ch7

Chapter 7



ISM: Linear Algebra



Chapter 7 7.1

1. If v is an eigenvector of A, then Av = λv. Hence A3 v = A2 (Av) = A2 (λv) = A(Aλv) = A(λAv) = A(λ2 v) = λ2 Av = λ3 v, so v is an eigenvector of A3 with eigenvalue λ3 .

1 2. We know Av = λv so v = A−1 Av = A−1 λv = λA−1 v, so v = λA−1 v or A−1 v = λ v.



Hence v is an eigenvector of A−1 with eigenvalue



1 λ.



3. We know Av = λv, so (A + 2In )v = Av + 2In v = λv + 2v = (λ + 2)v, hence v is an eigenvector of (A + 2In ) with eigenvalue λ + 2. 4. We know Av = λv, so 7Av = 7λv, hence v is an eigenvector of 7A with eigenvalue 7λ. 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. 6. Yes. If Av = λv and Bv = µv, then ABv = A(µv) = µ(Av) = µλv 7. We know Av = λv so (A − λIn )v = Av − λIn v = λv − λv = 0 so a nonzero vector v is in the kernel of (A − λIn ) so ker(A − λIn ) = {0} and A − λIn is not invertible. 8. We want all a c b d such that a b c d 5 b . 0 d 1 1 =5 0 0 hence 5 a , i.e. the desired = 0 c



matrices must have the form 9. We want a c b d 1 0 =λ 1 0



for any λ. Hence



a c



=



λ , i.e., the desired matrices 0



must have the form 10. We want 11. We want a c a c b d b d



λ b , they must be upper triangular. 0 d



5 − 2b b 1 1 . , i.e. the desired matrices must have the form =5 10 − 2d d 2 2 2 3 =

−2−2a , 3



−2 . So, 2a + 3b = −2 and 2c + 3d = −3. Thus, b = −3 a −2−2a 3 and d = −3−2c . So all matrices of the form will fit. 3 c −3−2c 3 2 0 3 4 v1 v2 =2 v1 v2 we get v1 v2 328 = t

3 −2t



12. Solving



(with t = 0) and



ISM: Linear Algebra 2 0 3 4 v1 v2 v1 v2 =4 v1 v2 0 t =



Section 7.1



solving 13. Solving



=4 v1 v2



we get



= v1 v2



(with t = 0).

3 5t



−6 6 −15 13



v1 , we get v2



t



(with t = 0).



14. We want to find all 4 × 4 matrices A such    0 a 0 λ  e λ must be of the form  , so A =  0 h 0 0 k 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. 16. Rotation by 180◦ is a flip about the origin so every nonzero vector is an eigenvector with the eigenvalue −1. Any basis for R2 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⊥ .



that Ae2 = λe2 , i.e. the second column of A  c d f g . i j l m



20. Any nonzero vector along the e3 -axis is unchanged, hence is an eigenvector with eigenvalue 1. No other (real) eigenvalues can be found. 21. Any nonzero vector in R3 is an eigenvector with eigenvalue 5. Any basis for R3 consists of eigenvectors. 22. Any nonzero scalar multiple of v is an eigenvector with eigenvalue 1.



23. a. Since S = [v1 · · · vn ], S −1 vi = S −1 (Sei ) = ei . b. ith column of S −1 AS = S −1 ASei = S −1 Avi (by definition of S) 329



Chapter 7 = S −1 λi vi (since vi is an eigenvector) = λi S −1 vi = λi ei (by part a) 



ISM: Linear Algebra



λ1  0 hence S −1 AS =  .  . . 0



0 λ2 0



0 ··· 0 ···



0 · · · λn



 0 0  . 



24. See Figure 7.1.



Figure 7.1: for Problem 7.1.24. 25. See Figure 7.2.



Figure 7.2: for Problem 7.1.25. 26. See Figure 7.3. 27. See Figure 7.4. 330



ISM: Linear Algebra



Section 7.1



Figure 7.3: for Problem 7.1.26.



Figure 7.4: for Problem 7.1.27.



Figure 7.5: for Problem 7.1.28. 28. See Figure 7.5. 29. See Figure 7.6. 30. Since the matrix is diagonal, e1 and e2 are eigenvectors. See Figure 7.7. 31. See Figure 7.8. 331



Chapter 7



ISM: Linear Algebra



Figure 7.6: for Problem 7.1.29.



Figure 7.7: for Problem 7.1.30.



Figure 7.8: for Problem 7.1.31.



32. Since the matrix is diagonal, e1 and e2 are eigenvectors. See Figure 7.9. 33. We are given that x(t) = 2t 1 −1 + 6t , hence we know that the eigenvalues are 2 1 1 332



ISM: Linear Algebra



Section 7.1



Figure 7.9: for Problem 7.1.32. 1 1 1 −1 we want a matrix A such that A 1 1 and 6 with corresponding eigenvectors 1 −1 1 1

−1



and =



−1 respectively (see Fact 7.1.3), so 1 2 −6 . Multiplying on the right by 2 6



, we get A =



4 −2 . −2 4



34. (A2 + 2A + 3In )v = A2 v + 2Av + 3In v = 42 v + 2 · 4v + 3v = (16 + 8 + 3)v = 27v so v is an eigenvector of A2 + 2A + 3In with eigenvalue 27. 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. Conversely, if α is an eigenvalue of A with eigenvector w, then Aw = αw, for some nonzero w. Therefore, S −1 AS(S −1 w) = S −1 Aw = S −1 αw = αS −1 w, so S −1 w is an eigenvector of S −1 AS with eigenvalue α. 36. We want A such that A A= 15 10 5 20 3 1 1 2

−1



15 10 3 1 10 1 15 3 , so = , i.e. A = and A = 5 20 1 2 20 2 5 1 = 4 3 . −2 11



37. a. A = 5



0.6 0.8 is a scalar multiple of an orthogonal matrix. By Fact 7.1.2, the 0.8 −0.6 possible eigenvalues of the orthogonal matrix are ±1, so that the possible eigenvalues of A are ±5. In part b we see that both are indeed eigenvalues. 2 −1 , v2 = . 1 2 333



b. Solve Av = ±5v to get v1 =



Chapter 7



ISM: Linear Algebra



      4 1 1 1 2 1 38.  −5 0 −3   −1  =  −2  = 2  −1 . The associated eigenvalue is 2. −1 −1 2 −1 −2 −1 39. We want 0 0 0 . So b = 0, and d = λ (for any λ). Thus, we need = =λ λ 1 1 0 0 0 0 1 0 a 0 . +d +c =a matrices of the form 0 1 1 0 0 0 c d a c b d 0 0 0 0 1 0 , , 0 1 1 0 0 0 is a basis of V , and dim(V )= 3. a c b d 1 1 λ =λ = . −3 −3 −3λ







So,



40. We need all matrices A such that



Thus, a − 3b = λ and c − 3d = −3λ. Thus, c − 3d = −3(a − 3b) = −3a + 9b, or c = −3a + 9b + 3d. So A must be of the form Thus, a basis of V is a c a −3a + 9b + 3d 0 0 1 1 0 , , 3 9 0 −3 0 a c b d b 1 0 0 1 0 0 = a +b +d . d −3 0 9 0 3 1 0 , and the dimension of V is 3. 1 1 1 = λ2 . So, a + b = λ1 = c + d and 2 2



41. We want



b 1 1 = λ1 , and d 1 1 a + 2b = λ2 and 2λ2 = c + 2d.



