assign10_soln

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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- I ) = det( P- 1 BP- I ) = det( P- 1 BP- P- 1 P ) = det( P- 1 ( B- I ) P ) = det P- 1 det( B- I ) det P = det( B- I ) . 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- I = 4- - 1- 2 5- . The characteristic polynomial is det( A- I ) = (4- )(5- )- 2 = 2- 9 + 18 = ( - 3)( - 6) . Thus, the eigenvalues of A are = 3 and = 6. Since all the eigenvalues have algebraic multiplicity 1, we know that A is diagonalizable. For = 3 we have A- I = 1- 1- 2 2 1- 1 . The general solution of ( A- I ) ~x = ~ 0 is x 2 (1 , 1), x 2 R , so an eigenvector corresponding to = 3 is (1 , 1). For = 6 we have A- I =- 2- 1- 2- 1 1 1 / 2 ....
View Full Document

Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online