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Unformatted text preview: Exam Four Review Problem Solutions THEORY 1. Define eigenvalue and eigenvector and show that is an eigenvalue for A if and only if det ( A- I ) = 0. (1) is an eigenvalue for A with corresponding eigenvector- v if- v 6 =- 0 and A- v = - v . is an eigenvalue for A if and only if there exists- v 6 = 0 such that A- v = v , that is, if and only if there exists- v 6 = 0 such that ( A- I )- v =- 0 , which is if and only if A- I is singular, which is if and only if det ( A- I ) = 0. 2. Define similar and show that similar matrices have the same characteristic polynomial. (2) A is similar to B if and only if there is a non-singular matrix S such that A = SBS- 1 . Suppose that A and B are similar and let A = SBS- 1 . Observe that A- I = SBS- 1- I = SBS- 1- S ( I ) S- 1 ) = S ( B- I ) S- 1 , so that det ( A- I ) = det ( S ) det ( B- I ) det ( S- 1 ) = det ( S ) det ( S- 1 ) det ( B- I ) = det ( SS- 1 ) det ( B- I ) = det ( I ) det ( B- I ) = det ( B- I ). 3. Suppose the n n matrix A has eigenvalues 1 ,..., n . Derive expressions for the sum and the product of the eigenvalues in terms of the matrix A . (3) The characteristic polynomial p ( ) = det ( A- I ) = c + c 1 + + c n n . Recalling that the determinant is the sum of ( ) all possible products with one term from each row and column, we see that the coefficients c n and c n- 1 can only arise from the product ( a 11- )( a 22- ) ( a nn- ), so that c n = (- 1) n and p ( ) = ( 1- )( 2- ) ( n- ) Thus we have (i) 1 2 n = c = p (0) = det ( A- I ) = det ( A ). Also, (- 1) n- 1 [ a 11 + a 22 + + a nn = c n- 1 = (- 1) n- 1 [ 1 + 2 + + n ], so that (ii) 1 + 2 + + n = (- 1) n- 1 c n- 1 = a 11 + a 22 + + a nn = trace ( A ). 4. Show that if is an eigenvalue of a non-singular matrix A, then 1 / is an eigenvalue of A- 1 . (4) Suppose that A- v = - v , with- v 6 =- 0 . Then- v = I- v = ( A- 1 A )- v = A- 1 ( A- v ) = A- 1 ( - v ) = A- 1- v , so that (1 / )- v = A- 1- v , which makes (1 / ) an eigenvalue of A . 5. Show that A and A T have the same eigenvalues. Do they always have the same eigenvectors? (5) Observe that ( A- I ) T = A T- I T = A T- I , so that det ( A T- I ) = det ([ A- I ] T ) = det ( A- I ), the last step since det ( B ) = det ( B T ) for any B . Then is an eigenvalue of A if and only if det ( A- I ) = 0, which is if and only if det ( A T- I ) = 0, which is if and only if is an eigenvalue of A T . They need not have the same eigenvectors. For example, A = 1 0 1 1 has eigenvector 1 , whereas A T has eigenvector 1 ....
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- Spring '09