So (a + 2b) − (a + b) = λ2 − λ1 = b, a = λ1 − b = 2λ1 − λ2 . Also, (c + 2d) − (c + d) = 2λ1 − λ2 λ2 − λ 1 = 2λ2 −λ1 = d, c = λ1 −d = 2λ1 −2λ2 . So A must be of the form: 2λ1 − 2λ2 2λ2 − λ1 2 −1 −1 1 λ1 + λ2 . 2 −1 −2 2 So a basis of V is 2 −1 −1 1 , , and dim(V )= 2. 2 −1 −2 2



  1 42. We will do this in a slightly simpler manner than Exercise 40. Since A  0  is simply the 0 first column of A, the first column must be a multiple of e1 . Similarly, the third column must be a multiple of e3 . There are no other restrictions on the form of A, meaning it can 334



ISM: Linear Algebra a be any matrix of the form  0 0     0 0 0 0 0 0 d  0 0 0  + e  0 0 0 . 0 0 1 0 1 0  1 0 Thus, a basis of V is  0 0 0 0 and the dimension of V is 5.  b c d



Section 7.1        0 1 0 0 0 1 0 0 0 0 0 = a0 0 0+b0 0 0+c0 1 0+ e 0 0 0 0 0 0 0 0 0



         0 0 1 0 0 0 0 0 0 0 0 0 0 0,0 0 0,0 1 0,0 0 0,0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1



43. A = AIn = A[ e1 . . . en ] = [ λ1 e1 . . . λn en ], where the eigenvalues λ1 , . . . , λn are arbitrary. Thus A can be any diagonal matrix, and dim(V ) = n. 44. We see that each of the columns 1 through m of A will have to be a multiple of its respective vector ei . Thus, there will be m free variables in the first m columns. The remaining n − m columns will each have n free variables. Thus, in total, the dimension of V is m + (n − m)n = m + n2 − nm. 45. Consider a vector w that is not parallel to v. We want A[v w] = [λv av + bw], where λ, a and b are arbitrary constants. Thus the matrices A in V are of the form A = [λv av + bw][v w]−1 . Using summary 4.1.6, we see that [v 0][v w]−1 , [0 v][v w]−1 , [0 w][v w]−1 is a basis of V , so that dim(V ) = 3.



46. a. We need all matrices A such that



a c



b d



k 1 1 . = =k 2k 2 2 = −2d + 2a + 4b and A 0 0 1 . So a +d −2 1 0 of V is 3.



Thus, a + 2b = k and c + 2d = 2k. So, c + 2d = 2a + 4b, or c 0 1 0 a b +b =a must be of the form 4 2 0 −2d + 2a + 4b d 1 0 0 1 0 0 basis of V is , , , and the dimension 2 0 4 0 −2 1 b. Clearly 1 2



is a basis of the image of T by definition of V , so that the rank of T is 1. a c



b a b 1 such that = 0, or a+2b = d c d 2 0 0 −2 1 −2b b . +d =b 0, c+2d = 0. These are the matrices of the form −2 1 0 0 −2d d −2 1 0 0 Thus a basis of the kernel of T is , . 0 0 −2 1 The kernel of T consists of all matrices 335



Chapter 7



ISM: Linear Algebra



c. Let’s find the kernel of L first. In part (a) we saw that the matrices in V are a b . A matrix A in V is in the kernel of L if of the form A = −2d + 2a + 4b d a b 1 = 0, or a + 3b = 0, 2a + 4b + d = 0. This system simpli−2d + 2a + 4b d 3 fies to a = −3b and d = 2b, so that the matrices in the kernel of L are of the form −3b b −3 1 −3 1 =b . The matrix forms a basis of the kernel of L. By −6b 2b −6 2 −6 2 the rank-nullity theorem, the rank of L is dim(V )−dim(ker L) = 3 − 1 = 2, and the image of L is all of R2 .



47. Suppose V is a one-dimensional A-invariant subspace of Rn , and v is a non-zero vector in V . Then Av will be in V, so that Av = λv for some λ, and v is an eigenvector of A. Conversely, if v is any eigenvector of A, then V = span(v) will be a one-dimensional A-invariant subspace. Thus the one-dimensional A-invariant subspaces V are of the form V = span(v), where v is an eigenvector of A.



48. a. Since span(e1 ) is an A-invariant subspace of R3 , it must be that e1 is an eigenvector   of a A, as revealed in Exercise 47. Thus, the first column of A must be of the form  0 . 0 Since span(e1 , e2 ) is also an A-invariant subspace, it must that Ae2 is in span(e1 , e2 ). be  b Thus, the second column of A must have the form  c . The third column may be 0  1 1 1 any vector in R3 . Thus, we can choose A =  0 1 1  to maximize the number of 0 0 1 non-zero entries. b. We see, from our construction above, that upper-triangular  a b tion. This space, V consists of all matrices of the form  0 c 0 0 of 6. matrices fit this descrip d e  and has a dimension f



49. The eigenvalues of the system are λ1 = 1.1, and λ2 = 0.9 and corresponding eigenvectors 100 200 100 are v1 = and v2 = , respectively. So if x0 = , we can see that 300 100 800 336



ISM: Linear Algebra x0 = 3v1 − v2 . Therefore, by Fact 7.1.3, we have x(t) = 3(1.1)t c(t) = 300(1.1)t − 200(0.9)t and r(t) = 900(1.1)t − 100(0.9)t . h(t) , and Av(t) = v(t + 1), where A = f (t) in the example worked on Pages 292 through 295.



Section 7.1 100 200 − (0.9)t , i.e. 300 100



50. Let v(t) =



4 −2 . Now we will proceed as 1 1



a. v(0) =



100 , and we see that Av(0) = 100 100 100 . = 2t v(t) = At v(0) = At 100 100 So c(t) = r(t) = 100(2)t .



4 −2 1 1



100 100



=



200 200



=2



100 . So, 100



b. v(0) =



200 , and we see that Av(0) = 100 200 200 v(t) = At v(0) = At = 3t . 100 100 So c(t) = 200(3)t and r(t) = 100(3)t .



4 −2 1 1



200 100



=



600 300



=3



200 . So, 100



c. v(0) =



600 . We can write this in terms of the previous eigenvectors as v(0) = 500 100 200 100 200 100 + = 4(2)t + At . So, v(t) = At v(0) = At 4 + 4 100 100 100 100 100 200 (3)t . 100 So c(t) = 400(2)t + 200(3)t and r(t) = 400(2)t + 100(3)t .



51. Let v(t) =



c(t) 0 .75 , and Av(t) = v(t + 1), where A = . Now we will proceed r(t) −1.5 2.25 as in the example worked on Pages 292 through 295.



a. v(0) =



100 0 .75 , and we see that Av(0) = 200 −1.5 2.25 100 100 . = (1.5)t So, v(t) = At v(0) = At 200 200 So c(t) = 100(1.5)t and r(t) = 200(1.5)t. 337



100 150 100 = = 1.5 . 200 300 200



Chapter 7 100 0 .75 , and we see that Av(0) = 100 −1.5 2.25 100 100 So, v(t) = At v(0) = At = (.75)t . 100 100 So c(t) = 100(.75)t and r(t) = 100(.75)t. c. v(0) = 100 100



ISM: Linear Algebra 75 75 100 . 100



b. v(0) =



=



= .75



500 . We can write this in terms of the previous eigenvectors as v(0) = 700 100 100 100 100 100 + = 3(.75)t + At 2 . So, v(t) = At v(0) = At 3 +2 3 100 200 100 200 100 100 . 2(1.5)t 200 So c(t) = 300(.75)t + 200(1.5)t and r(t) = 300(.75)t + 400(1.5)t .



52. a.



0.978 −0.006 0.004 0.992 and



−1 −0.99 −1 = = 0.99 , 2 1.98 2 3 2.94 3 = = 0.98 . The eigenvalues are λ1 = 0.99 −1 −0.98 −1



0.978 −0.006 0.004 0.992 and λ2 = 0.98. g0 l0 =



b. x0 = hence



100 −1 3 −1 3 = 20 +40 so x(t) = 20(0.99)t +40(0.98)t , 0 2 −1 2 −1



g(t) = −20(0.99)t + 120(0.98)t and h(t) = 40(0.99)t − 40(0.98)t.



Figure 7.10: for Problem 7.1.52b. h(t) first rises, then falls back to zero. g(t) falls a little below zero, then goes back up to zero. See Figure 7.10. c. We set g(t) = −20(0.99)t + 120(0.98)t = 0. 338



ISM: Linear Algebra



Section 7.1



Solving for t we get that g(t) = 0 for t ≈ 176 minutes. (After t = 176, g(t) 0, no real eigenvalues when k k > −2, two solutions if k = 2 or k = −2, and one solution if |k| > 2. 33. a. fA (λ) = det(A − λI3 ) = −λ3 + cλ2 + bλ + a 345



Chapter 7



ISM: Linear Algebra



(−1, 2)



g(λ) = λ3 − 3λ (1, − 2)



Figure 7.14: for Problem 7.2.32. 0 b. By part a, we have c = 17, b = −5 and a = π, so M =  0 π   1 0 0 1 . −5 17



34. Consider the possible graphs of fA (λ) assuming that it has 2 distinct real roots.



(1 – λ) (–λ – 2) (λ2 + 1)



Figure 7.15: for Problem 7.2.34.



