Question

A lo V20 2 0 - 1 1 1 (a) (b) Determine the singular values of A. Find the singular value decomposition of A. Your answer has to consist of three matrices U, E, V satisfying the appropriate properties and multiplied together to retrieve A.

Answer 1

Part (a)

To determine singular value of A we need to find non-zero eigenvalue(s) of A^{​​​​​​T}​​​​​A, say λ.

Therefore, singular value of A is √λ.

o - 1 A √2 2 and AT - 2 2 1 O O - 1 - ATA = 2 2 J2 252 om held ONIN 2. and LATA - NIL 2-7 O 2 / O 2 v2 6-7 2 2 2- to O > (2.2) [16–2)x= -1) -- 4) – 23 [2J7 (2-1)] → (2-1) [ 17-82+7²-4-8 ) 02-21 (22-8.2) A (2-1) (A-8) =0 O er

id= 0,2,8

Hence, singular values of A are $\sqrt\lambda = \sqrt2,2\sqrt2$

Part (b)

z = VE say o=87 - 0,=2&E give decemposition , we canaides singular values, E = 252 To 3= diagonal matrin ol 0 0 0 0 Now, we find au erthememel act of eigenelectory for AA. 252 0 2 252 21 22. 6 2 Xz 8 2 2 23 x3

Now, we we find au Arthonormal at of egenvectory for ATA. 2 x, 252 6 2 xi 22 X2 = 252 0 co 2 2 23 x3 6 3 A 2 y 2 2 2x, +252x2 = 881 252 x + 6+ 2043 = 822 2x2 + 2x₂ = 823 » [21, 22, 2] = [ 22,3,1] 252 2 JE 6 Y Y Y J 29, +252 42 = 2y. 2524, +69₂ +243 = 2yz 242 + 2y3 = 243 [y, yz yz ] = [5, 0,2 2 2 l yu 4 5 16 > » [y, yz yz "B

2 Zi 2 01 Zz o 2 2 Zz 2JZ Zi 2 6 Zz Zz 2Z, +255 Zz o 222, + 62, + 22-0 2 Z2+ 2 Zz =0. [Z,, Zz ,Zz] =[ J2,-1,1) ㅋ 8 >>

Normalising equ@ ® and ©g and putting them together, as 1/2 7/2 / 1456 VjB 3/12 152 -2/50 156 753 V 3/512 -1/2 7512 -2/J6 ½ or V.

Siusilarly. U can be determined of eigenvectors of LAAT will have same normall set ty calculating to the eigenvalues as of Ata) AA AATE 2 2 2 2 6 2 2 2 2 For 2=8, eigenvector be [ 1 2 3 To For 2=2, eigenvector be - - - eigenvertor be 1 1 1 1 9 J3 For 7=0, حل re

.. U= '/a -- 2/16 Y16 - मि O

To sum up,

$\Sigma = \begin{bmatrix} 2\sqrt2&0 &0 \\ 0& \sqrt2 & 0\\ 0& 0& 0 \end{bmatrix}$

$V = \begin{bmatrix} \frac{1}{\sqrt6}& \frac{1}{\sqrt3} & \frac{1}{\sqrt2}\\ \frac{3}{\sqrt{12}}& 0 & -\frac{1}{2}\\ \frac{1}{\sqrt{12}}& - \frac{2}{\sqrt6}& \frac{1}{2} \end{bmatrix}$

And $U = \begin{bmatrix} \frac{1}{\sqrt6}& -\frac{1}{\sqrt3} & \frac{1}{\sqrt2}\\ \frac{2}{\sqrt{6}}& \frac{1}{\sqrt3}& 0\\ \frac{1}{\sqrt{6}}& - \frac{1}{\sqrt3}& -\frac{1}{\sqrt2} \end{bmatrix}$

Hence, $A = U \Sigma V^T$