This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 32 immediately says that An D zero matrix. The
key example is a single n by n Jordan block (with n 1 ones above the diagonal):
Check directly that J n D zero matrix.
23 Certainly Q1 R1 is similar to R1 Q1 D Q1 1 .Q1 R1 /Q1 . Then A1 D Q1 R1 c s 2 I is
similar to A2 D R1 Q1 c s 2 I:
24 A could have eigenvalues D 2 and D 1 (A could be diagonal). Then A 1 has the
2
same two eigenvalues (and is similar to A).
22 If all roots are D 0, this means that det.A Problem Set 6.7, page 371
1 A D U †V T D u1 u2 1
0 v1 v2 T 32
3
3 2p
1
3 5 4 50 0 5 4 1
25
00 2
1
3
1
Dp
p
10
5
2
4
12
2 This A D
is a 2 by 2 matrix of rank 1. Its row space has basis v1 , its nullspace
36
has basis v2 , its column space has basis u1 , its left nullspace has basis u2 :
11
1
2
Row space p
Nullspace p
2
1
5
5
1
1
1
3
T
Column space p
; N .A / p
:
3
1
10
10
3 If A has rank 1 then so does AT A. The only nonzero eigenvalue of AT A is its trace, 2
2
which is the sum of all aij . (Each diagonal entry of AT A is the sum of aij down one
column, so the trace is the sum down all columns.) Then 1 D square root of this sum,
2
2
and 1 D this sum of all aij .
p
p
5 But A is
3C 5 2
3
21
T
T
2
4 A A D AA D
has eigenvalues 1 D
, 2 D
.
11
indeﬁnite
2
2
p
p
1 D .1 C 5/=2 D 1 .A/; 2 D . 5 1/=2 D 2 .A/; u1 D v1 but u2 D v2 .
5 A proof that eigshow ﬁnds the SVD. When V 1 D .1; 0/; V 2 D .0; 1/ the demo ﬁnds
AV 1 and AV 2 at some angle . A 90ı turn by the mouse to V 2 ; V 1 ﬁnds AV 2 and
AV 1 at the angle . Somewhere between, the constantly orthogonal v1 and v2
must produce Av1 and Av2 at angle =2. Those orthogonal directions give u1 and u2 .
p
p
21
1=p2
1=p2
2
2
6 AAT D
has 1 D 3 with u1 D
and 2 D 1 with u2 D
.
12
1= 2
1= 2
p3
2
2 p3
"
#
1=p6
1= 2
110
2
2
AT A D 1 2 1 has 1 D 3 with v1 D 4 2=p6 5, 2 D 1 with v2 D 4 0p 5;
011
1= 2
1= 6
p3
2
p
1=p3
110
300
and v3 D 4 1=p3 5. Then
D Œ u1 u2
Œ v1 v2 v3 T .
011
0 10
1= 3 Solutions to Exercises 75 p
3 and 2 D 1 in †. The smallest change to
rank 1 is to make 2 D 0. In the factorization 7 The matrix A in Problem 6 had 1 D A D U †V T D u1 1 vT C u2 2 vT
1
2
this change 2 ! 0 will leave the closest rank–1 matrix as u1 1 vT . See Problem 14
1
for the general case of this problem.
8 The number max .A 1 /max .A/ is the same as max .A/=min .A/. This is certainly 1.
It equals 1 if all ’s are equal, and A D U †V T is a multiple of an orthogonal matrix.
The ratio max =min is the important condition number of A studied in Section 9:2.
9 A D U V T since all j D 1, which means that † D I . 10 A rank–1 matrix with Av D 12u would have u in its column space, so A D uwT for some vector w. I intended (but didn’t say) that w is a multiple of the unit vector
v D 1 .1; 1; 1; 1/ in the problem. Then A D 12uvT to get Av D 12u when vT v D 1.
2 11 If A has orthogonal columns w1 ; : : : ; wn of lengths 1 ; : : : ; n , then AT A will be di2
2
agonal with entries 1 ; : : : ; n . So the ’s are deﬁnitely the singular values of A (as
expected). The eigenvalues of t...
View
Full
Document
 Spring '12
 Minki
 Mass

Click to edit the document details