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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
halflengths 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 halflengths 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.
 Spring '12
 Minki
 Mass

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