7. eigenvalues and eigenvectors

7. Eigenvalues and Eigenvectors
Download Document
Showing pages : 1 - 3 of 73
This preview has blurred sections. Sign up to view the full version! View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 7 ISM: Linear Algebra Chapter 7 7.1 1. If ~v is an eigenvector of A , then A~v = λ~v . Hence A 3 ~v = A 2 ( A~v ) = A 2 ( λ~v ) = A ( Aλ~v ) = A ( λA~v ) = A ( λ 2 ~v ) = λ 2 A~v = λ 3 ~v , so ~v is an eigenvector of A 3 with eigenvalue λ 3 . 2. We know A~v = λ~v so ~v = A- 1 A~v = A- 1 λ~v = λA- 1 ~v , so ~v = λA- 1 ~v or A- 1 ~v = 1 λ ~v . Hence ~v is an eigenvector of A- 1 with eigenvalue 1 λ . 3. We know A~v = λ~v , so ( A + 2 I n ) ~v = A~v + 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 A~v = λ~v , so 7 A~v = 7 λ~v , hence ~v is an eigenvector of 7 A with eigenvalue 7 λ . 5. Assume A~v = λ~v and B~v = β~v for some eigenvalues λ , β . Then ( A + B ) ~v = A~v + B~v = λ~v + β~v = ( λ + β ) ~v so ~v is an eigenvector of A + B with eigenvalue λ + β . 6. Yes. If A~v = λ~v and B~v = μ~v , then AB~v = A ( μ~v ) = μ ( A~v ) = μλ~v 7. We know A~v = λ~v so ( A- λI n ) ~v = A~v- λI n ~v = λ~v- λ~v = ~ 0 so a nonzero vector ~v is in the kernel of ( A- λI n ) so ker( A- λI n ) 6 = { ~ } and A- λI 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 = λ 1 for any λ . Hence a c = λ , i.e., the desired matrices must have the form λ 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 = λ~ e 2 , i.e. the second column of A must be of the form λ , so A = a c d e λ 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