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Math 136 assignment

Math 136 assignment - Math 136 Assignment 10 Solutions 1...

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Math 136 Assignment 10 Solutions 1. Determine the matrix of the linear operator L : R 3 R 3 with respect to the given basis B and determine [ L ( ~x )] B where B = { ~v 1 ,~v 2 ,~v 3 } , L ( ~v 1 ) = 2 ~v 1 - ~v 2 - 3 ~v 3 , L ( ~v 2 ) = 4 ~v 1 + 3 ~v 2 - ~v 3 , L ( ~v 3 ) = - ~v 2 and ( ~x ) B = (2 , - 1 , - 1). Solution: To determine the matrix of L with respect to B , we need the B -coordinates of the images of the basis vectors. We have [ L ( ~v 1 )] B = 2 - 1 - 3 , [ L ( ~v 2 )] B = 4 3 - 1 , [ L ( ~v 3 )] B = 0 - 1 0 . Hence, the matrix of L with respect to B is [ L ] B = [ L ( ~v 1 )] B [ L ( ~v 2 )] B [ L ( ~v 3 )] B = 2 4 0 - 1 3 - 1 - 3 - 1 0 . Thus [ L ( ~x )] B = [ L ] B [ ~x ] B = 2 4 0 - 1 3 - 1 - 3 - 1 0 2 - 1 - 1 = 0 - 4 - 5 . 2. Assume each of the following matrices is the standard matrix of a linear operator. Determine the matrix of the linear transformation with respect to the given basis B . a) 1 3 - 8 7 , B = 1 2 , 1 4 Solution: We have [ L ] S = 1 3 - 8 7 , P = 1 1 2 4 , and hence P - 1 = 1 2 4 - 1 - 2 1 . Thus, the matrix of the linear transformation with respect to B is [ L ] B = P - 1 [ L ] S P = 11 16 - 4 - 3 b) - 6 - 2 9 - 5 - 1 7 - 7 - 2 10 , B = 1 1 1 , 1 0 1 , 1 3 2 . Solution: We have [ L ] S = - 6 - 2 9 - 5 - 1 7 - 7 - 2 10 , P = 1 1 1 1 0 3 1 1 2 , thus P - 1 = 3 1 - 3 - 1 - 1 2 - 1 0 1 . Thus, the matrix of the linear transformation with respect to B is [ L ] B = P - 1 [ L ] S P = 1 2 3 0 1 2 0 0 1 1
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2 3. By checking whether columns of P are eigenvectors of A , determine whether P diagonalizes A . If so, determine P - 1 , and check that P - 1 AP is diagonal. a) A = 4 2 - 5 3 , P = 1 3 - 1 1 . Solution: Check whether the columns of P are eigenvectors of A by multiplying the column vectors with the matrix A : 4 2 - 5 3 1 - 1 = 2 - 8 4 2 - 5 3 3 1 = 14 - 12 We see that the columns of P are not eigenvectors of A . It follows that P does not diagonalize A . b) A = 1 3 3 1 , P = 1 1 1 - 1 . Solution: Check whether the columns of P are eigenvectors of A by multiplying the column vectors with the matrix A : 1 3 3 1 1 1 = 4 4 = 4 1 1 1 3 3 1 1 - 1 = - 2 2 = - 2 1 - 1 Thus 1 1 is an eigenvector of A with eigenvalue 4, and 1 - 1 is an eigenvector of A with eigenvalue - 2. Since the columns of P are not scalar multiples of each other, they are linearly independent and hence a basis for R 2 . Thus P diagonalizes A . We check by calculating: P - 1 = 1 2 1 1 1 - 1 and P - 1 AP = 1 2 1 1 1 - 1 1 3 3 1 1 1 1 - 1 = 4 0 0 - 2 . 4. Let A and B be similar matrices. Prove that: a) A and B have the same eigenvalues. Solution: To show A and B have the same eigenvalues, we will show that they have the same characteristic polynomial. Since A = P - 1 BP we have det( A - λI ) = det( P - 1 BP - λI ) = det( P - 1 BP - λP - 1 P ) = det( P - 1 ( B - λI ) P ) = det P - 1 det( B - λI ) det P = det( B - λI ) .
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3 b) tr A = tr B . Solution: Observe that tr AB = n X i =1 n X k =1 a ik b ki = tr BA. Hence, tr A = tr( P - 1 BP ) = tr( P ( P - 1 B )) = tr B. c) A n is similar to B n for all positive integers n .
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