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 Av = λv . Hence A 3 v = A 2 ( Av ) = A 2 ( λv ) = A ( Aλv ) = A ( λAv ) = A ( λ 2 v ) = λ 2 Av = λ 3 v , so v is an eigenvector of A 3 with eigenvalue λ 3 . 3. We know Av = λv , so ( A + 2 I n ) v = Av + 2 I n v = λv + 2 v = ( λ + 2) v , hence v is an eigenvector of ( A + 2 I n ) with eigenvalue λ + 2. 5. Assume Av = λv and Bv = βv for some eigenvalues λ , β . Then ( A + B ) v = Av + Bv = λv + βv = ( λ + β ) v so v is an eigenvector of A + B with eigenvalue λ + β . 7. We know Av = λv so ( A λI n ) v = Av λI n v = λv λv = 0 so a nonzero vector v is in the kernel of ( A λI n ) so ker( A λI n ) = { 0 } and A λI n is not invertible. 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. 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. 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). 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. 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. Form 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 ( Se i ) = e i . b. i th column of S 1 AS = S 1 ASe i = S 1 Av i (by definition 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 0 0 · · · 0 0 λ 2 0 · · · 0 . . . 0 0 0 · · · λ n . 25. See Figure 7.1. Figure 7.1: for Problem 7.1.25 . 27. See Figure 7.2. 29. See Figure 7.3. 31. See Figure 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 Fact 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 1 1 = 2 6 2 6 . Multiplying on the right by 1 1 1 1 1 , we get A = 4 2 2 4 . 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 ASv = λv , i.e., ASv = Sλv = λSv so λ is an eigenvalue of A with eigenvector Sv .
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