 1 0 0 0 0  0 −2 0 Algebraic multiplicity of each eigenvalue is 1. Example:  . See 0 0 0 −1 0 0 1 0 Figure 7.15.   −2 0 0 0  0 −2 0 0  Algebraic multiplicity of each eigenvalue is 2. Example:  . See 0 0 1 0 0 0 0 1 346







ISM: Linear Algebra



Section 7.2



(–λ – 2)2 (1 – λ)2



Figure 7.16: for Problem 7.2.34. Figure 7.16.



(–λ – 2) (1 – λ)3



Figure 7.17: for Problem 7.2.34. Algebraic multiplicity of λ1 is 1, and of λ2 is 3.   −2 0 0 0  0 1 0 0 Example:  . See Figure 7.17. 0 0 1 0 0 0 0 1  0 −1 0 0 0 0 0 1 35. A =  , with fA (λ) = (λ2 + 1)2 0 0 0 −1 0 0 1 0   B 0   B  where B = 0 −1 , fA (λ) = (λ2 + 1)n . 36. Let A =  ..   1 0 . 0 B 347 



Chapter 7



ISM: Linear Algebra



37. We can write fA (λ) = (λ − λ0 )2 g(λ), for some polynomial g. The product rule for derivatives tells us that fA (λ) = 2(λ − λ0 )g(λ) + (λ − λ0 )2 g (λ), so that fA (λ0 ) = 0, as claimed. 38. By Fact 7.2.4, the characteristic polynomial of A is fA (λ) = λ2 − 5λ − 14 = (λ − 7)(λ + 2), so that the eigenvalues are 7 and -2. 39. tr(AB) =tr tr(BA) =tr are equal. a b c d e g f h e g a c f h b d =tr =tr ae + bg −−− ea + f c −−− −−− cf + dh −−− gb + hd = ae + bg + cf + dh. = ea + f c + gb + hd. So they



40. Let the entries of A be aij and the entries of B be bij . Now, tr(AB) = (a11 b11 + a12 b21 + · · · + a1n bn1 )+(a21 b12 + · · · + a2n bn2 )+· · ·+(an1 b1n + · · · +ann bnn ). This is the sum of all products of the form aij bji . We see that tr(BA) = (b11 a11 + · · · + b1n an1 ) + · · · + (bn1 a1n + · · · + bnn ann ) , which also is the sum of all products of the form bji aij = aij bji . Thus, tr(AB) = tr(BA). 41. So there exists an invertible S such that B = S −1 AS, and tr(B) =tr(S −1 AS) =tr((S −1 A)S). By Exercise 40, this equals tr(S(S −1 A)) =tr(A). 42. tr (A + B)2 = tr(A2 + AB + BA + B 2 ) = tr(A2 ) + tr(AB) + tr(BA) + tr(B 2 ). By Exercise 40, tr(AB) = tr(BA). Thus, tr (A + B)2 = tr(A2 ) + 2tr(BA) + tr(B 2 ) = tr(A2 ) + tr(B 2 ), since BA = 0. 43. tr(AB − BA) =tr(AB)−tr(BA) =tr(AB)−tr(AB) = 0, but tr(In ) = n, so no such A, B exist. We have used Exercise 40. 44. No, there are no such matrices A and B. We will argue indirectly, assuming that invertible matrices A and B with AB − BA = A do exist. Then AB = BA + A = (B + In )A, and ABA−1 = B + In . Using Exercise 41, we see that tr(B) = tr(ABA−1 ) = tr(B + In ) = tr(B) + n, a contradiction. 45. fA (λ) = λ2 −tr(A)λ+det(A) = λ2 −2λ+(−3−4k). We want fA (5) = 25−10−3−4k = 0, or, 12 − 4k = 0, or k = 3.



46. a. λ2 +λ2 = (λ1 +λ2 )2 −2λ1 λ2 = (trA)2 −2 det(A) = (a+d)2 −2(ad−bc) = a2 +d2 +2bc. 1 2 348



ISM: Linear Algebra



Section 7.2



b. Based on part (a), we need to show that a2 +d2 +2bc ≤ a2 +b2 +c2 +d2 , or 2bc ≤ b2 +c2 , or 0 ≤ (b − c)2 . But the last inequality is obvious. c. By parts (a) and (b), the equality λ2 + λ2 = a2 + b2 + c2 + d2 holds if (and only if) 1 2 0 = (b − c)2 , or b = c. Thus equality holds for symmetric matrices A. 2 0 , or, [ Av1 0 3 Since v1 or v2 must be nonzero, 2 or 3 must be an eigenvalue of A. v2 ]. We want A[ v1 v2 ] = [ v 1 v2 ]



47. Let M = [ v1



Av2 ] = [ 2v1



3v2 ].



48. Let S = [v1 v2 ]. Then AS = [Av1 Av2 ] and SD = [2v1 3v2 ], so that v1 must be an eigenvector with eigenvalue 2, and v2 must be an eigenvector with eigenvalue 3. Thus, both 2 and 3 must be eigenvalues of A. 49. As in problem 47, such an M will exist if A has an eigenvalue 2, 3 or 4.



50. a. If f (x) = x3 + 6x − 20 then f (x) = 3x2 + 6 so f (x) > 0 for all x, i.e. f is always increasing, hence has only one real root. b. If v 3 − u3 = 20 and vu = 2 then (v − u)3 + 6(v − u) = v 3 − 3v 2 u + 3vu2 − u3 + 6(v − u) = v 3 − u3 − 3vu(v − u) + 6(v − u) = 20 − 6(v − u) + 6(v − u) = 20 Hence x = v − u satisfies the equation x3 + 6x = 20. c. The second equation tells us that u = we find that v3 −

8 v3 2 v



or u3 =



8 v3 .



Substituting into the first equation



= 20, or, (v 3 )2 − 8 = 20v 3 or (v 3 )2 − 20v 3 − 8 = 0, with solutions √ √ √ √ 3 v 3 = 20± 400+32 = 10 ± 108 = 10 ± 6 3 and v = 10 ± 108. 2 √ √ 3 Now u3 = v 3 − 20 = −10 ± 108 and u = −10 ± 108.

3



d. Let v =



q 2



+



q 2 2



+



p 3 3

3



and u =

q 2 2



3



q −2 +



q 2 2



+



p 3 3 .



Then v 3 − u3 = q and vu = Since x = v − u we have



+



p 3 3



−



q 2 2



= p. 3



x3 + px = v 3 − 3v 2 u + 3vu2 − u3 + p(v − u) = v 3 − u3 − 3vu(v − u) + p(v − u) 349



Chapter 7 = q − p(v − u) + p(v − u) = q, as claimed.



ISM: Linear Algebra



may be negative. Also, the equation If p is negative, the expression q + p 2 3 x3 + px = q may have more than one solution in this case. e. Setting x = t −

a 3



2



3



we get t −



a 3 3



+a t−



a 2 3



+b t−



a 3



+ c = 0 or



t3 −at2 +at2 +(linear and constant terms) = 0 or t3 +(linear and constant terms) = 0, as claimed (bring the constant terms to the right-hand side).



