Otto bretscher linear algebra with applications solutions

Download Linear Algebra Otto Bretscher Solution Manual and more Linear Algebra Exercises in PDF only on Docsity! Answers to Odd-Numbered Exercises CHAPTER I XI -2t 1.1 Answers to more theoretical questions are omitted. X2 t 7. I X3 = 0=1. (x. y) (-1, 1) 3. No solutions 5. (x, y) = (0,0) 7. No solutions 9. (x, y, z) = (t. ~ - 2t, t), where t is arbitrary 11. (x, y) = (4, 1) 13. No solutions 15. (x, y, z) = (0,0,0) 17. (x, y) = (-Sa + 2b, 3a - b) 9. 19.a.Ifk=7 b. If k = 7, there are infinitely many solutions. c. If k = 7, the solutions are (x, y, z) = (1 - t, 2t - 3, t). 21. 1l,13,and17 11. X4 0 X5 0 Xl I I t - s ­ 2r x2 I I r x3 -t +s + I X4 t-2s+2 X5 s X6 X1 1 r -2t 1;: ~ 31~:4 23. a. Products are competing. [ b. PI = 26, P2 = 46 25. a = 400, b = 300 13. No solutions 27. a. (x, y) = (t, 2t); 15 m~ m b. (x, y) = (t, -3t); x,c. (x,y)=(O,O) X231. f(t)=1-5t+3t 2 33. f(t)=2t 2 -3t+4 17. IX3 35. f(t) = at2 + (l - 4a)t + 3a. for arbitrary a X4 2e3t 2t37. f(t) = - e X5 39. -20 - 2x - 4v + x 2 + i = 0, the circle centered at (I, 2) with radius 5 -8221/4340 8591/8680 4695/434 -459/434 699/434 41. If a - 2b + c = 0 19. [1 aod [143. a. The intercepts of the line x + y = 1 are (1.0) and (0. I). The intercepts of the line x + &y = tare (t, 0) and (0,2). The lines intersect if 21. b=O,c= l,d=O,withabeingarbitrary t oj- 2. 23. 4 types t 2t - 2 b. x = ----, )I = - ­ t-2' t-2 27. Yes; perform the operations backwards. 45. There are many correct answers. Example: 29. No; you cannot make the last column zero by ele­ -41 mentary row operations. x -5z = y -3z =-2 I 31. a = 2, b = c = d = I 49. Twenty $1 bills, eight $5 bills, and four $10 bills. 33. f(t) = I - 5t + 4t2 + 3t 3 - 2t4 1.2 Answers to more theoretical questions are omitted. 35. f(t) = -5 + 13t - IOt2 + 3t3 IOt+13] -8t - 8 I. [ 6t[~] -t] .. t 37. _:~. where t IS arbitrary. X] [4 - 2s - 3t] r 3. y = s [ Z t XI] [500]39. X2 = 300 [ X3 400 41. a. Neither the manufacturing nor the energy sector makes demands on agriculture. 5. [} r~~1 b. XI :0:0 18.67, X2 :0:0 22.60, X3 :0:0 3.63 471 0 472 ANSWERS TO ODD-NUMBERED EXERCISES II. Undefined43. In\ = ~m2 45. a"'" 12.17, b "'" -1.15. c "'" 0.18. The longest day 13. 15. 70 is about 13.3 hours. [~~] 47. a. If k is neither I nor 2 17. Undefined 19.b. If k = I c. If k = 2 m49. a. Xl = 3X3 - 2X4, x2 = 2X3 - X4, for arbitrary X3 and x4 b. Yes, x) = I,X2 = 5,.q = 9,X4 = 13. I 21. 23. 51. C = 25 123 0P~1 r~ 0 r1 53. xv = 0, the union of the two coordinate axes 25. The system A.r = chas infinitely many solutions or 55. a(xy - y) + b(y2 - y) = 0, where a#-O or b #- 0 none. 57. a(x 2 - x) + b(i - y) = 0, where a#-O or b #- 0 0 0 rIO 059. 25 - lOx - lOy + x 2 + i = 0, the circle of radius 27. o .1 29. [~ 0 o 0 I5 centered at (5,5) 0 o 0 ~] J0 61. No solutions 63. Statistics: $86; Set Theory: $92; Psychology: $55 2 31. 365. Beginning: 120 liberal. ]40 conservative. [~ -il fO'mmpk0End: 140 liberal, 120 conservative. 