of the following statement:
Ifkxk
2
+2(x·y)0,thenkx+ykkyk.
Which one of these is logically equivalent to the original statement?
(6) Give the negation of the following statement:
There is a unit vectorxinR
3
such thatxis parallel to[−231].
(7) Use a proof by induction to show that if the×matricesA
1A are lower triangular, then
P
=1
Ais lower triangular.
(8) DecomposeA=
⎡
⎣
58−2
−63−3
94 1
⎤
⎦into the sum ofSandV,whereSis a symmetric matrix andVis
a skew-symmetric matrix.
(9) Given the following information about the amount of foods (in ounces) eaten by three cats each week,
and the percentage of certain nutrients in each food type,find the total intake of each type of nutrient
for each cat each week:
Cat 1
Cat 2
Cat 3
Food A Food B Food C Food D
⎡
⎣
94 78
63 104
46 97
⎤
⎦
Food A
Food B
Food C
Food D
Nutrient 1 Nutrient 2 Nutrient 3
⎡
⎢
⎢
⎣
5% 4% 8%
2% 3% 2%
9% 2% 6%
6% 0% 8%
⎤
⎥
⎥
⎦
(10) Prove that ifAandBare both×skew-symmetric matrices, then(AB)
=BA.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 192
Andrilli/Hecker - Chapter Tests Chapter 1 - Version C
Test for Chapter 1 – Version C
(1) Using Newton’s Second Law of Motion,find the acceleration vector on a10kg object in a three-
dimensional coordinate system when the following forces are simultaneously applied:
•a force of6newtons in the direction of the vector[−405],
•a force of4newtons in the direction of the vector[23−6],and
•a force of8newtons in the direction of the vector[−14−2].
(2) Use a calculator tofind the angle(to the nearest degree) between the vectorsx=[−24−3]and
y=[8−3−2].
(3) Without using the Triangle Inequality, prove that, for any vectorsxy∈R
,
kx+yk
2
≤(kxk+kyk)
2
.
(Hint: Use the Cauchy-Schwarz Inequality.)
(4) Letx=[−42−7]represent the force on an object in a three-dimensional coordinate system, and let
a=[6−51]be a given vector. Useproj
axto decomposexinto two component forces in directions
parallel and orthogonal toa. Verify that your answer is correct.
(5) Use a proof by contrapositive to prove the following statement:
Ify6 =proj
xy,theny6 =xfor all∈R.
(6) Give the negation of the following statement:
For every vectorx∈R
,thereissomey∈R
such thatkykkxk.
(7) DecomposeA=
⎡
⎣
−36 3
75 2
−24−4
⎤
⎦into the sum ofSandV,whereSis a symmetric matrix andVis
a skew-symmetric matrix.
(8) GivenA=
⎡
⎣
−4315
6−92 −4
87 −12
⎤
⎦andB=
⎡
⎢
⎢
⎣
64 −2
−1−34
239
−25 −8
⎤
⎥
⎥
⎦
,calculate,ifpossible,thethirdrow
ofABand the second column ofBA.
(9) Use a proof by induction to show that if the×matricesA
1A are diagonal, then the product
A
1···A is diagonal.
(10) Prove that ifAis a skew-symmetric matrix, thenA
3
is also skew-symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 193
Test for Chapter 1 – Version C
(1) Using Newton’s Second Law of Motion,find the acceleration vector on a10kg object in a three-
dimensional coordinate system when the following forces are simultaneously applied:
•a force of6newtons in the direction of the vector[−405],
•a force of4newtons in the direction of the vector[23−6],and
•a force of8newtons in the direction of the vector[−14−2].
(2) Use a calculator tofind the angle(to the nearest degree) between the vectorsx=[−24−3]and
y=[8−3−2].
(3) Without using the Triangle Inequality, prove that, for any vectorsxy∈R
,
kx+yk
2
≤(kxk+kyk)
2
.
(Hint: Use the Cauchy-Schwarz Inequality.)
(4) Letx=[−42−7]represent the force on an object in a three-dimensional coordinate system, and let
a=[6−51]be a given vector. Useproj
axto decomposexinto two component forces in directions
parallel and orthogonal toa. Verify that your answer is correct.
(5) Use a proof by contrapositive to prove the following statement:
Ify6 =proj
xy,theny6 =xfor all∈R.
(6) Give the negation of the following statement:
For every vectorx∈R
,thereissomey∈R
such thatkykkxk.
(7) DecomposeA=
⎡
⎣
−36 3
75 2
−24−4
⎤
⎦into the sum ofSandV,whereSis a symmetric matrix andVis
a skew-symmetric matrix.
(8) GivenA=
⎡
⎣
−4315
6−92 −4
87 −12
⎤
⎦andB=
⎡
⎢
⎢
⎣
64 −2
−1−34
239
−25 −8
⎤
⎥
⎥
⎦
,calculate,ifpossible,thethirdrow
ofABand the second column ofBA.
(9) Use a proof by induction to show that if the×matricesA
1A are diagonal, then the product
A
1···A is diagonal.
(10) Prove that ifAis a skew-symmetric matrix, thenA
3
is also skew-symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 193
Andrilli/Hecker - Chapter Tests Chapter 2 - Version A
Test for Chapter 2 – Version A
(1) Use Gaussian Elimination tofind the quadratic equation=
2
++that goes through the points
(37),(−212),and(−456).
(2) Use Gaussian Elimination or the Gauss-Jordan Method to solve the following linear system. Give the
full solution set.
⎧
⎨
⎩
5
1+10 2+2 3+14 4+2 5=−13
−7
1−14 2−2 3−22 4+4 5=47
3
1+6 2+ 3+9 4 =−13
(3) Solve the following homogeneous system using the Gauss-Jordan Method. Give the full solution set,
expressing the vectors in it as linear combinations of particular solutions.
⎧
⎨
⎩
2
1+5 2−16 3−9 4=0
1+ 2−2 3−2 4=0
−3
1+2 2−14 3−2 4=0
(4) Find the rank ofA=
⎡
⎣
2−18
−22 −10
−53 −21
⎤
⎦.IsArow equivalent toI
3?
(5) Solve the following two systems simultaneously:
⎧
⎨
⎩
2
1+5 2+11 3=8
2
1+7 2+14 3=6
3
1+11 2+22 3=9
and
⎧
⎨
⎩
2
1+5 2+11 3=25
2
1+7 2+14 3=30
3
1+11 2+22 3=47
(6) LetAbe a×matrix,Bbe an×matrix, and letbe the Type (I) row operation:hi←hi,
for some scalarand somewith1≤≤.Provethat(AB)=(A)B. (Since you are proving
part of Theorem 2.1 in the textbook, you may not use that theorem in your proof. However, you may
use the fact from Chapter 1 that (th row of(AB))=(th row ofA)B.)
(7) Determine whether or not[−1510−23]is in the row space of
A=
⎡
⎣
5−48
212
−1−51
⎤
⎦
(8) Find the inverse ofA=
⎡
⎣
3−12
8−93
4−
32
⎤
⎦.
(9) Without using row reduction,find the inverse of the matrixA=
∙
6−8
5−9
¸
.
(10) LetAandBbe nonsingular×matrices. Prove thatAandBcommute if and only if(AB)
2
=
A
2
B
2
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 194
Test for Chapter 2 – Version A
(1) Use Gaussian Elimination tofind the quadratic equation=
2
++that goes through the points
(37),(−212),and(−456).
(2) Use Gaussian Elimination or the Gauss-Jordan Method to solve the following linear system. Give the
full solution set.
⎧
⎨
⎩
5
1+10 2+2 3+14 4+2 5=−13
−7
1−14 2−2 3−22 4+4 5=47
3
1+6 2+ 3+9 4 =−13
(3) Solve the following homogeneous system using the Gauss-Jordan Method. Give the full solution set,
expressing the vectors in it as linear combinations of particular solutions.
⎧
⎨
⎩
2
1+5 2−16 3−9 4=0
1+ 2−2 3−2 4=0
−3
1+2 2−14 3−2 4=0
(4) Find the rank ofA=
⎡
⎣
2−18
−22 −10
−53 −21
⎤
⎦.IsArow equivalent toI
3?
(5) Solve the following two systems simultaneously:
⎧
⎨
⎩
2
1+5 2+11 3=8
2
1+7 2+14 3=6
3
1+11 2+22 3=9
and
⎧
⎨
⎩
2
1+5 2+11 3=25
2
1+7 2+14 3=30
3
1+11 2+22 3=47
(6) LetAbe a×matrix,Bbe an×matrix, and letbe the Type (I) row operation:hi←hi,
for some scalarand somewith1≤≤.Provethat(AB)=(A)B. (Since you are proving
part of Theorem 2.1 in the textbook, you may not use that theorem in your proof. However, you may
use the fact from Chapter 1 that (th row of(AB))=(th row ofA)B.)
(7) Determine whether or not[−1510−23]is in the row space of
A=
⎡
⎣
5−48
212
−1−51
⎤
⎦
(8) Find the inverse ofA=
⎡
⎣
3−12
8−93
4−
32
⎤
⎦.
(9) Without using row reduction,find the inverse of the matrixA=
∙
6−8
5−9
¸
.
(10) LetAandBbe nonsingular×matrices. Prove thatAandBcommute if and only if(AB)
2
=
A
2
B
2
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 194
Andrilli/Hecker - Chapter Tests Chapter 2 - Version B
Test for Chapter 2 – Version B
(1) Use Gaussian Elimination tofind the circle
2
+
2
+++=0that goes through the points
(5−1),(6−2),and(1−7).
(2) Use Gaussian Elimination or the Gauss-Jordan Method to solve the following linear system. Give the
full solution set.
⎧
⎨
⎩
3
1−2 2+19 3+11 4+9 5−27 6=31
−2
1+ 2−12 3−7 4−8 5+21 6=−22
2
1−2 2+14 3+8 4+3 5−14 6=19
(3) Solve the following homogeneous system using the Gauss-Jordan Method. Give the full solution set,
expressing the vectors in it as linear combinations of particular solutions.
⎧
⎨
⎩
−3
1+9 2+8 3+4 4=0
5
1−15 2+ 3+22 4=0
4
1−12 2+3 3+22 4=0
(4) Find the rank ofA=
⎡
⎣
3−97
1−22
−15 41−34
⎤
⎦.IsArow equivalent toI
3?
(5) Solve the following two systems simultaneously:
⎧
⎨
⎩
4
1−3 2− 3=−13
−4
1−2 2−3 3=14
5
1− 2+ 3=−17
and
⎧
⎨
⎩
4
1−3 2− 3=13
−4
1−2 2−3 3=−16
5
1− 2+ 3=18
(6) LetAbe a×matrix,Bbe an×matrix, and letbe the Type (II) row operation
:hi←hi+hi, for some scalarand some with1≤ ≤.Provethat(AB)=(A)B.
(Since you are proving part of Theorem 2.1 in the textbook, you may not use that theorem in your
proof. However, you may use the fact from Chapter 1 that (th row of(AB))=(th row ofA)B.)
(7) Determine whether[−231]is in the row space ofA=
⎡
⎣
2−1−5
−419
−113
⎤
⎦.
(8) Find the inverse ofA=
⎡
⎣
33 1
21 0
10 3−
1
⎤
⎦.
(9) Without using row reduction, solve the linear system
⎧
⎨
⎩
−5
1+9 2+4 3=−91
7
1−14 2−10 3= 146
−7
1+12 2+4 3=−120
where
⎡
⎣
32 6 −17
21 4 −11
−7−
3
2
7
2
⎤
⎦
is the inverse of the coefficient matrix
(10) (a) Prove that ifAis a nonsingular matrix and6 =0,then(A)
−1
=(
1
)A
−1
.
(b) Use the result from part (a) to prove that ifAis a nonsingular skew-symmetric matrix, thenA
−1
is also skew-symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 195
Test for Chapter 2 – Version B
(1) Use Gaussian Elimination tofind the circle
2
+
2
+++=0that goes through the points
(5−1),(6−2),and(1−7).
(2) Use Gaussian Elimination or the Gauss-Jordan Method to solve the following linear system. Give the
full solution set.
⎧
⎨
⎩
3
1−2 2+19 3+11 4+9 5−27 6=31
−2
1+ 2−12 3−7 4−8 5+21 6=−22
2
1−2 2+14 3+8 4+3 5−14 6=19
(3) Solve the following homogeneous system using the Gauss-Jordan Method. Give the full solution set,
expressing the vectors in it as linear combinations of particular solutions.
⎧
⎨
⎩
−3
1+9 2+8 3+4 4=0
5
1−15 2+ 3+22 4=0
4
1−12 2+3 3+22 4=0
(4) Find the rank ofA=
⎡
⎣
3−97
1−22
−15 41−34
⎤
⎦.IsArow equivalent toI
3?
(5) Solve the following two systems simultaneously:
⎧
⎨
⎩
4
1−3 2− 3=−13
−4
1−2 2−3 3=14
5
1− 2+ 3=−17
and
⎧
⎨
⎩
4
1−3 2− 3=13
−4
1−2 2−3 3=−16
5
1− 2+ 3=18
(6) LetAbe a×matrix,Bbe an×matrix, and letbe the Type (II) row operation
:hi←hi+hi, for some scalarand some with1≤ ≤.Provethat(AB)=(A)B.
(Since you are proving part of Theorem 2.1 in the textbook, you may not use that theorem in your
proof. However, you may use the fact from Chapter 1 that (th row of(AB))=(th row ofA)B.)
(7) Determine whether[−231]is in the row space ofA=
⎡
⎣
2−1−5
−419
−113
⎤
⎦.
(8) Find the inverse ofA=
⎡
⎣
33 1
21 0
10 3−
1
⎤
⎦.
(9) Without using row reduction, solve the linear system
⎧
⎨
⎩
−5
1+9 2+4 3=−91
7
1−14 2−10 3= 146
−7
1+12 2+4 3=−120
where
⎡
⎣
32 6 −17
21 4 −11
−7−
3
2
7
2
⎤
⎦
is the inverse of the coefficient matrix
(10) (a) Prove that ifAis a nonsingular matrix and6 =0,then(A)
−1
=(
1
)A
−1
.
(b) Use the result from part (a) to prove that ifAis a nonsingular skew-symmetric matrix, thenA
−1
is also skew-symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 195
Andrilli/Hecker - Chapter Tests Chapter 2 - Version C
Test for Chapter 2 – Version C
(1) Use Gaussian Elimination tofind the values of,,,andthat solve the following partial fractions
problem:
5
2
−18+1
(−2)
2
(+3)
=
−2
+
+
(−2)
2
+
+3
(2) Use Gaussian Elimination or the Gauss-Jordan Method to solve the following linear system. Give the
full solution set.
⎧
⎪
⎪
⎨
⎪
⎪
⎩
9
1+36 2−11 3+3 4+34 5−15 6=−6
−10
1−40 2+9 3−2 4−38 5−9 6=−89
4
1+16 2−4 3+ 4+15 5 =22
11
1+44 2−12 3+2 4+47 5+3 6=77
(3) Solve the following homogeneous system using the Gauss-Jordan Method. Give the full solution set,
expressing the vectors in it as linear combinations of particular solutions.
⎧
⎨
⎩
−2
1−2 2+14 3+ 4−7 5=0
6
1+9 2−27 3−2 4+21 5=0
−3
1−4 2+16 3+ 4−10 5=0
(4) Find the rank ofA=
⎡
⎢
⎢
⎣
56 −223
24 −113
−6−91 −27
46 −119
⎤
⎥
⎥
⎦
.IsArow equivalent toI
4?
(5) Find the reduced row echelon form matrixBfor the matrix
A=
⎡
⎣
223
−321
513
⎤
⎦
and list a series of row operations that convertsBtoA.
(6) Determine whether[2512−19]is a linear combination ofa=[735],b=[334],andc=[323].
(7) Find the inverse ofA=
⎡
⎢
⎢
⎣
−201 −1
4−1−13
31 −1−2
37 −2−16
⎤
⎥
⎥
⎦
.
(8) Without using row reduction, solve the linear system
⎧
⎨
⎩
−4
1+2 2−5 3=−3
−4
1+ 2−3 3=−7
6
1+3 2−4 3=26
where
⎡
⎣
5
2
−
7
2
−
1
2
−17 23 4
−912 2
⎤
⎦
is the inverse of the coefficient matrix.
(9) Prove by induction that ifA
1A are nonsingular×matrices, then
(A
1···A )
−1
=(A )
−1
···(A 1)
−1
.
(10) Suppose thatAandBare×matrices, and that
1are row operations such that
1(2(···( (AB))···)) =I .Provethat 1(2(···( (A))···)) =B
−1
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 196
Test for Chapter 2 – Version C
(1) Use Gaussian Elimination tofind the values of,,,andthat solve the following partial fractions
problem:
5
2
−18+1
(−2)
2
(+3)
=
−2
+
+
(−2)
2
+
+3
(2) Use Gaussian Elimination or the Gauss-Jordan Method to solve the following linear system. Give the
full solution set.
⎧
⎪
⎪
⎨
⎪
⎪
⎩
9
1+36 2−11 3+3 4+34 5−15 6=−6
−10
1−40 2+9 3−2 4−38 5−9 6=−89
4
1+16 2−4 3+ 4+15 5 =22
11
1+44 2−12 3+2 4+47 5+3 6=77
(3) Solve the following homogeneous system using the Gauss-Jordan Method. Give the full solution set,
expressing the vectors in it as linear combinations of particular solutions.
⎧
⎨
⎩
−2
1−2 2+14 3+ 4−7 5=0
6
1+9 2−27 3−2 4+21 5=0
−3
1−4 2+16 3+ 4−10 5=0
(4) Find the rank ofA=
⎡
⎢
⎢
⎣
56 −223
24 −113
−6−91 −27
46 −119
⎤
⎥
⎥
⎦
.IsArow equivalent toI
4?
(5) Find the reduced row echelon form matrixBfor the matrix
A=
⎡
⎣
223
−321
513
⎤
⎦
and list a series of row operations that convertsBtoA.
(6) Determine whether[2512−19]is a linear combination ofa=[735],b=[334],andc=[323].
(7) Find the inverse ofA=
⎡
⎢
⎢
⎣
−201 −1
4−1−13
31 −1−2
37 −2−16
⎤
⎥
⎥
⎦
.
(8) Without using row reduction, solve the linear system
⎧
⎨
⎩
−4
1+2 2−5 3=−3
−4
1+ 2−3 3=−7
6
1+3 2−4 3=26
where
⎡
⎣
5
2
−
7
2
−
1
2
−17 23 4
−912 2
⎤
⎦
is the inverse of the coefficient matrix.
(9) Prove by induction that ifA
1A are nonsingular×matrices, then
(A
1···A )
−1
=(A )
−1
···(A 1)
−1
.
(10) Suppose thatAandBare×matrices, and that
1are row operations such that
1(2(···( (AB))···)) =I .Provethat 1(2(···( (A))···)) =B
−1
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 196
Andrilli/Hecker - Chapter Tests Chapter 3 - Version A
Test for Chapter 3 – Version A
(1) Calculate the area of the parallelogram determined by the vectorsx=[−25]andy=[6−3].
(2) IfAis a5×5matrix with determinant4,whatis|6A|?Why?
(3) Calculate the determinant ofA=
⎡
⎣
5−6−3
461
510 2
⎤
⎦by row reducingAto upper triangular form.
(4) Prove that ifAandBare×matrices, then|AB
|=|A
B|.
(5) Use cofactor expansion along any row or column tofind the determinant of
A=
⎡
⎢
⎢
⎣
30 −46
4−252
−330 −5
700 −8
⎤
⎥
⎥
⎦
Be sure to use cofactor expansion tofind any3×3determinants needed as well.
(6) Prove that ifAandB
−1
are similar×matrices, then|A||B|=1.
(7) Use Cramer’s Rule to solve the following system:
⎧
⎨
⎩
5
1+ 2+2 3=3
−8
1+2 2− 3=17
6
1− 2+ 3=−10
(8) LetA=
⎡
⎣
01 −1
101
−1−10
⎤
⎦Find a nonsingular matrixPhaving all integer entries, and a diagonal
matrixDsuch thatD=P
−1
AP.
(9) Prove that an×matrixAis singular if and only if=0is an eigenvalue forA.
(10) Suppose that
1=2is an eigenvalue for an×matrixA.Provethat 2=8is an eigenvalue for
A
3
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 197
Test for Chapter 3 – Version A
(1) Calculate the area of the parallelogram determined by the vectorsx=[−25]andy=[6−3].
(2) IfAis a5×5matrix with determinant4,whatis|6A|?Why?
(3) Calculate the determinant ofA=
⎡
⎣
5−6−3
461
510 2
⎤
⎦by row reducingAto upper triangular form.
(4) Prove that ifAandBare×matrices, then|AB
|=|A
B|.
(5) Use cofactor expansion along any row or column tofind the determinant of
A=
⎡
⎢
⎢
⎣
30 −46
4−252
−330 −5
700 −8
⎤
⎥
⎥
⎦
Be sure to use cofactor expansion tofind any3×3determinants needed as well.
(6) Prove that ifAandB
−1
are similar×matrices, then|A||B|=1.
(7) Use Cramer’s Rule to solve the following system:
⎧
⎨
⎩
5
1+ 2+2 3=3
−8
1+2 2− 3=17
6
1− 2+ 3=−10
(8) LetA=
⎡
⎣
01 −1
101
−1−10
⎤
⎦Find a nonsingular matrixPhaving all integer entries, and a diagonal
matrixDsuch thatD=P
−1
AP.
(9) Prove that an×matrixAis singular if and only if=0is an eigenvalue forA.
(10) Suppose that
1=2is an eigenvalue for an×matrixA.Provethat 2=8is an eigenvalue for
A
3
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 197
Andrilli/Hecker - Chapter Tests Chapter 3 - Version B
Test for Chapter 3 – Version B
(1) Calculate the volume of the parallelepiped determined by the vectorsx=[27−1],y=[42−3],and
z=[−562].
(2) IfAis a4×4matrix with determinant7,whatis|5A|?Why?
(3) Calculate the determinant of
A=
⎡
⎢
⎢
⎣
2−4−12 15
61125 −32
41332 −41
3−16−45 57
⎤
⎥
⎥
⎦
by row reducingAto upper triangular form.
(4) Prove that ifAandBare×matrices withAB=−BA,andis odd, then eitherAorBis
singular.
(5) Use cofactor expansion along any row or column tofind the determinant of
A=
⎡
⎢
⎢
⎣
2152
43−10
−6800
17 0 −3
⎤
⎥
⎥
⎦
Be sure to use cofactor expansion tofind any3×3determinants needed as well.
(6) SupposeAis an×matrix,istheType(II)rowoperationhi←hi+hi,andis the Type
(II)columnoperationhcol.i←hcol.i+hcol.i.Provethat((A))
=
¡
A
¢
by showing that
theth row of((A))
=th row of
¡
A
¢
,first when6 =,andthenwhen=.
(7) Use Cramer’s Rule to solve the following system:
⎧
⎨
⎩
4
1−4 2−3 3=−10
6
1−5 2−10 3=−28
−2
1+2 2+2 3=6
(8) LetA=
⎡
⎣
21−3
12−3
11−2
⎤
⎦Find a nonsingular matrixPhaving all integer entries, and a diagonal
matrixDsuch thatD=P
−1
AP.
(9) Consider the matrixA=
⎡
⎢
⎢
⎣
2100
0210
0021
0002
⎤
⎥
⎥
⎦
.
(a) Show thatAhas only one eigenvalue. What is it?
(b) Show that Step 3 of the Diagonalization Method of Section 3.4 produces only one fundamental
eigenvector forA.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 198
Test for Chapter 3 – Version B
(1) Calculate the volume of the parallelepiped determined by the vectorsx=[27−1],y=[42−3],and
z=[−562].
(2) IfAis a4×4matrix with determinant7,whatis|5A|?Why?
(3) Calculate the determinant of
A=
⎡
⎢
⎢
⎣
2−4−12 15
61125 −32
41332 −41
3−16−45 57
⎤
⎥
⎥
⎦
by row reducingAto upper triangular form.
(4) Prove that ifAandBare×matrices withAB=−BA,andis odd, then eitherAorBis
singular.
(5) Use cofactor expansion along any row or column tofind the determinant of
A=
⎡
⎢
⎢
⎣
2152
43−10
−6800
17 0 −3
⎤
⎥
⎥
⎦
Be sure to use cofactor expansion tofind any3×3determinants needed as well.
(6) SupposeAis an×matrix,istheType(II)rowoperationhi←hi+hi,andis the Type
(II)columnoperationhcol.i←hcol.i+hcol.i.Provethat((A))
=
¡
A
¢
by showing that
theth row of((A))
=th row of
¡
A
¢
,first when6 =,andthenwhen=.
(7) Use Cramer’s Rule to solve the following system:
⎧
⎨
⎩
4
1−4 2−3 3=−10
6
1−5 2−10 3=−28
−2
1+2 2+2 3=6
(8) LetA=
⎡
⎣
21−3
12−3
11−2
⎤
⎦Find a nonsingular matrixPhaving all integer entries, and a diagonal
matrixDsuch thatD=P
−1
AP.
(9) Consider the matrixA=
⎡
⎢
⎢
⎣
2100
0210
0021
0002
⎤
⎥
⎥
⎦
.
(a) Show thatAhas only one eigenvalue. What is it?
(b) Show that Step 3 of the Diagonalization Method of Section 3.4 produces only one fundamental
eigenvector forA.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 198
Andrilli/Hecker - Chapter Tests Chapter 3 - Version B
(10) Consider the matrixA=
"
3
5
−
4
5
4
5
3
5
#
. It can be shown that, for every nonzero vectorX, the vectors
Xand(AX)form an angle with each other measuring= arccos(
3
5
)≈53
◦
.(Youmayassumethis
fact.) Show whyAis not diagonalizable,first from an algebraic perspective, and then from a geometric
perspective.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 199
(10) Consider the matrixA=
"
3
5
−
4
5
4
5
3
5
#
. It can be shown that, for every nonzero vectorX, the vectors
Xand(AX)form an angle with each other measuring= arccos(
3
5
)≈53
◦
.(Youmayassumethis
fact.) Show whyAis not diagonalizable,first from an algebraic perspective, and then from a geometric
perspective.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 199
Andrilli/Hecker - Chapter Tests Chapter 3 - Version C
Test for Chapter 3 – Version C
(1) Calculate the volume of the parallelepiped determined by the vectorsx=[5−23],y=[−31−2],
andz=[4−51].
(2) Calculate the determinant of
A=
⎡
⎢
⎢
⎣
431 2
190 2
832 −2
431 1
⎤
⎥
⎥
⎦
by row reducingAto upper triangular form.
(3) Use a cofactor expansion along any row or column tofind the determinant of
A=
⎡
⎢
⎢
⎢
⎢
⎣
0 0 005
012 094
0 0 083
0141172
15 13 10 6 1
⎤
⎥
⎥
⎥
⎥
⎦
Be sure to use cofactor expansion on each4×4,3×3,and2×2determinant you need to calculate as
well.