7.3

1. λ1 = 7, λ2 = 9, E7 = ker Eigenbasis: 1 , 0 4 1 1 −1 , E0 = span 1 1 0 8 1 −2 8 4 = span , E9 = ker = span 0 2 0 0 0 1



2. λ1 = 2, λ2 = 0, E2 = span Eigenbasis: 1 , 1 −1 1



3. λ1 = 4, λ2 = 9, E4 = span Eigenbasis: 3 , −2 1 1



3 1 , E9 = span −2 1



4. λ1 = λ2 = 1, E1 = span No eigenbasis



−1 1



5. No real eigenvalues as fA (λ) = λ2 − 2λ + 2. 6. λ1,2 =

√ 7± 57 2



Eigenbasis:



3 3 ≈ , λ1 − 2 5.27



3 3 ≈ λ2 − 2 −2.27



7. λ1 = 1, λ2 = 2, λ3 = 3, eigenbasis: e1 , e2 , e3 350



ISM: Linear Algebra       1 1 1 8. λ1 = 1, λ2 = 2, λ3 = 3, eigenbasis:  0  ,  1  ,  2  0 0 1       −1 0 1 9. λ1 = λ2 = 1, λ3 = 0, eigenbasis:  0  ,  1  ,  0  1 0 0     0 1 10. λ1 = λ2 = 1, λ3 = 0, E1 = span  0  , E0 = span  0  1 0 No eigenbasis      1 1 1 = λ2 = 0, λ3 = 3, eigenbasis:  −1  ,  0  ,  1  0 −1 1   1 = λ2 = λ3 = 1, E1 = span  0 , no eigenbasis 0       0 1 1 = 0, λ2 = 1, λ3 = −1, eigenbasis:  1  ,  −3  ,  −1  0 1 2       0 1 0 = 0, λ1 = λ3 = 1, eigenbasis:  1  ,  −5  ,  2  0 0 1   0 = 0, λ2 = λ3 = 1, E0 = span  1  . We can use Kyle Numbers to see that 0 −1 0 −1 0   2 1 1 = span  −1  . 1 2 2 



Section 7.3



11. λ1



12. λ1



13. λ1



14. λ1



15. λ1



There is no eigenbasis since the eigenvalue 1 has algebraic multiplicity 2, but the geometric multiplicity is only 1.   1 16. λ1 = 0 (no other real eigenvalues), with eigenvector  −1  1 No real eigenbasis 351



 1 −2 E1 = ker  −3 −4



Chapter 7 17. λ1 = λ2 = 0, λ3 = λ4 = 1         0 0 0 1  0   −1   1   0  with eigenbasis   ,  , ,  0 0 1 0 1 0 0 0 18. λ1 = λ2 = 0, λ3 = λ4 = 1, E0 = span(e1 , e3 ), E1 = span(e2 ) No eigenbasis



ISM: Linear Algebra



19. Since 1 is the only eigenvalue, with algebraic multiplicity 3, there exists an eigenbasis for A if (and only if) the geometric multiplicity of the eigenvalue 1 is 3 as well, that is, if   0 a b E1 = R3 . Now E1 = ker  0 0 c  is R3 if (and only if) a = b = c = 0. 0 0 0 If a = b = c = 0 then E1 is 3-dimensional with eigenbasis e1 , e2 , e3 . If a = 0 and c = 0 then E1 is 1-dimensional and otherwise E1 is 2-dimensional. The geometric multiplicity of the eigenvalue 1 is dim(E1 ).     0 a b 0 a 0 20. For λ1 = 1, E1 = ker  0 0 c  = ker  0 0 1  so if a = 0 then E1 is 2-dimensional, 0 0 1 0 0 0 otherwise it is 1-dimensional.   1 −a −b For λ2 = 2, E2 = ker  0 1 −c  so E2 is 1-dimensional. 0 0 0 Hence, there is an eigenbasis if a = 0. 1 1 2 2 4 1 2 1 4 = and A =2 = , i.e. A = 2 2 3 3 6 2 3 2 6

−1



21. We want A such that A so A = 1 4 2 6 1 2 2 3



=



5 −2 . 6 −2



The answer is unique. 22. We want A such that Ae1 = 7e1 and Ae2 = 7e2 hence A = 7 0 . 0 7



23. λ1 = λ2 = 1 and E1 = span(e1 ), hence there is no eigenbasis. The matrix represents a shear parallel to the x-axis. 352



ISM: Linear Algebra a b . First we want c d a c b d 2 1



Section 7.3



24. Let A =



2 , or 2a + b = 2, 2c + d = 1. This 1 a 2 − 2a condition is satisfied by all matrices of the form A = . Next, we want there c 1 − 2c to be no other eigenvalue, besides 1, so that 1 must have an algebraic multiplicity of 2. = We want the characteristic polynomial to be (λ − 1)2 = λ2 − 2λ + 1, so that the trace must be 2, and a + (1 − 2c) = 2, or, a = 1 + 2c. Thus we want a matrix of the form 1 + 2c −4c . A= c 1 − 2c



2 instead of E1 = R2 . This means 1 that we must exclude the case A = I2 . In order to ensure this, we state simply that 1 + 2c −4c A= , where c is any nonzero constant. c 1 − 2c   −λ 1 0 25. If λ is an eigenvalue of A, then Eλ = ker(A − λI3 ) = ker  0 −λ 1 . a b c−λ Finally, we have to make sure the E1 = span



The second and third columns of the above matrix aren’t parallel, hence Eλ is always 1-dimensional, i.e., the geometric multiplicity of λ is 1.

λ→∞



26. Note that fA (0) = det(A − 0I6 ) = det(A) is negative. Since lim fA (λ) = ∞, there must be a positive root, by the Intermediate Value Theorem (see Exercise 2.2.47c). Therefore, the matrix A has a positive eigenvalue. See Figure 7.18.



Figure 7.18: for Problem 7.3.26. 27. By Fact 7.2.4, we have fA (λ) = λ2 − 5λ + 6 = (λ − 3)(λ − 2) so λ1 = 2, λ2 = 3. 28. Since Jn (k) is triangular, its eigenvalues are its diagonal entries, hence its only eigenvalue is k. Moreover, 353



Chapter 7 0 1 0 0  . . . . Ek = ker(Jn (k) − kIn ) = ker  . . . . . . . . 0 0 0 ··· 0 . . 1 . . . 0 .  = span (e1 ).  .  . . 1 0 0



ISM: Linear Algebra



The geometric multiplicity of k is 1 while its algebraic multiplicity is n. 29. Note that r is the number of nonzero diagonal entries of A, since the nonzero columns of A form a basis of im(A). Therefore, there are n − r zeros on the diagonal, so that the algebraic multiplicity of the eigenvalue 0 is n − r. It is true for any n × n matrix A that the geometric multiplicity of the eigenvalue 0 is dim(ker(A)) = n − rank(A) = n − r. 30. Since A is triangular, fA (λ) = (a11 − λ)(a22 − λ) · · · (amm − λ)(0 − λ)n−m . Hence the algebraic multiplicity of λ = 0 is (n − m). Also note that the rank of A is at least m, since the first m columns of A are linearly independent. Therefore, the geometric multiplicity of the eigenvalue 0 is dim(ker(A)) = n − rank(A) ≤ n − m. 31. They must be the same. For if they are not, by Fact 7.3.7, the geometric multiplicities would not add up to n. 32. Recall that a matrix and its transpose have the same rank (Fact 5.3.9c). The geometric multiplicity of λ as an eigenvalue of A is dim(ker(A − λIn )) = n − rank(A − λIn ). The geometric multiplicity of λ as an eigenvalue of AT is dim(ker(AT − λIn )) = dim(ker(A − λIn )T ) = n − rank(A − λIn )T = n − rank(A − λIn ). We can see that the two multiplicities are the same. 33. If S −1 AS = B, then S −1 (A − λIn )S = S −1 (AS − λS) = S −1 AS − λS −1 S = B − λIn . 34. Note that SB = AS.



a. If x is in the kernel of B, then ASx = SBx = S 0 = 0, so that Sx is in ker(A). b. T is clearly linear, and the transformation R(x) = S −1 x is the inverse of T (if x is in the kernel of B, then S −1 x is in the kernel of A, by part (a)). c. The equation nullity(A) = nullity(B) follows from part (b); the equation rank(A) = rank(B) then follows from the rank-nullity theorem (Fact 3.3.7). 354



ISM: Linear Algebra



Section 7.3



35. No, since the two matrices have different eigenvalues (see Fact 7.3.6c). 36. No, since the two matrices have different traces (see Fact 7.3.6.d)



37. a. Av · w = (Av)T w = (v T AT )w = (v T A)w = v T (Aw) = v · Aw ↑ A symmetric b. Assume Av = λv and Aw = αw for λ = α, then (Av) · w = (λv) · w = λ(v · w), and v · Aw = v · αw = α(v · w). By part a, λ(v · w) = α(v · w) i.e., (λ − α)(v · w) = 0. Since λ = α, it must be that v · w = 0, i.e., v and w are perpendicular.



38. Note that fA (0) = det(A − 0I3 ) = det(A) = 1. Since lim fA (λ) = −∞, the polynomial fA (λ) must have a positive root λ0 , by the Intermediate Value Theorem. In other words, the matrix A will have a positive eigenvalue λ0 . Since A is orthogonal, this eigenvalue λ0 will be 1, by Fact 7.1.2. This means that there is a nonzero vector v in R3 such that Av = 1v = v, as claimed. See Figure 7.19.

