# Be the diagonal entries of d by fact 433 we know that

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Chapter 9 / Exercise 26
Applied Calculus for the Managerial, Life, and Social Sciences
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be the diagonal entries of D . By Fact 4.3.3., we know that [ T ( f i )] D = ( i th column of D ) = λ i e i , for i = 1 , 2 , . . . , n , so that T ( f i ) = λ i f i , by definition of coordinates. Thus f 1 , . . . , f n is an eigenbasis for T , as claimed. 54. Note that A 2 = 0, but B 2 = 0. Since A 2 fails to be similar to B 2 , matrix A isn’t similar to B (see Example 7 of Section 3.4). 55. Let A = 0 1 0 0 and B = 1 0 0 0 , for example. 56. The hint shows that matrix M = AB 0 B 0 is similar to N = 0 0 B BA ; thus matri- ces M and N have the same characteristic polynomial, by Fact 7.3.6a. Now f M ( λ ) = det AB - λ I n 0 B - λ I n = ( - λ ) n det( AB - λ I n ) = ( - λ ) n f AB ( λ ). To understand the second equality, consider Fact 6.1.8. Likewise, f N ( λ ) = ( - λ ) n f BA ( λ ). It follows that ( - λ ) n f AB ( λ ) = ( - λ ) n f BA ( λ ) and therefore f AB ( λ ) = f BA ( λ ), as claimed. 57. Modifying the hint in Exercise 56 slightly, we can write AB 0 B 0 I m A 0 I n = I m A 0 I n 0 0 B BA . Thus matrix M = AB 0 B 0 is similar to N = 0 0 B BA . By Fact 7.3.6a, matrices M and N have the same characteristic polynomial. Now f M ( λ ) = det AB - λ I m 0 B - λ I n = ( - λ ) n det( AB - λ I m ) = ( - λ ) n f AB ( λ ). To understand the second equality, consider Fact 6.1.8. Likewise, f N ( λ ) = det - λ I m 0 B BA - λ I n = ( - λ ) m f BA ( λ ). It follows that ( - λ ) n f AB ( λ ) = ( - λ ) m f BA ( λ ). Thus matrices AB and BA have the same nonzero eigenvalues, with the same algebraic multiplicities. If mult( AB ) and mult( BA ) are the algebraic multiplicities of 0 as an eigenvalue of AB and BA, respectively, then the equation ( - λ ) n f AB ( λ ) = ( - λ ) m f BA ( λ ) implies that 370
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Chapter 9 / Exercise 26
Applied Calculus for the Managerial, Life, and Social Sciences
Tan Expert Verified
ISM: Linear Algebra Section 7.4 n + mult( AB ) = m + mult( BA ) . 58. Let B i = A - λ i I n ; note that B i and B j commute for any two indices i and j . If v is an eigenvector of A with eigenvalue λ i , then B i v = 0 and B 1 B 2 . . . B i . . . B m v = B 1 . . . B i - 1 B i +1 . . . B m B i v = 0. Since A is diagonalizable, any vector x in R n can be written as a linear combination of eigenvectors, so that B 1 B 2 . . . B m x = 0 and therefore B 1 B 2 . . . B m = 0, as claimed. 59. If v is an eigenvector with eigenvalue λ , then f A ( A ) v = (( - A ) n + a n - 1 A n - 1 + · · · + a 1 A + a 0 I n ) v = ( - λ ) n v + a n - 1 λ n - 1 v + · · · + a 1 λ v + a 0 v = (( - λ ) n + a n - 1 λ n - 1 + · · · + a 1 λ + a 0 ) v = f A ( λ ) v = 0 v = 0. Since A is diagonalizable, any vector x in R n can be written as a linear combination of eigenvectors, so that f A ( A ) x = 0. Since this equation holds for all x in R n , we have f A ( A ) = 0, as claimed. 60. a. For a diagonalizable n × n matrix A with only two distinct eigenvalues, λ 1 and λ 2 , we have ( A - λ 1 I n )( A - λ 2 I n ) = 0 , by Exercise 58. Thus the column vectors of A - λ 2 I n are in the kernel of A - λ 1 I n , that is, they are eigenvectors of A with eigenvalue λ 1 (or else they are 0). Conversely, the column vectors of A - λ 1 I n are eigenvectors of A with eigenvalue λ 2 (or else they are 0). b. If A is a 2 × 2 matrix with distinct eigenvalues λ 1 and λ 2 , then the nonzero columns of A - λ 1 I 2 are eigenvectors of A with eigenvalue λ 2 , as we observed in part (a).
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