(4) Prove that ifAis an orthogonal matrix (that is,A
=A
−1
), then|A|=±1.
(5) Suppose thatAis an×matrix,Bis an×matrix, andis a Type (I), (II), or (III)column
operation. Prove that(AB)=A((B)). (Hint: You can assume the fact that ifis therow
operation corresponding to thecolumnoperation, then, for any matrixD,((D)) =
¡
¡
D
¢¢
.)
(6) Perform (only) the Base Step in the proof by induction of the following statement, which is part of
Theorem 3.3: LetAbe an×matrix, with≥2,andletbetheType(II)rowoperation
hi←hi+hi.Then|A|=|(A)|.
(7) Use Cramer’s Rule to solve the following system:
⎧
⎨
⎩
−9
1+6 2+2 3=−41
5
1−3 2− 3=22
−8
1+6 2+ 3=−40
(8) Use diagonalization to calculateA
9
ifA=
∙
5−6
3−4
¸
.
(9) LetA=
⎡
⎣
−46 −6
020
3−35
⎤
⎦Find a nonsingular matrixPhaving all integer entries, and a diagonal
matrixDsuch thatD=P
−1
AP.
(10) Letbe an eigenvalue for an×matrixAhaving algebraic multiplicity.Provethatis also an
eigenvalue with algebraic multiplicityforA
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 200
Test for Chapter 3 – Version C
(1) Calculate the volume of the parallelepiped determined by the vectorsx=[5−23],y=[−31−2],
andz=[4−51].
(2) Calculate the determinant of
A=
⎡
⎢
⎢
⎣
431 2
190 2
832 −2
431 1
⎤
⎥
⎥
⎦
by row reducingAto upper triangular form.
(3) Use a cofactor expansion along any row or column tofind the determinant of
A=
⎡
⎢
⎢
⎢
⎢
⎣
0 0 005
012 094
0 0 083
0141172
15 13 10 6 1
⎤
⎥
⎥
⎥
⎥
⎦
Be sure to use cofactor expansion on each4×4,3×3,and2×2determinant you need to calculate as
well.
(4) Prove that ifAis an orthogonal matrix (that is,A
=A
−1
), then|A|=±1.
(5) Suppose thatAis an×matrix,Bis an×matrix, andis a Type (I), (II), or (III)column
operation. Prove that(AB)=A((B)). (Hint: You can assume the fact that ifis therow
operation corresponding to thecolumnoperation, then, for any matrixD,((D)) =
¡
¡
D
¢¢
.)
(6) Perform (only) the Base Step in the proof by induction of the following statement, which is part of
Theorem 3.3: LetAbe an×matrix, with≥2,andletbetheType(II)rowoperation
hi←hi+hi.Then|A|=|(A)|.
(7) Use Cramer’s Rule to solve the following system:
⎧
⎨
⎩
−9
1+6 2+2 3=−41
5
1−3 2− 3=22
−8
1+6 2+ 3=−40
(8) Use diagonalization to calculateA
9
ifA=
∙
5−6
3−4
¸
.
(9) LetA=
⎡
⎣
−46 −6
020
3−35
⎤
⎦Find a nonsingular matrixPhaving all integer entries, and a diagonal
matrixDsuch thatD=P
−1
AP.
(10) Letbe an eigenvalue for an×matrixAhaving algebraic multiplicity.Provethatis also an
eigenvalue with algebraic multiplicityforA
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 200
Andrilli/Hecker - Chapter Tests Chapter 4 - Version A
Test for Chapter 4 – Version A
(1) Prove thatRis a vector space using the operations⊕and¯given byx⊕y=(
5
+
5
)
15
and¯x=
15
.
(2) Prove that the set of×skew-symmetric matrices is a subspace ofM
.
(3) Prove that the set{[5−2−1][2−1−1][8−21]}spansR
3
.
(4) Find a simplified general form for all the vectors inspan()inP
3if
={4
3
−20
2
−−116
3
−30
2
−2−18−5
3
+25
2
+2+16}
(5) Prove that the set{[2−1−11−4][9−122][40218][−2183]}inR
4
is linearly independent.
(6) Determine whether
½∙
413
−19 1
¸
∙
3−15
19−1
¸
∙
2−12
15−1
¸
∙
−47
−52
¸¾
is a basis forM
22.
(7) Find a subset of
={3
3
−2
2
+3−1−6
3
+4
2
−6+2
3
−3
2
+2−1
7
3
+5−111
3
−12
2
+13+3}
inP
3that is a basis forspan().Whatisdim(span())?
(8) Prove that the columns of a nonsingular×matrixAspanR
.
(9) Consider the matrixA=
⎡
⎣
5−65
100
−24 −3
⎤
⎦,whichhas=2as an eigenvalue. Find a basisforR
3
that contains a basis for the eigenspace 2forA.
(10) (a) Find the transition matrix from-coordinates to-coordinates if
=([−11−1912][716−2][−8−1211])and
=([24−1][−1−21][11−2])
are ordered bases forR
3
.
(b) Givenv=[−63−11362]∈R
3
,find[v] , and use your answer to part (a) tofind[v] .
Copyrightc°2016 Elsevier Ltd. All rights reserved. 201
Test for Chapter 4 – Version A
(1) Prove thatRis a vector space using the operations⊕and¯given byx⊕y=(
5
+
5
)
15
and¯x=
15
.
(2) Prove that the set of×skew-symmetric matrices is a subspace ofM
.
(3) Prove that the set{[5−2−1][2−1−1][8−21]}spansR
3
.
(4) Find a simplified general form for all the vectors inspan()inP
3if
={4
3
−20
2
−−116
3
−30
2
−2−18−5
3
+25
2
+2+16}
(5) Prove that the set{[2−1−11−4][9−122][40218][−2183]}inR
4
is linearly independent.
(6) Determine whether
½∙
413
−19 1
¸
∙
3−15
19−1
¸
∙
2−12
15−1
¸
∙
−47
−52
¸¾
is a basis forM
22.
(7) Find a subset of
={3
3
−2
2
+3−1−6
3
+4
2
−6+2
3
−3
2
+2−1
7
3
+5−111
3
−12
2
+13+3}
inP
3that is a basis forspan().Whatisdim(span())?
(8) Prove that the columns of a nonsingular×matrixAspanR
.
(9) Consider the matrixA=
⎡
⎣
5−65
100
−24 −3
⎤
⎦,whichhas=2as an eigenvalue. Find a basisforR
3
that contains a basis for the eigenspace 2forA.
(10) (a) Find the transition matrix from-coordinates to-coordinates if
=([−11−1912][716−2][−8−1211])and
=([24−1][−1−21][11−2])
are ordered bases forR
3
.
(b) Givenv=[−63−11362]∈R
3
,find[v] , and use your answer to part (a) tofind[v] .
Copyrightc°2016 Elsevier Ltd. All rights reserved. 201
Andrilli/Hecker - Chapter Tests Chapter 4 - Version B
Test for Chapter 4 – Version B
(1) Prove thatRwith the usual scalar multiplication, but with addition given byx⊕y=3(+)isnot
a vector space.
(2) Prove that the set of all polynomialspinP
4for which the coefficient of the second-degree term equals
the coefficient of the fourth-degree term is a subspace ofP
4.
(3) LetVbe a vector space with subspacesW
1andW 2. Prove that the intersection ofW 1andW 2is a
subspace ofV.
(4) Let={[5−15−4][−261][−9277]}. Determine whether[2−43]is inspan().
(5) Find a simplified general form for all the vectors inspan()inM
22if
=
½∙
2−6
−1−1
¸
∙
5−15
−11
¸
∙
8−24
−21
¸
∙
3−9
12
¸¾
(6) Letbe the set
{2
3
−11
2
−12−17
3
+35
2
+16−7
−5
3
+29
2
+33+2−5
3
−26
2
−12+5}
inP
3and letp()∈P 3. Prove that there is exactly one way to expressp()as a linear combination
of the elements of.
(7) Find a subset of
={[2−34−1][−69−123][31−22][28−123][76−104]}
that is a basis forspan()inR
4
.DoesspanR
4
?
(8) LetAbe a nonsingular×matrix. Prove that the rows ofAare linearly independent.
(9) Consider the matrixA=
⎡
⎢
⎢
⎣
−1−231
2−667
00 −20
2−465
⎤
⎥
⎥
⎦
, which has=−2as an eigenvalue. Find a basis
forR
4
that contains a basis for the eigenspace −2forA.
(10) (a) Find the transition matrix from-coordinates to-coordinates if
=(−21
2
+14−3817
2
−10+32−10
2
+4−23)and
=(−7
2
+4−142
2
−+4
2
−+1)
are ordered bases forP
2.
(b) Givenv=−100
2
+64−181∈P 3,find[v] , and use your answer to part (a) tofind[v] .
Copyrightc°2016 Elsevier Ltd. All rights reserved. 202
Test for Chapter 4 – Version B
(1) Prove thatRwith the usual scalar multiplication, but with addition given byx⊕y=3(+)isnot
a vector space.
(2) Prove that the set of all polynomialspinP
4for which the coefficient of the second-degree term equals
the coefficient of the fourth-degree term is a subspace ofP
4.
(3) LetVbe a vector space with subspacesW
1andW 2. Prove that the intersection ofW 1andW 2is a
subspace ofV.
(4) Let={[5−15−4][−261][−9277]}. Determine whether[2−43]is inspan().
(5) Find a simplified general form for all the vectors inspan()inM
22if
=
½∙
2−6
−1−1
¸
∙
5−15
−11
¸
∙
8−24
−21
¸
∙
3−9
12
¸¾
(6) Letbe the set
{2
3
−11
2
−12−17
3
+35
2
+16−7
−5
3
+29
2
+33+2−5
3
−26
2
−12+5}
inP
3and letp()∈P 3. Prove that there is exactly one way to expressp()as a linear combination
of the elements of.
(7) Find a subset of
={[2−34−1][−69−123][31−22][28−123][76−104]}
that is a basis forspan()inR
4
.DoesspanR
4
?
(8) LetAbe a nonsingular×matrix. Prove that the rows ofAare linearly independent.
(9) Consider the matrixA=
⎡
⎢
⎢
⎣
−1−231
2−667
00 −20
2−465
⎤
⎥
⎥
⎦
, which has=−2as an eigenvalue. Find a basis
forR
4
that contains a basis for the eigenspace −2forA.
(10) (a) Find the transition matrix from-coordinates to-coordinates if
=(−21
2
+14−3817
2
−10+32−10
2
+4−23)and
=(−7
2
+4−142
2
−+4
2
−+1)
are ordered bases forP
2.
(b) Givenv=−100
2
+64−181∈P 3,find[v] , and use your answer to part (a) tofind[v] .
Copyrightc°2016 Elsevier Ltd. All rights reserved. 202
Andrilli/Hecker - Chapter Tests Chapter 4 - Version C
Test for Chapter 4 – Version C
(1) The setR
2
with operations[ ]⊕[ ]=[++3+−4],and¯[ ]=[+3−3−4+4]
is a vector space. Find the zero vector0and the additive inverse ofv=[ ]for this vector space.
(2) Prove that the set of all 3-vectors orthogonal to[2−31]forms a subspace ofR
3
.
(3) Determine whether[−365]is inspan()inR
3
,if
={[3−6−1][4−8−3][5−10−1]}
(4) Find a simplified form for all the vectors inspan()inP
3if
={4
3
−
2
−11+18−3
3
+
2
+9−145
3
−
2
−13+22}
(5) Prove that the set
½∙
−363
9−46
¸
∙
−556
14−41
¸
∙
−444
13−32
¸
∙
514
−13−10
¸¾
inM
22is linearly independent.
(6) Find a subset of
=
½∙
3−1
−24
¸
∙
−62
4−8
¸
∙
21
−20
¸
∙
−1−3
24
¸
∙
4−3
−22
¸¾
inM
22that is a basis forspan().Whatisdim(span())?
(7) LetAbe an×singular matrix and letbe the set of columns ofA.Provethatdim(span()).
(8) Enlarge the linearly independent set
={2
3
−3
2
+3+1−
3
+4
2
−6−2}
ofP
3to a basis forP 3.
(9) LetA6 =I
be an×matrix having eigenvalue=1.Provethatdim( 1).
(10) (a) Find the transition matrix from-coordinates to-coordinates if
=([10−178][−410−5][29−3616])and
=([12−125][−53−1][1−21])
are ordered bases forR
3
.
(b) Givenv=[−109155−71]∈R
3
,find[v] , and use your answer to part (a) tofind[v] .
Copyrightc°2016 Elsevier Ltd. All rights reserved. 203
Test for Chapter 4 – Version C
(1) The setR
2
with operations[ ]⊕[ ]=[++3+−4],and¯[ ]=[+3−3−4+4]
is a vector space. Find the zero vector0and the additive inverse ofv=[ ]for this vector space.
(2) Prove that the set of all 3-vectors orthogonal to[2−31]forms a subspace ofR
3
.
(3) Determine whether[−365]is inspan()inR
3
,if
={[3−6−1][4−8−3][5−10−1]}
(4) Find a simplified form for all the vectors inspan()inP
3if
={4
3
−
2
−11+18−3
3
+
2
+9−145
3
−
2
−13+22}
(5) Prove that the set
½∙
−363
9−46
¸
∙
−556
14−41
¸
∙
−444
13−32
¸
∙
514
−13−10
¸¾
inM
22is linearly independent.
(6) Find a subset of
=
½∙
3−1
−24
¸
∙
−62
4−8
¸
∙
21
−20
¸
∙
−1−3
24
¸
∙
4−3
−22
¸¾
inM
22that is a basis forspan().Whatisdim(span())?
(7) LetAbe an×singular matrix and letbe the set of columns ofA.Provethatdim(span()).
(8) Enlarge the linearly independent set
={2
3
−3
2
+3+1−
3
+4
2
−6−2}
ofP
3to a basis forP 3.
(9) LetA6 =I
be an×matrix having eigenvalue=1.Provethatdim( 1).
(10) (a) Find the transition matrix from-coordinates to-coordinates if
=([10−178][−410−5][29−3616])and
=([12−125][−53−1][1−21])
are ordered bases forR
3
.
(b) Givenv=[−109155−71]∈R
3
,find[v] , and use your answer to part (a) tofind[v] .
Copyrightc°2016 Elsevier Ltd. All rights reserved. 203
Andrilli/Hecker - Chapter Tests Chapter 5 - Version A
Test for Chapter 5 – Version A
(1) Prove that the mapping:M →M given by(A)=BAB
−1
,whereBis somefixed nonsingular
×matrix, is a linear operator.
(2) Suppose:R
3
→R
3
is a linear transformation, and([−252]) = [21−1],([011]) = [1−10],
and([1−2−1]) = [03−1].Find([2−31]).
(3) Find the matrixA
for:R
2
→P 2given by
([ ]) = (−2+)
2
−(2)+(+2)
for the ordered bases=([3−7][2−5])forR
2
,
and=(9
2
+20−214
2
+9−910
2
+22−23)forP 2.
(4) For the linear transformation:R
4
→R
3
given by
⎛
⎜
⎜
⎝
⎡
⎢
⎢
⎣
1
2
3
4
⎤
⎥
⎥
⎦
⎞
⎟
⎟
⎠
=
⎡
⎣
4−8−1−7
−3616
−48210
⎤
⎦
⎡
⎢
⎢
⎣
1
2
3
4
⎤
⎥
⎥
⎦
find a basis forker(), a basis forrange(), and verify that the Dimension Theorem holds for.
(5) (a) Consider the linear transformation:P
3→R
3
given by
(p()) =
∙
p(3)p
0
(1)
Z
1
0
p()
¸
Prove thatker()is nontrivial.
(b) Use part (a) to prove that there is a nonzero polynomialp∈P
3such thatp(3) = 0,p
0
(1) = 0,
and
R
1
0
p()=0.
(6) Prove that the mapping:R
3
→R
3
given by
([ ]) =
⎡
⎣
−521
6−3−2
10−3−1
⎤
⎦
⎡
⎣
⎤
⎦
is one-to-one. Isan isomorphism?
(7) Show that:P
→P given by(p)=p−p
0
is an isomorphism.
(8) Consider the diagonalizable operator:M
22→M 22given by(K)=K−K
.LetAbe the matrix
representation ofwith respect to the standard basisforM
22.
(a) FindanorderedbasisofM
22consisting of fundamental eigenvectors for, and the diagonal
matrixDthat is the matrix representation ofwith respect to.
(b) Calculate the transition matrixPfromto,andverifythatD=P
−1
AP.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 204
Test for Chapter 5 – Version A
(1) Prove that the mapping:M →M given by(A)=BAB
−1
,whereBis somefixed nonsingular
×matrix, is a linear operator.
(2) Suppose:R
3
→R
3
is a linear transformation, and([−252]) = [21−1],([011]) = [1−10],
and([1−2−1]) = [03−1].Find([2−31]).
(3) Find the matrixA
for:R
2
→P 2given by
([ ]) = (−2+)
2
−(2)+(+2)
for the ordered bases=([3−7][2−5])forR
2
,
and=(9
2
+20−214
2
+9−910
2
+22−23)forP 2.
(4) For the linear transformation:R
4
→R
3
given by
⎛
⎜
⎜
⎝
⎡
⎢
⎢
⎣
1
2
3
4
⎤
⎥
⎥
⎦
⎞
⎟
⎟
⎠
=
⎡
⎣
4−8−1−7
−3616
−48210
⎤
⎦
⎡
⎢
⎢
⎣
1
2
3
4
⎤
⎥
⎥
⎦
find a basis forker(), a basis forrange(), and verify that the Dimension Theorem holds for.
(5) (a) Consider the linear transformation:P
3→R
3
given by
(p()) =
∙
p(3)p
0
(1)
Z
1
0
p()
¸
Prove thatker()is nontrivial.
(b) Use part (a) to prove that there is a nonzero polynomialp∈P
3such thatp(3) = 0,p
0
(1) = 0,
and
R
1
0
p()=0.
(6) Prove that the mapping:R
3
→R
3
given by
([ ]) =
⎡
⎣
−521
6−3−2
10−3−1
⎤
⎦
⎡
⎣
⎤
⎦
is one-to-one. Isan isomorphism?
(7) Show that:P
→P given by(p)=p−p
0
is an isomorphism.
(8) Consider the diagonalizable operator:M
22→M 22given by(K)=K−K
.LetAbe the matrix
representation ofwith respect to the standard basisforM
22.
(a) FindanorderedbasisofM
22consisting of fundamental eigenvectors for, and the diagonal
matrixDthat is the matrix representation ofwith respect to.
(b) Calculate the transition matrixPfromto,andverifythatD=P
−1
AP.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 204
Andrilli/Hecker - Chapter Tests Chapter 5 - Version A
(9) Indicate whether each given statement is true or false.
(a) Ifis a linear operator on a nontrivialfinite dimensional vector space for which every root of
()is real, thenis diagonalizable.
(b) Let:V→Vbe an isomorphism with eigenvalue.Then6 =0and1is an eigenvalue for
−1
.
(10) Prove that the linear operator onP
3given by(p()) =p(+1)is not diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 205
(9) Indicate whether each given statement is true or false.
(a) Ifis a linear operator on a nontrivialfinite dimensional vector space for which every root of
()is real, thenis diagonalizable.
(b) Let:V→Vbe an isomorphism with eigenvalue.Then6 =0and1is an eigenvalue for
−1
.
(10) Prove that the linear operator onP
3given by(p()) =p(+1)is not diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 205
Andrilli/Hecker - Chapter Tests Chapter 5 - Version B
Test for Chapter 5 – Version B
(1) Prove that the mapping:M →M given by(A)=A+A
is a linear operator.
(2) Suppose:R
3
−→P 3is a linear transformation, and([1−11]) = 2
3
−
2
+3,
([256]) =−11
3
+
2
+4−2,and([144]) =−
3
−3
2
+2+10.Find([6−47]).
(3) Find the matrixA
for:M 22→R
3
given by
µ∙
¸¶
=[−2+− −2+2−+]
for the ordered bases
=
µ∙
1−2
−4−1
¸
∙
−46
11 3
¸
∙
−21
31
¸
∙
14
51
¸¶
forM
22and=([−102][22−1][132])forR
3
.
(4) For the linear transformation:R
4
−→R
4
given by
⎛
⎜
⎜
⎝
⎡
⎢
⎢
⎣
1
2
3
4
⎤
⎥
⎥
⎦
⎞
⎟
⎟
⎠
=
⎡
⎢
⎢
⎣
25 48
1−1−5−1
−42161
41 −10 2
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
1
2
3
4
⎤
⎥
⎥
⎦
find a basis forker(), a basis forrange(), and verify that the Dimension Theorem holds for.
(5) LetAbe afixed×matrix, with6 =.Let
1:R
→R
be given by 1(x)=Ax,andlet
2:R
→R
be given by 2(y)=A
y.
(a) Prove or disprove:dim(range(
1)) = dim(range( 2))
(b) Prove or disprove:dim(ker(
1)) = dim(ker( 2))
(6) Prove that the mapping:M
22→M 22given by
µ∙
¸¶
=
∙
3−2−3−4−−−
−− 3−−−
¸
is one-to-one. Isan isomorphism?
(7) LetAbe afixed nonsingular×matrix. Show that:M
→M given by(B)=BA
−1
is
an isomorphism.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 206
Test for Chapter 5 – Version B
(1) Prove that the mapping:M →M given by(A)=A+A
is a linear operator.
(2) Suppose:R
3
−→P 3is a linear transformation, and([1−11]) = 2
3
−
2
+3,
([256]) =−11
3
+
2
+4−2,and([144]) =−
3
−3
2
+2+10.Find([6−47]).
(3) Find the matrixA
for:M 22→R
3
given by
µ∙
¸¶
=[−2+− −2+2−+]
for the ordered bases
=
µ∙
1−2
−4−1
¸
∙
−46
11 3
¸
∙
−21
31
¸
∙
14
51
¸¶
forM
22and=([−102][22−1][132])forR
3
.
(4) For the linear transformation:R
4
−→R
4
given by
⎛
⎜
⎜
⎝
⎡
⎢
⎢
⎣
1
2
3
4
⎤
⎥
⎥
⎦
⎞
⎟
⎟
⎠
=
⎡
⎢
⎢
⎣
25 48
1−1−5−1
−42161
41 −10 2
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
1
2
3
4
⎤
⎥
⎥
⎦
find a basis forker(), a basis forrange(), and verify that the Dimension Theorem holds for.
(5) LetAbe afixed×matrix, with6 =.Let
1:R
→R
be given by 1(x)=Ax,andlet
2:R
→R
be given by 2(y)=A
y.
(a) Prove or disprove:dim(range(
1)) = dim(range( 2))
(b) Prove or disprove:dim(ker(
1)) = dim(ker( 2))
(6) Prove that the mapping:M
22→M 22given by
µ∙
¸¶
=
∙
3−2−3−4−−−
−− 3−−−
¸
is one-to-one. Isan isomorphism?
(7) LetAbe afixed nonsingular×matrix. Show that:M
→M given by(B)=BA
−1
is
an isomorphism.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 206
Andrilli/Hecker - Chapter Tests Chapter 5 - Version B
(8) Consider the diagonalizable operator:M 22→M 22given by
µ∙
¸¶
=
∙
−712
−47
¸∙
¸
LetAbe the matrix representation ofwith respect to the standard basisforM
22.
(a) FindanorderedbasisofM
22consisting of fundamental eigenvectors for, and the diagonal
matrixDthat is the matrix representation ofwith respect to.
(b) Calculate the transition matrixPfromto,andverifythat
D=P
−1
AP.
(9) Indicate whether each given statement is true or false.
(a) If:V→Vis an isomorphism on a nontrivialfinite dimensional vector spaceVandis an
eigenvalue for,then=±1.
(b) A linear operatoron an-dimensional vector space is diagonalizable if and only ifhas
distinct eigenvalues.
(10) Letbe the linear operator onR
3
representing a counterclockwise rotation through an angle of6
radians around an axis through the origin that is parallel to[4−23]. Explain in words whyis not
diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 207
(8) Consider the diagonalizable operator:M 22→M 22given by
µ∙
¸¶
=
∙
−712
−47
¸∙
¸
LetAbe the matrix representation ofwith respect to the standard basisforM
22.
(a) FindanorderedbasisofM
22consisting of fundamental eigenvectors for, and the diagonal
matrixDthat is the matrix representation ofwith respect to.
(b) Calculate the transition matrixPfromto,andverifythat
D=P
−1
AP.
(9) Indicate whether each given statement is true or false.
(a) If:V→Vis an isomorphism on a nontrivialfinite dimensional vector spaceVandis an
eigenvalue for,then=±1.
(b) A linear operatoron an-dimensional vector space is diagonalizable if and only ifhas
distinct eigenvalues.
(10) Letbe the linear operator onR
3
representing a counterclockwise rotation through an angle of6
radians around an axis through the origin that is parallel to[4−23]. Explain in words whyis not
diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 207
Andrilli/Hecker - Chapter Tests Chapter 5 - Version C
Test for Chapter 5 – Version C
(1) LetAbe afixed×matrix. Prove that the function:P →M given by
(
+−1
−1
+···+ 1+ 0)
=
A
+−1A
−1
+···+ 1A+ 0I
is a linear transformation.
(2) Suppose:R
3
→M 22is a linear transformation such that([6−11]) =
∙
35−24
17 9
¸
([4−2−1]) =
∙
21−18
511
¸
and([−731]) =
∙
−38 31
−11−17
¸
Find([−31−4])
(3) Find the matrixA
for:R
3
−→R
4
given by
([ ]) = [3− −2+2−−−2]
for the ordered bases
=([−51−2][10−13][41−1220])forR
3
and
=([10−1−1][01−21][−1021][0−120])forR
4
(4) Consider:M
22→P 2given by
µ∙
1112
2122
¸¶
=(
11+22)
2
+( 11−21)+ 12
What isker()? What isrange()? Verify that the Dimension Theorem holds for.
(5) Suppose that
1:V→Wand 2:W→Yare linear transformations, whereVW,andYarefinite
dimensional vector spaces.
(a) Prove thatker(
1)⊆ker( 2◦1).
(b) Prove thatdim(range(
1))≥dim(range( 2◦1)).
(6) Prove that the mapping:P
2→R
4
given by
(
2
++)=[−5+−−7−2−−−2+5+]
is one-to-one. Isan isomorphism?
(7) Let=(v
1v )be an ordered basis for an-dimensional vector spaceV.Define:R
→Vby
([
1]) = 1v1+···+ v.Provethatis an isomorphism.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 208
Test for Chapter 5 – Version C
(1) LetAbe afixed×matrix. Prove that the function:P →M given by
(
+−1
−1
+···+ 1+ 0)
=
A
+−1A
−1
+···+ 1A+ 0I
is a linear transformation.
(2) Suppose:R
3
→M 22is a linear transformation such that([6−11]) =
∙
35−24
17 9
¸
([4−2−1]) =
∙
21−18
511
¸
and([−731]) =
∙
−38 31
−11−17
¸
Find([−31−4])
(3) Find the matrixA
for:R
3
−→R
4
given by
([ ]) = [3− −2+2−−−2]
for the ordered bases
=([−51−2][10−13][41−1220])forR
3
and
=([10−1−1][01−21][−1021][0−120])forR
4
(4) Consider:M
22→P 2given by
µ∙
1112
2122
¸¶
=(
11+22)
2
+( 11−21)+ 12
What isker()? What isrange()? Verify that the Dimension Theorem holds for.