67. Cow: 34121 liang; sheep: 20/21 liang 33. Ax = x 35. Aej is the ith column of A.69. Swallow: 24/19 liang; sparrow: 32/19 liang 71. A: 265; B: 191; C: ]48; D: 129; E: 76 37. x ~ [2: 2tl wh~e t i' "'bitr", 73. Gaussian elimination shows that Pigeons = -250 + ~(Swans) + 20(Peacocks) and I 0 0 Sarasas = 350 - .!J(Swans) - 21 (Peacocks). 39. 0 I 0 [ :]One solution (the one given by Mahavira) is 15 pi­ 001 geons, 28 sarasabirds, 45 swans, and 12 peacocks 41. One solution 43. No solutions (spending 9,20,35, and 36 panas, respectively). 47. a. x= 0 is a solution. 75. 53 sheep, 42 goats, and 5 hogs b. By part (a) and Theorem 1.3.3 77. Full Half Empty c. A(xj +X2) = Ax\ +AX2 = 0+0 = 0 d. A(k."i) = k(Ax) = kG = 0 1st Son p 10 - 2p p 49. a. Infinitely many solutions or none2nd Son q 10 - 2q q 3rd Son lO-p-q 2p+2q -10 lO-p-q b. One solution or none c. No solutions Here, p and q are integers between 0 and 5 such that d. Infinitely many solutions p + q 2: 5. 51. If In = rand s = p1.3 Answers to more theoretical questions are omitted. I. a. No solutions b. One solution 53. Yes c. Infinitely many solutions 3. Rank is 1. 55 Yo<; m~ -1 [~] +2 [~] 5. a. x l~] + y l~] = lIn b. x = 3, y = 2 57. [I~] [~]+[~] 7. One solution 59. c = 9, d = 11 61. For c = 2 and for c = 3 63. Line through the endpoint of vin the direction of in [0 -1] [ 0 I] 53. CD = I 0 and DC = -1 O' We com­ pose two reflections to obtain a rotation. [ -s2s -2t]55. X = t ,where sand t are arbitrary constants 57. X = [-~ - ~] 59. No such matrix X exists. [' +, -2+1] 61. X = -:s I ~ 2t , where sand t are arbitrary constants 63. No such matrix X exists. 65. X = [~ ~ ] , where t is an arbitrary constant 67a. [ I 1 I] A = [I I 1] 69a. Of the surfers who are on Page 1 initially, 25% will be on Page 3 after following two consecutive links. 69b. The ijth entry of A 2 is °if there is no path of length 2 from vertex j to vertex i in the graph of the mini-Web, meaning that a surfer cannot get from Page j to Page i by following two consecutive links. 71. A straightforward computation shows that the only nonzero entry of A4 is the first component of the third column. Thus there is no path of length 4 from Page 3 to Page I in the graph of the mini-Web, while there is a path of length 4 linking any other two pages. We cannot get from Page 3 to Page 1 by fol­ lowing four consecutive links. 73. lim (Amx) _ ( lim Am) X m~oo m~oo - (XI + ... + x ll ) Xequ[ i,~" X~"" ] X '-v-" components of X X equ 75. The ijth entry of Am+1 = Am A -is the dot product of the ith row tV or Am with the jth column i! of A. Since all components of tV and some components of i! are positive (with all components of i! being non­ negative), this dot product is positive, as claimed. 0.5003 OA985 0.5000] 77. We find A 10 ~ 0.0996 0.1026 0.0999 . This [ OA002 0.3989 OAOO1 shows that A is regular and it suggests then xequ = 0.5] [0'5]0.1 . To be sure, we verify that A 0.1 = [ 0.4 0.4 0.5] 0.1 . [ OA ANSWERS TO ODD-NUMBERED EXERCISES 475 79. An extreme example is the identity matrix In. In this case, Inx = Xfor all distribution vectors. 