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**Unformatted text preview: **MATH 152 L1 & L2 Applied Linear Algebra and Differential Equations Spring 2004-05 ¥ Week 12 Worksheet Solutions : Eigenvalue and Eigenvector, Differential Systems Q1. (Eigenvalues and eigenvectors) (a) Find an example of 3 × 3 matrices A and B , such that A and B have the same eigenvectors but distinct eigenvalues. Solution Let a , b , c be real, distinct. Consider the diagonal matrix diag( a,b,c ) def. = a b c . Then one can always choose e 1 = (1 , , 0), e 2 = (0 , 1 , 0), and e 3 = (0 , , 1) be the eigenvectors of the matrix, with distinct eigenvalues a , b , c . In particular, we may take A = diag(1 , 2 , 3), B = diag(4 , 5 , 6) for an example. (b) Find an example of 3 × 3 matrices A and B , such that A and B have the same eigenvalues but distinct eigenvectors. Solution Let A be a 3 × 3 matrix having three distinct real eigenvalues λ 1 , λ 2 , λ 3 . Also let v 1 , v 2 , v 3 be the corresponding (linearly independent) eigenvectors of A . Then A is diagonalizable and A = PBP- 1 , where P = £ v 1 v 2 v 3 / is invertible and B = diag( λ 1 ,λ 2 ,λ 3 ) is diagonal. Thus, A and B have the same eigenvalues (i.e., λ 1 , λ 2 , λ 3 ) but distinct eigenvectors (since one can always choose e 1 , e 2 , e 3 be the eigenvectors of B ). In particular, we may take the given matrix in Q4 as our matrix A . Then we may also take B = diag(- 1 , , 3). Q2. (Diagonalizable matrix) Consider A = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 , P = 1 . 1 . 01 . 001 . 0001 1 10 100 1000 e e 2 e 3 π √ π 1 . Is A diagonalizable? Find the eigenvalues of P- 1 A 2005 P . Solution A is a lower-triangular matrix of which the eigenvalues are given by the diagonal entries (i.e., 1, 2, 3, 4, 5 being the eigenvalues of A ). Since A has five distinct eigenvalues, A is diagonalizable. That is, there exist an invertible Q and a diagonal D such that A = QDQ- 1 , where D = diag(1 , 2 , 3 , 4 , 5). It follows that A 2005 = QD 2005 Q- 1 . Remark that the given matrix P is invertible. Multiplying P- 1 to the left and P to the right, the matrix P- 1 A 2005 P is diagonalizable because P- 1 A 2005 P = ¯ QD 2005 ¯ Q- 1 , where ¯ Q = P- 1 Q is invertible and D 2005 is diagonal. Therefore, P- 1 A 2005 P has the same eigenvalues of D 2005 , which are 1, 2 2005 , 3 2005 , 4 2005 , 5 2005 ....

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