Introduction to Linear algebra-Strang-Solutions-Manual_ver13

The nullspace of a 3 by 2 matrix with rank 2 is z

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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 rank-1 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...
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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.

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