81. Ami! = 5'lltj 83. By Definition 2.3.10, there exists a positive integer m such that Am is a positive transition matrix. Note that Am X = X. Thus, the ith component of x is the dot product of the ith row tV of Am with.r. Since all components of tV and some components of xare positive (with all components of xbeing nonnega­ tive), this dot product is positive, as claimed. 85. There is one and only one such X. 2.4 Answers to more theoretical questions are omitted. [8 -3]1. -5 2 3 [-: ~] 5. Not invertible 7. Not invertible [I ° -1]9. Not invertible 11. ° I ° ° ° 1 -2 °1 13 r-f o 1 I -2 ° ~1 -6 9 -5 -1 -5 15. 9 -5 -5 9 [ 1 2 -3 -~1 17. Not invertible 19. Xl = 3YI - 2.5Y2 + 0.5Y3 X2 = -3YI +4Y2 - Y3 X3 = YI - 1.5Y2 + 0.5Y3 21. Not invertible 23. Invertible 25. Invertible 27. Not invertible 29. For all k except k = land k = 2 31. It's never invertible. 33. If a2 + b2 = I 35. a. Invertible if a, d, f are all nonzero b. Invertible if all diagonal entries are nonzero c. Yes; use Theorem 2A.5. d. Invertible if all diagonal entries are nonzero I 37. (cA)-1 = -A-I C 39. M is invertible; if mij = k (where i =P j), then the ijth entry of M- 1 is -k; all other entries are the same. 41. The transformations in parts a, c, and d are invert­ ible, while the projection in part b is not. 43. Yes; x= B- 1(A -I y) 476 ANSWERS TO ODD·NUMBERED EXERCISES 45. a. 33 = 27 b. 17 3 c.\~J = 64 (seconds) 2 81.47. f (x) = x is not invertible, but the equation f (x) = 0 has the unique solution x = O. 51. a. Since rank A < n. all the entries in the last row of E will be zero. We can let c = e,,: the sys­ matrix ofT: [-~ ~~] o -1 0 matrix of L: [~ ~ ~] o 0 I 83. Yes; yes; each elementary row operation can be "un­tem Ex = e" will be inconsistent. Reversing done" by an elementary row operation. the row reduction from A to E, we can trans­ form the system gr = e into an inconsistent 85. a. Use Exercise 84: let S = EjE2···Ep' system A.r = b. 1l We have rank A ::s m < n. Now use part a. b. S = = b. [_~ ~] [t ~] [-t ~] 53. a. A] = 2. A2 = 6 b. We chose A = 2. A-Ah = [~ I] ~ [-I] 87. [ ~ ~], [~ ~], [~ ~l [~ ~]. [~ ~ ], 3 ,x = I' where k is nonzero and c is arbitrary. Cases 3 and 4 [3 I] ~ ~ represent shears, and case 5 is a reflection. c. Check that 3 5 x = 2x. 89. Yes; use Theorem 2.3.4. 55. detA = II [~] II sin(7f/2) II [~] II = 4, and A-I 2 [ 16 1~2 ] , a scaling by 1h A(m) A2], L __ [L(m)93. a. Write A = A3 A4 L3[ L~l -1, and A-I = A, a reflection VIm) VJ] V = 0 V~· Use Theorem 2.3.9. [ I, and c. Solve the equation A(1I-1) A = ~ v] [L~ 0] [VI y][ 0.8] [A-I = 0.6 . w k x t 0 I0.6 ' a rotatIOn -0.8 for .r, y, and t. 61. detA = II [_:] Ii sin(7f/2) II mII = 2, and A-I = 95. A is invertible if both A II and An are invertible. In this case, I [I -1]. ..:2 1 I' a rotatIOn through 7f /4 combmed with a scaling by ~/2 97. rank A = rank(A II) + rank(A23) 99. Only A = In A -I -_ 3 ' a reflection combined with a 215[-43 4] lOl. (ijth entry of AB) = Lk=l aikhkj ::s S Lk=1 bkj ::s srscaling by 1/5 107. g (J (x)) = x, for all x 65. detA = II[nllsin(7f/4)11[~]11 I, and A-I X . if x is even f ( g(x) ) = { x + 1 if x is odd [ _ ~ ~] , a vertical shear The functions f and g are not invertible. CHAPTER 3 67. False 69. False 71. True 3.1 Answers to more theoretical questions are omitted. I. kerA = {O} 3. el,e273. True 75. True 1 [ 977. A = BS- I 5. 