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InstructorsManualLeon7edPart3 - Chapter 6 SECTION 1 2 If A...

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Chapter 6 SECTION 1 2. If A is triangular then A - a ii I will be a triangular matrix with a zero entry in the ( i, i ) position. Since the determinant of a triangular matrix is the product of its diagonal elements it follows that det( A - a ii I ) = 0 Thus the eigenvalues of A are a 11 , a 22 , . . ., a nn . 3. A is singular if and only if det( A ) = 0. The scalar 0 is an eigenvalue if and only if det( A - 0 I ) = det( A ) = 0 Thus A is singular if and only if one of its eigenvalues is 0. 4. If A is a nonsingular matrix and λ is an eigenvalue of A , then there exists a nonzero vector x such that A x = λ x A - 1 A x = λA - 1 x It follows from Exercise 3 that λ = 0. Therefore A - 1 x = 1 λ x ( x = 0 ) and hence 1 is an eigenvalue of A - 1 . 110
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Section 1 111 5. The proof is by induction. In the case where m = 1, λ 1 = λ is an eigenvalue of A with eigenvector x . Suppose λ k is an eigenvalue of A k and x is an eigenvector belonging to λ k . A k +1 x = A ( A k x ) = A ( λ k x ) = λ k A x = λ k +1 x Thus λ k +1 is an eigenvalue of A k +1 and x is an eigenvector belonging to λ k +1 . It follows by induction that if λ an eigenvalue of A then λ m is an eigenvalue of A m , for m = 1 , 2 , . . . . 6. If A is idempotent and λ is an eigenvalue of A with eigenvector x , then A x = λ x A 2 x = λA x = λ 2 x and A 2 x = A x = λ x Therefore ( λ 2 - λ ) x = 0 Since x = 0 it follows that λ 2 - λ = 0 λ = 0 or λ = 1 7. If λ is an eigenvalue of A , then λ k is an eigenvalue of A k (Exercise 5). If A k = O , then all of its eigenvalues are 0. Thus λ k = 0 and hence λ = 0. 9. det( A - λI ) = det(( A - λI ) T ) = det( A T - λI ). Thus A and A T have the same characteristic polynomials and consequently must have the same eigenvalues. The eigenspaces however will not be the same. For example A = 1 1 0 1 and A T = 1 0 1 1 both have eigenvalues λ 1 = λ 2 = 1 The eigenspace of A corresponding to λ = 1 is spanned by (1 , 0) T while the eigenspace of A T is spanned by (0 , 1) T . Exercise 27 shows how the eigenvectors of A and A T are related. 10. det( A - λI ) = λ 2 - (2 cos θ ) λ + 1. The discriminant will be negative unless θ is a multiple of π . The matrix A has the effect of rotating a real vector x about the origin by an angle of θ . Thus A x will be a scalar multiple of x if and only if θ is a multiple of π . 12. Since tr( A ) equals the sum of the eigenvalues the result follows by solving n i =1 λ i = n i =1 a ii for λ j .
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112 Chapter 6 13. a 11 - λ a 12 a 21 a 22 - λ = λ 2 - ( a 11 + a 22 ) λ + ( a 11 a 22 - a 21 a 12 ) = λ 2 - (tr A ) λ + det( A ) 14. If x is an eigenvector of A belonging to λ , then any nonzero multiple of x is also an eigenvector of A belonging to λ . By Exercise 5 we know that A m x = λ m x , so A m x must be an eigenvector of A belonging to λ . Alternatively we could have proved the result by noting that A m x = λ m x = 0 and A ( A m x ) = A m +1 x = A m ( A x ) = A m ( λ x ) = λ ( A m x ) 15. If A - λ 0 I has rank k then N ( A - λ 0 I ) will have dimension n - k .
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