Eigenvalues and eigenvectors

Eigenvalues and eigenvectors

1. Eigenvalues and Eigenvectors
Consider multiplying nonzero vectors by a given square matrix, such as
We want to see what influence the multiplication of the given matrix has on the vectors.
In the first case, we get a totally new vector with a different direction and different length
when compared to the original vector. This is what usually happens and is of no interest
here. In the second case something interesting happens. The multiplication produces a
vector which means the new vector has the same direction as
the original vector. The scale constant, which we denote by is 10. The problem of
systematically finding such’s and nonzero vectors for a given square matrix will be the
theme of this chapter. It is called the matrix eigenvalueproblem or, more commonly, the
eigenvalueproblem.
We formalize our observation. Let be a given nonzero square matrix of
dimension Consider the following vector equation:
......(1)
.Ax lx
n n.
A[a
jk]
l
l
[30
40]
T
10 [3 4]
T
,
c
63
47
dc
5
1
dc
33
27
d, c
63
47
dc
3
4
dc
30
40
d.
c08.qxd 11/9/10 3:07 PM Page 323
(2))X0(
0


AI
AX
X
To have a non-zero solution of this set of homogeneous linear equation (2) | A-λI | must
.be equal to zero i.e
 (3)AI0
.The following procedure can find the eigen values & eigen vector of n order matrix A
1.
to find the characteristic polynomial P( λ) = det [A−λI
2.
to find the roots of the characteristic equations the roots are eigen
values that we required
P()0
3.
To solve the homogenous system
]
.To find n- eigen vectors
[Α−λΙ]Χ=0 wallaa alebady
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2. Example: Determine the eigen value and corresponding eigen vector at
the matrix







23
14
A
Solution:












X



1





a




 x
4 


x





x







 










 


X



 



 
 

xxxxx

















 
   
 (4

















0


1/ 2
1/ 2
3/ 10
1/ 10
the eigenvectors 
1/ 2
1/ 2
2
1
5the eigen vector corresponding to
,0523
05
5
23
14
5
3/ 10
1/ 10
3
1
10
31 for X

X is defind by
eigenvector may be normalized to unit length the normalized eigenvector
3
The eigen vector corresponding to
1
3 where a is arbitrary constant
0323
0341
23
14
To find the eigen vector for
1
51 ,&
05)(05
)(2)300
23
14
0
0
0
23
14
22
2
2121121
21121
2
1
2
1
2
1
1
21
21221
21121
2
1
2
1
2lengthaa
a
X
is
a xax xxxx
xxxxx
xx
AX
Xfor
a
a
lengtha
length
X
X
a
a
X
is
alet xa,x
xxxxx
x
x
x
x
AX
X
A
I





1)(6


 wallaa alebady 
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3. Example: Determine the eigen values and corresponding eigen vectors of
the matrix
A
P

P








X

3c










2 / 3



X
9
6























X
  
























 c 




























 




A










 




2 / 31/ 3
2 / 3 1/ 32 / 3
2 / 31/ 32 / 3
0(:
0(:
1/ 3
2 / 3
2 / 3
1
2
2
3144,
1
2
2
2
2
2
042
022
0223
0
402
022
223
0)(3
)(96 ,3 ,
9
18 )09)((01629918()
0
702
052
226
0(
)
702
052
226
333
222
1
1
213
31
21
321
3
2
1
11
321
223
I)XA

case
I)XAcase
c
clengthc
length
X
c
c
c
c
the eigenvecto rs is X
cxxlet

x
xx
xx
xxx
x
x
x
I XAcase
distinct

Simplifyin g we have
I








 wallaa alebady 
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4. .Find the eigenvalues. Find the corresponding eigenvectors
1. 2. 3. 4.D
13 5 2
27 8
547
TD
353
046
001
Tc
52
96
dc
3.0 0
00.6
d
5.E
3042
01 24
24 1 2
02 23
U
c08.qxd 10/30/10 10:56 AM Page 329
H.W4 wallaa alebady
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