Assignment 4 Solution

Assignment 4 Solution - D-20) = 0 This implies 4 C + 8 D =...

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CS3251, Homework 4 Solution 1. Problem Set 4.1, p. 191, # 5 (a) If Ax = b and A T y = 0 then y b because b C ( A ) and y N ( A T ) C ( A ). (b) If A T y = c and Ax = 0 then x c because c C ( A T ) and x N ( A ) C ( A T ). 2. Problem Set 4.2, p. 203, # 7 We have a 1 = ( - 1 , 2 , 2) T , a 2 = (2 , 2 , - 1) T , a 3 = (2 , - 1 , 2) T and P = aa T k a k 2 2 , is the projection on a vector a . Then P 1 = 1 9 1 - 2 - 2 - 2 4 4 - 2 4 4 , P 2 = 1 9 4 4 - 2 4 4 - 2 - 2 - 2 1 , P 3 = 1 9 4 - 2 4 - 2 1 - 2 4 - 2 4 . It’s easy to check that P 1 + P 2 + P 3 = I . 3. Problem Set 4.2, p. 204, # 17 ( I - P ) 2 = ( I - P )( I - P ) = I - 2 P + P 2 and since P 2 = P we have ( I - P ) 2 = I - P . When P projects on C ( A ), I - P projects on N ( A T ). 4. Problem Set 4.3, p. 205, # 29 B is a m × n matrix with full row rank, i.e., r = r ( B ) = m n . Therefore, B has m linearly independent rows. Recall that A has linearly independent columns if and only if A T A is non-singular, see 4G p. 200. The n × m matrix A = B T has m linearly independent columns. Thus N ( A ) = 0 = N ( A T A ). This implies that A T A = BB T is non-singular. 5. Problem Set 4.3, p. 215, # 4 t = 0 , 1 , 3 , 4, b = (0 , 8 , 8 , 20) T . E = k Ax - b k 2 2 = 4 i =1 ( C + Dt i - b i ) 2 = C 2 + ( C + D - 8) 2 + ( C + 3 D - 8) 2 + ( C + 4 D - 20) 2 . ∂F ∂C = 2 C + 2( C + D - 8) + 2( c + 3 D - 8) + 2( C + 4 D - 20) = 0 ∂F ∂D = 2( C + D - 8) + 2 · 3( C + 3 D - 8) + 2 · 4( C + 4
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Unformatted text preview: D-20) = 0 This implies 4 C + 8 D = 36 8 C + 26 D = 112 . 1 A T Ax = A t b is given by 4 8 8 26 C D = 36 112 . 6. Problem Set 4.5, p. 217, # 12 b = ( b 1 , . . . , b m ) and a = (1 , . . . , 1). Observe that there is a change in notation and b, a denote a row vectors. Thus b T , a T will denote column vectors. (a) a T a x = a T implies that m x = m i =1 b i and x = 1 m m i =1 b i . (b) e = b-a x = ( b 1- x, . . . , b m- x ). Then k e k 2 2 = m i =1 ( b i- x ) 2 and k e k 2 = { m i =1 ( b i- x ) 2 } 1 / 2 . (c) b = (1 , 2 , 6), a = (1 , 1 , 1), b = 3 and p = (3 , 3 , 3). Then for e = b-p = (-2 ,-1 , 3) we have pe T = 0 and P = a T a k a T k 2 2 = 1 3 1 1 1 1 1 1 1 1 1 . 2...
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This note was uploaded on 01/31/2012 for the course COMS 3251 taught by Professor Anargyrospapageorgiou during the Fall '11 term at Columbia.

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Assignment 4 Solution - D-20) = 0 This implies 4 C + 8 D =...

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