7. ker A = {O}79. A = 5" -2 H]I ~] 57. 59. 9. ker A = (O) 11. n~ -2 -3 0 I 0 0 13. 0 o ' -2 -1 ' 0 0 0 I 0 0 0 1 15. 17. AlloflR?2[:],[~] 19. The line spanned by [_~] 21. All oflR?3 23. kernel is (O), image is all of lR?2. 25. Same as Exercise 23 27. fix) = x 3 - x ¢ [Sin ¢ cos fI] 29. f [fI] = sin¢sinfl cos¢ (compare with spherical coordinates) [ -,., -3] 31.A= ~ ~ 33 T m~ x + 2y +3< 35. ker T is the plane with normal vector v; imT =lR?. 37. im A = Span(el, e2); ker A = span(el); im(A 2) = Span(el); ker(A 2 ) = Span(el, e); A 3 = 0 so ker(A 3 ) = lR?3 and im(A3) = {O} 39. a. ker B is contained in ker(AB), but they need not be equal. b. im(AB) is contained in im A, but they need not be equal. 41. a. im A is the line spanned by [~], and ker A is the perpendicular line, spanned by [ -~] . b. A 2 = A; if vis in im A, then Av= v. c. Orthogonal projection onto the line spanned by [~] 43. Suppose A is an n x m matrix of rank r. Let B be the matrix you get when you omit the first r rows and the first m columns of ITef [A ! Ill]' (What can you do when r = II?) ANSWERS TO ODD-NUMBERED EXERCISES 477 45. There are m - r nonleading variables, which can be chosen freely. The general vector in the kernel can be written as a linear combination of m - r vectors, with the nonleading variables as coefficients. 47. im T = L2 and kerT = LI 51. ker(AB) = (O) I 0 I 1 1 I 0 I 53. a. I II I ,0 I 0 0 0 o I 0 0 001 0 000 1 b. ker H span(vl, V2, V3, V4), by part a, and im M = span(v\, V2, V3, V4), by Theo­ rem 3.1.3. Thus ker H = im(A1). H(Mx) = O. since Mx is in im M = ker H. 3.2 Answers to more theoretical questions are omitted. 1. Not a subspace 3. W is a subspace. 7. Yes 9. Dependent II. Independent 13. Dependent 15. Dependent 17. Independent 19. r~1 ,," r~1 M'~""""L 21. V2 = vI, or, vI - V2 = 0, so that [_ ~] is in the kernel. ~ O~ [I] . . I23. VI = , so that 0 IS In the kerne. 25. '" "J, 0', '" '- "' ~ ii, '0 th" [_~] i' io the kernel. 27. 29. [~], m[ij,[i] 31. 33r~H!1 r~H~H!~ 35. Suppose there is a nontrivial relation CI VI + ... + ci Vi + ... + cm Vm = 0, with Ci i=- O. We can solve this equation for Vi and thus express Vi as a linear combination of the other vectors in the list. Conversely, if Vi is a linear combination of the other vectors, Vi = '" , then we can subtract Vi from both sides of the equation to obtain a nontri­ vial relation (the coefficient of Vi will be -1). 480 ANSWERS TO ODD·NUMBERED EXERCISES Each such cubic is the union of the y-axis, the 3.4 Answers to more theoretical questions are omitted. x-axis, and any line through the point (1, 1). 53. Plugging the points (1, 1), (2, I), (I. 2), and (3, 2) into the solutions of Problem 44, we find that Cs = C7 = Cs = C9 = 0, so that the solutions are of the form CIOY(Y - l)(y - 2) = 0, where CIO -=I- 0. Division by CIO produces the unique so­ lution y(y - 1)(y - 2) = 0, the union of the three horizontal lines y = 0, y = I, and y = 2. 