(5) Suppose that
1:V→Wand 2:W→Yare linear transformations, whereVW,andYarefinite
dimensional vector spaces.
(a) Prove thatker(
1)⊆ker( 2◦1).
(b) Prove thatdim(range(
1))≥dim(range( 2◦1)).
(6) Prove that the mapping:P
2→R
4
given by
(
2
++)=[−5+−−7−2−−−2+5+]
is one-to-one. Isan isomorphism?
(7) Let=(v
1v )be an ordered basis for an-dimensional vector spaceV.Define:R
→Vby
([
1]) = 1v1+···+ v.Provethatis an isomorphism.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 208
Andrilli/Hecker - Chapter Tests Chapter 5 - Version C
(8) Consider the diagonalizable operator:P 2→P 2given by
(
2
++)=(3+−)
2
+(−2+)+(4+2−)
LetAbe the matrix representation ofwith respect to the standard basisforP
2.
(a)FindanorderedbasisofP
2consisting of fundamental eigenvectors for, and the diagonal
matrixDthat is the matrix representation ofwith respect to.
(b) Calculate the transition matrixPfromto,andverifythat
D=P
−1
AP.
(9) Indicate whether each given statement is true or false.
(a) Ifis a diagonalizable linear operator on a nontrivialfinite dimensional vector space, then the
algebraic multiplicity ofequals the geometric multiplicity offor every eigenvalueof.
(b) A linear operatoron a nontrivialfinite dimensional vector spaceVis one-to-one if and only if
0is not an eigenvalue for.
(10) Prove that the linear operator onP
,for0,defined by(p)=p
0
is not diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 209
(8) Consider the diagonalizable operator:P 2→P 2given by
(
2
++)=(3+−)
2
+(−2+)+(4+2−)
LetAbe the matrix representation ofwith respect to the standard basisforP
2.
(a)FindanorderedbasisofP
2consisting of fundamental eigenvectors for, and the diagonal
matrixDthat is the matrix representation ofwith respect to.
(b) Calculate the transition matrixPfromto,andverifythat
D=P
−1
AP.
(9) Indicate whether each given statement is true or false.
(a) Ifis a diagonalizable linear operator on a nontrivialfinite dimensional vector space, then the
algebraic multiplicity ofequals the geometric multiplicity offor every eigenvalueof.
(b) A linear operatoron a nontrivialfinite dimensional vector spaceVis one-to-one if and only if
0is not an eigenvalue for.
(10) Prove that the linear operator onP
,for0,defined by(p)=p
0
is not diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 209
Andrilli/Hecker - Chapter Tests Chapter 6 - Version A
Test for Chapter 6 – Version A
(1) Use the Gram-Schmidt Process to enlarge the following set to an orthogonal basis forR
3
:{[4−32]}.
(2) Let:R
3
→R
3
be the orthogonal reflection through the plane−2+2=0. Use eigenvalues and
eigenvectors tofind the matrix representation ofwith respect to the standard basis forR
3
.(Hint:
[1−22]is orthogonal to the given plane.)
(3) For the subspaceW=span({[2−31][−322]})ofR
3
,find a basis forW
⊥
.Whatisdim(W
⊥
)?
(Hint: The two given vectors arenotorthogonal.)
(4) SupposeAis an×orthogonal matrix. Prove that for everyx∈R
,kAxk=kxk.
(5) Prove the following part of Corollary 6.14: LetWbe a subspace ofR
.ThenW⊆(W
⊥
)
⊥
.
(6) Let:R
4
−→R
4
be the orthogonal projection onto the subspace
W= span({[21−3−1][5233]})ofR
4
. Find the characteristic polynomial of.
(7) Explain why:R
3
→R
3
given by the orthogonal projection onto the plane3+5−6=0is
orthogonally diagonalizable.
(8) Consider:R
3
→R
3
given by
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
54−2
45 2
−22 8
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Find an ordered orthonormal basisof fundamental eigenvectors for, and the diagonal matrixD
that is the matrix forwith respect to.
(9) Use orthogonal diagonalization tofind a symmetric matrixAsuch that
A
3
=
1
5
∙
31 18
18 4
¸
(10) Letbe a symmetric operator onR
and let 1and 2be distinct eigenvalues forwith corresponding
eigenvectorsv
1andv 2.Provethatv 1⊥v2.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 210
Test for Chapter 6 – Version A
(1) Use the Gram-Schmidt Process to enlarge the following set to an orthogonal basis forR
3
:{[4−32]}.
(2) Let:R
3
→R
3
be the orthogonal reflection through the plane−2+2=0. Use eigenvalues and
eigenvectors tofind the matrix representation ofwith respect to the standard basis forR
3
.(Hint:
[1−22]is orthogonal to the given plane.)
(3) For the subspaceW=span({[2−31][−322]})ofR
3
,find a basis forW
⊥
.Whatisdim(W
⊥
)?
(Hint: The two given vectors arenotorthogonal.)
(4) SupposeAis an×orthogonal matrix. Prove that for everyx∈R
,kAxk=kxk.
(5) Prove the following part of Corollary 6.14: LetWbe a subspace ofR
.ThenW⊆(W
⊥
)
⊥
.
(6) Let:R
4
−→R
4
be the orthogonal projection onto the subspace
W= span({[21−3−1][5233]})ofR
4
. Find the characteristic polynomial of.
(7) Explain why:R
3
→R
3
given by the orthogonal projection onto the plane3+5−6=0is
orthogonally diagonalizable.
(8) Consider:R
3
→R
3
given by
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
54−2
45 2
−22 8
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Find an ordered orthonormal basisof fundamental eigenvectors for, and the diagonal matrixD
that is the matrix forwith respect to.
(9) Use orthogonal diagonalization tofind a symmetric matrixAsuch that
A
3
=
1
5
∙
31 18
18 4
¸
(10) Letbe a symmetric operator onR
and let 1and 2be distinct eigenvalues forwith corresponding
eigenvectorsv
1andv 2.Provethatv 1⊥v2.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 210
Andrilli/Hecker - Chapter Tests Chapter 6 - Version B
Test for Chapter 6 – Version B
(1) Use the Gram-Schmidt Process tofind an orthogonal basis forR
3
, starting with the linearly independent
set{[2−31][4−2−1]}.(Hint:Thetwogivenvectorsarenotorthogonal.)
(2) Let:R
3
→R
3
be the orthogonal projection onto the plane3+4=0. Use eigenvalues and
eigenvectors tofind the matrix representation ofwith respect to the standard basis forR
3
.(Hint:
[304]is orthogonal to the given plane.)
(3) LetW=span({[22−3][30−1]})inR
3
.Decomposev=[−12−36−43]intow 1+w2,where
w
1∈Wandw 2∈W
⊥
. (Hint: The two given vectors spanningWarenotorthogonal.)
(4) Suppose thatAandBare×orthogonal matrices. Prove thatABis an orthogonal matrix.
(5) Prove the following part of Corollary 6.14: LetWbe a subspace ofR
. AssumingW⊆(W
⊥
)
⊥
has
already been shown, prove thatW=(W
⊥
)
⊥
.
(6) Letv=[−3510],andletWbethesubspaceofR
4
spanned by={[036−2][−6203]}.Find
proj
W
⊥v.
(7) Is:R
2
→R
2
given by
µ∙
1
2
¸¶
=
∙
37
7−5
¸∙
1
2
¸
orthogonally diagonalizable? Explain.
(8) Consider:R
3
→R
3
given by
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
40 18 6
18 13 −12
6−12 45
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Find an ordered orthonormal basisof fundamental eigenvectors for, and the diagonal matrixD
that is the matrix forwith respect to.
(9) Use orthogonal diagonalization tofind a symmetric matrixAsuch that
A
2
=
∙
10−6
−610
¸
(10) LetAbe an×symmetric matrix. Prove that if all eigenvalues forAare±1,thenAis orthogonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 211
Test for Chapter 6 – Version B
(1) Use the Gram-Schmidt Process tofind an orthogonal basis forR
3
, starting with the linearly independent
set{[2−31][4−2−1]}.(Hint:Thetwogivenvectorsarenotorthogonal.)
(2) Let:R
3
→R
3
be the orthogonal projection onto the plane3+4=0. Use eigenvalues and
eigenvectors tofind the matrix representation ofwith respect to the standard basis forR
3
.(Hint:
[304]is orthogonal to the given plane.)
(3) LetW=span({[22−3][30−1]})inR
3
.Decomposev=[−12−36−43]intow 1+w2,where
w
1∈Wandw 2∈W
⊥
. (Hint: The two given vectors spanningWarenotorthogonal.)
(4) Suppose thatAandBare×orthogonal matrices. Prove thatABis an orthogonal matrix.
(5) Prove the following part of Corollary 6.14: LetWbe a subspace ofR
. AssumingW⊆(W
⊥
)
⊥
has
already been shown, prove thatW=(W
⊥
)
⊥
.
(6) Letv=[−3510],andletWbethesubspaceofR
4
spanned by={[036−2][−6203]}.Find
proj
W
⊥v.
(7) Is:R
2
→R
2
given by
µ∙
1
2
¸¶
=
∙
37
7−5
¸∙
1
2
¸
orthogonally diagonalizable? Explain.
(8) Consider:R
3
→R
3
given by
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
40 18 6
18 13 −12
6−12 45
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Find an ordered orthonormal basisof fundamental eigenvectors for, and the diagonal matrixD
that is the matrix forwith respect to.
(9) Use orthogonal diagonalization tofind a symmetric matrixAsuch that
A
2
=
∙
10−6
−610
¸
(10) LetAbe an×symmetric matrix. Prove that if all eigenvalues forAare±1,thenAis orthogonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 211
Andrilli/Hecker - Chapter Tests Chapter 6 - Version C
Test for Chapter 6 – Version C
(1) Use the Gram-Schmidt Process to enlarge the following set to an orthogonal basis forR
4
:
{[−31−12][2401]}.
(2) Let:R
3
→R
3
be the orthogonal projection onto the plane2−2−=0. Use eigenvalues and
eigenvectors tofind the matrix representation ofwith respect to the standard basis forR
3
.(Hint:
[2−2−1]is orthogonal to the given plane.)
(3) For the subspaceW=span({[51−2][−8−29]})ofR
3
,find a basis forW
⊥
.Whatisdim(W
⊥
)?
(Hint: The two given vectors arenotorthogonal.)
(4) LetAbe an×orthogonal skew-symmetric matrix. Show thatis even. (Hint: Assumeis odd
and calculate|A
2
|to reach a contradiction.)
(5) Prove that ifW
1andW 2are two subspaces ofR
such thatW 1⊆W 2,thenW
⊥
2
⊆W
⊥
1
.
(6) Let=(510−4),andlet
W= span({[2102][−2012][022−1]})
Find the minimum distance fromtoW.
(7) Explain why the matrix for:R
3
→R
3
given by the orthogonal reflection through the plane4−
3−=0with respect to the standard basis forR
3
is symmetric.
(8) Consider:R
3
→R
3
given by
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
49−84−72
−84 23 −84
−72−84 49
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Find an ordered orthonormal basisof fundamental eigenvectors for, and the diagonal matrixD
that is the matrix forwith respect to.
(9) Use orthogonal diagonalization tofind a symmetric matrixAsuch that
A
2
=
∙
10 30
30 90
¸
(10) LetAandBbe×symmetric matrices such that
A()= B(). Prove that there is an orthogonal
matrixPsuch thatA=PBP
−1
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 212
Test for Chapter 6 – Version C
(1) Use the Gram-Schmidt Process to enlarge the following set to an orthogonal basis forR
4
:
{[−31−12][2401]}.
(2) Let:R
3
→R
3
be the orthogonal projection onto the plane2−2−=0. Use eigenvalues and
eigenvectors tofind the matrix representation ofwith respect to the standard basis forR
3
.(Hint:
[2−2−1]is orthogonal to the given plane.)
(3) For the subspaceW=span({[51−2][−8−29]})ofR
3
,find a basis forW
⊥
.Whatisdim(W
⊥
)?
(Hint: The two given vectors arenotorthogonal.)
(4) LetAbe an×orthogonal skew-symmetric matrix. Show thatis even. (Hint: Assumeis odd
and calculate|A
2
|to reach a contradiction.)
(5) Prove that ifW
1andW 2are two subspaces ofR
such thatW 1⊆W 2,thenW
⊥
2
⊆W
⊥
1
.
(6) Let=(510−4),andlet
W= span({[2102][−2012][022−1]})
Find the minimum distance fromtoW.
(7) Explain why the matrix for:R
3
→R
3
given by the orthogonal reflection through the plane4−
3−=0with respect to the standard basis forR
3
is symmetric.
(8) Consider:R
3
→R
3
given by
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
49−84−72
−84 23 −84
−72−84 49
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Find an ordered orthonormal basisof fundamental eigenvectors for, and the diagonal matrixD
that is the matrix forwith respect to.
(9) Use orthogonal diagonalization tofind a symmetric matrixAsuch that
A
2
=
∙
10 30
30 90
¸
(10) LetAandBbe×symmetric matrices such that
A()= B(). Prove that there is an orthogonal
matrixPsuch thatA=PBP
−1
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 212
Andrilli/Hecker - Chapter Tests Chapter 7 - Version A
Test for Chapter 7 – Version A
Note that there are more than 10 problems in this test, with at least three problems from
each section of Chapter 7.
(1) (Section 7.1) Verify thata·b=b·afor the complex vectorsa=[22+]andb=[1+1−2+].
(2) (Section 7.1) SupposeHandPare×complex matrices and thatHis Hermitian. Prove that
P
∗
HPis Hermitian.
(3) (Section 7.1) Indicate whether each given statement is true or false.
(a) IfZandWare×complex matrices, thenZW=
(WZ).
(b) If an×complex matrix is normal, then it is either Hermitian or skew-Hermitian.
(4) (Section 7.2) Use Gaussian Elimination to solve the following system of linear equations:
½
1+(1+3) 2=6+2
(2 + 3)
1+(7+7) 2=20−2
(5) (Section 7.2) Give an example of a matrix having all real entries that is not diagonalizable when
thought of as a real matrix, but is diagonalizable when thought of as a complex matrix. Prove that
your example works.
(6) (Section 7.2) Find all eigenvalues and a basis of fundamental eigenvectors for each eigenspace for
:C
3
→C
3
given by
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
10 −1−
−1
−0−1+
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Isdiagonalizable?
(7) (Section 7.3) Prove or disprove: For≥1, the set of polynomials inP
C
with real coefficients is a
complex subspace ofP
C
.
(8) (Section 7.3) Find a basisforspan()that is a subset of
={[1 + 32+1−2][−2+41+33−][123+][5 + 63−1−2]}
(9) (Section 7.3) Find the matrix with respect to the standard bases for the complex linear transformation
:M
C
22
→C
2
given by
µ∙
¸¶
=[+ +]
Also,find the matrix forwith respect to the standard bases when thought of as a linear transformation
between real vector spaces.
(10) (Section 7.4) Use the Gram-Schmidt Process tofind an orthogonal basis for the complex vector space
C
3
starting with the linearly independent set{[11+][3−−1−]}.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 213
Test for Chapter 7 – Version A
Note that there are more than 10 problems in this test, with at least three problems from
each section of Chapter 7.
(1) (Section 7.1) Verify thata·b=b·afor the complex vectorsa=[22+]andb=[1+1−2+].
(2) (Section 7.1) SupposeHandPare×complex matrices and thatHis Hermitian. Prove that
P
∗
HPis Hermitian.
(3) (Section 7.1) Indicate whether each given statement is true or false.
(a) IfZandWare×complex matrices, thenZW=
(WZ).
(b) If an×complex matrix is normal, then it is either Hermitian or skew-Hermitian.
(4) (Section 7.2) Use Gaussian Elimination to solve the following system of linear equations:
½
1+(1+3) 2=6+2
(2 + 3)
1+(7+7) 2=20−2
(5) (Section 7.2) Give an example of a matrix having all real entries that is not diagonalizable when
thought of as a real matrix, but is diagonalizable when thought of as a complex matrix. Prove that
your example works.
(6) (Section 7.2) Find all eigenvalues and a basis of fundamental eigenvectors for each eigenspace for
:C
3
→C
3
given by
⎛
⎝
⎡
⎣
1
2
3
⎤
⎦
⎞
⎠=
⎡
⎣
10 −1−
−1
−0−1+
⎤
⎦
⎡
⎣
1
2
3
⎤
⎦
Isdiagonalizable?
(7) (Section 7.3) Prove or disprove: For≥1, the set of polynomials inP
C
with real coefficients is a
complex subspace ofP
C
.
(8) (Section 7.3) Find a basisforspan()that is a subset of
={[1 + 32+1−2][−2+41+33−][123+][5 + 63−1−2]}
(9) (Section 7.3) Find the matrix with respect to the standard bases for the complex linear transformation
:M
C
22
→C
2
given by
µ∙
¸¶
=[+ +]
Also,find the matrix forwith respect to the standard bases when thought of as a linear transformation
between real vector spaces.
(10) (Section 7.4) Use the Gram-Schmidt Process tofind an orthogonal basis for the complex vector space
C
3
starting with the linearly independent set{[11+][3−−1−]}.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 213
Andrilli/Hecker - Chapter Tests Chapter 7 - Version A
(11) (Section 7.4) Find a unitary matrixPsuch that
P
∗
∙
0
0
¸
P
is diagonal.
(12) (Section 7.4) Show that ifAandBare×unitary matrices, thenABis unitary and
¯
¯
¯|A|
¯
¯
¯=1.
(13) (Section 7.5) Prove that
hfgi=f(0)g(0) +f(1)g(1) +f(2)g(2)
is a real inner product onP
2.Forf()=2+1andg()=
2
−1,calculatehfgiandkfk.
(14) (Section 7.5) Find the distance betweenx=[412]andy=[335]in the real inner product space
consisting ofR
3
with inner product
hxyi=Ax·AywhereA=
⎡
⎣
40−1
33 1
12 1
⎤
⎦
(15) (Section 7.5) Prove that in a real inner product space, for any vectorsxandy,
kxk=kykif and only ifhx+yx−yi=0
(16) (Section 7.5) Use the Generalized Gram-Schmidt Process tofind an orthogonal basis forP
2starting
with the linearly independent set{
2
}using the real inner product given by
hfgi=
Z
1
−1
()()
(17) (Section 7.5) Decomposev=[01−3]inR
3
asw 1+w2,where
w
1∈W=span({[21−4][43−8]})andw 2∈W
⊥
, using the real inner product onR
3
given by
hxyi=Ax·Ay,with
A=
⎡
⎣
3−11
031
121
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 214
(11) (Section 7.4) Find a unitary matrixPsuch that
P
∗
∙
0
0
¸
P
is diagonal.
(12) (Section 7.4) Show that ifAandBare×unitary matrices, thenABis unitary and
¯
¯
¯|A|
¯
¯
¯=1.
(13) (Section 7.5) Prove that
hfgi=f(0)g(0) +f(1)g(1) +f(2)g(2)
is a real inner product onP
2.Forf()=2+1andg()=
2
−1,calculatehfgiandkfk.
(14) (Section 7.5) Find the distance betweenx=[412]andy=[335]in the real inner product space
consisting ofR
3
with inner product
hxyi=Ax·AywhereA=
⎡
⎣
40−1
33 1
12 1
⎤
⎦
(15) (Section 7.5) Prove that in a real inner product space, for any vectorsxandy,
kxk=kykif and only ifhx+yx−yi=0
(16) (Section 7.5) Use the Generalized Gram-Schmidt Process tofind an orthogonal basis forP
2starting
with the linearly independent set{
2
}using the real inner product given by
hfgi=
Z
1
−1
()()
(17) (Section 7.5) Decomposev=[01−3]inR
3
asw 1+w2,where
w
1∈W=span({[21−4][43−8]})andw 2∈W
⊥
, using the real inner product onR
3
given by
hxyi=Ax·Ay,with
A=
⎡
⎣
3−11
031
121
⎤
⎦
Copyrightc°2016 Elsevier Ltd. All rights reserved. 214
Andrilli/Hecker - Chapter Tests Chapter 7 - Version B
Test for Chapter 7 – Version B
Note that there are more than 10 problems in this test, with at least three problems from
each section of Chapter 7.
(1) (Section 7.1) CalculateA
∗
Bfor the complex matrices
A=
∙
3+2
5−2
¸
andB=
∙
1−3−2
3 7
¸
(2) (Section 7.1) Prove that ifHis a Hermitian matrix, thenHis a skew-Hermitian matrix.
(3) (Section 7.1) Indicate whether each given statement is true or false.
(a) Every skew-Hermitian matrix must have all zeroes on its main diagonal.
(b) Ifx,y∈C
,buteachentryofxandyis real, thenx·yproduces the same answer if the dot
product is thought of as taking place inC
as it would if the dot product were considered to be
taking place inR
.
(4) (Section 7.2) Use the an inverse matrix to solve the given linear system:
½
(4 +)
1+(2+14) 2=21+4
(2−)
1+(6+5 ) 2=10−6
(5) (Section 7.2) Explain why the sum of the algebraic multiplicities of a complex×matrix must
equal.
(6) (Section 7.2) Find all eigenvalues and a basis of fundamental eigenvectors for each eigenspace for
:C
2
→C
2
given by
µ∙
1
2
¸¶
=
∙
−1
43
¸∙
1
2
¸
Isdiagonalizable?
(7) (Section 7.3) Prove or disprove: The set of×Hermitian matrices is a complex subspace ofM
C
.
If it is a subspace, compute its dimension.
(8) (Section 7.3) If
={[−1−2+][2−2+25+2][022−2][1 + 3−2+2−3+5]}
find a basisforspan()that uses vectors having a simpler form than those in.
(9) (Section 7.3) Show that:M
C
22
→M
C
22
given by(Z)=Z
∗
is not a complex linear transformation.
(10) (Section 7.4) Use the Gram-Schmidt Process tofind an orthogonal basis for the complex vector space
C
3
containing{[1 +1−2]}.
(11) (Section 7.4) Prove thatZ=
∙
1+−1+
1−1+
¸
is unitarily diagonalizable.
(12) (Section 7.4) LetAbe an×unitary matrix. Prove thatA
2
=Iif and only ifAis Hermitian.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 215
Test for Chapter 7 – Version B
Note that there are more than 10 problems in this test, with at least three problems from
each section of Chapter 7.
(1) (Section 7.1) CalculateA
∗
Bfor the complex matrices
A=
∙
3+2
5−2
¸
andB=
∙
1−3−2
3 7
¸
(2) (Section 7.1) Prove that ifHis a Hermitian matrix, thenHis a skew-Hermitian matrix.
(3) (Section 7.1) Indicate whether each given statement is true or false.
(a) Every skew-Hermitian matrix must have all zeroes on its main diagonal.
(b) Ifx,y∈C
,buteachentryofxandyis real, thenx·yproduces the same answer if the dot
product is thought of as taking place inC
as it would if the dot product were considered to be
taking place inR
.
(4) (Section 7.2) Use the an inverse matrix to solve the given linear system:
½
(4 +)
1+(2+14) 2=21+4
(2−)
1+(6+5 ) 2=10−6
(5) (Section 7.2) Explain why the sum of the algebraic multiplicities of a complex×matrix must
equal.
(6) (Section 7.2) Find all eigenvalues and a basis of fundamental eigenvectors for each eigenspace for
:C
2
→C
2
given by
µ∙
1
2
¸¶
=
∙
−1
43
¸∙
1
2
¸
Isdiagonalizable?
(7) (Section 7.3) Prove or disprove: The set of×Hermitian matrices is a complex subspace ofM
C
.
If it is a subspace, compute its dimension.
(8) (Section 7.3) If
={[−1−2+][2−2+25+2][022−2][1 + 3−2+2−3+5]}
find a basisforspan()that uses vectors having a simpler form than those in.
(9) (Section 7.3) Show that:M
C
22
→M
C
22
given by(Z)=Z
∗
is not a complex linear transformation.
(10) (Section 7.4) Use the Gram-Schmidt Process tofind an orthogonal basis for the complex vector space
C
3
containing{[1 +1−2]}.
(11) (Section 7.4) Prove thatZ=
∙
1+−1+
1−1+
¸
is unitarily diagonalizable.
(12) (Section 7.4) LetAbe an×unitary matrix. Prove thatA
2
=Iif and only ifAis Hermitian.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 215
Andrilli/Hecker - Chapter Tests Chapter 7 - Version B
(13) (Section 7.5) LetAbe the3×3nonsingular matrix
⎡
⎣
431
−143
−111
⎤
⎦
Prove thathxyi=Ax·Ayis a real inner product onR
3
.Forx=[2−11]andy=[142],calculate
hxyiandkxkfor this inner product.
(14) (Section 7.5) Find the distance betweenw=[4+54+2+2]andz=[1+2−22+3]in the
complex inner product spaceC
3
using the usual complex dot product as an inner product.
(15) (Section 7.5) Prove that in a real inner product space, for any vectorsxandy,
hxyi=
1
4
(kx+yk
2
−kx−yk
2
)
(16) (Section 7.5) Use the Generalized Gram-Schmidt Process tofind an orthogonal basis forP
2containing
2
using the real inner product given by
hfgi=f(0)g(0) +f(1)g(1) +f(2)g(2)
(17) (Section 7.5) Decomposev=+3inP
2asw 1+w2,wherew 1∈W=span({
2
})andw 2∈W
⊥
,
using the real inner product onP
2given by
hfgi=
Z
1
0
()()
Copyrightc°2016 Elsevier Ltd. All rights reserved. 216
(13) (Section 7.5) LetAbe the3×3nonsingular matrix
⎡
⎣
431
−143
−111
⎤
⎦
Prove thathxyi=Ax·Ayis a real inner product onR
3
.Forx=[2−11]andy=[142],calculate
hxyiandkxkfor this inner product.
(14) (Section 7.5) Find the distance betweenw=[4+54+2+2]andz=[1+2−22+3]in the
complex inner product spaceC
3
using the usual complex dot product as an inner product.
(15) (Section 7.5) Prove that in a real inner product space, for any vectorsxandy,
hxyi=
1
4
(kx+yk
2
−kx−yk
2
)
(16) (Section 7.5) Use the Generalized Gram-Schmidt Process tofind an orthogonal basis forP
2containing
2
using the real inner product given by
hfgi=f(0)g(0) +f(1)g(1) +f(2)g(2)
(17) (Section 7.5) Decomposev=+3inP
2asw 1+w2,wherew 1∈W=span({
2
})andw 2∈W
⊥
,
using the real inner product onP
2given by
hfgi=
Z
1
0
()()
Copyrightc°2016 Elsevier Ltd. All rights reserved. 216
Andrilli/Hecker - Chapter Tests Chapter 7 - Version C
Test for Chapter 7 – Version C
Note that there are more than 10 problems in this test, with at least three problems from
each section of Chapter 7.
