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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 nonsingular 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 nonsingular 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
 RUDYAK

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