Question

a SVD for A and A2 if 2. Perform 1 3 2 -1 0 0 2 _ A 4 -2 1 -4 0 2

Answer 1

SVD means singular value decomposition.

SVD for the given Matrix A is as follows.

Here we have to find eigenvalues and eigenvectors.

Solution: 2 1 -1 3 0 0 -1 2 A = 1 -2 4 1 -4 2 +4.A 15 7 -1 12 7 5 -2 -1 -2 -13 22 12 8 -13 21 Find Eigen vector for A A A A A0 (15 -A) (5-A) 7 -1 12 -2 7 8 = 0 -2 -1 (22-A) -13 12 8 -13 (21 -A) 7 (5-A) -2 7 (5-A) |(5-A) 8 7 -2 8 -2 (22-A) -2 (22 A) -13 -1 (22-A) -13 -2 -13 -2 12 x1 : (15-A) x -7 x -1 x-1 0 = -13 (21 A) -13 8 (21 A) 12 -13 (21 A) 12 8 12 8 )-7x (313-2032-72 )-1 x(- 173 -255A 1212 (15 ) x (389-4401 4812-5 - 7 x 113 144A 12 x - )-( -1421A-491)+ -2076 3060A- 1441= 116022 - 6313 113 144A-2 5835 6989A 2191

(i-6313 +966 2-26521 1681) = 0 (24-6313 +9662-26521+ 1681 Roots can be found using newton raphson method +Newton Raphson method for x4- 63x3+966x2-2652x + 1681 0 x-0.93020384 x4-63x3966x- 2652x 1681 x3-62.06979616x2 908.26243745x 1807.13079578 Now, using long division x 0.93020384 Newton Raphson method for x -62.06979616x +908.26243745x 1807.13079578 0 = x-2.35397647 3-62.06979616x+908.26243745x-1807.130/95/8259.71581969x + 767.69280289 Now, using long division x 2.35397647 Now, x 59.71581969x 767.69280289 0 18.73127615 and x 40.98454354 The eigenvalues of the matrix A are given by A 0.93020384, 2.35397647, 18.73127615, 40.98454354 1. Eigenvectors for 40.98454354 0.59320416 0.37973154 -0.75601856 1

+2. Eigenvectors for A 18.73127615 = 6.14651506 2.62859506 7.46582039 1 +3. Eigenvectors for A 2.35397647 -1.09159899 0.3494441 0.64172219 1 +4. Eigenvectors for A 0.93020384 0.37977234 -2.4197119 = 0.40533608 1 For Eigenvector-1 (0.59320416, 0.37973154, 0.75601856, 1), Length L V0.593204162 + 0.379731542 (-0.75601856)2- 1.43793299 +1 0.59320416 0.37973154 -0.75601856 1 (0.4125395, 0.26408153, 0.52576759, 0.6954427) So, normalizing gives u1 1.43793299 1.43793299 1.43793299 1.43793299 - For Eigenvector-2 (6.14651506, 2.62859506, 7.46582039, 1), Length L v6.146515062 2.628595062 7.465820392 +12 10.07112871 6.14651506 2.62859506 746582039 1 So, normalizing gives u2 (0.61031045, 0.26100303, 0.7413092, 0.09929374) 10.07112871 10.07112871 10.07112871 10.07112871 For Eigenvector-3 (-1.09159899, 0.3494441, 0.64172219, 1), Length L (-1.09159899)20.34944412 +0.641722192 12 1.65091093 = = - 1.09159899 0.3494441 0.64172219 1 So, normalizing gives u (0.66121011, 0.21166745, 0.38870794, 0.6057262) 1.65091093 1.65091093 1.65091093 1.65091093 For Eigenvector-4 (0.37977234, 2.4197119, 0.40533608, 1), Length L 0.379772342(-2.4197119) 0.405336082 12-2.67647717 =

-1.67201863 0.3328001 0.26160317 1 So, normalizing gives v4 = ( - 0.8386512, 0.16692589, 0.13121493, 0.50158006) 1.99369969 1.99369969 1.99369969 1.99369969 1st Solution 40.98454354 0 0 0 6.40191718 0 V18.73127615 0 0 0 4.32796443 0 0 2.35397647 1.53426741 0 C 0 0 C 0.96447075 0 0 V0.93020384 0.4125395 0.61031045 -0.66121011 0.14189261 0.26408153 0.26100303 0.21166745 -0.90406596 U -0.52576759 0.7413092 0.38870794 0.15144388 0.6954427 0.09929374 0.6057262 0.37362545 1 -AT. V is found using formula v 0.15538378 047625711 -0.21377374 0.83865119 0.22869316 -0.20155155 -0.93766314 -0.1669259 0.86871823 0.39204305 -0.27278842 -0.1312149 0.41095492 0.76082744 0.02598303 -0.50158007 Or 2nd Solution 40.98454354 0 6.40191718 C 0 V18.73127615 0 0 0 4.32796443 0 0 2.35397647 0 0 1.53426741 0 0 C 0 0 0.96447075 V0.93020384 0 0 C

0.15538378 0.47625711 -0.21377373 -0.8386512 0.22869316 -0.20155155 -0.93766314 0.16692589 . V -0.86871823 0.39204305 -0.27278841 0.13121493 0.41095492 0.76082743 0.02598303 0.50158006 U is found using formula u =-A . V 0.4125395 0.61031045 -0.6612101 -0.1418926 0.26408153 0.26100303 0.21166745 0.90406597 . U= 0.52576759 0.7413092 0.38870795 -0.15144389 0.6954427 0.09929374 0.6057262 -0.37362543

Singular value matrix for the A^2.

