hw8solutions

hw8solutions - Homework 8 Solutions Joshua Hernandez...

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Homework 8 Solutions Joshua Hernandez November 18, 2009 6.2 - Gram-Schmidt Orthogonalization Process 2. Apply the Gram-Schmidt process to the given subset S of the inner product space V . Normalize the vectors in the resulting basis to obtain an orthonormal basis β , and compute the Fourier coefficients of the given vector relative to β . d. V = span( S ), where S = { v 1 = (1 ,i, 0) ,v 2 = (1 - i, 2 , 4 i ) } , and x = (3 + i, 4 i, -4). Solution: Orthogonalization: Define v 0 1 = v 1 . We compute the inner products k v 0 1 k 2 = k (1 ,i, 0) k 2 = | 1 | 2 + | 1 | 2 + | 0 | 2 = 2 h v 2 ,v 0 1 i = h (1 - i, 2 , 4 i ) , (1 ,i, 0) i = (1 - i ) 1 + 2 i + 4 i 0 = 1 - 3 i, and compute the next orthogonal vector v 0 2 = v 2 - proj v 0 1 ( v 2 ) = v 2 - h v 2 ,v 0 1 i h v 0 1 ,v 0 1 i v 0 1 = (1 - i, 2 , 4 i ) - 1 - 3 i 2 (1 ,i, 0) = 1 2 (1 + i, 1 - i, 8 i ) , where k v 0 2 k 2 = k 1 2 (1 + i, 1 - i, 8 i ) k 2 = 1 4 ( | 1 + i | 2 + | 1 - i | 2 + | 8 | 2 ) = 17 . Normalization: w 1 = v 0 1 k v 0 1 k = 2 2 (1 ,i, 0) , w 2 = v 0 2 k v 0 2 k = 17 34 (1 + i, 1 - i, 8 i ) . So β = { 2 2 (1 ,i, 0) , 17 34 (1+ i, 1 - i, 8 i ) } is an orthonormal basis of span( S ). Now we can compute the Fourier coefficients for x = (3 + i, 4 i, -4): c 1 = h x,w 1 i = D (3 + i, 4 i, -4) , 2 2 (1 ,i, 0) E = 2 2 ± (3 + i ) 1 + 4 i i - 4 0 ² = 2 2 (7 + i ) , c 2 = h x,w 2 i = D (3 + i, 4 i, -4) , 17 34 (1 + i, 1 - i, 8 i ) E = 17 34 ± (3 + i ) ( 1 + i ) + 4 i ( 1 - i ) - 4 8 i ² = 17 34 [(3 + i )(1 - i ) + 4 i (1 + i ) - 4(-8 i )] = 17 34 ((3 + 1 - 4) + (-3 + 1 + 4 + 32) i ) = 17 i. Checking that these coefficients work: x = c 1 w 1 + c 2 w 2 = 2 2 (7 + i ) 2 2 (1 ,i, 0) + 17 i 17 34 (1 + i, 1 - i, 8 i ) = 1 2 (7 + i, -1 + 7 i, 0) + 1 2 (-1 + i, 1 + i, -8) = (3 + i, 4 i, -4) . 1
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f. V = R 4 , S = { v 1 = (1 , -2 , -1 , 3) ,v 2 = (3 , 6 , 3 , -1) ,v 3 = (1 , 4 , 2 , 8) } , and x = (-1 , 2 , 1 , 1). Solution: Orthogonalization: Define v 0 1 = v 1 . We compute the inner products h v 0 1 ,v 0 1 i = h (1 , -2 , -1 , 3) , (1 , -2 , -1 , 3) i = 1 2 + (-2) 2 + (-1) 2 + 3 2 = 15 h v 2 ,v 0 1 i = h (3 , 6 , 3 , -1) , (1 , -2 , -1 , 3) i = 3(1) + 6(-2) + 3(-1) - 1(3) = -15 , and compute the next orthogonal vector v 0 2 = (3 , 6 , 3 , -1) - -15 15 (1 , -2 , -1 , 3) = (4 , 4 , 2 , 2) .
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This note was uploaded on 02/01/2010 for the course MATH math115a taught by Professor Shalom during the Spring '10 term at UCLA.

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hw8solutions - Homework 8 Solutions Joshua Hernandez...

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