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Unformatted text preview: Math 136 Assignment 10 Solutions 1. 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 . 2. 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 ) . 1 2 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 . Solution: Since A = P 1 BP , A 2 = ( P 1 BP )( P 1 BP ) = P 1 B 2 P , and A 2 is similar to B 2 . To prove the statement for all n , we use induction. We have proved the base case for n = 2, so we now assume it is true for n = k . Then A k +1 = AA k = ( P 1 BP )( P 1 B k P ) = P 1 B k +1 P, so the statement is true for n = k + 1. Hence the statement is true for all n . 3. For each of the following matrices, determine the eigenvalues and corresponding eigenvectors and hence determine if the matrix is diagonalizable. If it is, write the diagonalizing matrix P and the resulting matrix D . a) A = 4 1 2 5 Solution: A λI = 4 λ 1 2 5 λ . The characteristic polynomial is det( A λI ) = (4 λ )(5 λ ) 2 = λ 2 9 λ + 18 = ( λ 3)( λ 6) . Thus, the eigenvalues of A are λ = 3 and λ = 6. Since all the eigenvalues have algebraic multiplicity 1, we know that A is diagonalizable. For λ = 3 we have A λI = 1 1 2 2 ∼ 1 1 . The general solution of ( A λI ) ~x = ~ 0 is x 2 (1 , 1), x 2 ∈ R , so an eigenvector corresponding to λ = 3 is (1 , 1)....
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 Spring '08
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 Linear Algebra, Algebra, Eigenvectors, Vectors, Matrices, Eigenvalue, eigenvector and eigenspace, Orthogonal matrix, Det, λI

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