Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 2 1 2 2 4 4 2 1 1 2 4 4 p2 d 4 4 2 p1 and p2 are

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: nullspace and P ? as row space. ?  23 x in V ? is perpendicular to any vector in V . Since V contains all the vectors in S , x is also perpendicular to any vector in S . So every x in V ? is also in S ? . Solutions to Exercises 44 1 D I : Column 1 of A nth rows of A. 24 AA 1 is orthogonal to the space spanned by the 2nd, 3rd, : : :, 25 If the columns of A are unit vectors, all mutually perpendicular, then AT A D I . 26 A D " 2 1 2 # 1 This example shows a matrix with perpendicular columns. 2 , AT A D 9I is diagonal: .AT A/ij D .column i of A/  .column j of A/. 2 When the columns are unit vectors, then AT A D I . 2 2 1 27 The lines 3x C y D b1 and 6x C 2y D b2 are parallel. They are the same line if b2 D 2b1 . In that case .b1 ; b2 / is perpendicular to . 2 ; 1/. The nullspace of the 2 by 2 matrix is the line 3x C y D 0. One particular vector in the nullspace is . 1; 3/. 28 (a) .1; 1; 0/ is in both planes. Normal vectors are perpendicular, but planes still in- tersect! (b) Need three orthogonal vectors to span the whole orthogonal complement. (c) Lines can meet at the zero vector without being orthogonal. " # " # 123 1 1 1 A has v D .1; 2; 3/ in row space and column space 1 0 ; B has v in its column space and nullspace. 29 A D 2 1 0 ; B D 2 301 3 0 1 v can not be in the nullspace and row space, or in the left nullspace and column space. These spaces are orthogonal and vT v ¤ 0. 30 When AB D 0, the column space of B is contained in the nullspace of A. Therefore the dimension of C .B/  dimension of N .A/. This means rank.B/  4 rank.A/. 0 31 null.N / produces a basis for the row space of A (perpendicular to N.A/). 32 We need r T n D 0 and c T ` D 0. All possible examples have the form acr T with a ¤ 0. 33 Both r ’s orthogonal to both n’s, both c ’s orthogonal to both ` ’s, each pair independent. All A’s with these subspaces have the form Œc 1 c 2 M Œr 1 r 2 T for a 2 by 2 invertible M . Problem Set 4.2, page 214 1 (a) aT b=aT a D 5=3; p D 5a=3; e D . 2 ; 1; 1/=3 (b) aT b=aT a D 1; p D a; e D 0. 2 (a) The projection of b D .cos ; sin  / onto a D .1; 0/ is p D .cos ; 0/ (b) The projection of b D .1; 1/ onto a D .1; 1/ is p D .0; 0/ since aT b D 0. " # "# " # "# 1 1111 15 1131 1 1 1 and P1 b D 5 . P2 D 3 9 3 and P2 b D 3 . 3 P1 D 3111 35 11 1 3 1 1     11 10 1 P1 projects onto .1; 0/, P2 projects onto .1; 1/ 4 P1 D , P2 D . 00 1 1 P1 P2 ¤ 0 and P1 C P2 is not a projection matrix. 2 " # " # 1 2 2 4 4 2 1 1 2 4 4 , P2 D 4 4 2 . P1 and P2 are the projection 5 P1 D 9 9 2 4 4 2 2 1 matrices onto the lines through a1 D . 1; 2; 2/ and a2 D .2; 2; 1/ P1 P2 D zero matrix because a1 ? a2 . XXX Above solution does not ﬁt in 3 lines. 6 p1 D . 1 ; 9 2 ; 9 2 / 9 4 and p 2 D . 9 ; 4 ; 9 2 / 9 and p3 D . 4 ; 9 24 ; /. 99 So p1 C p2 C p3 D b. Solutions to Exercises 7 8 9 10 45 " # " # " # 1 2 2 4 4 2 4 2 4 1 1 1 2 4 4C 4 4 2C 2 1 2 D I. P1 C P2 C P3 D 9 9 9 2 4 4 2 2 1 4 2 4 We can add...
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.

Ask a homework question - tutors are online