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Unformatted text preview: 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 Bcoordinates 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 =  1 . 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 1 3 1 3 1 . Thus [ L ( ~x )] B = [ L ] B [ ~x ] B = 2 4 1 3 1 3 1 2 1 1 =  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 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 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 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 2 . 4. Let A and B be similar matrices. Prove that: a) A and B have the same eigenvalues....
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