(1) (Section 7.1) CalculateA
B
∗
for the complex matrices
A=
∙
23 −
1+2−
¸
andB=
∙
4+3
21−2
¸
(2) (Section 7.1) LetHandJbe×Hermitian matrices. Prove that ifHJis Hermitian, thenHJ=JH.
(3) (Section 7.1) Indicate whether each given statement is true or false.
(a) Ifx,y∈C
withx·y∈R,thenx·y=y·x.
(b) IfAis an×Hermitian matrix, andxy∈C
,thenAx·y=x·Ay.
(4) (Section 7.2) Use Cramer’s Rule to solve the following system of linear equations:
½
−
1+(1−) 2=−3−2
(1−2)
1+(4−) 2=−5−6
(5) (Section 7.2) Show that a complex×matrix that is not diagonalizable must have an eigenvalue
whose algebraic multiplicity is strictly greater than its geometric multiplicity.
(6) (Section 7.2) Find all eigenvalues and a basis of fundamental eigenvectors for each eigenspace for
:C
2
−→C
2
given by
µ∙
1
2
¸¶
=
∙
02
−22+2
¸∙
1
2
¸
Isdiagonalizable?
(7) (Section 7.3) Prove that the set of normal2×2matrices is not a complex subspace ofM
C
22
by showing
that it is not closed under matrix addition.
(8) (Section 7.3) Find a basisforC
4
that contains
{[13+ 1−][1 +3+4−1+2]}
(9) (Section 7.3) Letw∈C
be afixed nonzero vector. Show that:C
→Cgiven by(z)=w·zis
not a complex linear transformation.
(10) (Section 7.4) Use the Gram-Schmidt Process tofind an orthogonal basis for the complex vector space
C
3
, starting with the basis{[2−1+21][010][−102+]}.
(11) (Section 7.4) LetZbe an×complex matrix whose rows form an orthonormal basis forC
.Prove
that the columns ofZalso form an orthonormal basis forC
Copyrightc°2016 Elsevier Ltd. All rights reserved. 217
Test for Chapter 7 – Version C
Note that there are more than 10 problems in this test, with at least three problems from
each section of Chapter 7.
(1) (Section 7.1) CalculateA
B
∗
for the complex matrices
A=
∙
23 −
1+2−
¸
andB=
∙
4+3
21−2
¸
(2) (Section 7.1) LetHandJbe×Hermitian matrices. Prove that ifHJis Hermitian, thenHJ=JH.
(3) (Section 7.1) Indicate whether each given statement is true or false.
(a) Ifx,y∈C
withx·y∈R,thenx·y=y·x.
(b) IfAis an×Hermitian matrix, andxy∈C
,thenAx·y=x·Ay.
(4) (Section 7.2) Use Cramer’s Rule to solve the following system of linear equations:
½
−
1+(1−) 2=−3−2
(1−2)
1+(4−) 2=−5−6
(5) (Section 7.2) Show that a complex×matrix that is not diagonalizable must have an eigenvalue
whose algebraic multiplicity is strictly greater than its geometric multiplicity.
(6) (Section 7.2) Find all eigenvalues and a basis of fundamental eigenvectors for each eigenspace for
:C
2
−→C
2
given by
µ∙
1
2
¸¶
=
∙
02
−22+2
¸∙
1
2
¸
Isdiagonalizable?
(7) (Section 7.3) Prove that the set of normal2×2matrices is not a complex subspace ofM
C
22
by showing
that it is not closed under matrix addition.
(8) (Section 7.3) Find a basisforC
4
that contains
{[13+ 1−][1 +3+4−1+2]}
(9) (Section 7.3) Letw∈C
be afixed nonzero vector. Show that:C
→Cgiven by(z)=w·zis
not a complex linear transformation.
(10) (Section 7.4) Use the Gram-Schmidt Process tofind an orthogonal basis for the complex vector space
C
3
, starting with the basis{[2−1+21][010][−102+]}.
(11) (Section 7.4) LetZbe an×complex matrix whose rows form an orthonormal basis forC
.Prove
that the columns ofZalso form an orthonormal basis forC
Copyrightc°2016 Elsevier Ltd. All rights reserved. 217
Andrilli/Hecker - Chapter Tests Chapter 7 - Version C
(12) (Section 7.4)
(a) Show thatA=
∙
−2336
−36−2
¸
is normal, and hence unitarily diagonalizable.
(b) Find a unitary matrixPsuch thatP
−1
APis diagonal.
(13) (Section 7.5) Forx=[
12]andy=[ 12]inR
2
,provethathxyi=2 11−12−21+22
is a real inner product onR
2
.Forx=[23]andy=[4−4],calculatehxyiandkxkfor this inner
product.
(14) (Section 7.5) Find the distance betweenf()=sin
2
andg()=−cos
2
in the real inner product
space consisting of the set of all real-valued continuous functions definedontheinterval[0]with
inner product
hfgi=
Z
0
()()
(15) (Section 7.5) Prove that in a real inner product space, vectorsxandyare orthogonal if and only if
kx+yk
2
=kxk
2
+kyk
2
.
(16) (Section 7.5) Use the Generalized Gram-Schmidt Process tofind an orthogonal basis forR
3
,starting
with the linearly independent set{[−111][3−4−1]}using the real inner product given by
hxyi=Ax·AywhereA=
⎡
⎣
321
212
211
⎤
⎦
(17) (Section 7.5) Decomposev=
2
inP 2asw 1+w2,wherew 1∈W=span({−56−5})and
w
2∈W
⊥
, using the real inner product onP 2given by
hfgi=f(0)g(0) +f(1)g(1) +f(2)g(2)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 218
(12) (Section 7.4)
(a) Show thatA=
∙
−2336
−36−2
¸
is normal, and hence unitarily diagonalizable.
(b) Find a unitary matrixPsuch thatP
−1
APis diagonal.
(13) (Section 7.5) Forx=[
12]andy=[ 12]inR
2
,provethathxyi=2 11−12−21+22
is a real inner product onR
2
.Forx=[23]andy=[4−4],calculatehxyiandkxkfor this inner
product.
(14) (Section 7.5) Find the distance betweenf()=sin
2
andg()=−cos
2
in the real inner product
space consisting of the set of all real-valued continuous functions definedontheinterval[0]with
inner product
hfgi=
Z
0
()()
(15) (Section 7.5) Prove that in a real inner product space, vectorsxandyare orthogonal if and only if
kx+yk
2
=kxk
2
+kyk
2
.
(16) (Section 7.5) Use the Generalized Gram-Schmidt Process tofind an orthogonal basis forR
3
,starting
with the linearly independent set{[−111][3−4−1]}using the real inner product given by
hxyi=Ax·AywhereA=
⎡
⎣
321
212
211
⎤
⎦
(17) (Section 7.5) Decomposev=
2
inP 2asw 1+w2,wherew 1∈W=span({−56−5})and
w
2∈W
⊥
, using the real inner product onP 2given by
hfgi=f(0)g(0) +f(1)g(1) +f(2)g(2)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 218
Andrilli/Hecker - Answers to Chapter Tests Chapter 1 - Version A
Answers to Test for Chapter 1 – Version A
(1) The net velocity vector is[
5
√
2
2
2−
5
√
2
2
]≈[354−154]km/hr. The net speed is approximately385
km/hr.
(2) The angle (to the nearest degree) is137
◦
.
(3)kx+yk
2
+kx−yk
2
=(x+y)·(x+y)+(x−y)·(x−y)
=x·(x+y)+y·(x+y)+x·(x−y)−y·(x−y)
=x·x+x·y+y·x+y·y+x·x−x·y−y·x+y·y
=2(x·x)+2(y·y)=2
³
kxk
2
+kyk
2
´
(4) The vectorx=[−342]can be expressed as[−
65
35
13
35
−
39
35
]+[−
40
35
127
35
109
35
],wherethefirst vector
(proj
ax)of the sum is parallel toa(since it equals−
13
35
a), and the second vector(x−proj ax)of the
sum is easily seen to be orthogonal toa.
(5) Contrapositive: Ifxis parallel toy,thenkx+yk=kxk+kyk.
Converse: Ifxis not parallel toy,thenkx+yk6 =kxk+kyk.
Inverse: Ifkx+yk=kxk+kyk,thenxis parallel toy.
Only the contrapositive is logically equivalent to the original statement.
(6) The negation is:kxk6 =kykand(x−y)·(x+y)=0.
(7) Base Step: AssumeAandBare any two diagonal×matrices. Then,C=A+Bis also diagonal
because for
6 =, =+=0+0=0 ,sinceAandBare both diagonal.
Inductive Step: Assume that the sum of anydiagonal×matrices is diagonal. We must show for any
diagonal×matricesA
1A +1that
P
+1
=1
Ais diagonal. Now,
P
+1
=1
A=(
P
=1
A)+A +1.
LetB=
P
=1
A.ThenBis diagonal by the inductive hypothesis. Hence,
P
+1
=1
A=B+A +1is
the sum of two diagonal matrices, and so is diagonal by the Base Step.
(8)A=S+V,whereS=
⎡
⎢
⎢
⎣
−4
9
2
0
9
2
−1
11
2
0
11
2
−3
⎤
⎥
⎥
⎦
,V=
⎡
⎢
⎢
⎣
0−
3
2
−2
3
2
0
3
2
2−
3
2
0
⎤
⎥
⎥
⎦
,Sis symmetric, andVis skew-
symmetric.
(9)
TV Show1
TV Show2
TV Show3
TV Show4
Salary Perks
⎡
⎢
⎢
⎣
$1000000 $650000
$860000 $535000
$880000 $455000
$770000 $505000
⎤
⎥
⎥
⎦
(10) AssumeAandBare both×symmetric matrices. ThenABis symmetric if and only if(AB)
=AB
if and only ifB
A
=AB(by Theorem 1.18) if and only ifBA=AB(sinceAandBare symmetric)
if and only ifAandBcommute.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 219
Answers to Test for Chapter 1 – Version A
(1) The net velocity vector is[
5
√
2
2
2−
5
√
2
2
]≈[354−154]km/hr. The net speed is approximately385
km/hr.
(2) The angle (to the nearest degree) is137
◦
.
(3)kx+yk
2
+kx−yk
2
=(x+y)·(x+y)+(x−y)·(x−y)
=x·(x+y)+y·(x+y)+x·(x−y)−y·(x−y)
=x·x+x·y+y·x+y·y+x·x−x·y−y·x+y·y
=2(x·x)+2(y·y)=2
³
kxk
2
+kyk
2
´
(4) The vectorx=[−342]can be expressed as[−
65
35
13
35
−
39
35
]+[−
40
35
127
35
109
35
],wherethefirst vector
(proj
ax)of the sum is parallel toa(since it equals−
13
35
a), and the second vector(x−proj ax)of the
sum is easily seen to be orthogonal toa.
(5) Contrapositive: Ifxis parallel toy,thenkx+yk=kxk+kyk.
Converse: Ifxis not parallel toy,thenkx+yk6 =kxk+kyk.
Inverse: Ifkx+yk=kxk+kyk,thenxis parallel toy.
Only the contrapositive is logically equivalent to the original statement.
(6) The negation is:kxk6 =kykand(x−y)·(x+y)=0.
(7) Base Step: AssumeAandBare any two diagonal×matrices. Then,C=A+Bis also diagonal
because for
6 =, =+=0+0=0 ,sinceAandBare both diagonal.
Inductive Step: Assume that the sum of anydiagonal×matrices is diagonal. We must show for any
diagonal×matricesA
1A +1that
P
+1
=1
Ais diagonal. Now,
P
+1
=1
A=(
P
=1
A)+A +1.
LetB=
P
=1
A.ThenBis diagonal by the inductive hypothesis. Hence,
P
+1
=1
A=B+A +1is
the sum of two diagonal matrices, and so is diagonal by the Base Step.
(8)A=S+V,whereS=
⎡
⎢
⎢
⎣
−4
9
2
0
9
2
−1
11
2
0
11
2
−3
⎤
⎥
⎥
⎦
,V=
⎡
⎢
⎢
⎣
0−
3
2
−2
3
2
0
3
2
2−
3
2
0
⎤
⎥
⎥
⎦
,Sis symmetric, andVis skew-
symmetric.
(9)
TV Show1
TV Show2
TV Show3
TV Show4
Salary Perks
⎡
⎢
⎢
⎣
$1000000 $650000
$860000 $535000
$880000 $455000
$770000 $505000
⎤
⎥
⎥
⎦
(10) AssumeAandBare both×symmetric matrices. ThenABis symmetric if and only if(AB)
=AB
if and only ifB
A
=AB(by Theorem 1.18) if and only ifBA=AB(sinceAandBare symmetric)
if and only ifAandBcommute.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 219
Andrilli/Hecker - Answers to Chapter Tests Chapter 1 - Version B
Answers to Test for Chapter 1 – Version B
(1) Player B is pulling with a force of100
√
3lbs, or about57735lbs. Player C is pulling with a force of
200
√
3lbs, or about11547lbs.
(2) The angle (to the nearest degree) is152
◦
.
(3) See the proof in the textbook directly before Theorem 1.11 in Section 1.2.
(4) The vectorx=[5−2−4]can be expressed as[
8
22
−
12
22
12
22
]+[
102
22
−
32
22
−
100
22
],wherethefirst vector
(proj
ax)of the sum is parallel toa(since it equals−
4
22
a), and the second vector(x−proj
ax)
of the sum is easily seen to be orthogonal toa.
(5) Contrapositive: Ifkx+yk≤kyk,thenkxk
2
+2(x·y)≤0.
Converse: Ifkx+ykkyk,thenkxk
2
+2(x·y)0.
Inverse: Ifkxk
2
+2(x·y)≤0,thenkx+yk≤kyk.
Only the contrapositive is logically equivalent to the original statement.
(6) The negation is: No unit vectorx∈R
3
is parallel to[−231].
(7) Base Step: AssumeAandBare any two lower triangular×matrices. Then,C=A+Bis also
lower triangular because for,
=+=0+0=0,sinceAandBare both lower triangular.
Inductive Step: Assume that the sum of anylower triangular×matrices is lower triangular. We
mustshowforanylowertriangular×matricesA
1A +1that
P
+1
=1
Ais lower triangular.
Now,
P
+1
=1
A=(
P
=1
A)+A +1.LetB=
P
=1
A.ThenBis lower triangular by the inductive
hypothesis. Hence,
P
+1
=1
A=B+A +1is the sum of two lower triangular matrices, and so is lower
triangular by the Base Step.
(8)A=S+V,whereS=
⎡
⎢
⎢
⎣
51
7
2
13
1
2
7
2
1
2
1
⎤
⎥
⎥
⎦
V=
⎡
⎢
⎢
⎣
07 −
11
2
−70 −
7
2
11
2
7
2
0
⎤
⎥
⎥
⎦
Sis symmetric, andVis skew-
symmetric.
(9)
Cat 1
Cat 2
Cat 3
Nutrient 1 Nutrient 2 Nutrient 3
⎡
⎣
164 0 62 1 86
150 0 53 1 46
155 0 52 1 54
⎤
⎦(Allfigures are in ounces.)
(10) AssumeAandBare both×skew-symmetric matrices. Then(AB)
=B
A
(by Theorem 1.18)
=(−B)(−A)(sinceA,Bare skew-symmetric)=BA(by part (4) of Theorem 1.16).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 220
Answers to Test for Chapter 1 – Version B
(1) Player B is pulling with a force of100
√
3lbs, or about57735lbs. Player C is pulling with a force of
200
√
3lbs, or about11547lbs.
(2) The angle (to the nearest degree) is152
◦
.
(3) See the proof in the textbook directly before Theorem 1.11 in Section 1.2.
(4) The vectorx=[5−2−4]can be expressed as[
8
22
−
12
22
12
22
]+[
102
22
−
32
22
−
100
22
],wherethefirst vector
(proj
ax)of the sum is parallel toa(since it equals−
4
22
a), and the second vector(x−proj
ax)
of the sum is easily seen to be orthogonal toa.
(5) Contrapositive: Ifkx+yk≤kyk,thenkxk
2
+2(x·y)≤0.
Converse: Ifkx+ykkyk,thenkxk
2
+2(x·y)0.
Inverse: Ifkxk
2
+2(x·y)≤0,thenkx+yk≤kyk.
Only the contrapositive is logically equivalent to the original statement.
(6) The negation is: No unit vectorx∈R
3
is parallel to[−231].
(7) Base Step: AssumeAandBare any two lower triangular×matrices. Then,C=A+Bis also
lower triangular because for,
=+=0+0=0,sinceAandBare both lower triangular.
Inductive Step: Assume that the sum of anylower triangular×matrices is lower triangular. We
mustshowforanylowertriangular×matricesA
1A +1that
P
+1
=1
Ais lower triangular.
Now,
P
+1
=1
A=(
P
=1
A)+A +1.LetB=
P
=1
A.ThenBis lower triangular by the inductive
hypothesis. Hence,
P
+1
=1
A=B+A +1is the sum of two lower triangular matrices, and so is lower
triangular by the Base Step.
(8)A=S+V,whereS=
⎡
⎢
⎢
⎣
51
7
2
13
1
2
7
2
1
2
1
⎤
⎥
⎥
⎦
V=
⎡
⎢
⎢
⎣
07 −
11
2
−70 −
7
2
11
2
7
2
0
⎤
⎥
⎥
⎦
Sis symmetric, andVis skew-
symmetric.
(9)
Cat 1
Cat 2
Cat 3
Nutrient 1 Nutrient 2 Nutrient 3
⎡
⎣
164 0 62 1 86
150 0 53 1 46
155 0 52 1 54
⎤
⎦(Allfigures are in ounces.)
(10) AssumeAandBare both×skew-symmetric matrices. Then(AB)
=B
A
(by Theorem 1.18)
=(−B)(−A)(sinceA,Bare skew-symmetric)=BA(by part (4) of Theorem 1.16).
Copyrightc°2016 Elsevier Ltd. All rights reserved. 220
Andrilli/Hecker - Answers to Chapter Tests Chapter 1 - Version C
Answers to Test for Chapter 1 – Version C
(1) The acceleration vector is approximately[−04350870−0223]m/sec
2
. This can also be expressed
as1202[−03620724−0186]m/sec
2
, where the latter vector is a unit vector.
(2) The angle (to the nearest degree) is118
◦
.
(3) See the proof of Theorem 1.8 in Section 1.2 of the textbook.
(4) The vectorx=[−42−7]can be expressed as
£
−
246
62
205
62
−
41
62
¤
+
£
−
262
−
81
62
−
393
62
¤
where thefirst
vector(proj
ax)of the sum is parallel toa(since it equals−
41
62
a) and the second vector(x−proj ax)
of the sum is easily seen to be orthogonal toa.
(5) Assumey=x,forsome∈R.Wemustprovethaty=proj
xy.Now,ify=x,then
proj
xy=
³
x·y
kxk
2
´
x=
³
x·(x)
kxk
2
´
x=
³
(x·x)
kxk
2
´
x(by part (4) of Theorem 1.5)=x(by part (2) of
Theorem 1.5)=y.
(6) The negation is: For some vectorx∈R
, there is no vectory∈R
such thatkykkxk.
(7)A=S+V,whereS=
⎡
⎢
⎢
⎣
−3
13
2
1
2
13
2
53
1
2
3−4
⎤
⎥
⎥
⎦
V=
⎡
⎢
⎢
⎣
0−
1
2
5
2
1
2
0−1
−
5
2
10
⎤
⎥
⎥
⎦
Sis symmetric, andVis
skew-symmetric.
(8) The third row ofABis[3518−13], and the second column ofBAis[−325242−107].
(9) Base Step: SupposeAandBare two×diagonal matrices. LetC=AB. Then,
=
P
=1
.
But if6 =,then
=0.Thus, =. Then, if6 =, =0, hence =0=0. HenceCis
diagonal.
Inductive Step: Assume that the product of anydiagonal×matrices is diagonal. We must show
that for any diagonal×matricesA
1A +1that the productA 1···A +1is diagonal. Now,
A
1···A +1=(A 1···A )A+1.LetB=A 1···A .ThenBis diagonal by the inductive hypothesis.
Hence,A
1···A A+1=BA +1is the product of two diagonal matrices, and so is diagonal by the
Base Step.
(10) Assume thatAis skew-symmetric. Then(A
3
)
=(A
2
A)
=A
(A
2
)
(by Theorem 1.18)=
A
(AA)
=A
A
A
(again by Theorem 1.18)=(−A)(−A)(−A)(sinceAis skew-symmetric)
=−(A
3
)(by part (4) of Theorem 1.16). Hence,A
3
is skew-symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 221
Answers to Test for Chapter 1 – Version C
(1) The acceleration vector is approximately[−04350870−0223]m/sec
2
. This can also be expressed
as1202[−03620724−0186]m/sec
2
, where the latter vector is a unit vector.
(2) The angle (to the nearest degree) is118
◦
.
(3) See the proof of Theorem 1.8 in Section 1.2 of the textbook.
(4) The vectorx=[−42−7]can be expressed as
£
−
246
62
205
62
−
41
62
¤
+
£
−
262
−
81
62
−
393
62
¤
where thefirst
vector(proj
ax)of the sum is parallel toa(since it equals−
41
62
a) and the second vector(x−proj ax)
of the sum is easily seen to be orthogonal toa.
(5) Assumey=x,forsome∈R.Wemustprovethaty=proj
xy.Now,ify=x,then
proj
xy=
³
x·y
kxk
2
´
x=
³
x·(x)
kxk
2
´
x=
³
(x·x)
kxk
2
´
x(by part (4) of Theorem 1.5)=x(by part (2) of
Theorem 1.5)=y.
(6) The negation is: For some vectorx∈R
, there is no vectory∈R
such thatkykkxk.
(7)A=S+V,whereS=
⎡
⎢
⎢
⎣
−3
13
2
1
2
13
2
53
1
2
3−4
⎤
⎥
⎥
⎦
V=
⎡
⎢
⎢
⎣
0−
1
2
5
2
1
2
0−1
−
5
2
10
⎤
⎥
⎥
⎦
Sis symmetric, andVis
skew-symmetric.
(8) The third row ofABis[3518−13], and the second column ofBAis[−325242−107].
(9) Base Step: SupposeAandBare two×diagonal matrices. LetC=AB. Then,
=
P
=1
.
But if6 =,then
=0.Thus, =. Then, if6 =, =0, hence =0=0. HenceCis
diagonal.
Inductive Step: Assume that the product of anydiagonal×matrices is diagonal. We must show
that for any diagonal×matricesA
1A +1that the productA 1···A +1is diagonal. Now,
A
1···A +1=(A 1···A )A+1.LetB=A 1···A .ThenBis diagonal by the inductive hypothesis.
Hence,A
1···A A+1=BA +1is the product of two diagonal matrices, and so is diagonal by the
Base Step.
(10) Assume thatAis skew-symmetric. Then(A
3
)
=(A
2
A)
=A
(A
2
)
(by Theorem 1.18)=
A
(AA)
=A
A
A
(again by Theorem 1.18)=(−A)(−A)(−A)(sinceAis skew-symmetric)
=−(A
3
)(by part (4) of Theorem 1.16). Hence,A
3
is skew-symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 221
Andrilli/Hecker - Answers to Chapter Tests Chapter 2 - Version A
Answers to Test for Chapter 2 – Version A
(1) The quadratic equation is=3
2
−4−8.
(2) The solution set is{(−2−4−53+24)| ∈R}.
(3) The solution set is{(−2410)|∈R}.
(4) The rank ofAis2,andsoAis not row equivalent toI
3.
(5) The solution sets are, respectively,{(3−42)}and{(1−23)}.
(6) First,th row of(AB)
=(th row of(AB))
=(th row ofA)B(by the hint)
=(th row of(A))B
=th row of ((A)B)(by the hint).
Now, if6 =,th row of(AB)
=th row of(AB)
=(th row ofA)B(by the hint)
=(th row of(A))B
=th row of ((
A)B)(by the hint).
Hence,(AB)=(A)B, since they are equal on each row.
(7) Yes:[−1510−23] = (−2)(row1)+(−3)(row2)+(−1)(row3).
(8) The inverse ofAis
⎡
⎣
−9−415
−4−27
12 5 −19
⎤
⎦.
(9) The inverse ofAis−
1
14
∙
−98
−56
¸
=
" 9
14
−
4
7
5
14
−
3
7
#
.
(10) SupposeAandBare nonsingular×matrices. IfAandBcommute, thenAB=BA. Hence,
A(AB)B=A(BA)B,andsoA
2
B
2
=(AB)
2
.
Conversely, supposeA
2
B
2
=(AB)
2
.SincebothAandBare nonsingular, we know thatA
−1
and
B
−1
exist. Multiply both sides ofA
2
B
2
=ABABbyA
−1
on the left and byB
−1
on the right to
obtainAB=BA. Hence,AandBcommute.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 222
Answers to Test for Chapter 2 – Version A
(1) The quadratic equation is=3
2
−4−8.
(2) The solution set is{(−2−4−53+24)| ∈R}.
(3) The solution set is{(−2410)|∈R}.
(4) The rank ofAis2,andsoAis not row equivalent toI
3.
(5) The solution sets are, respectively,{(3−42)}and{(1−23)}.
(6) First,th row of(AB)
=(th row of(AB))
=(th row ofA)B(by the hint)
=(th row of(A))B
=th row of ((A)B)(by the hint).
Now, if6 =,th row of(AB)
=th row of(AB)
=(th row ofA)B(by the hint)
=(th row of(A))B
=th row of ((
A)B)(by the hint).
Hence,(AB)=(A)B, since they are equal on each row.
(7) Yes:[−1510−23] = (−2)(row1)+(−3)(row2)+(−1)(row3).
(8) The inverse ofAis
⎡
⎣
−9−415
−4−27
12 5 −19
⎤
⎦.
(9) The inverse ofAis−
1
14
∙
−98
−56
¸
=
" 9
14
−
4
7
5
14
−
3
7
#
.
(10) SupposeAandBare nonsingular×matrices. IfAandBcommute, thenAB=BA. Hence,
A(AB)B=A(BA)B,andsoA
2
B
2
=(AB)
2
.
Conversely, supposeA
2
B
2
=(AB)
2
.SincebothAandBare nonsingular, we know thatA
−1
and
B
−1
exist. Multiply both sides ofA
2
B
2
=ABABbyA
−1
on the left and byB
−1
on the right to
obtainAB=BA. Hence,AandBcommute.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 222
Andrilli/Hecker - Answers to Chapter Tests Chapter 2 - Version B
Answers to Test for Chapter 2 – Version B
(1) The equation of the circle is
2
+
2
−6+8+12=0,or,(−3)
2
+(+4)
2
=13.
(2) The solution set is{(−5−3++62+−3−22+1)| ∈R}.
(3) The solution set is{(3100) +(−40−21)| ∈R}.
(4) The rank ofAis3,andsoAis row equivalent toI
3.
(5) The solution sets are, respectively,{(−23−4)}and{(3−12)}.
