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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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 Spring '12
 Minki
 Mass

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