Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 2 0 1 1 0 1 1 11 11 2 1 1 10 11 110 k d and a d 1 2 1

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Unformatted text preview: ero trace also comes from adding the diagonal entries of A D uvT :     u1  u1 v1 u1 v2 v1 v2 D AD has trace u1 v1 C u2 v2 D vT u D 0 u2 u2 v1 u2 v2 I / D 0 gives the equation 4 D 1. This reﬂects the fact that P 4 D I . The solutions of 4 D 1 are  D 1; i; 1; i : The real eigenvector x 1 D .1; 1; 1; 1/ is not changed by the permutation P . Three more eigenvectors are .i; i 2 ; i 3 ; i 4 / and .1; 1; 1; 1/ and . i ; . i /2 ; . i /3 ; . i /4 /: 34 det.P 35 3 by 3 permutation matrices: Since P T P D I gives .det P /2 D 1, the determinant is 1 or 1. The pivots are always 1 (but there may be row exchanges). The trace of P can be 3 (for P D I ) or 1 (for row exchange) or 0 (for double exchange). The possible eigenvalues are 1 and 1 and e 2i=3 and e 2i=3 . Solutions to Exercises 61 36 1 D e 2i=3 and 2 D e   2i=3 give det 1 2 D 1 and trace 1 C 2 D 1. 2 sin  with  D has this trace and det. So does every M 1 AM ! cos  3 37 (a) Since the columns of A add to 1, one eigenvalue is  D 1 and the other is c :6 (to give the correct trace c C :4). cos  AD sin  (b) If c D 1:6 then both eigenvalues are 1, and all solutions to .A multiples of x D .1; 1/. I / x D 0 are (c) If c D :8, the eigenvectors for  D 1 are multiples of (1, 3). Since all powers An   111 n also have column sums D 1, A will approach D rank-1 matrix A1 with 433 eigenvalues 1; 0 and correct eigenvectors. .1; 3/ and .1; 1/. Problem Set 6.2, page 307     12 11 10 1 D 03 01 03 0   1 11 1 1 ; D 1 33 1   Put the eigenvectors in S 11 2 2 A D SƒS 1 D and eigenvalues in ƒ. 01 0 3 If A D SƒS 1 3 0 5 1   00 04 1 0 &quot; 3 4 1 4 1 4 1 4 # .   1 23 D . 1 05 then the eigenvalue matrix for A C 2I is ƒ C 2I and the eigenvector matrix is still S . A C 2I D S.ƒ C 2I /S 1 D SƒS 1 C S.2I /S 1 D A C 2I . 4 (a) False: don’t know ’s 5 With S D I; A D SƒS triangular, so SƒS 1 (b) True (c) True (d) False: need eigenvectors of S 1 1 D ƒ is a diagonal matrix. If S is triangular, then S is also triangular. 6 The columns of S are nonzero multiples of .2;1/ and .0;1/: either order. Same for A 7 A D SƒS  1 D  1 1 1 1  1 2   1 1 1 =2 D 1  1 C 2 1 2 3 3 3 1  . 1 2 =2 D 1 C 2  b for any a and b . a       1 11 1 2 1 0 1 2 1 8 A D SƒS D D . Sƒk S 10 1 1 1 1  2 1    0  2   k   1 2nd component is Fk 1 2 1 0 1 2 1 D . 1 1 1 0 .k k /=.1 2 / 0 k 1 2 1 1 2 2   :5 :5 9 (a) A D has 1 D 1, 2 D 1 with x 1 D .1; 1/, x 2 D .1; 2/ 2 10 # &quot; #   n &quot; 2 1 2 1 1 11 0 3 3 3 (b) An D ! A1 D 2 3 1 1 1 1 2 0 . :5/n a b is 1 D 3 10 The rule Fk C2 D Fk C1 C Fk produces the pattern: even, odd, odd, even, odd, odd, : : : (b) False (repeated  D 2 may have only one line of (c) False (repeated  may have a full set of eigenvectors) 11 (a) True (no zero eigenvalues) eigenvectors) Solutions to Exercises 62 12 (a) False: don’t know  13 A D  (b) True: an eigenvector is missing (c) True.     83 94 10 5 only eigenvectors (or other), A D , AD ; 32 41 50 are x D .c;...
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