(6) First,th row of(AB)
=(th row of(AB)) + (th row of(AB))
=(th row ofA)B+(th row ofA)B(by the hint)
=((th row ofA)+(th row ofA))B
=(th row of(A))B
=th row of ((A)B)(by the hint).
Now, if6 =,th row of(AB)
=th row of
(AB)
=(th row ofA)B(by the hint)
=(th row of(A))B
=th row of ((A)B)(by the hint).
Hence,(AB)=(A)B, since they are equal on each row.
(7) The vector[−231]isnotintherowspaceofA.
(8) The inverse ofAis
⎡
⎣
1−61
−213 −2
4−21 3
⎤
⎦.
(9) The solution set is{(4−7−2)}.
(10) (a) To prove that the inverse of(A)is(
1
)A
−1
, simply note that the product(A)times(
1
)A
−1
equalsI(by part (4) of Theorem 1.16).
(b) Now, assumeAis a nonsingular skew-symmetric matrix. Then,
(A
−1
)
=(A
)
−1
by part (4) of Theorem 2.12
=(−1A)
−1
sinceAis skew-symmetric
=(
1
−1
)A
−1
by part (a)
=−(A
−1
)
Hence,A
−1
is skew-symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 223
Answers to Test for Chapter 2 – Version B
(1) The equation of the circle is
2
+
2
−6+8+12=0,or,(−3)
2
+(+4)
2
=13.
(2) The solution set is{(−5−3++62+−3−22+1)| ∈R}.
(3) The solution set is{(3100) +(−40−21)| ∈R}.
(4) The rank ofAis3,andsoAis row equivalent toI
3.
(5) The solution sets are, respectively,{(−23−4)}and{(3−12)}.
(6) First,th row of(AB)
=(th row of(AB)) + (th row of(AB))
=(th row ofA)B+(th row ofA)B(by the hint)
=((th row ofA)+(th row ofA))B
=(th row of(A))B
=th row of ((A)B)(by the hint).
Now, if6 =,th row of(AB)
=th row of
(AB)
=(th row ofA)B(by the hint)
=(th row of(A))B
=th row of ((A)B)(by the hint).
Hence,(AB)=(A)B, since they are equal on each row.
(7) The vector[−231]isnotintherowspaceofA.
(8) The inverse ofAis
⎡
⎣
1−61
−213 −2
4−21 3
⎤
⎦.
(9) The solution set is{(4−7−2)}.
(10) (a) To prove that the inverse of(A)is(
1
)A
−1
, simply note that the product(A)times(
1
)A
−1
equalsI(by part (4) of Theorem 1.16).
(b) Now, assumeAis a nonsingular skew-symmetric matrix. Then,
(A
−1
)
=(A
)
−1
by part (4) of Theorem 2.12
=(−1A)
−1
sinceAis skew-symmetric
=(
1
−1
)A
−1
by part (a)
=−(A
−1
)
Hence,A
−1
is skew-symmetric.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 223
Andrilli/Hecker - Answers to Chapter Tests Chapter 2 - Version C
Answers to Test for Chapter 2 – Version C
(1)=3,=−2,=1,and=4.
(2) The solution set is{(−4−3+32−35−24)| ∈R}.
(3) The solution set is{(12−5100) +(−1−1031)| ∈R}.
(4) The rank ofAis3,andsoAis not row equivalent toI
4.
(5) The reduced row echelon form matrix forAisB=I
3. One possible sequence of row operations
convertingBtoAis:
(II):2←
11
10
3+2
(II):1←
2
5
3+1
(I):3←−
1
10
3
(II):3←−42+3
(II):1←12+1
(I):2←52
(II):3←51+3
(II):2←−31+2
(I):1←21
(6) Yes:[2512−19] = 4[735] + 2[334]−3[323].
(7) The inverse ofAis
⎡
⎢
⎢
⎣
13 −41
321−34 8
414−21 5
18 −13 3
⎤
⎥
⎥
⎦
.
(8) The solution set is{(4−6−5)}.
(9) Base Step: AssumeAandBare two×nonsingular matrices. Then(AB)
−1
=B
−1
A
−1
,bypart
(3) of Theorem 2.12.
Inductive Step: Assume that for any set ofnonsingular×matrices, the inverse of their product is
found by multiplying the inverses of the matrices in reverse order. We must show that ifA
1A +1
are nonsingular×matrices, then
(A
1···A +1)
−1
=(A +1)
−1
···(A 1)
−1
Now, letB=A
1···A . Then by the Base Step, or by part (3) of Theorem 2.12,
(A
1···A +1)
−1
=(BA +1)
−1
=(A +1)
−1
B
−1
=(A +1)
−1
(A1···A )
−1
=(A +1)
−1
(A)
−1
···(A 1)
−1
by the inductive hypothesis.
(10) Now,(
1(2(···( (A))···)))B= 1(2(···( (AB))···))(by part (2) of Theorem 2.1)=I
(given). Hence 1(2(···( (A))···)) =B
−1
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 224
Answers to Test for Chapter 2 – Version C
(1)=3,=−2,=1,and=4.
(2) The solution set is{(−4−3+32−35−24)| ∈R}.
(3) The solution set is{(12−5100) +(−1−1031)| ∈R}.
(4) The rank ofAis3,andsoAis not row equivalent toI
4.
(5) The reduced row echelon form matrix forAisB=I
3. One possible sequence of row operations
convertingBtoAis:
(II):2←
11
10
3+2
(II):1←
2
5
3+1
(I):3←−
1
10
3
(II):3←−42+3
(II):1←12+1
(I):2←52
(II):3←51+3
(II):2←−31+2
(I):1←21
(6) Yes:[2512−19] = 4[735] + 2[334]−3[323].
(7) The inverse ofAis
⎡
⎢
⎢
⎣
13 −41
321−34 8
414−21 5
18 −13 3
⎤
⎥
⎥
⎦
.
(8) The solution set is{(4−6−5)}.
(9) Base Step: AssumeAandBare two×nonsingular matrices. Then(AB)
−1
=B
−1
A
−1
,bypart
(3) of Theorem 2.12.
Inductive Step: Assume that for any set ofnonsingular×matrices, the inverse of their product is
found by multiplying the inverses of the matrices in reverse order. We must show that ifA
1A +1
are nonsingular×matrices, then
(A
1···A +1)
−1
=(A +1)
−1
···(A 1)
−1
Now, letB=A
1···A . Then by the Base Step, or by part (3) of Theorem 2.12,
(A
1···A +1)
−1
=(BA +1)
−1
=(A +1)
−1
B
−1
=(A +1)
−1
(A1···A )
−1
=(A +1)
−1
(A)
−1
···(A 1)
−1
by the inductive hypothesis.
(10) Now,(
1(2(···( (A))···)))B= 1(2(···( (AB))···))(by part (2) of Theorem 2.1)=I
(given). Hence 1(2(···( (A))···)) =B
−1
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 224
Andrilli/Hecker - Answers to Chapter Tests Chapter 3 - Version A
Answers for Test for Chapter 3 – Version A
(1) The area of the parallelogram is24square units.
(2) The determinant of6Ais6
5
(4) = 31104.
(3) The determinant ofAis−2.
(4) Note that|AB
|=|A||B
|(by Theorem 3.7)=|A
||B|(by Theorem 3.10)=|A
B|(again by
Theorem 3.7).
(5) The determinant ofAis1070.
(6) IfAandB
−1
are similar square matrices, there is some nonsingular matrixPsuch thatB
−1
=P
−1
AP.
Then,|B
−1
|=|P
−1
AP|.Since
|B
−1
|=
1
|B|
by Corollary 3.8, and since
|P
−1
AP|=|P
−1
||A||P|(by Theorem 3.7)=
1
|P|
|A||P|=(
1
|P|
|P|)|A|=|A|
we have
1
|B|
=|A|
which means|A||B|=1
(7) The solution set is{(−235)}.
(8) (Optional hint:
A()=
3
−.) Answer:P=
⎡
⎣
−1−1−1
011
101
⎤
⎦D=
⎡
⎣
100
0−10
000
⎤
⎦
(9)=0is an eigenvalue forA⇐⇒
A(0) = 0⇐⇒|0I −A|=0⇐⇒|−A|=0⇐⇒(−1)
|A|=0⇐⇒
|A|=0⇐⇒Ais singular.
(10) LetXbe an eigenvector forAcorresponding to the eigenvalue
1=2. Hence,AX=2X. Therefore,
A
3
X=A
2
(AX)=A
2
(2X)=2(A
2
X)=2A(AX)=2A(2X)=4AX=4(2X)=8X
This shows thatXis an eigenvector corresponding to the eigenvalue
2=8forA
3
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 225
Answers for Test for Chapter 3 – Version A
(1) The area of the parallelogram is24square units.
(2) The determinant of6Ais6
5
(4) = 31104.
(3) The determinant ofAis−2.
(4) Note that|AB
|=|A||B
|(by Theorem 3.7)=|A
||B|(by Theorem 3.10)=|A
B|(again by
Theorem 3.7).
(5) The determinant ofAis1070.
(6) IfAandB
−1
are similar square matrices, there is some nonsingular matrixPsuch thatB
−1
=P
−1
AP.
Then,|B
−1
|=|P
−1
AP|.Since
|B
−1
|=
1
|B|
by Corollary 3.8, and since
|P
−1
AP|=|P
−1
||A||P|(by Theorem 3.7)=
1
|P|
|A||P|=(
1
|P|
|P|)|A|=|A|
we have
1
|B|
=|A|
which means|A||B|=1
(7) The solution set is{(−235)}.
(8) (Optional hint:
A()=
3
−.) Answer:P=
⎡
⎣
−1−1−1
011
101
⎤
⎦D=
⎡
⎣
100
0−10
000
⎤
⎦
(9)=0is an eigenvalue forA⇐⇒
A(0) = 0⇐⇒|0I −A|=0⇐⇒|−A|=0⇐⇒(−1)
|A|=0⇐⇒
|A|=0⇐⇒Ais singular.
(10) LetXbe an eigenvector forAcorresponding to the eigenvalue
1=2. Hence,AX=2X. Therefore,
A
3
X=A
2
(AX)=A
2
(2X)=2(A
2
X)=2A(AX)=2A(2X)=4AX=4(2X)=8X
This shows thatXis an eigenvector corresponding to the eigenvalue
2=8forA
3
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 225
Andrilli/Hecker - Answers to Chapter Tests Chapter 3 - Version B
Answers for Test for Chapter 3 – Version B
(1) The volume of the parallelepiped is59cubic units.
(2) The determinant of5Ais5
4
(7) = 4375.
(3) The determinant ofAis−1.
(4) AssumeAandBare×matrices and thatAB=−BA,withodd. Then,|AB|=|−BA|.
Now,|−BA|=(−1)
|BA|(by Corollary 3.4)=−|BA|,sinceis odd. Hence,|AB|=−|BA|.By
Theorem 3.7, this means that|A|·|B|=−|B|·|A|.Since|A|and|B|are real numbers, this can only
be true if either|A|or|B|equals zero. Hence, eitherAorBis singular (by Theorem 3.5).
(5) The determinant ofAis−916.
(6) Case 1: Suppose6 =. Then, the(th row of((A))
)=th column of(A)=th column ofA
(since6 =,Cdoes not affect theth column of any matrix)=th row ofA
=th row of
¡
A
¢
(since6 =,does not affect theth row of any matrix).
Case 2: Consider theth row of((A))
.Theth row of((A))
=th column of(A)
=(th column ofA)+(th column ofA)=(th row ofA
)+(th row ofA
)=th row of
¡
A
¢
.
Hence, every row of((A))
equals the corresponding row of
¡
A
¢
, and so the matrices are equal.
(7) The solution set is{(−3−22)}.
(8) Optional hint:
A()=
3
−2
2
+.Answer:P=
⎡
⎣
−131
101
011
⎤
⎦D=
⎡
⎣
100
010
000
⎤
⎦
(9) (a) BecauseAis upper triangular, we can easily see that
A()=(−2)
4
. Hence=2is the only eigenvalue forA.
(b) The reduced row echelon form for2I
4−Ais
⎡
⎢
⎢
⎣
0100
0010
0001
0000
⎤
⎥
⎥
⎦
. There is only1non-pivot
column (column1), and so Step 3 of the Diagonalization Method will produce only1fundamental
eigenvector. That fundamental eigenvector is[1000].
(10) From the algebraic perspective,
A()=
2
−
6
5
+1, which has no roots, since its discriminant is−
64
25
.
Hence,Ahas no eigenvalues, and soAcan not be diagonalized. From the geometric perspective, for
any nonzero vectorX,AXwill be rotated away fromXthrough an angle. Hence,XandAXcan
not be parallel. Thus,Acan not have any eigenvectors. This makesAnondiagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 226
Answers for Test for Chapter 3 – Version B
(1) The volume of the parallelepiped is59cubic units.
(2) The determinant of5Ais5
4
(7) = 4375.
(3) The determinant ofAis−1.
(4) AssumeAandBare×matrices and thatAB=−BA,withodd. Then,|AB|=|−BA|.
Now,|−BA|=(−1)
|BA|(by Corollary 3.4)=−|BA|,sinceis odd. Hence,|AB|=−|BA|.By
Theorem 3.7, this means that|A|·|B|=−|B|·|A|.Since|A|and|B|are real numbers, this can only
be true if either|A|or|B|equals zero. Hence, eitherAorBis singular (by Theorem 3.5).
(5) The determinant ofAis−916.
(6) Case 1: Suppose6 =. Then, the(th row of((A))
)=th column of(A)=th column ofA
(since6 =,Cdoes not affect theth column of any matrix)=th row ofA
=th row of
¡
A
¢
(since6 =,does not affect theth row of any matrix).
Case 2: Consider theth row of((A))
.Theth row of((A))
=th column of(A)
=(th column ofA)+(th column ofA)=(th row ofA
)+(th row ofA
)=th row of
¡
A
¢
.
Hence, every row of((A))
equals the corresponding row of
¡
A
¢
, and so the matrices are equal.
(7) The solution set is{(−3−22)}.
(8) Optional hint:
A()=
3
−2
2
+.Answer:P=
⎡
⎣
−131
101
011
⎤
⎦D=
⎡
⎣
100
010
000
⎤
⎦
(9) (a) BecauseAis upper triangular, we can easily see that
A()=(−2)
4
. Hence=2is the only eigenvalue forA.
(b) The reduced row echelon form for2I
4−Ais
⎡
⎢
⎢
⎣
0100
0010
0001
0000
⎤
⎥
⎥
⎦
. There is only1non-pivot
column (column1), and so Step 3 of the Diagonalization Method will produce only1fundamental
eigenvector. That fundamental eigenvector is[1000].
(10) From the algebraic perspective,
A()=
2
−
6
5
+1, which has no roots, since its discriminant is−
64
25
.
Hence,Ahas no eigenvalues, and soAcan not be diagonalized. From the geometric perspective, for
any nonzero vectorX,AXwill be rotated away fromXthrough an angle. Hence,XandAXcan
not be parallel. Thus,Acan not have any eigenvectors. This makesAnondiagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 226
Andrilli/Hecker - Answers to Chapter Tests Chapter 3 - Version C
Answers for Test for Chapter 3 – Version C
(1) The volume of the parallelepiped is2cubic units.
(2) The determinant ofAis3.
(3) One possibility:|A|=5
¯
¯
¯
¯
¯
¯
¯
¯
012 09
0008
014117
15 13 10 6
¯
¯
¯
¯
¯
¯
¯
¯
=5(8)
¯
¯
¯
¯
¯
¯
012 0
01411
15 13 10
¯
¯
¯
¯
¯
¯
=5(8)(−12)
¯
¯
¯
¯
011
15 10
¯
¯
¯
¯
=5(8)(−12)(−11)|15|= 79200. Cofactor expansions were used along thefirst row in each case, except
for the4×4matrix, where the cofactor expansion was along the second row.
(4) Assume thatA
=A
−1
.Then|A
|=|A
−1
|. By Corollary 3.8 and Theorem 3.10, we have|A|=
1|A|. Hence,|A|
2
=1,andso|A|=±1.
(5) By the hint,(AB)=
³
³
(AB)
´´
=
¡
¡
B
A
¢¢
(by Theorem 1.18)=
¡
¡
B
¢
A
¢
(by
part (1) of Theorem 2.1)=A
¡
¡
B
¢¢
(by Theorem 1.18)=A((B))(by the hint).
(6) Base Step(=2):LetA=
∙
1112
2122
¸
Note that|A|=
1122−1221.SinceAhas only two
rows, there are only two possible forms for the row operation.
Case 1:=h1i←h2i+h1i.Then(A)=
∙
11+2112+22
21 22
¸
Hence|(A)|=
(
11+21)22−( 12+22)21=1122+2122−1221−2221=1122−1221=|A|.
Case 2:=h2i←h1i+h2i.Then(A)=
∙
11 12
21+1122+12
¸
Hence|(A)|=
11(22+12)− 12(21+11)= 1122+1112−1221−1211=1122−1221=|A|.
Therefore|A|=|(A)|in all possible cases when=2, thus completing the Base Step.
(7) The solution set is{(3−32)}.
(8) IfP=
∙
12
11
¸
,thenA=P
∙
−10
02
¸
P
−1
.Thus,
A
9
=P
∙
−10
02
¸
9
P
−1
=P
∙
−10
0 512
¸
P
−1
=
∙
1025−1026
513−514
¸
(9) (Optional hint:
A()=(+1)(−2)
2
.) Answer:P=
⎡
⎣
−21−1
01 0
10 1
⎤
⎦D=
⎡
⎣
−100
020
002
⎤
⎦
(10) We will show that
A()=
A
(). (Then the linear factor(−)will appear exactlytimes in
A
()since it appears exactlytimes in A().) Now,
A
()=|I −A
|=|(I )
−A
|(since
(I
)is diagonal, hence symmetric)=|(I −A)
|(by part (2) of Theorem 1.13)=|I −A|(by
Theorem 3.10)=
A().
Copyrightc°2016 Elsevier Ltd. All rights reserved. 227
Answers for Test for Chapter 3 – Version C
(1) The volume of the parallelepiped is2cubic units.
(2) The determinant ofAis3.
(3) One possibility:|A|=5
¯
¯
¯
¯
¯
¯
¯
¯
012 09
0008
014117
15 13 10 6
¯
¯
¯
¯
¯
¯
¯
¯
=5(8)
¯
¯
¯
¯
¯
¯
012 0
01411
15 13 10
¯
¯
¯
¯
¯
¯
=5(8)(−12)
¯
¯
¯
¯
011
15 10
¯
¯
¯
¯
=5(8)(−12)(−11)|15|= 79200. Cofactor expansions were used along thefirst row in each case, except
for the4×4matrix, where the cofactor expansion was along the second row.
(4) Assume thatA
=A
−1
.Then|A
|=|A
−1
|. By Corollary 3.8 and Theorem 3.10, we have|A|=
1|A|. Hence,|A|
2
=1,andso|A|=±1.
(5) By the hint,(AB)=
³
³
(AB)
´´
=
¡
¡
B
A
¢¢
(by Theorem 1.18)=
¡
¡
B
¢
A
¢
(by
part (1) of Theorem 2.1)=A
¡
¡
B
¢¢
(by Theorem 1.18)=A((B))(by the hint).
(6) Base Step(=2):LetA=
∙
1112
2122
¸
Note that|A|=
1122−1221.SinceAhas only two
rows, there are only two possible forms for the row operation.
Case 1:=h1i←h2i+h1i.Then(A)=
∙
11+2112+22
21 22
¸
Hence|(A)|=
(
11+21)22−( 12+22)21=1122+2122−1221−2221=1122−1221=|A|.
Case 2:=h2i←h1i+h2i.Then(A)=
∙
11 12
21+1122+12
¸
Hence|(A)|=
11(22+12)− 12(21+11)= 1122+1112−1221−1211=1122−1221=|A|.
Therefore|A|=|(A)|in all possible cases when=2, thus completing the Base Step.
(7) The solution set is{(3−32)}.
(8) IfP=
∙
12
11
¸
,thenA=P
∙
−10
02
¸
P
−1
.Thus,
A
9
=P
∙
−10
02
¸
9
P
−1
=P
∙
−10
0 512
¸
P
−1
=
∙
1025−1026
513−514
¸
(9) (Optional hint:
A()=(+1)(−2)
2
.) Answer:P=
⎡
⎣
−21−1
01 0
10 1
⎤
⎦D=
⎡
⎣
−100
020
002
⎤
⎦
(10) We will show that
A()=
A
(). (Then the linear factor(−)will appear exactlytimes in
A
()since it appears exactlytimes in A().) Now,
A
()=|I −A
|=|(I )
−A
|(since
(I
)is diagonal, hence symmetric)=|(I −A)
|(by part (2) of Theorem 1.13)=|I −A|(by
Theorem 3.10)=
A().
Copyrightc°2016 Elsevier Ltd. All rights reserved. 227
Andrilli/Hecker - Answers to Chapter Tests Chapter 4 - Version A
Answers for Test for Chapter 4 – Version A
(1) Clearly the operation⊕is commutative. Also,⊕is associative because
(x⊕y)⊕z=(
5
+
5
)
15
⊕z=(((
5
+
5
)
15
)
5
+
5
)
15
=((
5
+
5
)+
5
)
15
=(
5
+(
5
+
5
))
15
=(
5
+((
5
+
5
)
15
)
5
)
15
=x⊕(
5
+
5
)
15
=x⊕(y⊕z)
Clearly, the real number0acts as the additive identity, and, for any real number, the additive inverse
ofis−,because(
5
+(−)
5
)
15
=0
15
=0, the additive identity.
Also, thefirst distributive law holds because
¯(x⊕y)= ¯(
5
+
5
)
15
=
15
(
5
+
5
)
15
=((
5
+
5
))
15
=(
5
+
5
)
15
=((
15
)
5
+(
15
)
5
)
15
=(
15
)⊕(
15
)
=(¯x)⊕(¯y)
Similarly, the other distributive law holds because
(+)¯x=(+)
15
=(+)
15
(
5
)
15
=(
5
+
5
)
15
=((
15
)
5
+(
15
)
5
)
15
=(
15
)⊕(
15
)=(¯x)⊕(¯x)
Associativity of scalar multiplication holds because
()¯x=()
15
=
15
15
=
15
(
15
)=¯(
15
)=¯(¯x)
Finally,1¯x=1
15
=1=x.
(2) Since the set of×skew-symmetric matrices is nonempty, for any≥1,itisenoughtoprove
closure under addition and scalar multiplication. That is, we must show that ifAandBare any
two×skew-symmetric matrices, and ifis any real number, thenA+Bis skew-symmetric, and
Ais skew-symmetric. However,A+Bis skew-symmetric since(A+B)
=A
+B
(by part (2)
of Theorem 1.13)=−A−B(sinceAandBare skew-symmetric)=−(A+B). Similarly,Ais
skew-symmetric because(A)
=A
(by part (3) of Theorem 1.13) =(−A)=−A.
(3) Form the matrixAwhose rows are the given vectors:
⎡
⎣
5−2−1
2−1−1
8−21
⎤
⎦. It is easily shown thatA
row reduces toI
3. Hence, the given vectors spanR
3
. (Alternatively, the vectors spanR
3
since|A|is
nonzero.)
(4) A simplified form for the vectors inspan()is:
{(
3
−5
2
−2) +(+3)| ∈R}={
3
−5
2
++(−2+3)| ∈R}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 228
Answers for Test for Chapter 4 – Version A
(1) Clearly the operation⊕is commutative. Also,⊕is associative because
(x⊕y)⊕z=(
5
+
5
)
15
⊕z=(((
5
+
5
)
15
)
5
+
5
)
15
=((
5
+
5
)+
5
)
15
=(
5
+(
5
+
5
))
15
=(
5
+((
5
+
5
)
15
)
5
)
15
=x⊕(
5
+
5
)
15
=x⊕(y⊕z)
Clearly, the real number0acts as the additive identity, and, for any real number, the additive inverse
ofis−,because(
5
+(−)
5
)
15
=0
15
=0, the additive identity.
Also, thefirst distributive law holds because
¯(x⊕y)= ¯(
5
+
5
)
15
=
15
(
5
+
5
)
15
=((
5
+
5
))
15
=(
5
+
5
)
15
=((
15
)
5
+(
15
)
5
)
15
=(
15
)⊕(
15
)
=(¯x)⊕(¯y)
Similarly, the other distributive law holds because
(+)¯x=(+)
15
=(+)
15
(
5
)
15
=(
5
+
5
)
15
=((
15
)
5
+(
15
)
5
)
15
=(
15
)⊕(
15
)=(¯x)⊕(¯x)
Associativity of scalar multiplication holds because
()¯x=()
15
=
15
15
=
15
(
15
)=¯(
15
)=¯(¯x)
Finally,1¯x=1
15
=1=x.
(2) Since the set of×skew-symmetric matrices is nonempty, for any≥1,itisenoughtoprove
closure under addition and scalar multiplication. That is, we must show that ifAandBare any
two×skew-symmetric matrices, and ifis any real number, thenA+Bis skew-symmetric, and
Ais skew-symmetric. However,A+Bis skew-symmetric since(A+B)
=A
+B
(by part (2)
of Theorem 1.13)=−A−B(sinceAandBare skew-symmetric)=−(A+B). Similarly,Ais
skew-symmetric because(A)
=A
(by part (3) of Theorem 1.13) =(−A)=−A.
(3) Form the matrixAwhose rows are the given vectors:
⎡
⎣
5−2−1
2−1−1
8−21
⎤
⎦. It is easily shown thatA
row reduces toI
3. Hence, the given vectors spanR
3
. (Alternatively, the vectors spanR
3
since|A|is
nonzero.)
(4) A simplified form for the vectors inspan()is:
{(
3
−5
2
−2) +(+3)| ∈R}={
3
−5
2
++(−2+3)| ∈R}
Copyrightc°2016 Elsevier Ltd. All rights reserved. 228
Andrilli/Hecker - Answers to Chapter Tests Chapter 4 - Version A
(5) Use the Independence Test Method (in Section 4.4). Note that the matrix whose columns are the
given vectors row reduces toI
4. Thus, there is a pivot in every column, and the vectors are linearly
independent.
(6) The given set is a basis forM
22. Since the set contains four vectors, it is only necessary to check
either that it spansM
22or that it is linearly independent (seeTheorem 4.12). To check for linear
independence, form the matrix whose columns are the entries of the given matrices. (Be sure to take
the entries from each matrix in the same order.) Since this matrix row reduces toI
4, the set of vectors
is linearly independent, hence a basis forM
22.
(7) Using the Independence Test Method, as explained in Section 4.6, produces the basis
={3
3
−2
2
+3−1
3
−3
2
+2−111
3
−12
2
+13+3}
forspan().Sincehas3elements,dim(span()) = 3.
(8) Letbe the set of columns ofA. BecauseAis nonsingular, it row reduces toI
. Hence, the Inde-
pendence Test Method applied toshows thatis linearly independent. Thus,itself is a basis for
span().Sincehaselements,dim(span()) =,andsospan()=R
, by Theorem 4.13. Hence
spansR
.
(9) One possibility is={[210][100][001]}.
(10) (a) The transition matrix fromtois
⎡
⎣
−23 −1
4−32
−3−2−4
⎤
⎦.
