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Unformatted text preview: Exam Four Review Problem Solutions THEORY 1. Define “eigenvalue” and “eigenvector” and show that λ is an eigenvalue for A if and only if det ( A λI ) = 0. (1) λ is an eigenvalue for A with corresponding eigenvector→ v if→ v 6 =→ 0 and A→ v = λ→ v . λ is an eigenvalue for A if and only if there exists→ v 6 = 0 such that A→ v = λv , that is, if and only if there exists→ v 6 = 0 such that ( A λI )→ v =→ 0 , which is if and only if A λI is singular, which is if and only if det ( A λI ) = 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 λI = SBS 1 λI = SBS 1 S ( λI ) S 1 ) = S ( B λI ) S 1 , so that det ( A λI ) = det ( S ) det ( B λI ) det ( S 1 ) = det ( S ) det ( S 1 ) det ( B λI ) = det ( SS 1 ) det ( B λI ) = det ( I ) det ( B λI ) = det ( B λI ). 3. Suppose the n × n matrix A has eigenvalues λ 1 ,...,λ n . Derive expressions for the sum and the product of the eigenvalues in terms of the matrix A . (3) The characteristic polynomial p ( λ ) = det ( A λI ) = c + c 1 λ + ··· + c n λ 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 λ )( a 22 λ ) ··· ( a nn λ ), so that c n = ( 1) n and p ( λ ) = ( λ 1 λ )( λ 2 λ ) ··· ( λ n λ ) Thus we have (i) λ 1 λ 2 ··· λ 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 [ λ 1 + λ 2 + ··· + λ n ], so that (ii) λ 1 + λ 2 + ··· + λ n = ( 1) n 1 c n 1 = a 11 + a 22 + ··· + a nn = trace ( A ). 4. Show that if λ is an eigenvalue of a nonsingular matrix A, then 1 /λ is an eigenvalue of A 1 . (4) Suppose that A→ v = λ→ v , with→ v 6 =→ 0 . Then→ v = I→ v = ( A 1 A )→ v = A 1 ( A→ v ) = A 1 ( λ→ v ) = λA 1→ v , so that (1 /λ )→ v = A 1→ v , which makes (1 /λ ) 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 λI ) T = A T λI T = A T λI , so that det ( A T λI ) = det ([ A λI ] T ) = det ( A λI ), the last step since det ( B ) = det ( B T ) for any B . Then λ is an eigenvalue of A if and only if det ( A λI ) = 0, which is if and only if det ( A T λI ) = 0, which is if and only if λ 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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This note was uploaded on 12/27/2011 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.
 Spring '09
 RUDYAK

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