λ→∞



fA (λ) 1 λ0 = 1



Figure 7.19: for Problem 7.3.38.



39. a. There are two eigenvalues, λ1 = 1 (with E1 = V ) and λ2 = 0 (with E0 = V ⊥ ). 355



Chapter 7 Now geometric multiplicity(1) = dim(E1 ) = dim(V ) = m, and



ISM: Linear Algebra



geometric multiplicity(0) = dim(E0 ) = dim(V ⊥ ) = n − dim(V ) = n − m. Since geometric multiplicity(λ) ≤ algebraic multiplicity(λ), by Fact 7.3.7, and the algebraic multiplicities cannot add up to more than n, the geometric and algebraic multiplicities of the eigenvalues are the same here. b. Analogous to part a: E1 = V and E−1 = V ⊥ . geometric multiplicity(1) = algebraic multiplicity(1) = dim(V ) = m, and geometric multiplicity(−1) = algebraic multiplicity(−1) = dim(V ⊥ ) = n − m. a b b a



40. The matrix of the dynamical system is A =



so fA (λ) = (a − λ)2 − b2 . 1 1 and −1 . 1



Hence, λ1,2 = a ± b, and the respective eigenvectors are Since x(0) = 3

1 2



=



7 4



1 1 1 5 −1 7 5 −4 , by Fact 7.1.3, x(t) = 4 (a+b)t + 4 (a−b)t . 1 1 1 −1



Note that a−b is between 0 and 1, so that the second summand in the formula above goes to 0 as t goes to infinity. Qualitatively different outcomes occur depending on whether a + b exceeds 1, equals 1, or is less than 1. See Figure 7.20.



Figure 7.20: for Problem 7.3.40.



356



ISM: Linear Algebra



Section 7.3



      9 2 1 41. The eigenvalues of A are 1.2, −0.8, −0.4 with eigenvectors  6 ,  −2 ,  −2 . 2 1 2           2 9 1 2 9 Since x0 = 50  6 +50  −2 +50  −2  we have x(t) = 50(1.2)t  6 +50(−0.8)t  −2 + 1 2 2 1 2   1 50(−0.4)t  −2 , so, as t goes to infinity, j(t) : n(t) : a(t) approaches the proportion 2 9 : 6 : 2. 42. C(t+1) = 0.8C(t)+10 so if A 1 50 , 0 1 1 + −50 0 spectators. 0.8 10 C(t + 1) C(t) . A has eigenvectors ,A = = 0 1 1 1 C(0) 0 0 corresponding to λ1 = 0.8 and λ2 = 1. Since = , and = 1 1 1 50 , we have C(t) = −50(0.9)t + 50, hence in the long run, there will be 50 1 The graph of C(t) looks similar to the graph in Figure 7.21.



Figure 7.21: for Problem 7.3.42.



 0 1 1 43. a. A = 1  1 0 1  2 1 1 0







   7.6660156 7 b. After 10 rounds, we have A10  11  ≈  7.6699219 . 7.6640625 5     7.66666666667 7 After 50 rounds, we have A50  11  ≈  7.66666666667 . 7.66666666667 5  c. The eigenvalues of A are 1 and − 1 with 2       0 1 −1 E1 = span  1  and E− 1 = span  1  ,  −1  2 −1 1 2 357



Chapter 7   1  1  + −1 2 1   0  1 + −1 2 −1   −1  −1 . 2



ISM: Linear Algebra



so x(t) = 1 +



c0 3



t



t c0 3



, so that Carl After 1001 rounds, Alberich will be ahead of Brunnhilde by 1 2 needs to beat Alberich to win the game. A straightforward computation shows that 1001 (1 − c0 ); Carl wins if this quantity is positive, which is the c(1001) − a(1001) = 1 2 case if c0 is less than 1. Alternatively, observe that the ranking of the players is reversed in each round: Whoever is first will be last after the next round. Since the total number of rounds is odd (1001), Carl wants to be last initially to win the game; he wants to choose a smaller number than both Alberich and Brunnhilde. 44. a. a11 = 0.7 means that only 70% of the pollutant present in Lake Silvaplana at a given time is still there a week later; some is carried down to Lake Sils by the river Inn, and some is absorbed or evaporates. The other diagonal entries can be interpreted analogously. a21 = 0.1 means that 10% of the pollutant present in Lake Silvaplana at any given time can be found in Lake Sils a week later, carried down by the river Inn. The significance of the coefficient a32 = 0.2 is analogous; a31 = 0 means that no pollutant is carried down from Lake Silvaplana to Lake St. Moritz in just one week. The matrix is lower triangular since no pollutant is carried from Lake Sils to Lake Silvaplana, for example (the river Inn flows the other way). b. The eigenvalues of A are 0.8, 0.6, 0.7, with corresponding eigenvectors       0 0 1  0  ,  1  ,  1 . 1 −1 −2         1 0 0 100 x(0) =  0  = 100  0  − 100  1  + 100  1  , −2 −1 1 0       0 0 1 so x(t) = 100(0.8)t  0  − 100(0.6)t  1  + 100(0.7)t  1  or 1 −1 −2 x1 (t) = 100(0.7)t x2 (t) = 100(0.7)t − 100(0.6)t x3 (t) = 100(0.8)t + 100(0.6)t − 200(0.7)t . See Figure 7.22. 358



1001



ISM: Linear Algebra



Section 7.3



Figure 7.22: for Problem 7.3.44b. Using calculus, we find that the function x2 (t) = 100(0.7)t − 100(0.6)t reaches its maximum at t ≈ 2.33. Keep in mind, however, that our model holds for integer t only. 45. a. A = b. B = 0.1 0.2 1 ,b = 0.4 0.3 2 A b 0 1 1 2 and so 1 . −1 v 0 is



c. The eigenvalues of A are 0.5 and −0.1 with associated eigenvectors The eigenvalues of B are 0.5, −0.1, and 1. If Av = λv then B



v v =λ 0 0



an eigenvector of B.   2 Furthermore,  4  is an eigenvector of B corresponding to the eigenvalue 1. Note that 1 −(A − I2 )−1 b . this vector is 1         2 1 1 x1 (0) d. Write y(0) =  x2 (0)  = c1  2  + c2  −1  + c3  4 . 1 0 0 1 Note that c3 = 1.         1 1 2 2 t→∞ t→∞ Now y(t) = c1 (0.5)t  2  + c2 (−0.1)t  −1  +  4  −→  4  so that x(t) −→ 0 0 1 1 359 2 . 4



Chapter 7 46. a. T1 (t + 1) = 0.6T1 (t) + 0.1T2(t) + 20 T2 (t + 1) = 0.1T1 (t) + 0.6T2(t) + 0.1T3 (t) + 20 T3 (t + 1) = 0.1T2 (t) + 0.6T3(t) + 40     0.6 0.1 0 20 so A =  0.1 0.6 0.1  , b =  20  0 0.1 0.6 40 b. B = A b 0 1



ISM: Linear Algebra



    0 70.86  0   93.95  c. y(10) = B 10   ≈   0 120.56 1 1



    0 74.989  0   99.985  y(30) = B 30   ≈   0 124.989 1 1 



d. The eigenvalues of A are λ1 ≈ 0.45858, λ2 = 0.6, λ3 ≈ 0.74142 so the eigenvalues of B are λ1 ≈ 0.45858, λ2 = 0.6, λ3 ≈ 0.74142, λ4 = 1. v1 v v If v1 , v2 , v3 are eigenvectors of A (with Avi = λi vi ), then , 2 , 3 are 0 0 0   75  100  corresponding eigenvectors of B. Furthermore,   is an eigenvector if B with 125 1   75 eigenvalue 1. Since λ1 , λ2 , λ3 are all less than 1, lim x(t) =  100 , as in Exercise 45. t→∞ 125  1 1   0 r(t) 2 4  1  47. a. If x(t) =  p(t) , then x(t + 1) = Ax(t) with A =  1 1 2 . 2 2 w(t) 0 1 1

