This document discusses singular value decomposition (SVD) and provides an example to decompose the matrix A = [[2, -1], [2, 2]]. It finds the singular values σ1 = 3 and σ2 = 2 and constructs the matrices U, Σ, and V such that A = UΣV^T. It derives the eigenvalues and eigenvectors of A^TA to construct the diagonal matrix Σ and orthogonal matrix V, then uses the definition of U to construct it based on A and V.

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

2. Example 1
Find the matrices U Σ and V for the matrix
A =
"
2 −1
2 2
#
2/11

3. Finding the Singular Values
Solution
A =
"
2 −1
2 2
#
AT
=
"
2 2
−1 2
#
AT
A =
"
8 2
2 5
#
3/11

4. Characteristics Polynomial
det(AT
A) =

9. 8 − λ 2
2 5 − λ

14. (1)
P(AT
A) = (8 − λ)(5 − λ) − 4 (2)
= λ2
− 13λ + 36 (3)
= (λ − 9)(λ − 4) (4)
Eigenvalues
λ1 = σ2
1 = 9 (5)
λ2 = σ2
2 = 4 (6)
4/11

15. Singular Values
σ1 =
p
λ1 = 3 (7)
σ2 =
p
λ2 = 2 (8)
Singular Values Decompostion of A
Σ =
"
3 0
0 2
#
5/11

16. Constructing Matrix V
A = UΣV T
(9)
AT
A = (UΣV T
)T
UΣV T
(10)
= V ΣT
UT
UΣV T
(11)
= V Σ2
V T
(12)
= VDV T
(13)
AT
A =
"
8 2
2 5
#
(14)
6/11

17. Constructing Matrix V Cont’d
When λ = 9
"
8 − λ 2
2 5 − λ
#
=
"
−1 2
2 −4
#
(15)
By row reduction form, we have:
"
−1 2
2 −4
#
=>
"
−1 2
0 0
#
(16)
Forming equations with some variables:
−x + 2y = 0 (17)
Our eigenvector becomes: (2, 1)t
7/11

18. Constructing Matrix V Cont’d
When λ = 4
"
8 − λ 2
2 5 − λ
#
=
"
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
8/11

19. Let Z be the egigenvectors of the various eigenvalues.
Z =
"
2 1
1 −2
#
We normalized each column of Z to get V
V =
"
2
√
5
5
√
5
5
√
5
5 −2
√
5
5
#
9/11

20. Constructing U matrix
A = UΣV T
(21)
U = {u1, u2} (22)
u1 =
1
s1
Av1 (23)
u2 =
1
s2
Av2 (24)
10/11

21. Constructing U matrix
u1 =
1
3
"
2 −1
2 2
# "
2
√
5
5
√
5
5
#
=
" √
5
5
2
√
5
5
#
(25)
u2 =
1
2
"
2 −1
2 2
# " √
5
5
−2
√
5
5
#
=
"
2
√
5
5
−
√
5
5
#
(26)
A =
" √
5
5
2
√
5
5
2
√
5
5
−
√
5
5
# "
3 0
0 2
# "
2
√
5
5
√
5
5
√
5
5
−2
√
5
5
#
(27)
11/11