55. Plugging the points (1, 1), (2, I), (1, 2), (2, 2), and (3,3) into the solutions of Problem 44, we find that Cs = Cs = C9 = °and C7 + CIO = 0, so that the solutions are of the form C7X(X - l)(x - 2) ­ C7Y(y - I)(y - 2) = 0, where C7 -=I- O. Division by C7 produces the unique solution x (x - I) (x - 2) = y(y - 1)(y - 2). the union of the diagonal y = x with an ellipse. 57. See Exercises 41 and 56. Since the kernel of the 8 x IO matrix A is at least two-dimensional, and because everyone-dimensional subspace of ker A defines a unique cubic (compare with Exercise 40), there will be infinitely many such cubics. 59. There may be no such cubic [as in Exercise 49], ex­ actly one [take the 9 points in Exercise 47 and add (-1, -I)]. or infinitely many [as in Exercise 51]. 63. A basis of V is also a basis of W, by Theorem 3.3.4c. 65. dim(V+W) = dim V+dim W, by Exercise 3.2.51. 67. The first p columns of ITef A contain leading l' s be­ cause the Vi are linearly independent. Now apply Theorem 3.3.5. 71. [0 I 0 2 (I], [0 0 3 0] , [0 0 0 0 1] 73. a. A and E have the same row space, since ele­ mentary row operations leave the row space un­ changed. b. rank A = dim(rowspace(A»), by part a and Ex­ ercise 72. 77. Suppose rank A = n. The submatrix of A consisting of the n pivot columns of A is invertible, since the pivot columns are linearly independent. Conversely, if A has an invertible II x n subma­ trix, then the columns of that submatrix span lR", so im A = lR" and rank A = n. 79. Let m be the smallest number such that Am = O. By Exercise 78, there are m linearly independent A I1vectors in lR"; therefore, m :s II, and = Am A I1 m - = O. 83. a.3,4,or5 b. 0,I,or2 85. a. rank(AB).:s rank A b. rank(AB):S rank B I. [x]9.1 = m 3. [x] 9.1 = [~] 5. [x]9.1 = [-~] 7. [x] 9.1 = [~] 9. x isn't in V. 11. [xh; = [~j~] 13. [i]" ~ [-1] 15. [xb = H] 17. 19. B[xl" ~ [ :] = [~ -~] . -I 21. B 23. B= [~ ~] = [~ -~] 25. B = [-~ -~] 27. B = 0[~ ° ~]° 29. B = 1 31. B = 0[~ 0 ~] U 0 ~]0 0 33. B = I 35. B = 1[~ 0 ~] 0 ~]0 H° 45. If V is any vector in the plane that is not parallel to x, then V, ~(x - 2v) is a basis with the desired pmp«". Fo'"=ple. "~ mgi", th' b"", m·~ [=;j 47. A = [~ ~] 49. [=:] 53. x= [~~] 55. ~ [-: ~] ANSWERS TO ODD-NUMBERED EXERCISES 481 57. Consider a basis with two vectors parallel to the b . . [I27. A aSls IS 0 0] [0 ~ ] . so that the dimension plane, and one vector perpendicular. o ' 0 is 2. 59. Yes 29. A basis is [-~ ~] , so that the dimen­ 61. [-~], [~],forexamPle ~J. [-~ sion is 2. ooB [~ b~ ~ :d] [163. Yes 67. = 31. A baSIS IS I - ~ ] , so that the dimension ~J. [~ is 2. [I 2] [3 0]-I69. lfS= 2 I ,thenS AS= ° -I' 33. Only the zero matrix has this property, so that the basis is 0, and the dimension is 0. 71. a. If Bx = S-1 ASx = 0, then A(S.~) = O. b. Since we have the p linearly independent vec­ 1 ° 0] [0 ° 0]35. A basis is ° ° ° , ° 1 ° ,tors Sv 1, SV2, ... , Svp in ker A, we know that [ dim(ker B) = p :s dim(ker A), by Theo­ ° ° ° ° ° ° rem 3.3.4a. Reversing the roles of A and B, 0 ° 0] we find that dim(ker A) :s dim(ker B). Thus, ° ° ° .and the dimension is 3.