4 2



 75  100  y(t) seems to approach   as t → ∞ 125 1



360



ISM: Linear Algebra 



Section 7.3



     1 1 1 The eigenvalues of A are 0, 1 , 1 with eigenvectors  −2  ,  0  ,  2 . 2 1 −1 1           1 1 1 1 1 t Since x(0) =  0  = 1  −2  + 1  0  + 1  2  , x(t) = 1 1  0  + 4 2 4 2 2 1 −1 1 −1 0 t > 0. b. As t → ∞ the ratio is 1 : 2 : 1 (since the first term of x(t) drops out). 48. a. We are told that a(t + 1) = a(t) + j(t) j(t + 1) = a(t), so that A = 1 1 . 1 0

√ 1± 5 2



  1 1   2 for 4 1



b. fA (λ) = λ(λ − 1) − 1 = λ2 − λ − 1 so λ1,2 = Since x(0) = i.e. a(t) = j(t) =

1 √ ((λ1 )t+1 5 1 √ ((λ1 )t 5 a(t) j(t)



with eigenvectors

1 √ (λ1 )t 5



λ1 1



and



λ2 . 1



1 = 0



1 √ 5



λ1 λ2 1 − √5 we have x(t) = 1 1



λ1 λ 1 − √5 (λ2 )t 2 , 1 1



− (λ2 )t+1 )

√ 1+ 5 2 ,



− (λ2 )t ). → λ1 = since |λ2 | 2. Thus A is diagonalizable if and only if |a| 0 since b, c, d > 0. Therefore, λ1 λ2 λ3 > 0. Hence two of the eigenvalues must be negative, and the largest one (in absolute value) must be positive.



n



30. a. The ith entry of Ax is

k=1 n n n n



aik xk , so that the sum of all the entries of Ax is

n n n



aik xk =

i=1 k=1 k=1 i=1



aik xk =

k=1 i=1



aik ↑ 1



xk =

k=1



xk = 1.



b. As we do some computer experiments, At appears to approach a matrix with identical columns, with column sum 1. Let v1 , v2 , . . . , vn be an eigenbasis with λ1 = 1 and |λj | |λj | for 2 ≤ j ≤ 5.

5



Let v1 , . . . , v5 be corresponding eigenvectors. For a fixed i, write ei =

j=1



cj vj ; then



(ith column of At ) = At ei = c1 λt v1 + · · · + c5 λt v5 . 1 5 But in the last expression, for large t, the first term is dominant, so the ith column of At is almost parallel to v1 , the eigenvector corresponding to the dominant eigenvalue. 379



Chapter 7



ISM: Linear Algebra



d. By part c, the columns of B and C are almost eigenvectors of A associated with the largest eigenvalue, λ1 . Since the first row of C consists of 1’s, the entries in the first row of AC will be close to λ1 . 34. a. The eigenvalues of A − λIn are λ1 − λ, λ2 − λ, . . . , λn − λ, and we were told that |λ1 − λ| 0, so that A is invertible. c. Yes; if A = w z −z , then A−1 = w

1 |w|2 +|z|2



w −z



z w



is in H.



381



Chapter 7 i 0 0 −i



ISM: Linear Algebra 0 −1 , then AB = 1 0 0 −i −i 0



d. For example, if A = BA =  0 i i . 0 0 0 0 1 1 0 0 0



and B =



and



0 0 2 38. a. C4 =  1 0



Figure 7.31 illustrates how C4 acts on the basis vectors ei .



  0 0 1 1 3 0 0  , C4 =  0 0 0 0 1 0



0 1 0 0



 0 0 4 4+k k  , C4 = I4 , then C4 = C4 . 1 0



Figure 7.31: for Problem 7.5.38a. b. The eigenvalues are λ1 = 1, λ2 = −1, λ3 = i, and λ4 = −i, and for each eigenvalue  3 λk  λ2  λk , vk =  k  is an associated eigenvector. λk 1

2 3 c. M = aI4 + bC4 + cC4 + dC4



If v is an eigenvector of C4 with eigenvalue λ, then M v = av + bλv + cλ2 v + dλ3 v = (a + bλ + cλ2 + dλ3 )v, so that v is an eigenvector of M as well, with eigenvalue a + bλ + cλ2 + dλ3 . The eigenbasis for C4 we found in part b is an eigenbasis for all circulant 4×4 matrices.



39. Figure 7.32 illustrates how Cn acts on the standard basis vectors e1 , e2 , . . . , en of Rn .



k a. Based on Figure 7.9, we see that Cn takes ei to ei+k “modulo n,” that is, if i + k k exceeds n then Cn takes ei to ei+k−n (for k = 1, . . . , n − 1). k To put it differently: Cn is the matrix whose ith column is ei+k if i + k ≤ n and ei+k−n if i + k > n (for k = 1, . . . , n − 1).



382



ISM: Linear Algebra



Section 7.5



Figure 7.32: for Problem 7.5.39. b. The characteristic polynomial is 1 − λn , so that the eigenvalues are the n distinct solutions of the equation λn = 1 (the so-called nth roots of unity), equally spaced points along the unit circle, λk = cos 2πk +i sin 2πk , for k = 0, 1, . . . , n−1 (compare n n with Exercise 5 and Figure 7.7.). For each eigenvalue λk ,

n−1  λk  .   .   .  vk =  λ2  is an associated eigenvector.  k   λk  1







c. The eigenbasis v0 , v1 , . . . , vn−1 for Cn we found in part b is in fact an eigenbasis for all circulant n × n matrices.



40. In Exercise 7.2.50 we derived the formula x =



3



q 2



+



q 2 2 q 2 2



+



p 3 3 + p 3 3



3



q 2



−



q 2 2



+



p 3 3



for the solution of the equation x3 + px = q. Here write x=

3



+



is negative, and we can



q 2



+i



−



q 2 2



+



−p 3 3



+



3



q 2



−i



−



q 2 2



+



−p 3 . 3



Let us write this solution in polar coordinates: x= = =2

3



−p 3



3/2



(cos α + i sin α) +



3



p −3 −p 3



3/2



(cos α − i sin α)



− p cos α+2πk + i sin α+2πk + 3 3 3

p − 3 cos α+2πk 3



cos α+2πk − i sin α+2πk 3 3



, k = 0, 1, 2. See Figure 7.33.



Answer: 383



Chapter 7



ISM: Linear Algebra



Figure 7.33: for Problem 7.5.40.

−p 3

q 2 p 3/2 −3



x1,2,3 = 2



cos



α+2πk 3



, k = 0, 1, 2, where α = arccos

−p 3 ,2 −p 3



(



)



.

−p 3 ,− −p 3



Note that x is on the interval k = 1 and on 41. Substitute ρ =

14 x2



when k = 0, on



−2



when



−

1 x



−p 3 ,



−p 3



when k = −1 (Think about it!).



into 14ρ2 + 12ρ3 − 1 = 0;



+



12 x3



−1=0



14x + 12 − x3 = 0 x3 − 14x = 12 Now use the formula derived in Exercise 40 to find x, with p = −14 and q = 12. There 1 is only one positive solution, x ≈ 4.114, so that ρ = x ≈ 0.243.



42. a. We will use the fact that for any two complex numbers z and w, z + w = z + w and zw = zw.

p p p



The ij th entry of AB is

k=1



aik bkj =

k=1



aik bkj =

k=1



aik bkj , which is the ij th entry of



AB, as claimed. b. Use part a, where B is the n × 1 matrix v + iw. We are told that AB = λB, where ¯ ¯ ¯¯ λ = p + iq. Then AB = A B = AB = λB = λ B, or A(v − iw) = (p − iq)(v − iw).



