Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# It drops to zero 11 1 for a2 d 1 1 1 18 if we know

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Unformatted text preview: he column number) &quot; 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 &quot; 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 # &quot; 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 : &quot; 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. Multiplyingby 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 right-inverse B is also a left-inverse: 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! # # &quot; &quot;  10 110 110 D110C 22 A D .pivot columns/.nonzero rows of R/ D 1 4 001 110 18 &quot; #       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 &quot; 33 1 0 0 122 000 000 # has x2 ; x3 ; x4 free. If c ¤ 1; R D 2 &quot; 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.

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