2 1 -1 3 -1 2 0 2 1 -1 3 6 4 -19 13 -1 2 0 -2 0 0 1 2 -12 3 4 1 1 -2 4 7 -7 13 -25 3 1 1 -4 2 0 -4 2 1 0 1

Now to find the required singular value decomposition for A^2.

Is given below.

In this step, we are finding the eigenvalues and eigenvectors.

Solution: 2 1 -1 3 0 0 -1 2 A = 1 -2 4 1 -4 2 +4.A 15 7 -1 12 7 5 -2 -1 -2 -13 22 12 8 -13 21 Find Eigen vector for A A A A A0 (15 -A) (5-A) 7 -1 12 -2 7 8 = 0 -2 -1 (22-A) -13 12 8 -13 (21 -A) 7 (5-A) -2 7 (5-A) |(5-A) 8 7 -2 8 -2 (22-A) -2 (22 A) -13 -1 (22-A) -13 -2 -13 -2 12 x1 : (15-A) x -7 x -1 x-1 0 = -13 (21 A) -13 8 (21 A) 12 -13 (21 A) 12 8 12 8 )-7x (313-2032-72 )-1 x(- 173 -255A 1212 (15 ) x (389-4401 4812-5 - 7 x 113 144A 12 x - )-( -1421A-491)+ -2076 3060A- 1441= 116022 - 6313 113 144A-2 5835 6989A 2191

Roots can be found using newton raphson method +Newton Raphson method for x -1747x 3 + 368269x2- 2080803x+2825761 0 . x - 2.22482285 Now, using long division 1747x +368269x- 2080803x+ 28257613-1744.77517715x364387.18430937x- 1270106.06434531 x 2.22482285 +Newton Raphson method for x3 -1744.77517715x2+364387.18430937x 1270106.06434531 0 x-3.54566916 3-1744.77517715x+364387.18430937x 1270106.06434531 x-1741.22950799x+358213.36054652 Now, using long division x 3.54566916 Now, x-1741.22950799x+358213.36054652 0 = 238.3516052 and x 1502.87790279 .x The eigenvalues of the matrix A are given by A 2.22482285, 3.54566916, 238.3516052, 1502.87790279 = +1. Eigenvectors for A 1502.87790279 0.82797844 0.46899229 V1 = -0.48023937 1 2. Eigenvectors for A 238.3516052 -3.11360194 -0.27398885 -3.55342366 3. Eigenvectors for A 3.54566916 -1.37964247 1.69560727 1.35955643 1

4. Eigenvectors for 2 2.22482285 -0.20567786 -1.20154187 V 0.55428461 1 = 0.827978442 +0.468992292+(-048023937)2+12 1.4615512 For Eigenvector-1 (0.82797844, 0.46899229, 0.48023937, 1), Length L 0.82797844 0.46399229 -0.48023937 1 So, normalizing gives u ( 0.56650662, 0.32088666, 0.32858197, 0.68420456) 14615512 1.4615512 14615512 1.4615512 -3.11360194)2(-0.27398885)2 +(-3.55342366) 12-4.83698322 For Eigenvector-2 (-3.11360194, 0.27398885, -3.55342366, 1), Length L -3.11360194 -0.27398885 -3.55342366 So, normalizing gives u (0.64370741, 0.05664457, 0.73463634,0.20674043) 4.83698322 4.83698322 4.83698322 4.83698322 For Eigenvector-3 (- 1.37964247, 1.69560727, 1.35955643, 1), Length L (- 137964247)+1.695607272 1.35955643+12-2.76168265 = = -137964247 1.69560727 135955643 = ( -0.4995659 , 0.61397615, 0.49229278, 0.36209808) So, normalizing gives u = 2.76168265 2.76168265 2.76168265 2.76168265 -0.20567786)+(-1.20154187)2 0.55428461212 1.6712982 For Eigenvector-4 (-0.20567786, 1.20154187, 0.55428461, 1), Length L = -0.20567786 -1.20154187 0.55428461 So, normalizing gives u ( -0.12306473 , 0.71892728, 0.33164914, 0.59833727) 1.6712982 1.6712982 1.6712982 1.6712982

1st Solution 1502.87790279 0 0 0 38.76696922 0 238.3516052 0 0 0 15.43864 0 0 ,Σ= V3.54566916 0 1.88299473 0 C 0 0 1.49158401 2482285 0 0 0 0.56650662 -0.64370741 0.4995659 -0.12306473 0.32088666 -0.05664457 0.61397615 -0.71892728 .U -0.32858197 -0.73463634 0.49229278 0.33164914 0.68420456 0.20674043 0.36209808 059833727 Vis found using formula v,A u 0.03662532-0.58692627 0.56432989 0.57940305 0.29318038 0.2794935 -0.50848684 0.75984857 -0.92839203 -0.11714811-0.28072705 021344041 0.22537098-0.75078442 -0.58665598 0.203385

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