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hw6solutions - Homework 6 Solutions Joshua Hernandez...

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Homework 6 Solutions Joshua Hernandez November 11, 2009 2.5 - The Change of Coordinate Matrix 2. For each of the following pairs of ordered bases β and β 0 for R 2 , find the change of coordinate matrix that changes β 0 -coordinates into β coordinates. b. β = { (-1 , 3) , (2 , -1) } and β 0 = { (0 , 10) , (5 , 0) } . Solution: We want to find Q = [ I R 2 ] β β 0 . The usual procedure: I R 2 (0 , 10) = (0 , 10) = a (-1 , 3) + b (2 , -1) , I R 2 (5 , 0) = (5 , 0) = c (-1 , 3) + d (2 , -1) . We can write these two systems as matrix equations ± -1 2 3 -1 ²± a b ² = ± 0 10 ² , ± -1 2 3 -1 ²± c d ² = ± 5 0 ² , or together as ± -1 2 3 -1 ²± a c b d ² = ± 0 5 10 0 ² . The matrix ( a c b d ) will be our change-of-coordinate matrix. We solve by taking inverses: Q = ± a c b d ² = ± -1 2 3 -1 ² -1 ± 0 5 10 0 ² = 1 -5 ± -1 -2 -3 -1 ²± 0 5 10 0 ² = 1 -5 ± -20 -5 -10 -15 ² = ± 4 1 2 3 ² . Note: In general, let V be some finite-dimensional vector space, let α the standard basis of V , and let β and β 0 be two other bases of V . Then [ I V ] β β 0 = [ I V ] β α [ I V ] α β 0 = ([ I V ] α β ) -1 [ I V ] α β 0 The matrices [ I V ] α β 0 and [ I V ] α β are easily computed; one can simply read off the coefficients from the basis vectors. d. β = { (-4 , 3) , (2 , -1) } and β 0 = { (2 , 1) , (-4 , 1) } . Solution: Proceeding as above, Q = ± -4 2 3 -1 ² -1 ± 2 -4 1 1 ² = 1 -2 ± -1 -2 -3 -4 ²± 2 -4 1 1 ² = 1 -2 ± -4 2 -10 8 ² = ± 2 -1 5 -4 ² . 3. For each of the following pairs of ordered bases β and β 0 for P 2 ( R ), find the change of coordinate matrix that changes β 0 -coordinates into β -coordinates. b. β = { 1 ,x,x 2 } , and β 0 = { a 2 x 2 + a 1 x + a 0 ,b 2 x 2 + b 1 x + b 0 ,c 2 x 2 + c 1 x + c 0 } 1
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Solution: We can easily write vectors in β 0 as linear combinations in β : a 2 x 2 + a 1 x + a 0 = a 0 1 + a 1 x + a 2 x 2 b 2 x 2 + b 1 x + b 0 = b 0 1 + b 1 x + b 2 x 2 c 2 x 2 + c 1 x + c 0 = c 0 1 + c 1 x + c 2 x 2 Thus Q = a 0 b 0 c 0 a 1 b 1 c 1 a 2 b 2 c 2 . d. β = { x 2 - x + 1 ,x + 1 ,x 2 + 1 } and β 0 = { x 2 + x + 4 , 4 x 2 - 3 x + 2 , 2 x 2 + 3 } Solution: Let γ = { 1 ,x,x 2 } , the standard ordered basis of P 2 ( R ). We compute Q = [ I P 2 ( R ) ] β β 0 using the identity Q = ([ I P 2 ( R ) ] γ β ) -1 [ I P 2 ( R ) ] γ β 0 : Q = 1 1 1 -1 1 0 1 0 1 -1 4 2 3 1 -3 0 1 4 2 = 1 -1 -1 1 0 -1 -1 1 2 4 2 3 1 -3 0 1 4 2 = 2 1 1 3 -2 1 1 3 1 (I found the inverse matrix using Cramer’s rule, but you could also find it by solving the systems a i (1 , -1 , 1) + b i (1 , 1 , 0) + c i (1 , 0 , 1) = e i , (where e i is the i th standard basis vector of R 3 ) and writing those coefficients down the i th column. 6. For each matrix A and ordered basis β , find [ L A ] β . Also, find an invertible matrix such that [ L A ] β = Q -1 AQ . Solution:
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hw6solutions - Homework 6 Solutions Joshua Hernandez...

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