Introduction to Linear algebra-Strang-Solutions-Manual_ver13

I i 2 i 3 i 4 34 detp 35 3 by 3 permutation

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Unformatted text preview: e P 2 D P leads to 2 D  so  D 0 or 1. 13 (a) P u D .uuT /u D u.uT u/ D u so  D 1 (b) P v D .uuT /v D u.uT v/ D 0 (c) x 1 D . 1; 1; 0; 0/, x 2 D . 3; 0; 1; 0/, x 3 D . 5; 0; 0; 1/ all have P x D 0x D 0. 14 Two eigenvectors of this rotation matrix are x 1 D .1; i / and x 2 D .1; i / (more generally c x 1 , and d x 2 with cd ¤ 0). 15 The other two eigenvalues are  D 1 . 2 16 Set  D 0 in det.A p 1 ˙ i 3/; the three eigenvalues are 1; 1; 1. I / D .1 / : : : .n / to find det A D .1 /.2 /    .n /. p p 17 1 D C d C .a d /2 C 4bc/ and 2 D 1 .a C d / add to a C d . 2 2 If A has 1 D 3 and 2 D 4 then det.A I / D . 3/. 4/ D  7 C 12.     40 32 22 18 These 3 matrices have  D 4 and 5, trace 9, det 20: ; ; . 05 16 37 1 .a 2 19 (a) rank D 2 (d) eigenvalues of .B 2 C I / 1 1 are 1; 2 ; 1 . 5  01 has trace 11 and determinant 28, so  D 4 and 7. Moving to a 3 by 3 28 11 " # 0 10 0 1 has det.C I / D 3 C 62 11 C 6 D companion matrix, C D 0 6 11 6 .1 /.2 /.3 /. Notice the trace 6 D 1 C 2 C 3, determinant 6 D .1/.2/.3/, and also 11 D .1/.2/ C .1/.3/ C .2/.3/. 20 A D  (b) det.B T B/ D 0 Solutions to Exercises 60     I / has the same determinant as .A I /T 10 1 1 have different and 0 0 eigenvectors. because every square matrix has det M D det M T . 1 0 21 .A 22  D 1 (for Markov), 0 (for singular), 23  0 1 1 2 (so sum of eigenvalues D trace D 1 /. 2    0 01 11 Always A2 is the zero matrix if  D 0 and 0, , , . 0 00 11 by the Cayley-Hamilton Theorem in Problem 6.2.32. 24  D 0; 0; 6 (notice rank 1 and trace 6) with x 1 D .0; 2 ; 1/, x 2 D .1; 2 ; 0/, x 3 D .1; 2; 1/. 25 With the same n ’s and x ’s, Ax D c1 1 x 1 C    C cn n x n equals B x D c1 1 x 1 C    C cn n x n for all vectors x . So A D B . 26 The block matrix has  D 1, 2 from B and 5, 7 from D . All entries of C are multiplied by zeros in det.A I /, so C has no effect on the eigenvalues. 27 A has rank 1 with eigenvalues 0; 0; 0; 4 (the 4 comes from the trace of A). C has rank 2 (ensuring two zero eigenvalues) and .1; 1; 1; 1/ is an eigenvector with  D 2. With trace 4, the other eigenvalue is also  D 2, and its eigenvector is .1; 1; 1; 1/. 28 B has  D 1, 1, 1, 3 and C has  D 1; 1; 1; 3. Both have det D 3. p p 3; Rank-1 matrix: .C / D 29 Triangular matrix: .A/ D 1; 4; 6; .B/ D 2, 3, 0; 0; 6.       ab 1 aCb 1 30 D D .a C b / ; 2 D d b to produce the correct trace cd 1 cCd 1 .a C b/ C .d b / D a C d . 31 Eigenvector .1; 3; 4/ for A with  D 11 and eigenvector .3; 1; 4/ for PAP T . Eigenvec- tors with  ¤ 0 must be in the column space since Ax is always in the column space, and x D Ax =. 32 (a) u is a basis for the nullspace, v and w give a basis for the column space (b) x D .0; 1 ; 1 / is a particular solution. Add any c u from the nullspace 35 (c) If Ax D u had a solution, u would be in the column space: wrong dimension 3. 33 If vT u D 0 then A2 D u.vT u/vT is the zero matrix and 2 D 0; 0 and  D 0; 0 and trace .A/ D 0. This z...
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