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

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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...
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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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