Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# I believe this is a unifying concept from linear

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Unformatted text preview: 12 are never both positive). B is never positive deﬁnite (determinants d   15 13 A D is an example with a C c > 2b but ac < b 2 , so not positive deﬁnite. 5 10 14 The eigenvalues of A 1 are positive because they are 1=.A/. And the entries of A pass the determinant tests. And x T A 1 x D .A 1 x /T A.A 1 x / > 0 for all x ¤ 0. 1 15 Since x T Ax > 0 and x T B x > 0 we have x T .A C B/x D x T Ax C x T B x > 0 for all x ¤ 0. Then A C B is a positive deﬁnite matrix. The second proof uses the test T T A D RT R (independent   columns in R): If A D R R and B D S S pass this test, then  T R ACB D R S also passes, and must be positive deﬁnite. S 16 x T Ax is zero when .x1 ; x2 ; x3 / D .0; 1; 0/ because of the zero on the diagonal. Actu- ally x T Ax goes negative for x D .1; 10; 0/ because the second pivot is negative. 17 If ajj were smaller than all ’s, A deﬁnite). But A ajj I would have all eigenvalues > 0 (positive ajj I has a zero in the .j; j / position; impossible by Problem 16. 18 If Ax D x then x T Ax D x T x . If A is positive deﬁnite this leads to  D x T Ax =x T x > 0 (ratio of positive numbers). So positive energy ) positive eigenvalues. 19 All cross terms are x T x j D 0 because symmetric matrices have orthogonal eigenveci tors. So positive eigenvalues ) positive energy. 20 (a) The determinant is positive; all  > 0 (b) All projection matrices except I are singular (c) The diagonal entries of D are its eigenvalues (d) A D I has det D C1 when n is even. 21 A is positive deﬁnite when s > 8; B is positive deﬁnite when t > 5 by determinants. 22 R D 2 4 1 1 p 2 3 2p 15 4 9 1 p 1 32 54 3 1 15       21 40 31 11 T D ;RDQ QD . p 12 02 13 2 23 x 2 =a2 C y 2 =b 2 p x T Ax when A p diag.1=a2 ; 1=b 2 /. Then 1 D 1=a2 and 2 D is D 2 2 2 1=b so a D 1= 1 and b D 1= 2 . The ellipse 9x C 16y D 1 has axes with 1 half-lengths a D 3 and b D 1 . The points . 1 ; 0/ and .0; 1 / are at the ends of the axes. 4 3 4 p p p 24 The ellipse x 2 C xy C y 2 D 1 has axes with half-lengths 1=  D 2 and 2=3.         93 48 10 40 12 24 T 25 A D C C D ; D and C D 35 8 25 21 09 01 03 Solutions to Exercises 71 2 # 1 300 0 1 2 and C D 4 0 002 0 square roots of the pivots from D . Note again C T C D LDLT D A.  p T 26 The Cholesky factors C D L D D " 3 11 1 p 5 have 1 0 5 2 b 27 Writing out x T Ax D x T LDLT x gives ax 2 C 2bxy C cy 2 D a.x C a y/2 C ac a b y 2 . 28 29 30 31 32 33 So the LDLT from elimination is exactly the same as completing the square. The example 2x 2 C 8xy C 10y 2 D 2.x C 2y/2 C 2y 2 with pivots 2; 2 outside the squares and multiplier 2 inside. det A D .1/.10/.1/ D 10;  D 2 and 5; x 1 D .cos ; sin  /, x 2 D . sin ; cos  /; the ’s are positive. So A is positive deﬁnite. 2  6x 2x 1 H1 D is semideﬁnite; f1 D . 2 x 2 C y/2 D 0 on the curve 1 x 2 C y D 0; 2 2x 2    6x 1 01 H2 D D is indeﬁnite at .0; 1/ where 1st derivatives D 0. This is a 10 10 saddle point of the function f2 .x; y/. ax 2 C 2bxy C cy 2 has a saddle point if ac < b 2 . The matrix is...
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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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