Introduction to Linear algebra-Strang-Solutions-Manual_ver13

So positive eigenvalues positive energy 20 a the

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Unformatted text preview:  sin  cos  1 0 0 1 cos  sin  sin  cos 2 D cos  sin 2 26 Eigenvectors .1; 0/ and .1; 1/ give a 45ı angle even with AT very close to A. sin 2 . cos 2 Solutions to Exercises 69 p b 2 4 ac/. Then 1 2 is b 2 4 c . For det.A C tB I / we have b D 3 8t andp D 2 C 16t t 2 . The minimum of c b 2 4 c is 1=17 at t D 2=17. Then 2 1 D 1= 17.   T 4 2Ci 28 A D D A has real eigenvalues  D 5 and 1 with trace D 4 and 2i 0 1 27 The roots of 2 C b C c D 0 are 2 . b ˙ p T det D 5. The solution to 20 proves that  is real when A D A is Hermitian; I did not intend to repeat this part. 29 (a) A D QƒQ T times A T D Qƒ T Q T equals A T times A because ƒƒ T D ƒ T ƒ (diagonal!) (b) step 2: The 1; 1 entries of T T T and T T T are jaj2 and jaj2 C jb j2 . This makes b D 0 and T D ƒ.   T  1 q 11 : : : n q 1n  max jq11 j2 C    C jq1n j2 D max . 30 a11 is q11 : : : q1n x T Ax : (b) zT Az is pure imaginary, its real part is x Ax C y Ay D 0 C 0 (c) det A D 1 : : : n  0 W pairs of ’s D i b; i b . 32 Since A is diagonalizable with eigenvalue matrix ƒ D 2I , the matrix A itself has to be SƒS 1 D S.2I /S 1 D 2I . (The unsymmetric matrix Œ2 1 I 0 2 also has  D 2; 2.) 31 (a) x T .Ax / D .Ax /T x D x T AT x D T T Problem Set 6.5, page 350 1 Suppose a > 0 and ac > b 2 so that also c > b 2 =a > 0. (i) The eigenvalues have the same sign because 1 2 D det D ac b 2 > 0. (ii) That sign is positive because 1 C 2 > 0 (it equals the trace a C c > 0).   1 10 2 2 2 Only A4 D has two positive eigenvalues. x T A1 x D 5x1 C 12x1 x2 C 7x2 10 101 is negative for example when x1 D 4 and x2 D 3: A1 is not positive definite as its determinant confirms.      b 0 Positive definite 10 1 10 1 1b 3 D D LDLT for 3 < b < 3 b 1 0 9 b2 b 1 0 9 b2 01       Positive definite 10 2 4 10 2 0 12 D D LDLT . for c > 8 21 0c8 21 0c8 01 4 f .x; y/ D x 2 C 4xy C 9y 2 D .x C 2y/2 C 5y 2 ; x 2 C 6xy C 9y 2 D .x C 3y/2 . 5 x 2 C 4xy C 3y 2 D .x C 2y/2 y 2 D difference of squares is negative at x D 2, y D 1, where the first square is zero.       01 x 01 6 AD produces f .x; y/ D x y D 2xy . A has  D 1 and 10 10 y 1. Then A is an indefinite matrix and f .x; y/ D 2xy has a saddle point. " #     233 12 65 7 RT R D and RT R D are positive definite; RT R D 3 5 4 is 2 13 56 345 singular (and positive semidefinite). The first two R’s have independent columns. The 2 by 3 R cannot have full column rank 3, with only 2 rows.      Pivots 3; 4 outside squares, `ij inside. 10 30 12 36 .T D 8 AD 6 16 21 04 01 x Ax D 3.x C 2y/2 C 4y 2 Solutions to Exercises 70 9 AD " 4 4 8 4 4 8 8 8 16 # has only one pivot D 4, rank A D 1, eigenvalues are 24; 0; 0; det A D 0. 10 A D " 2 1 0 1 2 1 0 1 2 # has pivots BD 2; 3 ; 4 ; 23 " 2 1 1 1 2 1 # "# "# 1 1 0 1 is singular; B 1 D 0 . 2 1 0 11 Corner determinants jA1 j D 2, jA2 j D 6, jA3 j D 30. The pivots are 2=1; 6=2; 30=6. 12 A is positive definite for c > 1; determinants c; c 2 1, and .c 1/2 .c C 2/ > 0. 4 and 4 d C...
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