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View Full DocumentChapter 7 ISM: Linear Algebra Chapter 7 7.1 1. If ~v is an eigenvector of A , then A~v = ~v . Hence A 3 ~v = A 2 ( A~v ) = A 2 ( ~v ) = A ( A~v ) = A ( A~v ) = A ( 2 ~v ) = 2 A~v = 3 ~v , so ~v is an eigenvector of A 3 with eigenvalue 3 . 2. We know A~v = ~v so ~v = A- 1 A~v = A- 1 ~v = A- 1 ~v , so ~v = A- 1 ~v or A- 1 ~v = 1 ~v . Hence ~v is an eigenvector of A- 1 with eigenvalue 1 . 3. We know A~v = ~v , so ( A + 2 I n ) ~v = A~v + 2 I n ~v = ~v + 2 ~v = ( + 2) ~v , hence ~v is an eigenvector of ( A + 2 I n ) with eigenvalue + 2. 4. We know A~v = ~v , so 7 A~v = 7 ~v , hence ~v is an eigenvector of 7 A with eigenvalue 7 . 5. Assume A~v = ~v and B~v = ~v for some eigenvalues , . Then ( A + B ) ~v = A~v + B~v = ~v + ~v = ( + ) ~v so ~v is an eigenvector of A + B with eigenvalue + . 6. Yes. If A~v = ~v and B~v = ~v , then AB~v = A ( ~v ) = ( A~v ) = ~v 7. We know A~v = ~v so ( A- I n ) ~v = A~v- I n ~v = ~v- ~v = ~ 0 so a nonzero vector ~v is in the kernel of ( A- I n ) so ker( A- I n ) 6 = { ~ } and A- I n is not invertible. 8. We want all a b c d such that a b c d 1 = 5 1 hence a c = 5 , i.e. the desired matrices must have the form 5 b d . 9. We want a b c d 1 = 1 for any . Hence a c = , i.e., the desired matrices must have the form b 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 3 4 v 1 v 2 = 2 v 1 v 2 we get v 1 v 2 = t- 3 2 t (with t 6 = 0) and 328 ISM: Linear Algebra Section 7.1 solving 2 3 4 v 1 v 2 = 4 v 1 v 2 we get v 1 v 2 = t (with t 6 = 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 6 = 0). 14. We want to find all 4 4 matrices A such that A~ e 2 = ~ e 2 , i.e. the second column of A must be of the form , so A = a c d e f g h i j k l m . 15. Any vector on L is unaffected 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.... View Full Document
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