Introduction to Linear algebra-Strang-Solutions-Manual_ver13

1 0 0 1 and b is 6 8 8 6 59 29 divide

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Unformatted text preview: When the x average of .b Ax /.b Ax /T is  2 I , the average of .b x /.b x /T will be the x x output covariance matrix .AT A/ 1 AT  2 A.AT A/ 1 which simplifies to  2 .AT A/ 1 . x /2 as D  =m. By taking m measurements, the variance drops from  2 to 15 When A has 1 column of ones, Problem 14 gives the expected error .b x 2 T 1 2  .A A/  2 =m. 1 9 1 16 b10 C b9 D x .b1 C    C b10 /. Knowing b9 avoids adding all b ’s. x 10 10 10 " # "#       1 1 7 C 9 32 C 35 1 7 . The solution b D 17 1 D x comes from D . D 4 26 D 42 1 2 21 18 p D Ab D .5; 13; 17/ gives the heights of the closest line. The error is b x .2; 6; 4/. This error e has P e D P b Pp D p p D 0. pD 19 If b D error e then b is perpendicular to the column space of A. Projection p D 0. 20 If b D Ab D .5; 13; 17/ then b D .9; 4/ and e D 0 since b is in the column space x x of A. 21 e is in N.AT /; p is in C.A/; b is in C.AT /; N.A/ D f0g D zero vector only. x Solutions to Exercises 48      50 C 5 22 The least squares equation is D . Solution: C D 1, D D 0 10 D 10 T Line 1 t . Symmetric t ’s ) diagonal A A 23 e is orthogonal to p ; then ke k2 D e T .b p/ D e T b D bT b 1. bT p. b k2 D x T AT Ax 2bT Ax C bT b (this term is constant) are zero when 2A Ax D 2A b, or x D .AT A/ 1 AT b. 24 The derivatives of kAx T T 25 3 points on a line: Equal slopes .b2 b1 /=.t2 t1 / D .b3 b2 /=.t3 t2 /. Linear algebra: Orthogonal to .1; 1; 1/ and .t1 ; t2 ; t3 / is y D .t2 t3 ; t3 t1 ; t1 t2 / in the left nullspace. b is in the column space. Then y T b D 0 is the same equal slopes condition written as .b2 b1 /.t3 t2 / D .b3 b2 /.t2 t1 /. 2 3 23 " # " #" # 1 1 0" # 0 C 400 8 C 0 17 61 617 2;DD 26 4 5 D D 4 3 5 has AT A D 0 2 0 ; AT b D 1 1 0 E 002 3 E 1 0 1 4 " # 2 1 . At x; y D 0; 0 the best plane 2 x 3 y has height C D 2 D average of 2 3=2 0; 1; 3; 4. 27 The shortest link connecting two lines in space is perpendicular to those lines. 28 Only 1 plane contains 0; a1 ; a2 unless a1 ; a2 are dependent. Same test for a1 ; : : : ; an . 29 There is exactly one hyperplane containing the n points 0; a1 ; : : : ; an 1 When the n 1 vectors a1 ; : : : ; an 1 are linearly independent. (For n D 3, the vectors a1 and a2 must be independent. Then the three points 0; a1 ; a2 determine a plane.) The equation of the plane in Rn will be aT x D 0. Here an is any nonzero vector on the line (it is only a n line!) perpendicular to a1 ; : : : ; an 1 . Problem Set 4.4, page 239 1 (a) Independent (b) Independent and orthogonal (c) Independent and orthonormal. For orthonormal vectors, (a) becomes .1; 0/, .0; 1/ and (b) is .:6; :8/, .:8; :6/. "   5=9 2=9 Divide by length 3 to get 10 T T 2=9 8=9 2 Q QD but QQ D 1 01 q 1 D . 2 ; 2 ; 1 /. q 2 D . 3 ; 2 ; 2 /: 33 3 33 4 =9 2=9 3 (a) AT A will be 16I (b) AT A will be diagonal with entries 1, 4, 9. " # " # 10 100 T 4 (a) Q D 0 1 , QQ D 0 1 0 ¤ I . Any Q with n < m has QQT ¤ 00 000 I . (b) .1; 0/ and .0; 0/ are orthogonal, not independent. Nonzero...
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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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