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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 non-singular 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 non-singular 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