(b) Forv=[−63−11362],[v]
=[3−22],and[v] =[−1422−13].
Copyrightc°2016 Elsevier Ltd. All rights reserved. 229
(5) Use the Independence Test Method (in Section 4.4). Note that the matrix whose columns are the
given vectors row reduces toI
4. Thus, there is a pivot in every column, and the vectors are linearly
independent.
(6) The given set is a basis forM
22. Since the set contains four vectors, it is only necessary to check
either that it spansM
22or that it is linearly independent (seeTheorem 4.12). To check for linear
independence, form the matrix whose columns are the entries of the given matrices. (Be sure to take
the entries from each matrix in the same order.) Since this matrix row reduces toI
4, the set of vectors
is linearly independent, hence a basis forM
22.
(7) Using the Independence Test Method, as explained in Section 4.6, produces the basis
={3
3
−2
2
+3−1
3
−3
2
+2−111
3
−12
2
+13+3}
forspan().Sincehas3elements,dim(span()) = 3.
(8) Letbe the set of columns ofA. BecauseAis nonsingular, it row reduces toI
. Hence, the Inde-
pendence Test Method applied toshows thatis linearly independent. Thus,itself is a basis for
span().Sincehaselements,dim(span()) =,andsospan()=R
, by Theorem 4.13. Hence
spansR
.
(9) One possibility is={[210][100][001]}.
(10) (a) The transition matrix fromtois
⎡
⎣
−23 −1
4−32
−3−2−4
⎤
⎦.
(b) Forv=[−63−11362],[v]
=[3−22],and[v] =[−1422−13].
Copyrightc°2016 Elsevier Ltd. All rights reserved. 229
Andrilli/Hecker - Answers to Chapter Tests Chapter 4 - Version B
Answers for Test for Chapter 4 – Version B
(1) The operation⊕is not associative in general because
(x⊕y)⊕z=(3(+))⊕z=3(3(+)) +)=9+9+3
while
x⊕(y⊕z)=x⊕(3(+)) = 3(+3(+)) = 3+9+9
instead. For a particular counterexample,(1⊕2)⊕3=9⊕3=30, but1⊕(2⊕3) = 1⊕15 = 48.
(2) The set of all polynomials inP
4for which the coefficient of the second-degree term equals the coefficient
of the fourth-degree term has the form{
4
+
3
+
2
++}. To show this set is a subspace
ofP
4, it is enough to show that it is closed under addition and scalar multiplication, as it is clearly
nonempty. Clearly, this set is closed under addition because
(
4
+
3
+
2
++)+(
4
+
3
+
2
++)
=(+)
4
+(+)
3
+(+)
2
+(+)+(+)
which has the correct form because this latter polynomial has the coefficients of its second-degree and
fourth-degree terms equal. Similarly,
(
4
+
3
+
2
++)=()
4
+()
3
+()
2
+()+()
which also has the coefficients of its second-degree and fourth-degree terms equal.
(3) LetVbe a vector space with subspacesW
1andW 2.LetW=W 1∩W2.First,Wis clearly nonempty
because0∈Wsince0must be in bothW
1andW 2due to the fact that they are subspacesThus,
to show thatWis a subspace ofV, it is enough to show thatWis closed under addition and scalar
multiplication. Letxandybe elements ofW,andletbe any real number. Then,xandyare
elements of bothW
1andW 2, and sinceW 1andW 2are both subspaces ofV, hence closed under
addition, we know thatx+yis in bothW
1andW 2.Thusx+yis inW. Similarly, sinceW 1andW 2
are both subspaces ofV,thenxis in bothW 1andW 2, and hencexis inW.
(4) The vector[2−43]isnotinspan().Notethat
⎡
⎣
5−2−9
−15 6 27
−417
¯
¯
¯
¯
¯
¯
2
−4
3
⎤
⎦row reduces to
⎡
⎢
⎣
10−
5
3
01
1
3
00 0
¯
¯
¯
¯
¯
¯
¯
−
8
3
−
23
3
2
⎤
⎥
⎦.
(5) A simplified form for the vectors inspan()is:
½
∙
1−3
00
¸
+
∙
00
10
¸
+
∙
00
01
¸¯
¯
¯
¯
∈R
¾
=
½∙
−3
¸¯
¯
¯
¯
∈R
¾
(6) Form the matrixAwhose columns are the coefficients of the polynomials in. Row reducingAshows
that there is a pivot in every column. Henceis linearly independent by the Independence Test
Method in Section 4.4. Hence, sincehasfourelementsanddim(P
3)=4, Theorem 4.12 shows that
is a basis forP
3.Therefore,spansP 3, showing thatp()can be expressed as a linear combination of
the elements in. Finally, sinceis linearly independent, the uniqueness assertion is true by Theorem
4.9.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 230
Answers for Test for Chapter 4 – Version B
(1) The operation⊕is not associative in general because
(x⊕y)⊕z=(3(+))⊕z=3(3(+)) +)=9+9+3
while
x⊕(y⊕z)=x⊕(3(+)) = 3(+3(+)) = 3+9+9
instead. For a particular counterexample,(1⊕2)⊕3=9⊕3=30, but1⊕(2⊕3) = 1⊕15 = 48.
(2) The set of all polynomials inP
4for which the coefficient of the second-degree term equals the coefficient
of the fourth-degree term has the form{
4
+
3
+
2
++}. To show this set is a subspace
ofP
4, it is enough to show that it is closed under addition and scalar multiplication, as it is clearly
nonempty. Clearly, this set is closed under addition because
(
4
+
3
+
2
++)+(
4
+
3
+
2
++)
=(+)
4
+(+)
3
+(+)
2
+(+)+(+)
which has the correct form because this latter polynomial has the coefficients of its second-degree and
fourth-degree terms equal. Similarly,
(
4
+
3
+
2
++)=()
4
+()
3
+()
2
+()+()
which also has the coefficients of its second-degree and fourth-degree terms equal.
(3) LetVbe a vector space with subspacesW
1andW 2.LetW=W 1∩W2.First,Wis clearly nonempty
because0∈Wsince0must be in bothW
1andW 2due to the fact that they are subspacesThus,
to show thatWis a subspace ofV, it is enough to show thatWis closed under addition and scalar
multiplication. Letxandybe elements ofW,andletbe any real number. Then,xandyare
elements of bothW
1andW 2, and sinceW 1andW 2are both subspaces ofV, hence closed under
addition, we know thatx+yis in bothW
1andW 2.Thusx+yis inW. Similarly, sinceW 1andW 2
are both subspaces ofV,thenxis in bothW 1andW 2, and hencexis inW.
(4) The vector[2−43]isnotinspan().Notethat
⎡
⎣
5−2−9
−15 6 27
−417
¯
¯
¯
¯
¯
¯
2
−4
3
⎤
⎦row reduces to
⎡
⎢
⎣
10−
5
3
01
1
3
00 0
¯
¯
¯
¯
¯
¯
¯
−
8
3
−
23
3
2
⎤
⎥
⎦.
(5) A simplified form for the vectors inspan()is:
½
∙
1−3
00
¸
+
∙
00
10
¸
+
∙
00
01
¸¯
¯
¯
¯
∈R
¾
=
½∙
−3
¸¯
¯
¯
¯
∈R
¾
(6) Form the matrixAwhose columns are the coefficients of the polynomials in. Row reducingAshows
that there is a pivot in every column. Henceis linearly independent by the Independence Test
Method in Section 4.4. Hence, sincehasfourelementsanddim(P
3)=4, Theorem 4.12 shows that
is a basis forP
3.Therefore,spansP 3, showing thatp()can be expressed as a linear combination of
the elements in. Finally, sinceis linearly independent, the uniqueness assertion is true by Theorem
4.9.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 230
Andrilli/Hecker - Answers to Chapter Tests Chapter 4 - Version B
(7) A basis forspan()is
{[2−34−1][31−22][28−123]}
Hence,dim(span()) = 3.Sincedim(R
4
)=4,does not spanR
4
.
(8) SinceAis nonsingular,A
is also nonsingular (by part (4) of Theorem 2.12). ThereforeA
row
reduces toI
, giving a pivot in every column. The Independence Test Method from Section 4.4 now
showsthecolumnsofA
, and hence the rows ofA, are linearly independent.
(9) One possible answer:={[2100][−3010][1000],[0001]}
(10) (a) The transition matrix fromtois
⎡
⎣
3−11
24 −3
−423
⎤
⎦.
(b) Forv=−100
2
+64−181,[v] =[2−4−1],and[v] =[9−9−19].
Copyrightc°2016 Elsevier Ltd. All rights reserved. 231
(7) A basis forspan()is
{[2−34−1][31−22][28−123]}
Hence,dim(span()) = 3.Sincedim(R
4
)=4,does not spanR
4
.
(8) SinceAis nonsingular,A
is also nonsingular (by part (4) of Theorem 2.12). ThereforeA
row
reduces toI
, giving a pivot in every column. The Independence Test Method from Section 4.4 now
showsthecolumnsofA
, and hence the rows ofA, are linearly independent.
(9) One possible answer:={[2100][−3010][1000],[0001]}
(10) (a) The transition matrix fromtois
⎡
⎣
3−11
24 −3
−423
⎤
⎦.
(b) Forv=−100
2
+64−181,[v] =[2−4−1],and[v] =[9−9−19].
Copyrightc°2016 Elsevier Ltd. All rights reserved. 231
Andrilli/Hecker - Answers to Chapter Tests Chapter 4 - Version C
Answers for Test for Chapter 4 – Version C
(1) From part (2) of Theorem 4.1, we know0v=0in any general vector space. Thus,
0¯[ ]=[0+3(0)−30−4(0) + 4] = [−34]
is the additive identity0in this given vector space. Similarly, by part (3) of Theorem 4.1,(−1)vis
the additive inverse ofv. Hence, the additive inverse of[ ]is
(−1)¯[ ]=[−−3−3−+4+4]=[−−6−+8]
(2) LetWbe the set of all 3-vectors orthogonal to[2−31].First,[000]∈Wbecause
[000]·[2−31] = 0
HenceWis nonempty. Thus, to show thatWis a subspace of
R
3
, it is enough to show thatWis
closed under addition and scalar multiplication. Letxy∈W;thatis,letxandyboth be orthogonal
to[2−31]. Then,x+yis also orthogonal to[2−31],because
(x+y)·[2−31] = (x·[2−31]) + (y·[2−31]) = 0 + 0 = 0
Similarly, ifis any real number, andxis inW,thenxis also inWbecause
(x)·[2−31] =(x·[2−31])(by part (4) of Theorem 1.5)
=(0) = 0
(3) The vector
[−365]is inspan()since it equals
11
5
[3−6−1]−
12
5
[4−8−3] + 0[5−10−1]
(4) A simplified form for the vectors inspan()is:
{(
3
−2+4)+(
2
+3−2)| ∈R}={
3
+
2
+(−2+3)+(4−2)| ∈R}
(5) Form the matrixAwhose columns are the entries of each given matrix (taking the entries in the same
order each time). Noting that every column in the reduced row echelon form forAcontains a pivot,
the Independence Test Method from Section 4.4 shows that the given set is linearly independent.
(6) Using the Independence Test Method, as explained in Section 4.6, produces the basis
=
½∙
3−1
−24
¸
∙
21
−20
¸
∙
4−3
−22
¸¾
forspan().Sincehas3elements,dim(span()) = 3.
(7) Note thatspan()istherowspaceofA
.NowA
is also singular (use Theorem 2.12), and so
rank(A
)by Theorem 2.15. Hence, the Simplified Span Method as explained in Section 4.6 for
finding a basis forspan()will produce fewer thanbasiselements(thenonzerorowsofthereduced
row echelon form ofA
), showing thatdim(span()).
(8) One possible answer is{2
3
−3
2
+3+1−
3
+4
2
−6−2
3
}. (This is obtained by enlarging
with the standard basis forP
3and then eliminating
2
and1for redundancy.)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 232
Answers for Test for Chapter 4 – Version C
(1) From part (2) of Theorem 4.1, we know0v=0in any general vector space. Thus,
0¯[ ]=[0+3(0)−30−4(0) + 4] = [−34]
is the additive identity0in this given vector space. Similarly, by part (3) of Theorem 4.1,(−1)vis
the additive inverse ofv. Hence, the additive inverse of[ ]is
(−1)¯[ ]=[−−3−3−+4+4]=[−−6−+8]
(2) LetWbe the set of all 3-vectors orthogonal to[2−31].First,[000]∈Wbecause
[000]·[2−31] = 0
HenceWis nonempty. Thus, to show thatWis a subspace of
R
3
, it is enough to show thatWis
closed under addition and scalar multiplication. Letxy∈W;thatis,letxandyboth be orthogonal
to[2−31]. Then,x+yis also orthogonal to[2−31],because
(x+y)·[2−31] = (x·[2−31]) + (y·[2−31]) = 0 + 0 = 0
Similarly, ifis any real number, andxis inW,thenxis also inWbecause
(x)·[2−31] =(x·[2−31])(by part (4) of Theorem 1.5)
=(0) = 0
(3) The vector
[−365]is inspan()since it equals
11
5
[3−6−1]−
12
5
[4−8−3] + 0[5−10−1]
(4) A simplified form for the vectors inspan()is:
{(
3
−2+4)+(
2
+3−2)| ∈R}={
3
+
2
+(−2+3)+(4−2)| ∈R}
(5) Form the matrixAwhose columns are the entries of each given matrix (taking the entries in the same
order each time). Noting that every column in the reduced row echelon form forAcontains a pivot,
the Independence Test Method from Section 4.4 shows that the given set is linearly independent.
(6) Using the Independence Test Method, as explained in Section 4.6, produces the basis
=
½∙
3−1
−24
¸
∙
21
−20
¸
∙
4−3
−22
¸¾
forspan().Sincehas3elements,dim(span()) = 3.
(7) Note thatspan()istherowspaceofA
.NowA
is also singular (use Theorem 2.12), and so
rank(A
)by Theorem 2.15. Hence, the Simplified Span Method as explained in Section 4.6 for
finding a basis forspan()will produce fewer thanbasiselements(thenonzerorowsofthereduced
row echelon form ofA
), showing thatdim(span()).
(8) One possible answer is{2
3
−3
2
+3+1−
3
+4
2
−6−2
3
}. (This is obtained by enlarging
with the standard basis forP
3and then eliminating
2
and1for redundancy.)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 232
Andrilli/Hecker - Answers to Chapter Tests Chapter 4 - Version C
(9) First, by Theorem 4.4, 1is a subspace ofR
, and so by Theorem 4.13,dim( 1)≤dim(R
)=.But,
ifdim(
1)=, then Theorem 4.13 shows that 1=R
. Hence, for every vectorX∈R
,AX=X.
In particular, this is true for each vectore
1e . Using these vectors as columns forms the matrix
I
.ThusweseethatAI =I. implyingA=I . This contradicts the given conditionA6 =I . Hence
dim(
1)6 =,andsodim( 1).
(10) (a) The transition matrix fromtois
⎡
⎣
2−13
3−22
1−23
⎤
⎦.
(b) Forv=[−109155−71],[v]
=[−13−3],and[v] =[−14−15−16].
Copyrightc°2016 Elsevier Ltd. All rights reserved. 233
(9) First, by Theorem 4.4, 1is a subspace ofR
, and so by Theorem 4.13,dim( 1)≤dim(R
)=.But,
ifdim(
1)=, then Theorem 4.13 shows that 1=R
. Hence, for every vectorX∈R
,AX=X.
In particular, this is true for each vectore
1e . Using these vectors as columns forms the matrix
I
.ThusweseethatAI =I. implyingA=I . This contradicts the given conditionA6 =I . Hence
dim(
1)6 =,andsodim( 1).
(10) (a) The transition matrix fromtois
⎡
⎣
2−13
3−22
1−23
⎤
⎦.
(b) Forv=[−109155−71],[v]
=[−13−3],and[v] =[−14−15−16].
Copyrightc°2016 Elsevier Ltd. All rights reserved. 233
Andrilli/Hecker - Answers to Chapter Tests Chapter 5 - Version A
Answers to Test for Chapter 5 – Version A
(1) To showis a linear operator, we must show that(A 1+A2)=(A 1)+(A 2), for allA 1A2∈M ,
and that(A)=(A), for all∈Rand allA∈M
. However, thefirst equation holds since
B(A
1+A2)B
−1
=BA 1B
−1
+BA 2B
−1
by the distributive laws of matrix multiplication over addition. Similarly, the second equation holds
true since
B(A)B
−1
=BAB
−1
by part (4) of Theorem 1.16.
(2)([2−31]) = [−1−114].
(3) The matrixA
=
⎡
⎣
167 117
46 32
−170−119
⎤
⎦.
(4) The kernel of={[2+ −3 ]| ∈R}. Hence, a basis forker()={[2100][10−31]}.
Abasisforrange()is the set{[4−34][−112]},thefirst and third columns of the original matrix.
Thus,dim(ker()) + dim(range())=2+2=4=dim( R
4
), and the Dimension Theorem is verified.
(5) (a) Now,dim(range())≤dim(R
3
)=3. Hencedim(ker()) = dim(P 3)−dim(range())(by the
Dimension Theorem)=4−dim(range())≥4−3=1.Thus,ker()is nontrivial.
(b) Any nonzero polynomialpinker()satisfies the given conditions.
(6) The3×3matrixgiveninthedefinition ofis clearly the matrix forwith respect to the standard
basis forR
3
. Since the determinant of this matrix is−1, it is nonsingular. Theorem 5.16 thus shows
thatis an isomorphism, implying also that it is one-to-one.
(7) First,is a linear operator, because
(p
1+p2)=(p 1+p2)−(p 1−p2)
0
=(p 1−p1
0)+(p 2−p2
0)=(p 1)+(p 2)
and because
(p)=p−(p)
0
=p−p
0
=(p−p
0
)=(p)
Since the domain and codomain ofhavethesamedimension,inordertoshowis an isomorphism,
it is only necessary to show eitheris one-to-one, oris onto (by Corollary 5.13). We showis
one-to-one. Suppose(p)=0(the zero polynomial). Then,p−p
0
=0,andsop=p
0
.Butthese
polynomials have different degrees unlesspis constant, in which casep
0
=0. Therefore,p=p
0
=0.
Hence,ker()={0},andis one-to-one.
(8)=
µ∙
10
00
¸
∙
01
10
¸
∙
00
01
¸
∙
0−1
10
¸¶
,P=
⎡
⎢
⎢
⎣
100 0
010 −1
010 1
001 0
⎤
⎥
⎥
⎦
,D=
⎡
⎢
⎢
⎣
0000
0000
0000
0002
⎤
⎥
⎥
⎦
Other answers forandPare possible since the eigenspace
0is three-dimensional.
(9) (a) False
(b) True
Copyrightc°2016 Elsevier Ltd. All rights reserved. 234
Answers to Test for Chapter 5 – Version A
(1) To showis a linear operator, we must show that(A 1+A2)=(A 1)+(A 2), for allA 1A2∈M ,
and that(A)=(A), for all∈Rand allA∈M
. However, thefirst equation holds since
B(A
1+A2)B
−1
=BA 1B
−1
+BA 2B
−1
by the distributive laws of matrix multiplication over addition. Similarly, the second equation holds
true since
B(A)B
−1
=BAB
−1
by part (4) of Theorem 1.16.
(2)([2−31]) = [−1−114].
(3) The matrixA
=
⎡
⎣
167 117
46 32
−170−119
⎤
⎦.
(4) The kernel of={[2+ −3 ]| ∈R}. Hence, a basis forker()={[2100][10−31]}.
Abasisforrange()is the set{[4−34][−112]},thefirst and third columns of the original matrix.
Thus,dim(ker()) + dim(range())=2+2=4=dim( R
4
), and the Dimension Theorem is verified.
(5) (a) Now,dim(range())≤dim(R
3
)=3. Hencedim(ker()) = dim(P 3)−dim(range())(by the
Dimension Theorem)=4−dim(range())≥4−3=1.Thus,ker()is nontrivial.
(b) Any nonzero polynomialpinker()satisfies the given conditions.
(6) The3×3matrixgiveninthedefinition ofis clearly the matrix forwith respect to the standard
basis forR
3
. Since the determinant of this matrix is−1, it is nonsingular. Theorem 5.16 thus shows
thatis an isomorphism, implying also that it is one-to-one.
(7) First,is a linear operator, because
(p
1+p2)=(p 1+p2)−(p 1−p2)
0
=(p 1−p1
0)+(p 2−p2
0)=(p 1)+(p 2)
and because
(p)=p−(p)
0
=p−p
0
=(p−p
0
)=(p)
Since the domain and codomain ofhavethesamedimension,inordertoshowis an isomorphism,
it is only necessary to show eitheris one-to-one, oris onto (by Corollary 5.13). We showis
one-to-one. Suppose(p)=0(the zero polynomial). Then,p−p
0
=0,andsop=p
0
.Butthese
polynomials have different degrees unlesspis constant, in which casep
0
=0. Therefore,p=p
0
=0.
Hence,ker()={0},andis one-to-one.
(8)=
µ∙
10
00
¸
∙
01
10
¸
∙
00
01
¸
∙
0−1
10
¸¶
,P=
⎡
⎢
⎢
⎣
100 0
010 −1
010 1
001 0
⎤
⎥
⎥
⎦
,D=
⎡
⎢
⎢
⎣
0000
0000
0000
0002
⎤
⎥
⎥
⎦
Other answers forandPare possible since the eigenspace
0is three-dimensional.
(9) (a) False
(b) True
Copyrightc°2016 Elsevier Ltd. All rights reserved. 234
Andrilli/Hecker - Answers to Chapter Tests Chapter 5 - Version A
(10) The matrix forwith respect to the standard basis forP 3is lower triangular with all1’s on the main
diagonal. (Theth column of this matrix consists of the coefficients of(+1)
4−
in order of descending
degree.) Thus,
()=(−1)
4
,and=1is the only eigenvalue for.Now,ifwere diagonalizable,
the geometric multiplicity of=1would have to be4, and we would have
1=P3. This would imply
that(p)=pfor allp∈P
3. But this is clearly false since the image of the polynomialis+1.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 235
(10) The matrix forwith respect to the standard basis forP 3is lower triangular with all1’s on the main
diagonal. (Theth column of this matrix consists of the coefficients of(+1)
4−
in order of descending
degree.) Thus,
()=(−1)
4
,and=1is the only eigenvalue for.Now,ifwere diagonalizable,
the geometric multiplicity of=1would have to be4, and we would have
1=P3. This would imply
that(p)=pfor allp∈P
3. But this is clearly false since the image of the polynomialis+1.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 235
Andrilli/Hecker - Answers to Chapter Tests Chapter 5 - Version B
Answers to Test for Chapter 5 – Version B
(1) To showis a linear operator, we must show that(A 1+A2)=(A 1)+(A 2), for allA 1A2∈M ,
and that(A)=(A), for all∈Rand allA∈M
. However, thefirst equation holds since
(A
1+A2)+(A 1+A2)
=A 1+A2+A1
+A2
(by part (2) of Theorem 1.13)
=(A
1+A2)+(A 1+A2)
=A 1+A2+A1
+A2
=(A 1+A1
)+(A 2+A2
)(by commutativity of matrix addition).
Similarly,
(A)=(A)+(A)
=A+A
=(A+A
)=(A)
(2)([6−47]) =−
2
+2+3
(3) The matrixA
=
⎡
⎣
−55 165 49 56
−46 139 42 45
32−97−29−32
⎤
⎦.
(4) The kernel of={[3−2 0]|∈R}. Hence, a basis forker()={[3−210]}. The range ofis
spanned by columns1,2,and4of the original matrix. Since these columns are linearly independent, a
basis forrange()={[21−44][5−121][8−112]}. The Dimension Theorem is verified because
dim(ker())
=1,dim(range()) = 3, and the sum of these dimensions equals the dimension ofR
4
the
domain of.
(5) (a) This statement is true. For,dim(range(
1)) = rank(A)(by part (1) of Theorem 5.9)=rank(A
)
(by Corollary 5.11)= dim(range(
2))(by part (1) of Theorem 5.9).
(b) This statement is false. For a particular counterexample, useA=O
,inwhichcase
dim(ker(
1)) =6 ==dim(ker( 2))
In general, using the Dimension Theorem,
dim(ker(
1)) =−dim(range( 1))
and
dim(ker(
2)) =−dim(range( 1))
But, by part (a),
dim(range(
1)) = dim(range( 2))
and since6 =,dim(ker(
1))can never equaldim(ker( 2)).
(6) Solving the appropriate homogeneous system tofind the kernel ofgivesker()=
½∙
00
00
¸¾
.
Hence,is one-to-one by part (1) of Theorem 5.12. Since the domain and codomain ofhave the
same dimension,is also an isomorphism by Corollary 5.13.
(7) First,is a linear operator, since
(B
1+B2)=(B 1+B2)A
−1
=B1A
−1
+B2A
−1
=(B 1)+(B 2)
and since
(B)=(B)A
−1
=(BA
−1
)=(B)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 236
Answers to Test for Chapter 5 – Version B
(1) To showis a linear operator, we must show that(A 1+A2)=(A 1)+(A 2), for allA 1A2∈M ,
and that(A)=(A), for all∈Rand allA∈M
. However, thefirst equation holds since
(A
1+A2)+(A 1+A2)
=A 1+A2+A1
+A2
(by part (2) of Theorem 1.13)
=(A
1+A2)+(A 1+A2)
=A 1+A2+A1
+A2
=(A 1+A1
)+(A 2+A2
)(by commutativity of matrix addition).
Similarly,
(A)=(A)+(A)
=A+A
=(A+A
)=(A)
(2)([6−47]) =−
2
+2+3
(3) The matrixA
=
⎡
⎣
−55 165 49 56
−46 139 42 45
32−97−29−32
⎤
⎦.
(4) The kernel of={[3−2 0]|∈R}. Hence, a basis forker()={[3−210]}. The range ofis
spanned by columns1,2,and4of the original matrix. Since these columns are linearly independent, a
basis forrange()={[21−44][5−121][8−112]}. The Dimension Theorem is verified because
dim(ker())
=1,dim(range()) = 3, and the sum of these dimensions equals the dimension ofR
4
the
domain of.
(5) (a) This statement is true. For,dim(range(
1)) = rank(A)(by part (1) of Theorem 5.9)=rank(A
)
(by Corollary 5.11)= dim(range(
2))(by part (1) of Theorem 5.9).
(b) This statement is false. For a particular counterexample, useA=O
,inwhichcase
dim(ker(
1)) =6 ==dim(ker( 2))
In general, using the Dimension Theorem,
dim(ker(
1)) =−dim(range( 1))
and
dim(ker(
2)) =−dim(range( 1))
But, by part (a),
dim(range(
1)) = dim(range( 2))
and since6 =,dim(ker(
1))can never equaldim(ker( 2)).
(6) Solving the appropriate homogeneous system tofind the kernel ofgivesker()=
½∙
00
00
¸¾
.
Hence,is one-to-one by part (1) of Theorem 5.12. Since the domain and codomain ofhave the
same dimension,is also an isomorphism by Corollary 5.13.
