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assign10_soln - Math 136 Assignment 10 Solutions 1 By...

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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 0 0 - 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
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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 0 0 . 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). For λ = 6 we have A - λI = - 2 - 1 - 2 - 1 1 1 / 2 0 0 . The general solution of ( A - λI ) ~x = ~ 0 is x 2 ( - 1 / 2 , 1), x 2 R , so an eigenvector corresponding to λ = 6 is (1 , - 2). It follows that A is diagonalized by P = 1 1 1 - 2 . The resulting diagonal matrix D has the eigenvalues of A as its diagonal entries, so D = 3 0 0 6 .
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3 b) B = 2 1 - 1 4 Solution: B - λI = 2 - λ 1 - 1 4 - λ . The characteristic polynomial is det( B - λI ) = (2 - λ )(4 - λ ) + 1 = λ 2 - 6 λ + 9 = ( λ - 3) 2 .
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