Singular Value Decompostion (SVD): Worked example 2

It is about a general understanding of SVD with 3x2 worked Example.

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

2. Example 2
Find the matrices Uand V for the matrix
A=
2
6
6
4
2 1
1 0
0 1
3
7
7
5
2/13

3. Finding the Singular Values
Solution
A=
2
6
6
4
2 1
1 0
0 1
3
7
7
5
A
T
=
"
2 1 0
1 0 1
#
A
T
A=
"
5 2
2 2
#
3/13

4. Characteristics Polynomial
det(A
T
A) =





52
2 2





(1)
P(A
T
A) = (5)(2)4 (2)
=
2
7+6 (3)
= (6)(1) (4)
Eigenvalues
1=
2
1=6 (5)
2=
2
2=1 (6)
4/13

5. Singular Values
1=
p
1=
p
6 (7)
2=
p
2=1 (8)
Singular Values Decompostion of A
=
2
6
6
4
p
6 0
0 1
0 0
3
7
7
5
5/13

6. Constructing Matrix V
A=UV
T
(9)
A
T
A= (UV
T
)
T
UV
T
(10)
=V
T
U
T
UV
T
(11)
=V
2
V
T
(12)
=VDV
T
(13)
A
T
A=
"
5 2
2 2
#
(14)
6/13

7. Constructing Matrix V Cont'd
When=6
"
52
2 2
#
=
"
1 2
24
#
(15)
By row reduction form, we have:
"
1 2
24
#
=>
"
1 2
0 0
#
(16)
Forming equations with some variables:
x+2y=0 (17)
Our eigenvector becomes:(2;1)
t
7/13

8. Constructing Matrix V Cont'd
When=1
"
52
2 2
#
=
"
4 2
2 1
#
(18)
By row reduction form, we have:
"
2 1
4 2
#
=>
"
2 1
0 0
#
(19)
Forming equations with some variables:
2x+y=0 (20)
Our eigenvector becomes:(1;2)
t
8/13

9. Let Z be the egigenvectors of the various eigenvalues.
Z=
"
2 1
12
#
We normalized each column of Z to get V
V=
"
2
p
5
5
p
5
5p
5
5

2
p
5
5
#
9/13

10. Constructing U matrix
A=UV
T
(21)
U=fu1;u2g (22)
u1=
1
s1
Av1 (23)
u2=
1
s2
Av2 (24)
10/13

11. Constructing U matrix
u1=
1
p
6
2
6
6
4
2 1
1 0
0 1
3
7
7
5
"
2
p
5
5p
5
5
#
=
2
6
6
4
p
30
6p
30
15p
30
30
3
7
7
5
(25)
u2=
2
6
6
4
2 1
1 0
0 1
3
7
7
5
"p
5
5
2
p
5
5
#
=
2
6
6
4
0
p
5
5
2
p
5
5
3
7
7
5
(26)
11/13

12. Find the remaining eigenvector
A
T
X=0 (27)
"
2 1 0
1 0 1
#
2
6
6
4
x
y
z
3
7
7
5
=
"
0
0
#
(28)
Forming equations with the variables:
2x+y=0 (29)
x+z=0 (30)
Our eigenvector becomes:(1;2;1)
t
12/13

13. Singular Value Decomposition
U=
2
6
6
4
p
30
6
0
p
6
6p
30
15
p
5
5

p
6
3p
30
30
2
p
5
5

p
6
6
3
7
7
5
(31)
A=
2
6
6
4
p
30
6
0
p
6
6p
30
15
p
5
5

p
6
3p
30
30
2
p
5
5

p
6
6
3
7
7
5
2
6
6
4
p
6 0
0 1
0 0
3
7
7
5
"
2
p
5
5
p
5
5p
5
5

2
p
5
5
#
(32)
13/13