Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Problem set 51 page 251 1 det2a d 8i det a d 14 det

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Unformatted text preview: solutions to .1; 1; 1; 1/T x D 0, a basis for S ? is .1; 1; 1; 1/ (c) Split .1; 1; 1; 1/ D b1 C b2 11 3 by projection on S ? and S : b2 D . 1 ; 1 ; 1 ; 1 / and b1 D . 2 ; 2 ; 1 ; 2 /. 222 2 2 This questionshows 2  2 formulas for QR breakdown R22 D 0 when A is sinby ;        12 1 53 11 21 1 11 1 gular. Dp p . Singular Dp  11 2 11 1 51 501 21   1 22 p . The Gram-Schmidt process breaks down when ad b c D 0. 200 T .q T C  /q 2 D BT c B because q 2 D B and the extra q 1 in C  is orthogonal to q 2 . 2 kB k BB When a and b are not orthogonal, the projections onto these lines do not add to the projection onto the plane of a and b. We must use the orthogonal A and B (or orthonormal q 1 and q 2 ) to be allowed to add 1D projections. 28 There are mn multiplications in (11) and 1 m2 n multiplications in each part of (12). 2 29 q 1 D 1 .2; 2; 3 1 1/, q 2 D 1 .2; 1; 2/, q 3 D 3 .1; 2 ; 2/. 3 31 32 33 34 35 1 D W T . See Section 7.2 for more about wavelets : a useful orthonormal basis with many zeros. 1 (a) c D 2 normalizes all the orthogonal columns to have unit length (b) The pro1 jection .aT b=aT a/a of b D .1; 1; 1; 1/ onto the first column is p1 D 2 . 1; 1; 1; 1/. (Check e D 0.) To project onto the plane, add p2 D 1 .1; 1; 1; 1/ to get .0; 0; 1; 1/. 2 " #   1 0 0 1 0 0 1 across plane y C z D 0. Q1 D reflects across x axis, Q2 D 0 0 1 0 1 0 Orthogonal and lower triangular ) ˙1 on the main diagonal and zeros elsewhere. (a) Qu D .I 2uuT /u D u 2uuT u. This is u, provided that uT u equals 1 (b) Qv D .I 2uuT /v D u 2uuT v D u, provided that uT v D 0. Starting from A D .1; 1; 0; 0/, the orthogonal (not orthonormal) vectors B D .1; 1; 2 ; 0/ and C D .1; 1; 1; 3/ and D D .1; 1; 1; 1/ are in the directions of q 2 ; q 3 ; q 4 . The 4 by 4 and 5 by 5 matrices with integer orthogonal columns (not orthogonal3 rows, 2 2 3 1 1 11 1 1 17 61 since not orthonormal Q!) are 4 A B C D 5 D 4 and 0 2 1 15 0 0 31 2 3 1 1 1 11 61 1 1 1 17 6 7 2 1 1 17 60 40 0 3 1 15 0 0 0 41 30 The columns of the wavelet matrix W are orthonormal. Then W Solutions to Exercises 51 36 ŒQ; R D q r .A/ produces from A (m by n of rank n) a “full-size” square Q D Œ Q1 Q2   R . The columns of Q1 are the orthonormal basis from Gram-Schmidt of the 0 column space of A. The m n columns of Q2 are an orthonormal basis for the left nullspace of A. Together the columns of Q D Œ Q1 Q2  are an orthonormal basis for Rm . and 37 This question describes the next q nC1 in Gram-Schmidt using the matrix Q with the columns q 1 ; : : : ; q n (instead of using those q ’s separately). Start from a, subtract its projection p D QT a onto the earlier q ’s, divide by the length of e D a QT a to get q nC1 D e =ke k. Problem Set 5.1, page 251 1 det.2A/ D 8I det. A/ D . 1/4 det A D 2 det. 1 A/ 2 1 D /D det.A . 1 /3 2 1. det A D 1 8 1 I 2 det.A2 / D 1 I det.A 4 3 1 / D 2 D det.AT / 1 2 and det. A/ D . 1/ det A D 1; det.A / D 1; 3 (a) False: det.I C I / is not 1 C 1 (b) True: The product rule extends...
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