Math 136 assignment

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

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

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