384



ISM: Linear Algebra



Section 7.5



43. Note that f (z) is not the zero polynomial, since f (i) = det(S1 + iS2 ) = det(S) = 0, as S is invertible. A nonzero polynomial has only finitely many zeros, so that there is a real number x such that f (x) = det(S1 + xS2 ) = 0, that is, S1 + xS2 is invertible. Now SB = AS or (S1 + iS2 )B = A(S1 + iS2 ). Considering the real and the imaginary part, we can conclude that S1 B = AS1 and S2 B = AS2 and therefore (S1 + xS2 )B = A(S1 + xS2 ). Since S1 + xS2 is invertible, we have B = (S1 + xS2 )−1 A(S1 + xS2 ), as claimed. 44. Let A be a complex 2 × 2 matrix. Let λ be a complex eigenvalue of A, and consider an associated eigenvector v, so that Av = λv. Now let P be an invertible 2 × 2 matrix of the form P = [v w] (the first column of P is our eigenvector v). Then P −1 AP will be of the λ ∗ , so that we have found an upper triangular matrix similar to A (compare form 0 ∗ with the proof of Fact 7.4.1). Yes, any complex square matrix is similar to an upper triangular matrix, although the proof is challenging at this stage of the course. Following the hint, we will assume that the claim holds for n × n matrices, and we will prove it for an (n + 1) × (n + 1) matrix A. As in the case of a 2 × 2 matrix discussed above, we can find an invertible P such that λ w P −1 AP is of the form for some scalar λ, a row vector w with n components, and 0 B an n×n matrix B (just make the first column of P an eigenvector of A). By the induction hypothesis, B is similar to some upper triangular matrix T , that is, R −1 BR = T for some invertible R. Now let S = P S −1 AS = 1 0 , an invertible (n + 1) × (n + 1) matrix. Then 0 R



1 0 λ w 1 0 1 0 1 0 λ wR P −1 AP = = , an 0 R 0 B 0 R−1 0 R−1 0 R 0 T upper triangular matrix, showing that A is indeed similar to an upper triangular matrix. You will see an analogous proof in Section 8.1 (proof of Fact 8.1.1, Page 368). √ 45. If a = 0, then there are two distinct eigenvalues, 1 ± a, so that the matrix is diagonal1 1 1 1 fails to be diagonalizable. = izable. If a = 0, then 0 1 a 1 46. If a = 0, then there are two distinct eigenvalues, ±ia, so that the matrix is diagonaliz0 −a 0 0 able. If a = 0, then = is diagonalizable as well. Thus the matrix is a 0 0 0 diagonalizable for all a. √ 47. If a = 0, then there are  three distinct eigenvalues, 0, ± a, so that the matrix is diago   0 0 0 0 0 0 nalizable. If a = 0, then  1 0 a  =  1 0 0  fails to be diagonalizable. 0 1 0 0 1 0 385



Chapter 7



ISM: Linear Algebra



48. The characteristic polynomial is f (λ) = −λ3 + 3λ + a. We need to find the values a such that this polynomial has multiple roots. Now λ is a multiple root if (and only if) f (λ) = f (λ) = 0 (see Exercise 7.2.37). Since f (λ) = −3λ2 + 3 = −3(λ − 1)(λ + 1), the only possible multiple roots are 1 and −1. Now 1 is a multiple root if f (1) = 2 + a = 0, or, a = −2, and −1 is a multiple root if a = 2. Thus, if a is neither 2 nor −2, then the matrix is diagonalizable. Conversely, if a = 2 or a = −2, then the matrix fails to be diagonalizable, since all the eigenspaces will be one-dimensional (verify this!). 49. The eigenvalues are 0, 1, a − 1. If a is neither 1 nor 2, then there are three distinct eigenvalues, so that the matrix is diagonalizable. Conversely, if a = 1 or a = 2, then the matrix fails to be diagonalizable, since all the eigenspaces will be one-dimensional (verify this!). 50. The eigenvalues are 0, 0, 1. Since the kernel is always two-dimensional, with basis     1 0  1  ,  1 , the matrix is diagonalizable for all values of constant a. 0 1



7.6

1. λ1 = 0.9, λ2 = 0.8, so, by Fact 7.6.2, 0 is a stable equilibrium. 2. λ1 = −1.1, λ2 = 0.9, so by Fact 7.6.2, 0 is not a stable equilibrium. (|λ1 | > 1) √ 3. λ1,2 = 0.8 ± (0.7)i so |λ1 | = |λ2 | = 0.64 + 0.49 > 1 so 0 is not a stable equilibrium. √ 4. λ1,2 = −0.9 ± (0.4)i so |λ1 | = |λ2 | = 0.81 + 0.16 1 and 0 is not a stable equilibrium. 8. λ1 = 0.9, λ2 = 0.8 so 0 is a stable equilibrium. 9. λ1,2 = 0.8 ± (0.6)i, λ3 = 0.7, so |λ1 | = |λ2 | = 1 and 0 is not a stable equilibrium. 10. λ1,2 = 0, λ3 = 0.9 so 0 is a stable equilibrium. 11. λ1 = k, λ2 = 0.9 so 0 is a stable equilibrium if |k| 1. Thus, the zero state isn’t a stable equilibrium for any real k. √ √ 1+30k 1 16. λ1,2 = 2± 10 so |2 ± 1 + 30k| must be less than 10. λ1,2 are real if k ≥ − 30 . In this √ √ case √ it is required that 2 + 1 + 30k − 97 . 30 21 Overall, 0 is a stable equilibrium if − 97 1.



26. Stable since A and AT have the same eigenvalues. 27. Stable since if λ is an eigenvalue of −A, then −λ is an eigenvalue of −A and | − λ| = |λ|. 28. Not stable, since if λ is an eigenvalue of A, then (λ − 2) is an eigenvalue of (A − 2I n ) and |λ − 2| > 1. 391



Chapter 7

1 2



ISM: Linear Algebra 0

1 2 3 2



29. Cannot tell; for example, if A = not stable, but if A = −1 2 0 −1 2 0



0



, then A + I2 is

1 2



0

3 2



0



and the zero state is



then A + I2 =



0

1 2



0



and the zero state is stable.



30. Consider the dynamical systems x(t + 1) = A2 x(t) and y(t + 1) = Ay(t) with equal initial values, x(0) = y(0). Then x(t) = y(2t) for all positive integers t. We know that lim y(t) = 0; thus lim x(t) = 0, proving that the zero state is a stable equilibrium of the system x(t + 1) = A2 x(t). 31. We need to determine for which values of det(A) and tr(A) the modulus of both eigenvalues is less than 1. We will first think about the border line case and examine when one of the moduli is exactly 1: If one of the eigenvalues is 1 and the other is λ, then tr(A) = λ + 1 and det(A) = λ, so that det(A) = tr(A) − 1. If one of the eigenvalues is −1 and the other is λ, then tr(A) = λ − 1 and det(A) = −λ, so that det(A) = −tr(A) − 1. If the eigenvalues are complex conjugates with modulus 1, then det(A) = 1 and |tr(A)| tr(A) − 1 and det(A) > −tr(A) − 1. This means that |trA| − 1 1 then at least one eigenvalue is greater than one in modulus and the zero state fails to be stable. b. If | det A| = |λ1 λ2 | · · · |λn | 0, detA = αγ αγ − α αγ



Use Exercise 31; stable if det(A) = αγ < 1 and trA − 1 = αγ + γ − 1 < αγ. The second condition is satisfied since γ < 1. Stable if γ <

1 α



394



ISM: Linear Algebra eigenvalues are real if γ ≥

4α (1+α)2



Section 7.6



38. a. T (v) = Av + b = v if v − Av = b or (In − A)v = b. In − A is invertible since 1 is not an eigenvalue of A. Therefore, v = (In − A)−1 b is the only solution. b. Let y(t) = x(t) − v be the deviation of x(t) from the equilibrium v. Then y(t + 1) = x(t + 1) − v = Ax(t) + b− v = A(y(t) + v) + b− v = Ay(t) + Av + b− v = Ay(t), so that y(t) = At y(0), or x(t) = v + At (x0 − v).

t→∞



lim x(t) = v for all x0 if lim At (x0 − v) = 0. This is the case if the modulus of all t→∞ the eigenvalues of A is less than 1.



39. Use Exercise 38: v = (I2 − A)−1 b = 2 4



0.9 −0.2 −0.4 0.7



−1



1 2 = . 2 4



is a stable equilibrium since the eigenvalues of A are 0.5 and −0.1.