(7) First,is a linear operator, since
(B
1+B2)=(B 1+B2)A
−1
=B1A
−1
+B2A
−1
=(B 1)+(B 2)
and since
(B)=(B)A
−1
=(BA
−1
)=(B)
Copyrightc°2016 Elsevier Ltd. All rights reserved. 236
Andrilli/Hecker - Answers to Chapter Tests Chapter 5 - Version B
Since the domain and codomain ofhave the same dimension, in order to show thatis an isomor-
phism, by Corollary 5.13 it is enough to show either thatis one-to-one, or thatis onto. We show
is one-to-one. Suppose(B
1)=(B 2).ThenB 1A
−1
=B 2A
−1
. Multiplying both sides on the right
byAgivesB
1=B 2. Hence,is one-to-one.
(8) (Optional Hint:
()=
4
−2
2
+1=(−1)
2
(+1)
2
.)
Answer:
=
µ∙
30
20
¸
∙
03
02
¸
∙
20
10
¸
∙
02
01
¸¶
A=
⎡
⎢
⎢
⎣
−70120
0−7012
−4 070
0−407
⎤
⎥
⎥
⎦
,P=
⎡
⎢
⎢
⎣
3020
0302
2010
0201
⎤
⎥
⎥
⎦
,D=
⎡
⎢
⎢
⎣
10 0 0
01 0 0
00−10
00 0 −1
⎤
⎥
⎥
⎦
Other answers forandPare possible since the eigenspaces
1and −1are both two-dimensional.
(9) (a) False
(b) False
(10) All vectors parallel to[4−23]are mapped to themselves under,andso=1is an eigenvalue for
. Now, all other nonzero vectors inR
3
are rotated around the axis through the origin parallel to
[4−23], and so their images underare not parallel to themselves. Hence
1is the one-dimensional
subspace spanned by[4−23], and there are no other eigenvalues. Therefore, the sum of the geometric
multiplicities of all eigenvalues is1, which is less than3, the dimension ofR
3
. Thus Theorem 5.28
shows thatis not diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 237
Since the domain and codomain ofhave the same dimension, in order to show thatis an isomor-
phism, by Corollary 5.13 it is enough to show either thatis one-to-one, or thatis onto. We show
is one-to-one. Suppose(B
1)=(B 2).ThenB 1A
−1
=B 2A
−1
. Multiplying both sides on the right
byAgivesB
1=B 2. Hence,is one-to-one.
(8) (Optional Hint:
()=
4
−2
2
+1=(−1)
2
(+1)
2
.)
Answer:
=
µ∙
30
20
¸
∙
03
02
¸
∙
20
10
¸
∙
02
01
¸¶
A=
⎡
⎢
⎢
⎣
−70120
0−7012
−4 070
0−407
⎤
⎥
⎥
⎦
,P=
⎡
⎢
⎢
⎣
3020
0302
2010
0201
⎤
⎥
⎥
⎦
,D=
⎡
⎢
⎢
⎣
10 0 0
01 0 0
00−10
00 0 −1
⎤
⎥
⎥
⎦
Other answers forandPare possible since the eigenspaces
1and −1are both two-dimensional.
(9) (a) False
(b) False
(10) All vectors parallel to[4−23]are mapped to themselves under,andso=1is an eigenvalue for
. Now, all other nonzero vectors inR
3
are rotated around the axis through the origin parallel to
[4−23], and so their images underare not parallel to themselves. Hence
1is the one-dimensional
subspace spanned by[4−23], and there are no other eigenvalues. Therefore, the sum of the geometric
multiplicities of all eigenvalues is1, which is less than3, the dimension ofR
3
. Thus Theorem 5.28
shows thatis not diagonalizable.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 237
Andrilli/Hecker - Answers to Chapter Tests Chapter 5 - Version C
Answers to Test for Chapter 5 – Version C
(1) To show thatis a linear transformation, we must prove that(p+q)=(p)+(q)and that(p)=
(p)for allpq∈P
and all∈R.Letp=
+···+ 1+ 0and letq=
+···+ 1+ 0.
Then
(p+q)= (
+···+ 1+ 0+
+···+ 1+ 0)
=((
+)
+···+( 1+1)+( 0+0))
=(
+)A
+···+( 1+1)A+( 0+0)I
= A
+···+ 1A+ 0I+A
+···+ 1A+ 0I
=(p)+(q)
Similarly,
(p)= ((
+···+ 1+ 0))
=(
+···+ 1+ 0)
=
A
+···+ 1A+ 0I
=( A
+···+ 1A+ 0I)
=(p)
(2)([−31−4]) =
∙
−29 21
−26
¸
.
(3) The matrixA
=
⎡
⎢
⎢
⎣
−44 91 350
−13 23 118
−28 60 215
−10 16 97
⎤
⎥
⎥
⎦
.
(4) The matrix forwith respect to the standard bases forM
22andP 2isA=
⎡
⎣
10 01
10−10
01 00
⎤
⎦,which
row reduces to
⎡
⎣
1001
0100
0011
⎤
⎦. Henceker()=
½
∙
−10
−11
¸¯
¯
¯
¯
∈R
¾
, which has dimension1.A
basis forrange()consists of thefirst three “columns” ofA, and hencedim(range()) = 3, implying
range()=P
2.Notethatdim(ker()) + dim(range())=1+3=4=dim( M 22),thusverifyingthe
Dimension Theorem.
(5) (a)v∈ker(
1)=⇒ 1(v)=0 W=⇒ 2(1(v)) = 2(0W)=⇒( 2◦1)(v)=0 Y
=⇒v∈ker( 2◦1). Hence,ker( 1)⊆ker( 2◦1).
(b) By part (a) and Theorem 4.13,dim(ker(
1))≤dim(ker( 2◦1)). Hence, by the Dimension
Theorem,dim(range(
1)) = dim(V)−dim(ker( 1))≥dim(V)−dim(ker( 2◦1))
= dim(range(
2◦1)).
(6) Solving the appropriate homogeneous system tofind the kernel ofshows thatker()consistsonly
of the zero polynomial. Hence,is one-to-one by part (1) of Theorem 5.12. Also,dim(ker())=0.
However,is not an isomorphism, because by the Dimension Theorem,dim(range())=3,whichis
less than the dimension of the codomainR
4
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 238
Answers to Test for Chapter 5 – Version C
(1) To show thatis a linear transformation, we must prove that(p+q)=(p)+(q)and that(p)=
(p)for allpq∈P
and all∈R.Letp=
+···+ 1+ 0and letq=
+···+ 1+ 0.
Then
(p+q)= (
+···+ 1+ 0+
+···+ 1+ 0)
=((
+)
+···+( 1+1)+( 0+0))
=(
+)A
+···+( 1+1)A+( 0+0)I
= A
+···+ 1A+ 0I+A
+···+ 1A+ 0I
=(p)+(q)
Similarly,
(p)= ((
+···+ 1+ 0))
=(
+···+ 1+ 0)
=
A
+···+ 1A+ 0I
=( A
+···+ 1A+ 0I)
=(p)
(2)([−31−4]) =
∙
−29 21
−26
¸
.
(3) The matrixA
=
⎡
⎢
⎢
⎣
−44 91 350
−13 23 118
−28 60 215
−10 16 97
⎤
⎥
⎥
⎦
.
(4) The matrix forwith respect to the standard bases forM
22andP 2isA=
⎡
⎣
10 01
10−10
01 00
⎤
⎦,which
row reduces to
⎡
⎣
1001
0100
0011
⎤
⎦. Henceker()=
½
∙
−10
−11
¸¯
¯
¯
¯
∈R
¾
, which has dimension1.A
basis forrange()consists of thefirst three “columns” ofA, and hencedim(range()) = 3, implying
range()=P
2.Notethatdim(ker()) + dim(range())=1+3=4=dim( M 22),thusverifyingthe
Dimension Theorem.
(5) (a)v∈ker(
1)=⇒ 1(v)=0 W=⇒ 2(1(v)) = 2(0W)=⇒( 2◦1)(v)=0 Y
=⇒v∈ker( 2◦1). Hence,ker( 1)⊆ker( 2◦1).
(b) By part (a) and Theorem 4.13,dim(ker(
1))≤dim(ker( 2◦1)). Hence, by the Dimension
Theorem,dim(range(
1)) = dim(V)−dim(ker( 1))≥dim(V)−dim(ker( 2◦1))
= dim(range(
2◦1)).
(6) Solving the appropriate homogeneous system tofind the kernel ofshows thatker()consistsonly
of the zero polynomial. Hence,is one-to-one by part (1) of Theorem 5.12. Also,dim(ker())=0.
However,is not an isomorphism, because by the Dimension Theorem,dim(range())=3,whichis
less than the dimension of the codomainR
4
.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 238
Andrilli/Hecker - Answers to Chapter Tests Chapter 5 - Version C
(7) First, we must show thatis a linear transformation. Now
([
1]+[ 1]) =([ 1+1+])
=(
1+1)v1+···+( +)v
= 1v1+1v1+···+ v+v
= 1v1+···+ v+1v1+···+ v
=([ 1]) +([ 1])
Also,
([
1]) =([ 1])
=
1v1+···+ v
=( 1v1+···+ v)
=([
1])
In order to prove thatis an isomorphism, it is enough to show thatis one-to-one or thatis onto
(by Corollary 5.13), since the dimensions of the domain and the codomain are the same. We show
thatis one-to-one. Suppose([
1]) =0 V.Then 1v1+2v2+···+ v=0V, implying
1=2=···= =0becauseis a linearly independent set (sinceis a basis forV). Hence
[
1]=[00]. Therefore,ker()={[00]},andis one-to-one by part (1) of Theorem
5.12.
(8)=(
2
−+2
2
−2
2
+2),P=
⎡
⎣
111
−1−20
202
⎤
⎦,D=
⎡
⎣
000
010
001
⎤
⎦
Other answers forandPare possible since the eigenspace
1is two-dimensional.
(9) (a) True
(b) True
(10) Ifpis a nonconstant polynomial, thenp
0
is nonzero and has a degree lower than that ofp. Hence
p
0
6=p, for any∈R. Hence, nonconstant polynomials can not be eigenvectors. All constant
polynomials, however, are in the eigenspace
0of=0.Thus, 0is one-dimensional. Therefore, the
sum of the geometric multiplicities of all eigenvalues is1,whichislessthandim(P
)=+1,since
0. Henceis not diagonalizable by Theorem 5.28.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 239
(7) First, we must show thatis a linear transformation. Now
([
1]+[ 1]) =([ 1+1+])
=(
1+1)v1+···+( +)v
= 1v1+1v1+···+ v+v
= 1v1+···+ v+1v1+···+ v
=([ 1]) +([ 1])
Also,
([
1]) =([ 1])
=
1v1+···+ v
=( 1v1+···+ v)
=([
1])
In order to prove thatis an isomorphism, it is enough to show thatis one-to-one or thatis onto
(by Corollary 5.13), since the dimensions of the domain and the codomain are the same. We show
thatis one-to-one. Suppose([
1]) =0 V.Then 1v1+2v2+···+ v=0V, implying
1=2=···= =0becauseis a linearly independent set (sinceis a basis forV). Hence
[
1]=[00]. Therefore,ker()={[00]},andis one-to-one by part (1) of Theorem
5.12.
(8)=(
2
−+2
2
−2
2
+2),P=
⎡
⎣
111
−1−20
202
⎤
⎦,D=
⎡
⎣
000
010
001
⎤
⎦
Other answers forandPare possible since the eigenspace
1is two-dimensional.
(9) (a) True
(b) True
(10) Ifpis a nonconstant polynomial, thenp
0
is nonzero and has a degree lower than that ofp. Hence
p
0
6=p, for any∈R. Hence, nonconstant polynomials can not be eigenvectors. All constant
polynomials, however, are in the eigenspace
0of=0.Thus, 0is one-dimensional. Therefore, the
sum of the geometric multiplicities of all eigenvalues is1,whichislessthandim(P
)=+1,since
0. Henceis not diagonalizable by Theorem 5.28.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 239
Andrilli/Hecker - Answers to Chapter Tests Chapter 6 - Version A
Answers for Test for Chapter 6 – Version A
(1) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis:{[4−32][1312−8][023]}.
(2) Matrix for=
1
9
⎡
⎣
74−4
41 8
−48 1
⎤
⎦.
(3) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis forW:{[2−31][−11−119]}. Continuing the process leads to
the following basis forW
⊥
:{[875]}.Also,dim(W
⊥
)=1.
(4)kAxk
2
=Ax·Ax=x·x(by Theorem 6.9)=kxk
2
. Taking square roots yieldskAxk=kxk.
(5) Letw∈W. We need to show thatw∈(W
⊥
)
⊥
. To do this, we need only prove thatw⊥vfor all
v∈W
⊥
. But, by the definition ofW
⊥
,w·v=0,sincew∈W, which completes the proof.
(6) The characteristic polynomial is
()=
2
(−1)
2
=
4
−2
3
+
2
. (This is the characteristic
polynomial for every orthogonal projection onto a2-dimensional subspace ofR
4
.)
(7) Ifv
1andv 2are unit vectors in the given plane withv 1⊥v2, then the matrix forwith respect to
the ordered orthonormal basis
µ
1
√
70
[35−6]v
1v2
¶
is
⎡
⎣
000
010
001
⎤
⎦
(8) (Optional Hint:
()=
3
−18
2
+81.)
Answer:=(
1
3
[2−21]
1
3
[122]
1
3
[21−2]),D=
⎡
⎣
000
090
009
⎤
⎦
Other answers forare possible since the eigenspace
9is two-dimensional. Another likely answer
foris(
1
3
[2−21]
1
√
2
[110]
1
√
18
[−114]).
(9)A
3
=
µ
1
√
5
∙
−12
21
¸¶∙
−10
08
¸µ
1
√
5
∙
−12
21
¸¶
.
Hence, one possible answer is
A=
µ
1
√
5
∙
−12
21
¸¶∙
−10
02
¸µ
1
√
5
∙
−12
21
¸¶
=
1
5
∙
76
6−2
¸
(10) (Optional Hint: Use the definition of a symmetric operator to show that(
2−1)(v1·v2)=0.)
Answer:symmetric=⇒v
1·(v 2)=(v 1)·v2=⇒v 1·(2v2)=( 1v1)·v2=⇒ 2(v1·v2)= 1(v1·v2)
=⇒(
2−1)(v1·v2)=0=⇒(v 1·v2)=0(since 26 =1)=⇒v 1⊥v2.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 240
Answers for Test for Chapter 6 – Version A
(1) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis:{[4−32][1312−8][023]}.
(2) Matrix for=
1
9
⎡
⎣
74−4
41 8
−48 1
⎤
⎦.
(3) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis forW:{[2−31][−11−119]}. Continuing the process leads to
the following basis forW
⊥
:{[875]}.Also,dim(W
⊥
)=1.
(4)kAxk
2
=Ax·Ax=x·x(by Theorem 6.9)=kxk
2
. Taking square roots yieldskAxk=kxk.
(5) Letw∈W. We need to show thatw∈(W
⊥
)
⊥
. To do this, we need only prove thatw⊥vfor all
v∈W
⊥
. But, by the definition ofW
⊥
,w·v=0,sincew∈W, which completes the proof.
(6) The characteristic polynomial is
()=
2
(−1)
2
=
4
−2
3
+
2
. (This is the characteristic
polynomial for every orthogonal projection onto a2-dimensional subspace ofR
4
.)
(7) Ifv
1andv 2are unit vectors in the given plane withv 1⊥v2, then the matrix forwith respect to
the ordered orthonormal basis
µ
1
√
70
[35−6]v
1v2
¶
is
⎡
⎣
000
010
001
⎤
⎦
(8) (Optional Hint:
()=
3
−18
2
+81.)
Answer:=(
1
3
[2−21]
1
3
[122]
1
3
[21−2]),D=
⎡
⎣
000
090
009
⎤
⎦
Other answers forare possible since the eigenspace
9is two-dimensional. Another likely answer
foris(
1
3
[2−21]
1
√
2
[110]
1
√
18
[−114]).
(9)A
3
=
µ
1
√
5
∙
−12
21
¸¶∙
−10
08
¸µ
1
√
5
∙
−12
21
¸¶
.
Hence, one possible answer is
A=
µ
1
√
5
∙
−12
21
¸¶∙
−10
02
¸µ
1
√
5
∙
−12
21
¸¶
=
1
5
∙
76
6−2
¸
(10) (Optional Hint: Use the definition of a symmetric operator to show that(
2−1)(v1·v2)=0.)
Answer:symmetric=⇒v
1·(v 2)=(v 1)·v2=⇒v 1·(2v2)=( 1v1)·v2=⇒ 2(v1·v2)= 1(v1·v2)
=⇒(
2−1)(v1·v2)=0=⇒(v 1·v2)=0(since 26 =1)=⇒v 1⊥v2.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 240
Andrilli/Hecker - Answers to Chapter Tests Chapter 6 - Version B
Answers for Test for Chapter 6 – Version B
(1) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis:{[2−31][3011−27][568]}.
(2) Matrix for=
1
25
⎡
⎣
16 0−12
025 0
−12 0 9
⎤
⎦.
(3) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis forW:{[22−3][33−1810]}. Continuing the process leads to
the following basis forW
⊥
:{[276]}. Using this information, we can show that[−12−36−43] =
[067] + [−12−42−36],wherethefirst vector of this sum is inW,andthesecondisinW
⊥
.
(4) BecauseAandBare orthogonal,AA
=BB
=I. Hence,
(AB)(AB)
=(AB)(B
A
)
=A(BB
)A
=AI A
=AA
=I
Thus,ABis orthogonal.
(5) By Corollary 6.13,dim(W)=−dim(W
⊥
)=−(−dim((W
⊥
)
⊥
)) = dim((W
⊥
)
⊥
). Thus, by
Theorem 4.13,W=(W
⊥
)
⊥
.
(6) Nowis an orthogonal basis forW,andso
proj
Wv=
v·[036−2]
[036−2]·[036−2]
[036−2] +
v·[−6203]
[−6203]·[−6203]
[−6203] =
1
7
[−2417186]
Thus,proj
W
⊥v=v−proj
Wv=
1
7
[318−11−6].
(7) Yes, by Theorems 6.19 and 6.22, because the matrix forwith respect to the standard basis is
symmetric.
(8) (Optional Hint:
1=0, 2= 49)
Answer:=(
1
7
[−362]
1
7
[623]
1
7
[23−6]),D=
⎡
⎣
000
049 0
0049
⎤
⎦
Other answers forarepossiblesincetheeigenspace
49is two-dimensional. Another likely answer
foris(
1
7
[−362]
1
√
5
[210]
1
√
245
[2−415]).
(9)A
2
=
µ
1
√
2
∙
1−1
11
¸¶∙
40
016
¸µ
1
√
2
∙
11
−11
¸¶
.
Hence, one possible answer is
A=
µ
1
√
2
∙
1−1
11
¸¶∙
20
04
¸µ
1
√
2
∙
11
−11
¸¶
=
∙
3−1
−13
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 241
Answers for Test for Chapter 6 – Version B
(1) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis:{[2−31][3011−27][568]}.
(2) Matrix for=
1
25
⎡
⎣
16 0−12
025 0
−12 0 9
⎤
⎦.
(3) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis forW:{[22−3][33−1810]}. Continuing the process leads to
the following basis forW
⊥
:{[276]}. Using this information, we can show that[−12−36−43] =
[067] + [−12−42−36],wherethefirst vector of this sum is inW,andthesecondisinW
⊥
.
(4) BecauseAandBare orthogonal,AA
=BB
=I. Hence,
(AB)(AB)
=(AB)(B
A
)
=A(BB
)A
=AI A
=AA
=I
Thus,ABis orthogonal.
(5) By Corollary 6.13,dim(W)=−dim(W
⊥
)=−(−dim((W
⊥
)
⊥
)) = dim((W
⊥
)
⊥
). Thus, by
Theorem 4.13,W=(W
⊥
)
⊥
.
(6) Nowis an orthogonal basis forW,andso
proj
Wv=
v·[036−2]
[036−2]·[036−2]
[036−2] +
v·[−6203]
[−6203]·[−6203]
[−6203] =
1
7
[−2417186]
Thus,proj
W
⊥v=v−proj
Wv=
1
7
[318−11−6].
(7) Yes, by Theorems 6.19 and 6.22, because the matrix forwith respect to the standard basis is
symmetric.
(8) (Optional Hint:
1=0, 2= 49)
Answer:=(
1
7
[−362]
1
7
[623]
1
7
[23−6]),D=
⎡
⎣
000
049 0
0049
⎤
⎦
Other answers forarepossiblesincetheeigenspace
49is two-dimensional. Another likely answer
foris(
1
7
[−362]
1
√
5
[210]
1
√
245
[2−415]).
(9)A
2
=
µ
1
√
2
∙
1−1
11
¸¶∙
40
016
¸µ
1
√
2
∙
11
−11
¸¶
.
Hence, one possible answer is
A=
µ
1
√
2
∙
1−1
11
¸¶∙
20
04
¸µ
1
√
2
∙
11
−11
¸¶
=
∙
3−1
−13
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 241
Andrilli/Hecker - Answers to Chapter Tests Chapter 6 - Version B
(10) (Optional Hint: Consider the orthogonal diagonalization ofAand the fact thatA
2
=AA
.)
Answer: SinceAis symmetric, it is orthogonally diagonalizable. So there is an orthogonal matrixP
and a diagonal matrixDsuch thatD=P
−1
AP,orequivalently,
A=PDP
−1
=PDP
Since all eigenvalues ofAare±1,Dhas only these values on its main diagonal, and henceD
2
=I.
Therefore,
AA
=PDP
(PDP
)
=PDP
((P
)
D
P
)
=PDP
(PDP
)
=PD(P
P)DP
=PDI DP
=PD
2
P
=PI P
=PP
=I
Thus,A
=A
−1
,andAis orthogonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 242
(10) (Optional Hint: Consider the orthogonal diagonalization ofAand the fact thatA
2
=AA
.)
Answer: SinceAis symmetric, it is orthogonally diagonalizable. So there is an orthogonal matrixP
and a diagonal matrixDsuch thatD=P
−1
AP,orequivalently,
A=PDP
−1
=PDP
Since all eigenvalues ofAare±1,Dhas only these values on its main diagonal, and henceD
2
=I.
Therefore,
AA
=PDP
(PDP
)
=PDP
((P
)
D
P
)
=PDP
(PDP
)
=PD(P
P)DP
=PDI DP
=PD
2
P
=PI P
=PP
=I
Thus,A
=A
−1
,andAis orthogonal.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 242
Andrilli/Hecker - Answers to Chapter Tests Chapter 6 - Version C
Answers for Test for Chapter 6 – Version C
(1) (Optional Hint: The two given vectors are orthogonal. Therefore, only two additional vectors need
to be found.) Performing the Gram-Schmidt Process as outlined in the text and using appropriate
multiples to avoid fractions leads to the following basis:
{[−31−12][2401][22−19−2132][01−7−4]}
(2) Matrix for=
1
9
⎡
⎣
5−42
−452
228
⎤
⎦.
(3) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis forW:{[51−2][205]}. Continuing the process leads to the
following basis forW
⊥
:{[5−29−2]}.Also,dim(W
⊥
)=1.
(4) Supposeis odd. Then
|A|
2
=|A
2
|=|AA|=|A(−A
)|=|−AA
|=(−1)
|AA
|=(−1)|I |=−1
But|A|
2
can not be negative. This contradiction shows thatis even.
(5) Letw
2∈W
⊥
2
. We must show thatw 2∈W
⊥
1
.Todothis,weshowthatw 2·w1=0for allw 1∈W 1.
So, letw
1∈W 1⊆W 2.Thus,w 1∈W 2. Hence,w 1·w2=0,becausew 2∈W
⊥
2
. This completes the
proof.
(6) Letv=[510−4]. Then,proj
Wv=
1
3
[145−2−12];proj
W
⊥v=
1
3
[1−220];minimumdistance
=
°
°
1
3
[1−220]
°
°
=1.
(7) Ifv
1andv 2are unit vectors in the given plane withv 1⊥v2, then the matrix forwith respect to
the ordered orthonormal basis
µ
1
√
26
[4−3−1]v
1v2
¶
is
⎡
⎣
−100
010
001
⎤
⎦
Hence,is orthogonally diagonalizable. Therefore,is a symmetric operator, and so its matrix with
respect to any orthonormal basis is symmetric. In particular, this is true of the matrix forwith
respect to the standard basis.
(8) (Optional Hint:
1= 121, 2=−121)Answer:
=
¡
1
11
[26−9]
1
11
[−962]
1
11
[676]
¢
;D=
⎡
⎣
121 0 0
0 121 0
00 −121
⎤
⎦
Other answers forare possible since the eigenspace
121is two-dimensional. Another likely answer
foris µ
1
√
85
[−760]
1
11
√
85
[−36−4285]
1
11
[676]
¶
Copyrightc°2016 Elsevier Ltd. All rights reserved. 243
Answers for Test for Chapter 6 – Version C
(1) (Optional Hint: The two given vectors are orthogonal. Therefore, only two additional vectors need
to be found.) Performing the Gram-Schmidt Process as outlined in the text and using appropriate
multiples to avoid fractions leads to the following basis:
{[−31−12][2401][22−19−2132][01−7−4]}
(2) Matrix for=
1
9
⎡
⎣
5−42
−452
228
⎤
⎦.
(3) Performing the Gram-Schmidt Process as outlinedin the text and using appropriate multiples to avoid
fractions leads to the following basis forW:{[51−2][205]}. Continuing the process leads to the
following basis forW
⊥
:{[5−29−2]}.Also,dim(W
⊥
)=1.
(4) Supposeis odd. Then
|A|
2
=|A
2
|=|AA|=|A(−A
)|=|−AA
|=(−1)
|AA
|=(−1)|I |=−1
But|A|
2
can not be negative. This contradiction shows thatis even.
(5) Letw
2∈W
⊥
2
. We must show thatw 2∈W
⊥
1
.Todothis,weshowthatw 2·w1=0for allw 1∈W 1.
So, letw
1∈W 1⊆W 2.Thus,w 1∈W 2. Hence,w 1·w2=0,becausew 2∈W
⊥
2
. This completes the
proof.
(6) Letv=[510−4]. Then,proj
Wv=
1
3
[145−2−12];proj
W
⊥v=
1
3
[1−220];minimumdistance
=
°
°
1
3
[1−220]
°
°
=1.
(7) Ifv
1andv 2are unit vectors in the given plane withv 1⊥v2, then the matrix forwith respect to
the ordered orthonormal basis
µ
1
√
26
[4−3−1]v
1v2
¶
is
⎡
⎣
−100
010
001
⎤
⎦
Hence,is orthogonally diagonalizable. Therefore,is a symmetric operator, and so its matrix with
respect to any orthonormal basis is symmetric. In particular, this is true of the matrix forwith
respect to the standard basis.
