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hw6solutions

# hw6solutions - Homework 6 Solutions Joshua Hernandez 2.5...

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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 a b = 0 10 , -1 2 3 -1 c d = 5 0 , or together as -1 2 3 -1 a c b d = 0 5 10 0 . The matrix ( a c b d ) will be our change-of-coordinate matrix. We solve by taking inverses: Q = a c b d = -1 2 3 -1 -1 0 5 10 0 = 1 -5 -1 -2 -3 -1 0 5 10 0 = 1 -5 -20 -5 -10 -15 = 4 1 2 3 . 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 -1 2 -4 1 1 = 1 -2 -1 -2 -3 -4 2 -4 1 1 = 1 -2 -4 2 -10 8 = 2 -1 5 -4 . 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: Let α be the standard ordered basis of the relevant R n (i.e. the basis in which [ L A ] α α = A ). We can write [ L A ] β β = [ I R n L A I R n ] β β = [ I R n ] β α [ L A ] α α [ I R n ] α β = ([ I R n ] α β ) -1 A [ I R n ] α β Letting Q = [ I R n ] α β (this is just the matrix whose column vectors are the corresponding elements of β ), we have our matrix such that [ L A ] β = Q -1 AQ . b. A = 1 2 2 1 and β = 1 1 , 1 -1 Solution: Here Q = ( 1 1 1 -1 ) and [ L A ] β = 1 1 1 -1 -1 1 2 2 1 1 1 1 -1 = 1 -2 -1 -1 -1 1 3 -1 3 1 = 1 -2 -6 0 0 2 = 3 0 0 -1 . d. A = 13 1 4 1 13 4 4 4 10 and β = 1 1 -2 , 1 -1 0 , 1 1 1 Solution: Now, to avoid another instance of Cramer’s rule, I’ll try to compute [ L A ] β directly
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