This preview shows page 1. Sign up to view the full content.
Unformatted text preview: transpose of AT y D 0).
(a) u and w
(b) v and z
(c) rank < 2 if u and w are dependent or if v and z
are dependent
(d) The rank of uvT C wzT is 2.
"
#
" 3 2 # has column space spanned
12
T
10
T
AD u w v z D 2 2
D 4 2 by u and w, row space
11
41
5 1 spanned by v and z: Solutions to Exercises 42 23 As in Problem 22: Row space basis .3; 0; 3/; .1; 1; 2/; column space basis .1; 4; 2/, 24
25 26
27
28 29
30
31 32 .2; 5; 7/; the rank of (3 by 2) times (2 by 3) cannot be larger than the rank of either
factor, so rank 2 and the 3 by 3 product is not invertible.
AT y D d puts d in the row space of A; unique solution if the left nullspace (nullspace
of AT ) contains only y D 0.
(a) True (A and AT have the same rank)
(b) False A D Œ 1 0 and AT have very
different left nullspaces
(c) False (A can be invertible and unsymmetric even if
C .A/ D C .AT /)
(d) True (The subspaces for A and A are always the same. If
AT D A or AT D A they are also the same for AT )
The rows of C D AB are combinations of the rows of B . So rank C rank B . Also
rank C rank A, because the columns of C are combinations of the columns of A.
b
Choose d D bc=a to make a d a rank1 matrix. Then the row space has basis .a; b/
c
and the nullspace has basis . b; a/. Those two vectors are perpendicular !
B and C (checkers and chess) both have rank 2 if p ¤ 0. Row 1 and 2 are a basis for the
row space of C , B T y D 0 has 6 special solutions with 1 and 1 separated by a zero;
N.C T / has . 1; 0; 0; 0; 0; 0; 0; 1/ and .0; 1; 0; 0; 0; 0; 1; 0/ and columns 3; 4; 5; 6 of
I ; N.C / is a challenge.
a11 D 1; a12 D 0; a13 D 1; a22 D 0; a32 D 1; a31 D 0; a23 D 1; a33 D 0; a21 D 1.
The subspaces for A D uvT are pairs of orthogonal lines (v and v? , u and u? ).
If B has those same four subspaces then B D cA with c ¤ 0.
(a) AX D 0 if each column of X is a multiple of .1; 1; 1/; dim.nullspace/ D 3.
(b) If AX D B then all columns of B add to zero; dimension of the B ’s D 6.
(c) 3 C 6 D dim.M 33 / D 9 entries in a 3 by 3 matrix.
The key is equal row spaces. First row of A D combination of the rows of B : only
possible combination (notice I ) is 1 (row 1 of B ). Same for each row so F D G . Problem Set 4.1, page 202
1 Both nullspace vectors are orthogonal to the row space vector in R3 . The column space
2 3 4 5 is perpendicular to the nullspace of AT (two lines in R2 because rank D 1).
The nullspace of a 3 by 2 matrix with rank 2 is Z (only zero vector) so x n D 0, and
row space D R2 . Column space D plane perpendicular to left nullspace D line in R3 .
"
#
"#
"#
"#
"#
1
2
3
2
1
1
1
2
3
1 (b) Impossible, 3 not orthogonal to 1 (c) 1 and 0 in
(a)
3
5
2
5
1
1
0
C .A/ and N .AT / is impossible: not perpendicular (d) Need A2 D 0; take A D 1 1
11
(e) .1; 1; 1/ in the nullspace (columns add to 0) and also row space; no such matrix.
If AB D 0, the columns of B are in the nullspace of A. The rows of A are in the left
nullspace of B . If rank D 2, those four subsp...
View
Full
Document
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

Click to edit the document details