Linear Algebra by otto brestscher chapter 07 solutions

# Linear Algebra with Applications (3rd Edition)

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SSM: Linear Algebra Section 7.1 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 . 3. We know A~v = λ~v ,so( A +2 I n ) ~v = A~v I n ~v = λ~v ~v =( λ +2) ~v , hence ~v is an eigenvector of ( A I n ) with eigenvalue λ +2. 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 . 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 = { ~ 0 } and A λI n is not invertible. 9. We want · ab cd ¸· 1 0 ¸ = λ · 1 0 ¸ for any λ . Hence · a c ¸ = · λ 0 ¸ , i.e., the desired matrices must have the form · λb 0 d ¸ , they must be upper triangular. 11. We want · 2 3 ¸ = · 2 3 ¸ .So ,2 a +3 b = 2 and 2 c 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 Ft. 13. Solving · 66 15 13 v 1 v 2 ¸ =4 · v 1 v 2 ¸ ,weget · v 1 v 2 ¸ = · 3 5 t t ¸ (with t 6 = 0). 15. Any vector on L is una±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. 17. No (real) eigenvalues 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. ²orm a basis consisting of eigenvectors by picking any nonzero vector in L and any two nonparallel vectors in L . 21. Any nonzero vector in R 3 is an eigenvector with eigenvalue 5. Any basis for R 3 consists of eigenvectors. 167

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Chapter 7 SSM: Linear Algebra 23. a. Since S =[ ~v 1 ··· ~v n ], S 1 ~v i = S 1 ( S~e i )= ~e i . b. i th column of S 1 AS = S 1 AS~e i = S 1 A~v i (by deFnition of S ) = S 1 λ i ~v i (since ~v i is an eigenvector) = λ i S 1 ~v i = λ i ~e i (by part a) hence S 1 AS = λ 1 00 0 0 λ 2 0 0 . . . 0 λ n . 25. See ±igure 7.1. ±igure 7.1: for Problem 7.1.25 . 27. See ±igure 7.2. 29. See ±igure 7.3. 31. See ±igure 7.4. 33. We are given that ~x ( t )=2 t · 1 1 ¸ +6 t · 1 1 ¸ , hence we know that the eigenvalues are 2 and 6 with corresponding eigenvectors · 1 1 ¸ and · 1 1 ¸ respectively (see ±act 7.1.3), so 168
SSM: Linear Algebra Section 7.1 Figure 7.2: for Problem 7.1.27 . Figure 7.3: for Problem 7.1.29 . Figure 7.4: for Problem 7.1.31 . we want a matrix A such that A · 1 1 11 ¸ = · 2 6 26 ¸ . Multiplying on the right by · 1 1 ¸ 1 ,weget A = · 4 2 24 ¸ . 169

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Chapter 7 SSM: Linear Algebra 35. Let λ be an eigenvalue of S 1 AS . Then for some nonzero vector ~v , S 1 AS~v = λ~v , i.e., AS~v = Sλ~v = λS~v so λ is an eigenvalue of A with eigenvector S~v .
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Linear Algebra by otto brestscher chapter 07 solutions -...

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