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Unformatted text preview: he column number) " 0
is 2 for this matrix A and 1 for AT : A D 0
0 7 Special solutions in N D Œ 2 4 1 0I 3
"
1
2
8 The new entries keep rank 1: A D
4
a
b
MD
.
c bc=a #
10
0 0.
00 5 0 1 and Œ 1 0 0I
#
"
2
24
1
4 8; BD
8 16
2 0 2 1 .
#
6
3
3
3=2 ;
6
3 Solutions to Exercises 32 9 If A has rank 1, the column space is a line in Rm . The nullspace is a plane in Rn (given by one equation). The nullspace matrix N is n by n 1 (with n 1 special solutions
in its columns). The column space of AT is a line in Rn .
2
3 2 3
366
3
122
2
2
6
4
2 1132
41 2 25D415
10
and
D
1
1
3
2
1
488
4
11 A rank one matrix has one pivot. (That pivot is in row 1 after possible row exchange; it could come in any column.) The second row of U is zero.
Invertible r by r submatrices
13
10
12
SD
and S D Œ 1 and S D
.
Use pivot rows and columns
14
01
13 P has rank r (the same as A) because elimination produces the same pivot columns.
14 The rank of RT is also r . The example matrix A has rank 2 with invertible S : " 1
PD 2
2 3
6
7 # PT D 12
36 2
7 ST D 1
3 2
7 SD
13
:
27 15 The product of rank one matrices has rank one or zero. These particular matrices have rank.AB/ D 1; rank.AM / D 1 except AM D 0 if c D 1=2. 16 .uvT /.wzT / D u.vT w/zT has rank one unless the inner product is vT w D 0.
17 (a) By matrix multiplication, each column of AB is A times the corresponding column of B . So if column j of B is a combination of earlier columns, then column j of AB
is the same combination of earlier columns of AB . Then rank .AB/ rank .B/. No
new pivot columns! (b) The rank of B is r D 1. Multiplyingby A cannot increase
this rank.
The rank of AB stays the same for A1 D I and B D 1 1 . It drops to zero
11
1
for A2 D 1 1 .
1 18 If we know that rank.B T AT / rank.AT /, then since rank stays the same for transposes, (apologies that this fact is not yet proved), we have rank.AB/ rank.A/. 19 We are given AB D I which has rank n. Then rank.AB/ rank.A/ forces rank.A/ D n. This means that A is invertible. The rightinverse B is also a leftinverse: BA D I
and B D A 1 . 20 Certainly A and B have at most rank 2. Then their product AB has at most rank 2. Since BA is 3 by 3, it cannot be I even if AB D I .
21 (a) A and B will both have the same nullspace and row space as the R they share. (b) A equals an invertible matrix times B , when they share the same R. A key fact!
#
#
"
"
10
110
110
D110C
22 A D .pivot columns/.nonzero rows of R/ D 1 4
001
110
18
"
#
000
22 10
columns
20
02
0 0 4 . BD
D
D
C
23 01
times rows
20
03
008 Solutions to Exercises 23 If c D 1; R D " 33 1
0
0 122
000
000 # has x2 ; x3 ; x4 free. If c ¤ 1; R D
2 " 10
01
00 22
00
00 # 3
1
2
2
0
07
61
has x3 ; x4 free. Special solutions in N D 4
(for c D 1) and N D
0
1
05
0
0
1
2
3
2
2
07
1
2
01
60
(for c ¤ 1). If c D 1; R D
and x1 free; if c D 2; R D
41
05
00
0
0
0
1
1
and x2 free; R D I if c ¤ 1; 2. Special solutions in N D
.c D...
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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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