Find an SVD of the matrix. - 2 3 A= 0 -2 A=Find an SVD of the matrix. 4 7 A= 0 0 1 4 Give an SVD of matrix A below. A= (Type an exact answer, using radicals as needed.)Use technology to compute the singular values of the following 4 x 4 matrix and compute the condition number o /64. 4 0 -3 -7 - 6 9 9 9 7 -5 10 19 -1 2 4 -1 Write the singular values in decreasing order. 02 = 04 =\f10 Answer - 2 A = 4 7 4 Here we are trying to find out Solution using AAT & AT. A 1 Solution using A.A for normalized vectors v. A.AT = 65 0 32 32 0 17 - Find for : A.AT I A. AT - 21 = D 65- 2 0. 32 0 - 2 32 0 17-k . - 23 + 9272 + 812 = 0 .: 2 =0 2 : +1 1 : 81 -7 Feigenvectors . for 1: 81 65 0 32 0 0 O 0 O 81 0 10 32 0 . 17 0 0 111 - 16 0 32 O - 81 O 32 0 - 64 Using Row tounsfromation we give 1 -2 O O O O = The system associated with 2:81 . (A. A - 81 1 ) X 1 10 - 2 X 1 O * 2 O X 2 X3 0 0.0 X3 0 . X - 2 X 3 - 6 2 : 2x3 X2-6 . eigen vectors corresponding to the eigenvalue 2 = 81 150 V= 2x 3 ; Let x 3 = 1 O . Length= 14+0+1 - 2.236 2 : 4 , = (0. 894 0, 0.4:47) 012 = > feigenvectors for 2=1 65 0 32 - L 32 0 17 0 0 1 64 0 32 O -1 0 32 0 16 After Row transformation we give 1 0 0.5 0 1 . 0 - The system associated with eigenvalue 20=1 CAAT - 12 ) 1 0 .0.5 0 O 1 O 0 O 0 * 2: -0.513 . eigenvector Corresponding to the eigenvalue 2:1 - 1/2 23 VE O Let 23= 1 Length = 16 0.5 ) to+ 1 = 1.11913 = Feigen vectors for = 2:0 (AAT - 21 ) : 65 0 32 10 0 O O O - o 32 0 17 65 0 32 O O O 32 0 17 After Row transformation we give O -7 The system associated with eigenvalue 1=0 CA.AT - 02 ) x, 10 02 O 0 0 0 x3 eigenvectors corresponding to the eigenvalue 2=0. V - O Let 2q = 1 Length = 10 + 1to = 1 : 43 ( 0. 1, 0 )14 and Solution using AT. A for normalized Vector Vi ALA = 17 . 32 32 65 AT A - 21 = 0 172 32 32 65- 2 (17- 2 ) ( 65 -2 ) - 38*32 -0 2 :1 0 fa : 81 1 Eigenvector for 2 = 81 . A . A - 21 : 17 32 32. 65 O - 64 32 32 -16 After xoco transformation we will give. 1- 112 O - The system associated with 2 : 81(ATA - 812) = cigenvectors corresponding to eigenvalue 2: 81 15 V : 1/2 2 2 Let azel . U . = 1 /2 No Length. = ( 1/2 ) 2 + 12 = 1. 118 " V = ( 0 . 447 0 . 894 )/ 2 Eigenvectors for 2=1 AT. A - 21 = 17 32 32 65 16 32 32 64 After gow transformation we will tiny 0 0 The system associated with 2=l16 1x - - 2 12 eigenvectors corresponding to the eigenvalue 2 = 1 is V : - 2x2 " Let (2= 1 V. = - 2 Length: (-2)2 + 12 = 9.236 (- 0. 849 , 0.497 ) Now ist SUD solution using A AT E = 181 9 . 0 O 0 O O O O 0.844 - 0.447 O 1 0.447 0.894 V is found by using Uj = A. U; 0.447 - 0. 894 V = 0. 894 0.4 47\f2 - Now, eigenvector for 2= 16 A'A - 21 : 4 - 6 - 16 1 -6 13 O 1 4 - 6 16 0 - 6 13 O 16 - 12" - 6 -6 -3 Now , R. = R, / - 12 1 12 -C -3 Now R2 : Re'+ 6 R , o The system associated with the . eigenvalue 2 = 16 ( A !A - 16 7 ) 1 1/2 D X2 O = > 24 + 0. 5 012 = 0 : 2 : - 0.5 ( 2 . . eigen vectors . corresponding to the eigenvalue 2= 16 ip : V = -0.5 X 2 Let or 2 = . .: V. = -0.5 DC 2 13 Now . Ecignvector for 2 = 1 . "(AT. A - 21 ) - 4 -6 -1 1 0 3 -6 - 6 13 0 1 - 6 12 Now , R, = RI/ 3 1 - 2 - 6 12 Now R 2 = R2 + ( +6 R, ) 1 - 2 O . O The system associated with the eigenvalue 2 =1 (A'A - aI ) 1 -1 O O 21 -212=0" . eigen vectors Corresponding to the eigenvalue . 2 = 1 is V = let bel X 2 24) for eigen vectors - 1 ( - 0.5 , 1) Length L = (- 0.5 ) 2 + 12 2 LIL8 so , normalizing gives V. = / - 0.5 1 ) : ( 0 .4 4 72 , 0 84 46) 1418 For eigenvaluetor - ( & ), Length L = 22 + 12 = 2.2361 So normalizing gives , - 60 8949 , 0.4672 ( 2.2361 2.236.1 Sup Solution E = 016 0 4 0 - 0.4472 0. 84 4 4 0. 8449 0. 4 472 0. 89 49 - 0. 4672 6; -0.4672 - 0.84 49\f6 Now, Second Solution Using A . AT for normalized vetos u ; ( A . AT = - 2 - 2 0 13 - 6 0 - 2 .3 -2 - 6 4 Find cigen vector for A.AT AA - 21 =0 13-2 - 6 - 6 4-2 . (13 - 2 ) ( 4 -2 ) - 36 :0 . Ca - 12 ( 2 - 16) = 0 2 : 16 Now, for 2 = 16 (A AT - 21) : 13 - 6 - 16 b -3 -6 L- 6 4 1 - 6 - 12 Now , R . = RI / - 3 = 1 2 -6 - 12 Now R2 = Rat 6 R = 0 CA.AT - 167 1 2 2 2 0\f-> for eigen vector - 1 (-2.1) , Length L= (- 2 )2 + 12 : 2 .236 1 So, normalizing gives 2 1 2. 2361 2. 2361 2 , = ( - 0 . 8 944, 6. 4 472 ) - for eigenvector - 2 (as + 1 ), Length: 1= (0.5)2 + (172 = 1.118 So normalizing gives 0.5 1 ( 0.44 72, 0.89 94 1.118 1.118 JUL : C0.4472, 0.8944 O 4 0 6 JI O 1 -0. 8949 0.4472 0. 4 472 0. 894 4: V is were found using formula U;= L.AT.y . Gi V = 0. 4472 -0. 8944 - 0. 8944 -0.4472 verify god Solution U X E : - 0.89443 0.44921 4 0 0.45721 0. 89443 0 1 - 3.57772 0.49721 1. 78 884 0. 89443 ( U X E X ( V - 3:57772 0.4972) 1. 78884 0.894 43 0.4472 .:- 0.89 4b -0.8949 - 0.4972 - 1.99 99 4 9. 99 4 4 2 O - 1. 99 943 Solution is possible