40. Note that A can be partitioned as A =



B −C T , where B and C are rotation-scaling C BT matrices. Also note that BC = CB, B T B = (p2 + q 2 )I2 , and C T C = (r2 + s2 )I2 .



a. AT A =



BT −C



CT B



B C



−C T BT



= (p2 + q 2 + r2 + s2 )I4 if A = 0.



b. By part a, A−1 =



1 T p2 +q 2 +r 2 +s2 A



c. (det A)2 = (p2 + q 2 + r2 + s2 )4 , by part a, so that det A = ±(p2 + q 2 + r2 + s2 )2 . Laplace Expansion along the first row produces the term +p4 , so that det(A) = (p2 + q 2 + r2 + s2 )2 . d. Consider det(A − λI4 ). Note that the matrix A − λI4 has the same “format” as A, with p replaced by p − λ and q, r, s remaining unchanged. By part c, det(A − λI4 ) = ((p − λ)2 + q 2 + r2 + s2 )2 = 0 when (p − λ)2 = −q 2 − r2 − s2 p − λ = ±i q 2 + r2 + s2 395



Chapter 7 λ=p±i q 2 + r 2 + s2



ISM: Linear Algebra



Each of these eigenvalues has algebraic multiplicity 2 (if q = r = s = 0 then λ = p has algebraic multiplicity 4). e. By part a we can write A = p2 + q 2 + r 2 + s2 1 p2 + q 2 + r 2 + s2

S



A , where S is



orthogonal.



p2 + q 2 + r2 + s2 (Sx) = p2 + q 2 + r2 + s2 x . Therefore, Ax =       3 −3 −4 −5 1 −39 3 5 −4  3 2  13  f. Let A =   and x =  ; then Ax =  . 4 −5 3 3 4 18 5 4 −3 3 4 13 By part e, Ax

2



= (32 + 32 + 42 + 52 ) x 2 , or



392 + 132 + 182 + 132 = (32 + 32 + 42 + 52 )(12 + 22 + 42 + 42 ), as desired. g. Any positive integer m can be written as m = p1 p2 . . . pn . Using part f repeatedly we see that the numbers p1 , p1 p2 , p1 p2 p3 , . . . , p1 p2 p3 · · · pn−1 , and finally m = p1 · · · pn can be expressed as the sums of four squares.



41. Find the 2 × 2 matrix A that transforms A 8 −3 −3 −8 = 6 4 4 −6 and A =



8 6



into



−3 4



and

−1



−3 4

1 50



into



−8 : −6



−3 −8 4 −6



8 −3 6 4



=



36 −73 . 52 −36



There are many other correct answers.



42. a. x(t + 1) = x(t) − ky(t) y(t + 1) = kx(t) + y(t) = kx(t) + (1 − k 2 )y(t) so b. fA (λ) = λ2 − (2 − k 2 )λ + 1 = 0 The discriminant is (2 − k 2 )2 − 4 = −4k 2 + k 4 = k 2 (k 2 − 4), which is negative if k is a small positive number (k < 2). Therefore, the eigenvalues are complex. By Fact 7.6.4 the trajectory will be an ellipse, since det(A) = 1. 396 1 x(t + 1) = k y(t + 1) −k 1 − k2 x(t) . y(t)



ISM: Linear Algebra



True or False



True or False

1. T, by Fact 7.2.2 2. T, by Definition 7.2.3 3. F; If 1 1 , then eigenvalue 1 has geometric multiplicity 1 and algebraic multiplicity. 2. 0 1



4. T, by Fact 7.4.3 5. T; A = AIn = A[e1 . . . en ] = [λ1 e1 . . . λn en ] is diagonal. 6. T; If Av = λv, then A3 v = λ3 v. 7. T; Consider a diagonal 5 × 5 matrix with only two distinct diagonal entries. 8. F, by Fact 7.2.7. 9. T, by Summary 7.1.5 10. T, by Fact 7.2.4 11. F; Consider A = 12. F; Let A = 1 1 . 0 1 4 0 , β = 5, for example. Then αβ = 10 isn’t an 0 5



2 0 , α = 2, B = 0 3 8 0 eigenvalue of AB = . 0 15



13. T; If Av = 3v, then A2 v = 9v. 14. T; Construct an eigenbasis by concatenating a basis of V with a basis of V ⊥ . 15. T, by Fact 7.5.5 16. F; Let A = 1 1 , for example. 0 −1



17. T, by Example 6 of Section 7.5 18. T; The geometric multiplicity of eigenvalue 0 is dim(kerA) = n − rank(A). 19. T; If S −1 AS = D, then S T AT (S T )−1 = D. 397



Chapter 7    2 0 0 1 0 0 20. F; Let A =  0 3 0  and B =  0 4 0 , for example. 0 0 0 0 0 0 21. F; Consider A = 22. T, by Fact 7.5.5 23. F; Let A = 24. F; Let A = 1 0 , for example. 1 1 1 1 0 0 and B = 0 0 0 1 . , with A2 = 0 0 0 0 



ISM: Linear Algebra



0 2 0 1 , for example. , with AB = 0 0 0 1



25. T; If S −1 AS = D, then S −1 A−1 S = D−1 is diagonal. 26. F; the equation det(A) = det(AT ) holds for all square matrices, by Fact 6.2.7. 27. T; The sole eigenvalue, 7, must have geometric multiplicity 3. 28. F; Let A = 1 1 0 0 and B = 1 2 0 1 , for example. , with A + B = 0 1 0 1



29. F; Consider the zero matrix. 30. T; If Av = αv and Bv = βv, then (A + B)v = Av + Bv = αv + βv = (α + β)v. 31. F; Consider the identity matrix.  1 0 0 32. T; Both A and B are similar to  0 2 0 , by Fact 7.4.1 0 0 3  33. F; Let A = 34. F; Consider 35. F; Let A = 1 1 0 1 and v = 1 , for example. 0



1 1 0 1 0 1 2 0 , for example. , and w = ,v = 1 0 0 3



36. T; A nonzero vector on L and a nonzero vector on L⊥ form an eigenbasis. 37. T; The eigenvalues are 3 and −2. 38. T, We will us Fact 7.3.7 throughout: The geometric multiplicity of an eigenvalue is ≤ its algebraic multiplicity. 398



ISM: Linear Algebra



True or False



Now let’s show the contrapositive of the given statement: If the geometric multiplicity of some eigenvalue is less than its algebraic multiplicity, then the matrix A fails to be diagonalizable. Indeed, in this case the sum of the geometric multiplicities of all the eigenvalues is less than the sum of their algebraic multiplicities, which in turn is ≤ n (where A is an n × n matrix). Thus the geometric multiplicities do not add up to n, so that A fails to be diagonalizable, by Fact 7.3.4b. 39. T, Consider the proof of Fact 7.3.4a. 40. T; An eigenbasis for A is an eigenbasis for A + 4I4 as well. 41. F; Consider a rotation through π/2. 42. T; Suppose v λv A(v + w) v A A . If w for a nonzero vector = = w λw Aw w 0 A is nonzero, then it is an eigenvector of A with eigenvalue λ; otherwise v is such an eigenvector. 1 0 0 1 and 1 1 . 0 1



43. F; Consider



44. T; Note that S −1 AS = D, so that D4 = S −1 A4 S = S −1 0S = 0, and therefore D = 0 (since D is diagonal) and A = SDS −1 = 0. 45. T; There is an eigenbasis v1 , . . . , vn , and we can write v = c1 v1 + · · · + cn vn . The vectors ci vi are either eigenvectors or zero. 46. T; If Av = αv and Bv = βv, then ABv = αβv. 47. T, by Fact 7.3.6a 48. F; Let A = 1 0 , for example. 0 0



49. T; Recall that the rank is the dimension of the image. If v is in the image of A, then Av is in the image of A as well, so that Av is parallel to v. 50. F; Consider 0 1 . 0 0



51. T; If Av = λv for a nonzero v, then A4 v = λ4 v = 0, so that λ4 = 0 and λ = 0. 52. F; Let A = 1 1 0 0 and B = 0 1 , for example. 0 1



53. T; If the eigenvalue associated with v is λ = 0, then Av = 0, so that v is in the kernel of 1 A; otherwise v = A λ v , so that v is in the image of A. 399



Chapter 7



ISM: Linear Algebra



54. T; either there are two distinct real eigenvalues, or the matrix is of the form kI 2 . 55. T; Either Au = 3u or Au = 4u. 56. T; Note that (uuT )u = u 2 u. 57. T; Suppose Avi = αi vi and Bvi = βi vi , and let S = [v1 . . . vn ]. Then ABS = BAS = [α1 β1 v1 . . . αn βn vn ], so that AB = BA. 58. T; Note that a nonzero vector v = a b p if (and only if) is an eigenvector of A = c d q p ap + bq p ap + bq = 0. Check that this , that is, if det is parallel to v = Av = q cp + dq q cp + dq is the case if (and only if) v is an eigenvector of adj(A) (use the same criterion).



400