Singular Value Decompostion (SVD): Worked example 3

It is about a general understanding of SVD with 2x3 worked Example.

1. Singular Value Decomposition (SVD)
Isaac Amornortey Yowetu
December 31, 2021

2. Example 3
Find the matrices Uand V for the matrix
A=
"
3 1 1
1 3 1
#
2/18

3. Finding the Singular Values
Solution
A=
"
3 1 1
1 3 1
#
A
T
=
2
6
6
4
31
1 3
1 1
3
7
7
5
AA
T
=
"
11 1
1 11
#
3/18

4. Characteristics Polynomial
det(A
T
A) =





11 1
1 11





(1)
P(A
T
A) = (11)(11)1 (2)
=
2
22+120 (3)
= (12)(10) (4)
Eigenvalues
1=
2
1=12 (5)
2=
2
2=10 (6)
4/18

5. Singular Values
1=
p
1=
p
12 (7)
2=
p
2=
p
10 (8)
Singular Values Decompostion of A
=
"p
12 0 0
0
p
10 0
#
5/18

6. Constructing Matrix U
A=UV
T
(9)
AA
T
=UV
T
(UV
T
)
T
(10)
=U
T
V
T
VU
T
(11)
=U
2
U
T
(12)
=UDU
T
(13)
AA
T
=
"
11 1
1 11
#
(14)
6/18

7. Constructing Matrix U Cont'd
When=12
"
11 1
1 11
#
=
"
1 1
11
#
(15)
By row reduction form, we have:
"
1 1
11
#
=>
"
1 1
0 0
#
(16)
Forming equations with some variables:
x+y=0 (17)
Our eigenvector becomes:(1;1)
t
7/18

8. Constructing Matrix U Cont'd
When=10
"
11 1
1 11
#
=
"
1 1
1 1
#
(18)
By row reduction form, we have:
"
1 1
1 1
#
=>
"
1 1
0 0
#
(19)
Forming equations with some variables:
x+y=0 (20)
Our eigenvector becomes:(1;1)
t
8/18

9. Constructing Matrix U Cont'd
Let Z be the egigenvectors of the various eigenvalues.
Z=
"
1 1
11
#
We normalized each column of Z to get U
U=
"p
2
2
p
2
2p
2
2

p
2
2
#
9/18

10. Constructing Matrix V
A
T
A=
2
6
6
4
31
1 3
1 1
3
7
7
5
"
3 1 1
1 3 1
#
(21)
=
2
6
6
4
10 0 2
0 10 4
2 4 2
3
7
7
5
(22)
10/18

11. Constructing Matrix V Cont'd
Characteristics Polynomial
det(AA
T
) =








10 0 2
0 104
2 4 2








(23)
P(AA
T
) =(12)(10) (24)
Eigenvalues
1=
2
1=12 (25)
2=
2
2=10 (26)
3=
2
3=0 (27)
11/18

12. Constructing Matrix V Cont'd
When=12
2
6
6
4
10 0 2
0 104
2 4 2
3
7
7
5
=
2
6
6
4
2 0 2
02 4
2 4 10
3
7
7
5
(28)
By row reduction form, we have:
2
6
6
4
2 0 2
02 4
2 4 10
3
7
7
5
=>
2
6
6
4
2 0 2
02 4
0 4 8
3
7
7
5
=>
2
6
6
4
2 0 2
02 4
0 0 0
3
7
7
5
(29)
12/18

13. Constructing Matrix V Cont'd
Forming equations with some variables:
2x+2z=0 (30)
2y+4z=0 (31)
Our eigenvector becomes:(1;2;1)
t
13/18

14. Constructing Matrix V Cont'd
When=10
2
6
6
4
10 0 2
0 104
2 4 2
3
7
7
5
=
2
6
6
4
0 0 2
0 0 4
2 48
3
7
7
5
(32)
By row reduction form, we have:
2
6
6
4
2 48
0 0 2
0 0 4
3
7
7
5
=>
2
6
6
4
2 48
0 0 2
0 0 0
3
7
7
5
(33)
14/18

15. Constructing Matrix V Cont'd
Forming equations with some variables:
2x+4y=0 (34)
z=0 (35)
Our eigenvector becomes:(2;1;0)
t
15/18

16. Find the remaining eigenvector
AX=0 (36)
"
3 1 1
1 3 1
#
2
6
6
4
x
y
z
3
7
7
5
=
"
0
0
#
(37)
By row reduction form, we have:
"
3 1 1
1 3 1
#
=>
"
3 1 1
0 10 4
#
=>
"
3 1 1
0 5 2
#
(38)
Forming equations with the variables:
3x+y+z=0 (39)
5y+2z=0 (40)
Our eigenvector becomes:(1;2;5)
t
16/18

17. Constructing Matrix V Cont'd
Z=
2
6
6
4
1 2 1
21 2
1 0 5
3
7
7
5
(41)
We normalize each column vector of Z to obtain matrix V
V=
2
6
6
4
p
6
6
2
p
5
5
p
30
30p
6
3

p
5
5
p
30
15p
6
6
0

p
30
6
3
7
7
5
(42)
17/18

18. Singular Value Decomposition
A=
"p
2
2
p
2
2p
2
2

p
2
2
# "p
12 0 0
0
p
10 0
#
2
6
6
4
p
6
6
p
6
3
p
6
6
2
p
5
5

p
5
5
0
p
30
30
p
30
15

p
30
6
3
7
7
5
(43)
18/18