(8) (Optional Hint:
1= 121, 2=−121)Answer:
=
¡
1
11
[26−9]
1
11
[−962]
1
11
[676]
¢
;D=
⎡
⎣
121 0 0
0 121 0
00 −121
⎤
⎦
Other answers forare possible since the eigenspace
121is two-dimensional. Another likely answer
foris µ
1
√
85
[−760]
1
11
√
85
[−36−4285]
1
11
[676]
¶
Copyrightc°2016 Elsevier Ltd. All rights reserved. 243
Andrilli/Hecker - Answers to Chapter Tests Chapter 6 - Version C
(9)A
2
=
µ
1
√
10
∙
31
−13
¸¶∙
00
0 100
¸µ
1
√
10
∙
3−1
13
¸¶
.
Hence, one possible answer is
A=
µ
1
√
10
∙
31
−13
¸¶∙
00
010
¸µ
1
√
10
∙
3−1
13
¸¶
=
∙
13
39
¸
(10) (Optional Hint: First show thatAandBorthogonally diagonalize to the same matrix.)
Answer: First, becauseAandBare symmetric, they are orthogonally diagonalizable by Corollary
6.23. But, since the characteristic polynomials ofAandBare equal, these matrices have the same
eigenvalues with the same algebraic multiplicities. Thus, there is a single diagonal matrixDhaving
these eigenvalues on its main diagonal, such thatD=P
−1
1
AP1andD=P
−1
2
BP2, for some orthogonal
matricesP
1andP 2. HenceP
−1
1
AP1=P
−1
2
BP2,orA=P 1P
−1
2
BP2P
−1
1
.LetP=P 1P
−1
2
Then
A=PBP
−1
,wherePis orthogonal by parts (2) and (3) of Theorem 6.6.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 244
(9)A
2
=
µ
1
√
10
∙
31
−13
¸¶∙
00
0 100
¸µ
1
√
10
∙
3−1
13
¸¶
.
Hence, one possible answer is
A=
µ
1
√
10
∙
31
−13
¸¶∙
00
010
¸µ
1
√
10
∙
3−1
13
¸¶
=
∙
13
39
¸
(10) (Optional Hint: First show thatAandBorthogonally diagonalize to the same matrix.)
Answer: First, becauseAandBare symmetric, they are orthogonally diagonalizable by Corollary
6.23. But, since the characteristic polynomials ofAandBare equal, these matrices have the same
eigenvalues with the same algebraic multiplicities. Thus, there is a single diagonal matrixDhaving
these eigenvalues on its main diagonal, such thatD=P
−1
1
AP1andD=P
−1
2
BP2, for some orthogonal
matricesP
1andP 2. HenceP
−1
1
AP1=P
−1
2
BP2,orA=P 1P
−1
2
BP2P
−1
1
.LetP=P 1P
−1
2
Then
A=PBP
−1
,wherePis orthogonal by parts (2) and (3) of Theorem 6.6.
Copyrightc°2016 Elsevier Ltd. All rights reserved. 244
Andrilli/Hecker - Answers to Chapter Tests Chapter 7 - Version A
Answers to Test for Chapter 7 – Version A
(1)a·b=b·a=[2−2−1+5]
(2)(P
∗
HP)
∗
=P
∗
H
∗
(P
∗
)
∗
(by part (5) of Theorem 7.2)=P
∗
H
∗
P(by part (1) of Theorem 7.2)=
P
∗
HP(sinceHis Hermitian). HenceP
∗
HPis Hermitian.
(3) (a) False
(b) False
(4)
1=1+, 2=1−2
(5) LetA=
∙
01
10
¸
.Then
A()=
2
+1,whichhasnorealroots. HenceAhas no real eigenvalues,
and so is not diagonalizable as a real matrix. However,
A()has complex rootsand−.Thus,this
2×2matrix has two complex distinct eigenvalues, and thus must be diagonalizable.
(6) (Optional Hint:
()=
3
−2
2
−.)
Answer:
1=0, 2=;basisfor 0={[1 +−1]},basisfor ={[01][010]}.
is diagonalizable.
(7) The statement is false. The polynomialis in the given subset, butis not. Hence, the subset is not
closed under scalar multiplication.
(8)={[1 + 32+1−2][123+]}
(9)
As a complex linear
transformation:
As a real linear transformation:
∙
100
001
¸
⎡
⎢
⎢
⎣
100 −1000 0
011 0000 0
000 0100 −1
000 0011 0
⎤
⎥
⎥
⎦
(10) (Optional Hint: The two given vectors are already orthogonal.)
Answer:{[11+][3−−1−][02−1+]}
(11)
P=
√
2
2
∙
11
1−1
¸
(12)(AB)
∗
=B
∗
A
∗
=B
−1
A
−1
=(AB)
−1
,andsoABis unitary. Also,
¯
¯
¯|A|
¯
¯
¯
2
=|A|
|A|=|A||A
∗
|(by
part (3) of Theorem 7.5)=|AA
∗
|=|I |=1.
(13) Proofs for thefive properties of an inner product:
(1)hffi=f
2
(0) +f
2
(1) +f
2
(2)≥0, since the sum of squares is always nonnegative.
(2) Supposef()=
2
++.Thenhffi=0
=⇒f(0) = 0f(1) = 0f(2) = 0(since a sum of squares of real numbers can only be zero if
each number is zero)
=⇒( )satisfies
⎧
⎨
⎩
=0
++=0
4+2+=0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 245
Answers to Test for Chapter 7 – Version A
(1)a·b=b·a=[2−2−1+5]
(2)(P
∗
HP)
∗
=P
∗
H
∗
(P
∗
)
∗
(by part (5) of Theorem 7.2)=P
∗
H
∗
P(by part (1) of Theorem 7.2)=
P
∗
HP(sinceHis Hermitian). HenceP
∗
HPis Hermitian.
(3) (a) False
(b) False
(4)
1=1+, 2=1−2
(5) LetA=
∙
01
10
¸
.Then
A()=
2
+1,whichhasnorealroots. HenceAhas no real eigenvalues,
and so is not diagonalizable as a real matrix. However,
A()has complex rootsand−.Thus,this
2×2matrix has two complex distinct eigenvalues, and thus must be diagonalizable.
(6) (Optional Hint:
()=
3
−2
2
−.)
Answer:
1=0, 2=;basisfor 0={[1 +−1]},basisfor ={[01][010]}.
is diagonalizable.
(7) The statement is false. The polynomialis in the given subset, butis not. Hence, the subset is not
closed under scalar multiplication.
(8)={[1 + 32+1−2][123+]}
(9)
As a complex linear
transformation:
As a real linear transformation:
∙
100
001
¸
⎡
⎢
⎢
⎣
100 −1000 0
011 0000 0
000 0100 −1
000 0011 0
⎤
⎥
⎥
⎦
(10) (Optional Hint: The two given vectors are already orthogonal.)
Answer:{[11+][3−−1−][02−1+]}
(11)
P=
√
2
2
∙
11
1−1
¸
(12)(AB)
∗
=B
∗
A
∗
=B
−1
A
−1
=(AB)
−1
,andsoABis unitary. Also,
¯
¯
¯|A|
¯
¯
¯
2
=|A|
|A|=|A||A
∗
|(by
part (3) of Theorem 7.5)=|AA
∗
|=|I |=1.
(13) Proofs for thefive properties of an inner product:
(1)hffi=f
2
(0) +f
2
(1) +f
2
(2)≥0, since the sum of squares is always nonnegative.
(2) Supposef()=
2
++.Thenhffi=0
=⇒f(0) = 0f(1) = 0f(2) = 0(since a sum of squares of real numbers can only be zero if
each number is zero)
=⇒( )satisfies
⎧
⎨
⎩
=0
++=0
4+2+=0
Copyrightc°2016 Elsevier Ltd. All rights reserved. 245
Andrilli/Hecker - Answers to Chapter Tests Chapter 7 - Version A
=⇒===0(since the coefficient matrix of the homogeneous system has determinant−2).
Conversely, it is clear thath00i=0.
(3)hfgi=f(0)g(0) +f(1)g(1) +f(2)g(2)
=g(0)f(0) +g(1)f(1) +g(2)f(2) =hgfi
(4)hf+ghi=(f(0) +g(0))h(0) + (f(1) +g(1))h(1) + (f(2) +g(2))h(2)
=f(0)h(0) +g(0)h(0) +f(1)h(1) +g
(1)h(1) +f(2)h(2) +g(2)h(2)
=f(0)h(0) +f(1)h(1) +f(2)h(2) +g(0)h(0) +g(1)h(1) +g(2)h(2)
=hfhi+hghi
(5)hfgi=(f)(0)g(0) + (f)(1)g(1) + (f)(2)g(2)
=f(0)g(0) +f(1)g(1) +f(2)g(2)
=(f(0)g(0) +f(1)g(1) +f(2)g
(2))
=hfgi
Also,hfgi=14andkfk=
√
35.
(14) Thedistancebetweenxandyis11.
(15) Note thathx+yx−yi=hxxi+hx−yi+hyxi+hy−yi=kxk
2
−kyk
2
.So,ifkxk=kyk,then
hx+yx−yi=kxk
2
−kyk
2
=0.
Conversely,hx+yx−yi=0=⇒kxk
2
−kyk
2
=0=⇒kxk
2
=kyk
2
=⇒kxk=kyk.
(16){
2
3−5
2
}
(17)w
1=[−8−516],w 2=[86−19]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 246
=⇒===0(since the coefficient matrix of the homogeneous system has determinant−2).
Conversely, it is clear thath00i=0.
(3)hfgi=f(0)g(0) +f(1)g(1) +f(2)g(2)
=g(0)f(0) +g(1)f(1) +g(2)f(2) =hgfi
(4)hf+ghi=(f(0) +g(0))h(0) + (f(1) +g(1))h(1) + (f(2) +g(2))h(2)
=f(0)h(0) +g(0)h(0) +f(1)h(1) +g
(1)h(1) +f(2)h(2) +g(2)h(2)
=f(0)h(0) +f(1)h(1) +f(2)h(2) +g(0)h(0) +g(1)h(1) +g(2)h(2)
=hfhi+hghi
(5)hfgi=(f)(0)g(0) + (f)(1)g(1) + (f)(2)g(2)
=f(0)g(0) +f(1)g(1) +f(2)g(2)
=(f(0)g(0) +f(1)g(1) +f(2)g
(2))
=hfgi
Also,hfgi=14andkfk=
√
35.
(14) Thedistancebetweenxandyis11.
(15) Note thathx+yx−yi=hxxi+hx−yi+hyxi+hy−yi=kxk
2
−kyk
2
.So,ifkxk=kyk,then
hx+yx−yi=kxk
2
−kyk
2
=0.
Conversely,hx+yx−yi=0=⇒kxk
2
−kyk
2
=0=⇒kxk
2
=kyk
2
=⇒kxk=kyk.
(16){
2
3−5
2
}
(17)w
1=[−8−516],w 2=[86−19]
Copyrightc°2016 Elsevier Ltd. All rights reserved. 246
Andrilli/Hecker - Answers to Chapter Tests Chapter 7 - Version B
Answers to Test for Chapter 7 – Version B
(1)A
∗
B=
∙
−1+1433−3
−5−55+2
¸
(2)(H)
∗
=
H
∗
(by part (3) of Theorem 7.2)=(−)H
∗
=(−)H(sinceHis Hermitian)=−(H). Hence,
His skew-Hermitian.
(3) (a) False
(b) True
(4) The determinant of the matrix of coefficients is1, and so the inverse matrix is
∙
6+5−2−14
−2+ 4+
¸
.
Using this gives
1=2+and 2=−.
(5) LetAbe an×complex matrix. By the Fundamental Theorem of Algebra,
A()factors into
linear factors. Hence,
A()=(− 1)
1
···(− )
,where 1are the eigenvalues ofA,
and
1are their corresponding algebraic multiplicities. But, since the degree of A()is,
P
=1
=.
(6)=;basisfor
={[12]}.is not diagonalizable.
(7) The set of Hermitian×matrices is not closed under scalar multiplication, and so is not a subspace
ofM
C
.Toseethis,notethat
∙
10
00
¸
=
∙
0
00
¸
.However,
∙
10
00
¸
is Hermitian and
∙
0
00
¸
is not. Since the set of Hermitian×matrices is not a vector space, it does not have a dimension.
(8)={[10][011−]}
(9) Note that
µ
∙
10
00
¸¶
=
µ∙
0
00
¸¶
=
∙
0
00
¸
∗
=
∙
−0
00
¸
,but
µ∙
10
00
¸¶
=
∙
10
00
¸
∗
=
∙
10
00
¸
=
∙
0
00
¸
.
(10){[1 +1−2][3−1+][02−1−]}
(11) Now,Z
∗
=
∙
1−1+
−1−1−
¸
andZZ
∗
=
∙
44
−44
¸
=Z
∗
Z. Hence,Zis normal. Thus, by Theorem
7.9,Zis unitarily diagonalizable.
(12) Now,Aunitary impliesAA
∗
=I.SupposeA
2
=I. HenceAA
∗
=AA. Multiplying both sides
byA
−1
(=A
∗
)on the left yieldsA
∗
=A,andAis Hermitian. Conversely, ifAis Hermitian, then
A=A
∗
. HenceAA
∗
=IyieldsA
2
=I.
(13) Proofs for thefive properties of an inner product:
(1)hxxi=Ax·Ax≥0,since·is an inner product.
(2)hxxi=0 =⇒Ax·Ax=0 =⇒Ax=0(since·is an inner product)
=⇒x=0(sinceAis nonsingular)
Conversely,h00i=A0·A0=0·0=0.
(3)hxyi=Ax·Ay=Ay·Ax=hyxi
Copyrightc°2016 Elsevier Ltd. All rights reserved. 247
Answers to Test for Chapter 7 – Version B
(1)A
∗
B=
∙
−1+1433−3
−5−55+2
¸
(2)(H)
∗
=
H
∗
(by part (3) of Theorem 7.2)=(−)H
∗
=(−)H(sinceHis Hermitian)=−(H). Hence,
His skew-Hermitian.
(3) (a) False
(b) True
(4) The determinant of the matrix of coefficients is1, and so the inverse matrix is
∙
6+5−2−14
−2+ 4+
¸
.
Using this gives
1=2+and 2=−.
(5) LetAbe an×complex matrix. By the Fundamental Theorem of Algebra,
A()factors into
linear factors. Hence,
A()=(− 1)
1
···(− )
,where 1are the eigenvalues ofA,
and
1are their corresponding algebraic multiplicities. But, since the degree of A()is,
P
=1
=.
(6)=;basisfor
={[12]}.is not diagonalizable.
(7) The set of Hermitian×matrices is not closed under scalar multiplication, and so is not a subspace
ofM
C
.Toseethis,notethat
∙
10
00
¸
=
∙
0
00
¸
.However,
∙
10
00
¸
is Hermitian and
∙
0
00
¸
is not. Since the set of Hermitian×matrices is not a vector space, it does not have a dimension.
(8)={[10][011−]}
(9) Note that
µ
∙
10
00
¸¶
=
µ∙
0
00
¸¶
=
∙
0
00
¸
∗
=
∙
−0
00
¸
,but
µ∙
10
00
¸¶
=
∙
10
00
¸
∗
=
∙
10
00
¸
=
∙
0
00
¸
.
(10){[1 +1−2][3−1+][02−1−]}
(11) Now,Z
∗
=
∙
1−1+
−1−1−
¸
andZZ
∗
=
∙
44
−44
¸
=Z
∗
Z. Hence,Zis normal. Thus, by Theorem
7.9,Zis unitarily diagonalizable.
(12) Now,Aunitary impliesAA
∗
=I.SupposeA
2
=I. HenceAA
∗
=AA. Multiplying both sides
byA
−1
(=A
∗
)on the left yieldsA
∗
=A,andAis Hermitian. Conversely, ifAis Hermitian, then
A=A
∗
. HenceAA
∗
=IyieldsA
2
=I.
(13) Proofs for thefive properties of an inner product:
(1)hxxi=Ax·Ax≥0,since·is an inner product.
(2)hxxi=0 =⇒Ax·Ax=0 =⇒Ax=0(since·is an inner product)
=⇒x=0(sinceAis nonsingular)
Conversely,h00i=A0·A0=0·0=0.
(3)hxyi=Ax·Ay=Ay·Ax=hyxi
Copyrightc°2016 Elsevier Ltd. All rights reserved. 247
Andrilli/Hecker - Answers to Chapter Tests Chapter 7 - Version B
(4)hx+yzi=A(x+y)·Az=(Ax+Ay)·Az=Ax·Az+Ay·Az=hxzi+hyzi
(5)hxyi=A(x)·Ay=(Ax)·(Ay)=((Ax)·(Ay)) =hxyi
Also,hxyi=35andkxk=7.
(14) Thedistancebetweenwandzis8.
(15)
1
4
(kx+yk
2
−kx−yk
2
)=
1
4
(hx+yx+yi−hx−yx−yi)
=
1
4
(hx+yxi+hx+yyi−hx−yxi+hx−yyi)
=
1
4
(hxxi+hyxi+hxyi+hyyi−hxxi+hyxi+hxyi−hyyi)
=
1
4
(4hxyi)=hxyi.
(16){
2
17−9
2
2−3+
2
}
(17)w
1=
25
4
2
,w2=+3−
25
4
2
Copyrightc°2016 Elsevier Ltd. All rights reserved. 248
(4)hx+yzi=A(x+y)·Az=(Ax+Ay)·Az=Ax·Az+Ay·Az=hxzi+hyzi
(5)hxyi=A(x)·Ay=(Ax)·(Ay)=((Ax)·(Ay)) =hxyi
Also,hxyi=35andkxk=7.
(14) Thedistancebetweenwandzis8.
(15)
1
4
(kx+yk
2
−kx−yk
2
)=
1
4
(hx+yx+yi−hx−yx−yi)
=
1
4
(hx+yxi+hx+yyi−hx−yxi+hx−yyi)
=
1
4
(hxxi+hyxi+hxyi+hyyi−hxxi+hyxi+hxyi−hyyi)
=
1
4
(4hxyi)=hxyi.
(16){
2
17−9
2
2−3+
2
}
(17)w
1=
25
4
2
,w2=+3−
25
4
2
Copyrightc°2016 Elsevier Ltd. All rights reserved. 248
Andrilli/Hecker - Answers to Chapter Tests Chapter 7 - Version C
Answers to Test for Chapter 7 – Version C
(1)A
B
∗
=
∙
14−51+4
8−78−3
¸
(2)HJ=(HJ)
∗
=J
∗
H
∗
=JH, becauseH,J,andHJare all Hermitian.
(3) (a) True
(b) True
(4)
1=4−3, 2=−1+
(5) LetAbe an×complex matrix that is not diagonalizable. By the Fundamental Theorem of Algebra,
A()factors intolinear factors. Hence, A()=(− 1)
1
···(− )
,where 1are
the eigenvalues ofA,and
1are their corresponding algebraic multiplicities. Now, since the
degree of
A()is,
P
=1
=. Hence, if each geometric multiplicity actually equaled each algebraic
multiplicity for each eigenvalue, then the sum of the geometric multiplicities would also be, implying
thatAis diagonalizable. However, sinceAis not diagonalizable, some geometric multiplicity must be
less than its corresponding algebraic multiplicity.
(6)
1=2, 2=2;basisfor 2={[11]},basisfor 2={[1]}.is diagonalizable.
(7) We need tofind normal matricesZandWsuch that(Z+W)is not normal. Note thatZ=
∙
11
10
¸
andW=
∙
0
0
¸
are normal matrices (Zis Hermitian andWis skew-Hermitian). LetY=
Z+W=
∙
11+
1+0
¸
.ThenYY
∗
=
∙
11+
1+0
¸∙
11−
1−0
¸
=
∙
21−
1+2
¸
, while
Y
∗
Y=
∙
11−
1−0
¸∙
11+
1+0
¸
=
∙
21+
1−2
¸
,andsoYis not normal.
(8) One possibility:={[13+ 1−][1 +3+4−1+2][1000][0010]}
(9) Letbe such that
,theth entry ofw, does not equal zero. Then,(e )= (
)= 1(−).
However,(e
)=( )(1). Setting these equal gives− =, which implies0=2 , and thus
=0. But this contradicts the assumption that 6 =0.
(10) (Optional Hint: The last vector in thegiven basis is already orthogonal to thefirst two.)
Answer:{[2−1+21][56−1+2][−102+]}
(11) Because the rows ofZform an orthonormal basis forC
, by part (1) of Theorem 7.7,Zis a unitary
matrix. Hence, by part (2) of Theorem 7.7, the columns ofZform an orthonormal basis forC
.
(12) (a) Note thatAA
∗
=A
∗
A=
∙
1825 900
−9001300
¸
.
(b) (Optional Hint:
A()=
2
−25+ 1250,or, 1=25and 2=−50.)
Answer:P=
1
5
∙
34
4−3
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 249
Answers to Test for Chapter 7 – Version C
(1)A
B
∗
=
∙
14−51+4
8−78−3
¸
(2)HJ=(HJ)
∗
=J
∗
H
∗
=JH, becauseH,J,andHJare all Hermitian.
(3) (a) True
(b) True
(4)
1=4−3, 2=−1+
(5) LetAbe an×complex matrix that is not diagonalizable. By the Fundamental Theorem of Algebra,
A()factors intolinear factors. Hence, A()=(− 1)
1
···(− )
,where 1are
the eigenvalues ofA,and
1are their corresponding algebraic multiplicities. Now, since the
degree of
A()is,
P
=1
=. Hence, if each geometric multiplicity actually equaled each algebraic
multiplicity for each eigenvalue, then the sum of the geometric multiplicities would also be, implying
thatAis diagonalizable. However, sinceAis not diagonalizable, some geometric multiplicity must be
less than its corresponding algebraic multiplicity.
(6)
1=2, 2=2;basisfor 2={[11]},basisfor 2={[1]}.is diagonalizable.
(7) We need tofind normal matricesZandWsuch that(Z+W)is not normal. Note thatZ=
∙
11
10
¸
andW=
∙
0
0
¸
are normal matrices (Zis Hermitian andWis skew-Hermitian). LetY=
Z+W=
∙
11+
1+0
¸
.ThenYY
∗
=
∙
11+
1+0
¸∙
11−
1−0
¸
=
∙
21−
1+2
¸
, while
Y
∗
Y=
∙
11−
1−0
¸∙
11+
1+0
¸
=
∙
21+
1−2
¸
,andsoYis not normal.
(8) One possibility:={[13+ 1−][1 +3+4−1+2][1000][0010]}
(9) Letbe such that
,theth entry ofw, does not equal zero. Then,(e )= (
)= 1(−).
However,(e
)=( )(1). Setting these equal gives− =, which implies0=2 , and thus
=0. But this contradicts the assumption that 6 =0.
(10) (Optional Hint: The last vector in thegiven basis is already orthogonal to thefirst two.)
Answer:{[2−1+21][56−1+2][−102+]}
(11) Because the rows ofZform an orthonormal basis forC
, by part (1) of Theorem 7.7,Zis a unitary
matrix. Hence, by part (2) of Theorem 7.7, the columns ofZform an orthonormal basis forC
.
(12) (a) Note thatAA
∗
=A
∗
A=
∙
1825 900
−9001300
¸
.
(b) (Optional Hint:
A()=
2
−25+ 1250,or, 1=25and 2=−50.)
Answer:P=
1
5
∙
34
4−3
¸
Copyrightc°2016 Elsevier Ltd. All rights reserved. 249
Andrilli/Hecker - Answers to Chapter Tests Chapter 7 - Version C
(13) Proofs for thefive properties of an inner product:
Property 1:hxxi=2
11−12−21+22=
2
1
+
2
1
−2 12+
2
2
=
2
1
+( 1−2)
2
≥0.
Property 2:hxxi=0precisely when
1=0and 1−2=0,whichiswhen 1=2=0.
Property 3:hyxi=2
11−12−21+22=2 11−12−21+22=hxyi
Property 4: Letz=[
12]. Then,hx+yzi
=2(
1+1)1−( 1+1)2−( 2+2)1+( 2+2)2
=2 11+2 11−12−12−21−21+22+21
=2 11−12−21+22+2 11−12−21+21
=hxzi+hyzi
Property 5:hxyi=2(
1)1−( 1)2−( 2)1+( 2)2
=(2 11−12−21+22)=hxyi
Also,hxyi=0andkxk=
√
5.
(14) Distance betweenfandgis
√
.
(15)x⊥y⇐⇒hxyi=0⇐⇒2hxyi=0⇐⇒kxk
2
+2hxyi+kyk
2
=kxk
2
+kyk
2
⇐⇒hxxi+hxyi+hyxi+hyyi=kxk
2
+kyk
2
⇐⇒hxx+yi+hyx+yi=kxk
2
+kyk
2
⇐⇒hx+yx+yi=kxk
2
+kyk
2
⇐⇒kx+yk
2
=kxk
2
+kyk
2
(16) (Optional Hint:h[−111][3−4−1]i=0)Answer:{[−111][3−4−1][−120]}
(17)w
1=2−
1
3
,w2=
2
−2+
1
3
Copyrightc°2016 Elsevier Ltd. All rights reserved. 250
Elementary Linear Algebra 5th Edition Larson Solutions Manual
Full Download: http://alibabadownload.com/product/elementary-linear-algebra-5th-edition-larson-solutions-manual/
This sample only, Download all chapters at: alibabadownload.com
(13) Proofs for thefive properties of an inner product:
Property 1:hxxi=2
11−12−21+22=
2
1
+
2
1
−2 12+
2
2
=
2
1
+( 1−2)
2
≥0.
Property 2:hxxi=0precisely when
1=0and 1−2=0,whichiswhen 1=2=0.
Property 3:hyxi=2
11−12−21+22=2 11−12−21+22=hxyi
Property 4: Letz=[
12]. Then,hx+yzi
=2(
1+1)1−( 1+1)2−( 2+2)1+( 2+2)2
=2 11+2 11−12−12−21−21+22+21
=2 11−12−21+22+2 11−12−21+21
=hxzi+hyzi
Property 5:hxyi=2(
1)1−( 1)2−( 2)1+( 2)2
=(2 11−12−21+22)=hxyi
Also,hxyi=0andkxk=
√
5.
(14) Distance betweenfandgis
√
.
(15)x⊥y⇐⇒hxyi=0⇐⇒2hxyi=0⇐⇒kxk
2
+2hxyi+kyk
2
=kxk
2
+kyk
2
⇐⇒hxxi+hxyi+hyxi+hyyi=kxk
2
+kyk
2
⇐⇒hxx+yi+hyx+yi=kxk
2
+kyk
2
⇐⇒hx+yx+yi=kxk
2
+kyk
2
⇐⇒kx+yk
2
=kxk
2
+kyk
2
(16) (Optional Hint:h[−111][3−4−1]i=0)Answer:{[−111][3−4−1][−120]}
(17)w
1=2−
1
3
,w2=
2
−2+
1
3
Copyrightc°2016 Elsevier Ltd. All rights reserved. 250
Elementary Linear Algebra 5th Edition Larson Solutions Manual
Full Download: http://alibabadownload.com/product/elementary-linear-algebra-5th-edition-larson-solutions-manual/
This sample only, Download all chapters at: alibabadownload.com
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