[nullity A = dim(ker A) = dim(ker B) = 001 nullity B. n n(n + 1) 37. 3,5,or9 39. Lk =[ 0.36 0.48 0.8] 2 73. 0.48 0.64 -0.6 k=1 -0.8 0.6 0 41. 0,3,6,or9 43. 2 45. dim(V) = 3 75. [~ -~] 47. Yes and yes 49. Yes 3x + be4x51. f(x) = ae 77. bij = an+l-i,ll+l-j 4.2 Answers to more theoretical questions are omitted. 79. By Theorem 3.4.7, we seek a basis VI, V2 such that 1. NonlinearAVI = VI and AV2 = -V2. Solving the linea~ sys­ tems Ax = x and Ax = -.~ [or (A - h)"x = °and 3. Linear, not an isomorphism (A + h)x = OJ, we find VI = [n and V2 = [~], 5. Nonlinear 7. Isomorphism for example. 9. Isomorphism 11. Isomorphism 13. Linear, not an isomorphism CHAPTER 4 15. Isomorphism 4.1 Answers to more theoretical questions are omitted. 17. Linear, not an isomorphism I. Not a subspace 19. Isomorphism 21. Isomorphism 3. Subspace with basis I - t, 2 - t 2 23. Linear, not an isomorphism 5. Subspace with basis t 7. Subspace 25. Linear, not an isomorphism 9. Not a subspace II. Not a subspace 27. Isomorphism 13. Not a subspace 15. Subspace 29. Linear, not an isomorphism 17. Matrices with one entry equal to I and all other en­ 31. Linear, not an isomorphism tries equal to 0. The dimension is mn. 33. Linear, not an isomorphism 19. A basis is [~J. [~], [~], [~] ,so that the dimension 35. Linear, not an isomorphism 37. Linear, not an isomorphism is 4. 39. Linear, not an isomorphism . . [I 0] [0 0] h d' .21. A baSIS IS ° ' ° I ' so that t e ImenSlOn 41. Nonlinear 43. Isomorphism is 2. 45. Linear. not an isomorphism 47. Linear, not an isomorphism 23. A basis is [~ ~], [~ ~], [~ ~], so that the 49. Linear, not an isomorphism dimension is 3. 51. ker T consists of all matrices of the form [~ ~ ] , 25. A basis is I - t, I - t 2 , so that the dimension is 2. so that the nullity is 2. 482 ANSWERS TO ODD-NUMBERED EXERCISES 53. The image consists of all linear functions, of the form mt +b, so that the rank is 2. The kernel consists of the constant functions, so that the nullity is 1. 55. The image consists of all infinite sequences, and the kernel consists of all sequences of the form (0, XI, 0, X3, 0, xs, ... ). 57. The kernel consists of all functions of the form 2r + be3rae , so that the nullity is 2. 59. The kernel has the basis t - 7, (t - 7)2, so that the nullity is 2. The image is all of JR, so that the rank is 1. 61. The kernel consists of the zero function alone, and the image consists of all polynomials get) whose constant tenn is zero [that is, g(O) = 0]. 63. Impossible, since dim(P3) oF dim(JR3). 65. b. ker T consists of the zero matrix alone. d. This dimension is mn. 67. For all k except k = 2 and k = 4 69. No; if B = S-l AS, then T(S) = 0. 71. Yes, there is exactly one such polynomial. 73. Yes 77. Yes and yes 79. Surprisingly, yes 83. The transfoffimtion T induces a transformation T from ker(L a T) to ker L, with ker T = ker T. Applying the rank-nullity theorem as stated in Exercise 82 to T, we find that dim (ker(L a T)) = dim (ker T) + dim (im T) :s dim (ker T) + dim (ker L), since im T is a subspace of the kernel of L. 4.3 Answers to more theoretical questions are omitted. I. Yes 3. Yes 5 7[~ ~ ~] [H ~] 9 11.