Chapter 7 Notes - Chapter 7 ISM Linear Algebra Chapter 7...

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Chapter 9 / Exercise 26
Applied Calculus for the Managerial, Life, and Social Sciences
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Chapter 7 ISM: Linear Algebra Chapter 7 7.1 1. If v is an eigenvector of A , then Av = λ v . Hence A 3 v = A 2 ( Av ) = A 2 ( λ v ) = A ( A λ v ) = A ( λ Av ) = A ( λ 2 v ) = λ 2 Av = λ 3 v , so v is an eigenvector of A 3 with eigenvalue λ 3 . 2. We know Av = λ v so v = A - 1 Av = A - 1 λ v = λ A - 1 v , so v = λ A - 1 v or A - 1 v = 1 λ v . Hence v is an eigenvector of A - 1 with eigenvalue 1 λ . 3. We know Av = λ v , so ( A + 2 I n ) v = Av + 2 I n v = λ v + 2 v = ( λ + 2) v , hence v is an eigenvector of ( A + 2 I n ) with eigenvalue λ + 2. 4. We know Av = λ v , so 7 Av = 7 λ v , hence v is an eigenvector of 7 A with eigenvalue 7 λ . 5. Assume Av = λ v and Bv = β v for some eigenvalues λ , β . Then ( A + B ) v = Av + Bv = λ v + β v = ( λ + β ) v so v is an eigenvector of A + B with eigenvalue λ + β . 6. Yes. If Av = λ v and Bv = μv , then ABv = A ( μv ) = μ ( Av ) = μ λ v 7. We know Av = λ v so ( A - λ I n ) v = Av - λ I n v = λ v - λ v = 0 so a nonzero vector v is in the kernel of ( A - λ I n ) so ker( A - λ I n ) = { 0 } and A - λ I n is not invertible. 8. We want all a b c d such that a b c d 1 0 = 5 1 0 hence a c = 5 0 , i.e. the desired matrices must have the form 5 b 0 d . 9. We want a b c d 1 0 = λ 1 0 for any λ . Hence a c = λ 0 , i.e., the desired matrices must have the form λ b 0 d , they must be upper triangular. 10. We want a b c d 1 2 = 5 1 2 , i.e. the desired matrices must have the form 5 - 2 b b 10 - 2 d d . 11. We want a b c d 2 3 = - 2 - 3 . So, 2 a + 3 b = - 2 and 2 c + 3 d = - 3. Thus, b = - 2 - 2 a 3 , and d = - 3 - 2 c 3 . So all matrices of the form a - 2 - 2 a 3 c - 3 - 2 c 3 will fit. 12. Solving 2 0 3 4 v 1 v 2 = 2 v 1 v 2 we get v 1 v 2 = t - 3 2 t (with t = 0) and 328
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Chapter 9 / Exercise 26
Applied Calculus for the Managerial, Life, and Social Sciences
Tan
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ISM: Linear Algebra Section 7.1 solving 2 0 3 4 v 1 v 2 = 4 v 1 v 2 we get v 1 v 2 = 0 t (with t = 0). 13. Solving - 6 6 - 15 13 v 1 v 2 = 4 v 1 v 2 , we get v 1 v 2 = 3 5 t t (with t = 0). 14. We want to find all 4 × 4 matrices A such that Ae 2 = λ e 2 , i.e. the second column of A must be of the form 0 λ 0 0 , so A = a 0 c d e λ f g h 0 i j k 0 l m . 15. Any vector on L is una ff ected by the reflection, so that a nonzero vector on L is an eigenvector with eigenvalue 1. Any vector on L is flipped about L , so that a nonzero vector on L is an eigenvector with eigenvalue - 1. Picking a nonzero vector from L and one from L , we obtain a basis consisting of eigenvectors. 16. Rotation by 180 is a flip about the origin so every nonzero vector is an eigenvector with the eigenvalue - 1. Any basis for R 2 consists of eigenvectors. 17. No (real) eigenvalues 18. Any nonzero vector in the plane is unchanged, hence is an eigenvector with the eigenvalue 1. Since any nonzero vector in V is flipped about the origin, it is an eigenvector with eigenvalue - 1. Pick any two non-collinear vectors from V and one from V to form a basis consisting of eigenvectors. 19. Any nonzero vector in L is an eigenvector with eigenvalue 1, and any nonzero vector in the plane L is an eigenvector with eigenvalue 0. Form a basis consisting of eigenvectors by picking any nonzero vector in L and any two nonparallel vectors in L .