Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Those orthogonal directions give u1 and u2 p p 21 1p2

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Unformatted text preview: C 5e t . 2 3 2 3 m21 m22 m23 m24 0 m12 m13 0 60 6 0 m22 m23 0 7 0 0 07 7 6 7 If M 1 JM D K then JM D6 4 m41 m42 m43 m44 5 D MKD4 0 m32 m33 0 5. 0 0 0 0 0 m42 m43 0 That means m21 D m22 D m23 D m24 D 0. M is not invertible, J not similar to K . The five 4 by 4 Jordan forms with  D 0; 0; 0; 0 are J1 D zero matrix and 2 3 2 3 0100 0100 60 0 0 07 60 0 1 07 J2 D 4 J3 D 4 0 0 0 05 0 0 0 05 0000 0000 2 3 2 3 0100 0100 60 0 0 07 60 0 1 07 J4 D 4 J5 D 4 0 0 0 15 0 0 0 15 0000 0000 7 (a) .M 8 9 10 11 12 13 Solutions to Exercises 73 Problem 12 showed that J3 and J4 are not similar, even with the same rank. Every matrix with all  D 0 is “nilpotent” (its nth power is An D zero matrix). You see J 4 D 0 for these matrices. How many possible Jordan forms for n D 5 and all  D 0? 1 Ji Mi D MiT in each block (2) M0 has those diagonal blocks Mi to get M0 JM0 D J T . (3) AT D .M 1 /T J T M T equals .M 1 /T M0 1 JM0 M T D .MM0 M T / 1 A.MM0 M T /, and AT is similar to A. 14 (1) Choose Mi D reverse diagonal matrix to get Mi 1 AM I / D det.M 1 AM M 1 IM /. This is det.M 1 .A I /M /. By the product rule, the determinants of M and M 1 cancel to leave det.A I /.        ab dc ba cd 16 is similar to ; is similar to . So two pairs of similar cd ba dc ab     10 01 matrices but is not similar to : different eigenvalues! 01 10 15 det.M 1 17 (a) False: Diagonalize a nonsymmetric A D SƒS 1 . Then ƒ is symmetric and similar     01 0 1 and are similar 10 1 0 (d) True: Adding I increases all eigenvalues by 1 (b) True: A singular matrix has  D 0. (c) False: (they have  D ˙1) 18 AB D B 1 .BA/B so AB is similar to BA. If AB x D x then BA.B x / D .B x /. 19 Diagonal blocks 6 by 6, 4 by 4; AB has the same eigenvalues as BA plus 6 20 (a) A D M to B 2 .  3 (c) 0  3 (d) 0 1 2 1 1 1 2 4 zeros. 2 BM ) A D .M BM /.M BM / D M B M . So A is similar (b) A2 equals . A/2 but A may not be similar to B D A (it could be!).    30 1 because 1 ¤ 2 , sothesematrices are similar. is diagonalizableto 04 4  1 has only one eigenvector, sonot diagonalizable (e) PAP T is similar to A. 3 21 J 2 has three 1’s down the second superdiagonal, and two independent eigenvectors for #   010 01 .  D 0. Its 5 by 5 Jordan form is with J3 D 0 0 1 and J2 D 00 J2 000 Note to professors: An interesting question: Which matrices A have (complex) square roots R2 D A? If A is invertible, no problem. But any Jordan blocks for  D 0 must have sizes n1  n2  : : :  nk  nk C1 D 0 that come in pairs like 3 and 2 in this example: n1 D (n2 or n2 C 1) and n3 D (n4 or n4 C 1) and so on. #" # " a10 a00 A list of all 3 by 3 and 4 by 4 Jordan forms could be 0 b 0 , 0 a 0 , 00c 00b 3 2 " # a1 a10 (for any numbers a; b; c ) a 6 7 0a1 5, b with 3; 2; 1 eigenvectors; diag.a; b; c; d / and 4 00a c 3 32 32 2 a1 a1 a1 a1 a1 a 76 7 6 76 with 4; 3; 2; 1 eigenvectors. , 5, 4 4 a 15 a b 15 4 b a b  J3  " Solutions to Exercises 74 I / must be just n . The CayleyHamilton Theorem in Problem 6.2....
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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