Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Tb 12 a a d 1 1 has a a d m a b d b1 c c bm

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Unformatted text preview: The column space is the line through a D 31 The error e D b p must be perpendicular to all the a’s. 32 Since P1 b is in C .A/; P2 .P1 b/ equals P1 b. So P2 P1 D P1 D aaT =aT a where a D .1; 2; 0/. 33 If P1 P2 D P2 P1 then S is contained in T or T is contained in S . 34 BB T is invertible as in Problem 29. Then .AT A/.BB T / D product of r by r invertible matrices, so rank r . AB can’t have rank < r , since AT and B T cannot increase the rank. Conclusion: A (m by r of rank r ) times B (r by n of rank r ) produces AB of rank r . Problem Set 4.3, page 226 3 23 0 0     17 48 36 687 T T and b D 4 5 give A A D and A b D . 35 8 8 26 112 4 20 23 23 1 1  1 657 6 37 T T A Ab D A b gives b D x x and p D Ab D 4 5 and e D b p D 4 5 x 4 13 5 17 E D ke k2 D 44 3 2 3 23 23 10 0 1  1 61 17 C 6 8 7 This Ax D b is unsolvable 6 5 7 24 x 5 D D 4 8 5. Change b to p D P b D 4 13 5; b D 4 exactly solves 13 14 20 17 Ab D p. x 2 1 61 1 AD4 1 1 3 In Problem 2, p D A.AT A/ 1 AT b D .1; 5; 13; 17/ and e D b p D . 1; 3; 5; 3/. p e is perpendicular to both columns of A. This shortest distance ke k is 44. 4 E D .C C 0D/2 C .C C 1D 8/2 C .C C 3D 8/2 C .C C 4D 20/2 . Then @[email protected] D 2C C 2.C C D 8/ C 2.C C 3D 8/ C 2.C C 4D 20/ D 0 and @[email protected] D 1  2.C C D 8 C 3  2.C C  8/ C 4  2.C C 4D 20/ D 0. These /   3D  48 C 36 normal equations are again D . 8 26 D 112 Solutions to Exercises 47 0/2 C .C 8/2 C .C 8/2 C .C 20/2 . AT D Œ 1 1 1 1  and AT A D Œ 4 . AT b D Œ 36  and .AT A/ 1 AT b D 9 D best height C . Errors e D . 9; 1; 1; 11/. 5 E D .C 6 a D .1; 1; 1; 1/ and b D .0; 8; 8; 20/ give b D aT b=aT a D 9 and the projection is x ba D p D .9; 9; 9; 9/. Then e T a D . 9; 1; 1; 11/T .1; 1; 1; 1/ D 0 and ke k D x p 204. 7 A D Œ 0 1 3 4 T , AT A D Œ 26  and AT b D Œ 112 . Best D D 112=26 D 56=13. 8 b D 56=13, p D .56=13/.0; 1; 3; 4/. .C; D/ D .9; 56=13/ don’t match .C; D/ D .1; 4/. x Columns of A were not perpendicular so we can’t project separately to find C and D . 2 3 23 " #" # " # 10 0"# 0 Parabola C 4 8 26 C 36 61 1 17 6 87 T D D 4 5. A Ab D 8 26 92 D D 112 . 9 Project b 4 x 1 3 95 8 4D to 3D E 26 92 338 E 400 1 4 16 20 2 32 3 2 3 23 2 3 10 0 0 C 0 C 0 Exact cubic so p D b, e D 0. 6 1 1 1 1 76 D 7 6 8 7 6 D 7 1 6 47 7 This Vandermonde matrix . 10 4 D . Then 4 5 D 4 1 3 9 27 54 E 5 4 8 5 E 28 5 gives exact interpolation 3 1 4 16 64 F 20 F 5 by a cubic at 0; 1; 3; 4 11 (a) The best line x D 1 C 4t gives the center point b D 9 when b D 2. b t (b) The first equation C m C D T P ti D bi divided by m gives C C Db D b. tb 12 (a) a D .1; : : : ; 1/ has a a D m, a b D b1 C    C bm . Therefore b D aT b=m is the x P ba b D .1; 2; b/ ke k2 D x " # 1 111 p D .3; 3; 3/ T 111 . p e D 0. P D e D . 2 ; 1; 3/ 3 111 mean of the b ’s (c) T P 13 .AT A/ 1 A T .b 14 The matrix .b x (b) e D b Ax / D b x T x . When e D b T 1 T m i D1 .bi b/2 D variance x Ax averages to 0, so does b x T T x. 1 x /.b x / is .A A/ A .b Ax /.b Ax / A.A A/ ....
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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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