[H n [~ ! ~] 13 r ~ ~1~ ~ 17. [~ -~] 19. [~ -;] 21. [-~ -~ ~] 23 [H ~]° 0-3 25 [~ - ~] 27. [~ -i -:]~ 1 °-1 ° 0]-1 1 -1 1[1 c. S = 2: I 29 [~ ~ Sf] 33 [~ ! ~] cose - sin e] 2/9 -14/9]53. [ sine cose 55. [-1/9 7/9 57. [=~ ~] 59. T(J(t)) = t· f(t) from P to P, for example 61. a. S = 4] b. S-l ~[-3 = [-3 43]4 3 25 4 63 a. hI ~ [~l h, ~ [~l foc ",mp" -1 b. S = 1 -~ ][ ANSWERS TO ODD-NUMBERED EXERCISES 485 59. The kernel consists of the symmetric n x n matri­ 9. Ilxoll < Ilxll for all other vectors.r in S ces, and the image consists of the skew-symmetric 1 l. b. L (L + G)) = y matrices. c. L+(L(x») = projvx, where V = (kerA)J. = im(A T)o ° ° 0] 61 ° ° ° () d. im(L+) = im(A T) and ker(L +) = {O} . ° ° ° ° f° ° ° 2 , /+(Yl ~ [~ !] J63. If A = LDU, then AT = UT DL T is the LDU 0 factorization of AT. Since A = AT the two factor­ 13. b. L+(L(;») = projvx, where V = (ker A)J. = izations are identical, so that U = L T , as claimed. im(A T ) 65 [0.6 -0.8] 0.8 -0.6] [0.6 0.8] c. L(L+G)) = projwY, where W = imA . 0.8 [ -0.6 ' and0.6 ' 0.6 0.8' 0.8 (ker(A T ))1. 0.8 0.6] d. im(L+) = im(A T) and ker(L+) = ker(A T )[0.6 -0.8 I I -I 0] I I 2 0-1 e. I' (Yl ~ [~ ~] J69. Pw - ­- 3 -I () 2-1[ ° -I -I I 15. Let B = (A T A)-I AT. 71. dim V = 2n - I n-I 17. Yes; note that ker A = ker(A T A). 73. a. Note that ab = -I. Now a.h = L (ah)k k=O 19. U] I - (ah)fI ---- = °for even n. 1 - ah 21. ;* = [-~] , lib - Ax*11 = 42b. By Theorem 5.3.10 and the preceding para­ graph, P is a linear combination of the matrices M = aa T and N = bhT . It suffices to show that 23. [~] M and N are Hankel matrices. Indeed, m i i = . . -2 .+. :1 a l +.I- = mi+l,j-l and nij = hl.l- = [I -3t] .25. t ' for arbItrary t ni+l,j-1 for all i = 1, .... n - 1 and all j = 2, ... , n. * * ~ I27. L~] 29. XI = x2 2:5.4 Answers to more theoretical questions are omitted. I. im A = span [~] and ker(A T ) = span [-~] 31. 3 + 1.5t 33. approximately 1.5 +0.1 sin(t) - IAI cos(t) 3. The vectors form a basis of JF!?n. co + 35CJ = log(35) 5. VJ. = (kerA)J. = im(A T ), where Co + 46CJ = log(46) 37. a. Try to solve the system I 59 I (77)"Co + CI = ogA= [1 1 I 1] 1 2 5 4 Co + 69c1 = log(l33) [c*] [0915]Least-squares solution cr ~ 0:017' Use B,,;, of V" fl f~] approximation log (d) = 0.915 + 0.017t. b. Exponentiate the equation in part (a): d = IOlogd = 100.915+0.0171 ~ 8.221. 10°.0171 ~ 7. im A = (ker A)J. 8.221 . 1.041 . im(A 7) = (ker A).1. c. Predicts 259 displays for the A320; there are much fewer since the A320 is highly comput­ erized. 39. a. Try to solve the system s: parallel to ker A Co + 10g(600,000)c1 = log(250) Co + log(200,000)CJ = log(60) Co + log(60. ooo)c 1 = log(25),. Co + 10g(IO,OOO)CJ = log(12) Co + log(2,500)Cj = log(5) 486 ANSWERS TO ODD·NUMBERED EXERCISES 5.5 Answers to more theoretical questions are omitted. 45. They are the same. 3. a. If S is invertible 49. The kernel is the plane span(v, w) and the image b. If S is orthogonal is R. 5. Yes 7. For positive k 51. Let au be the first diagonal entry that does not be­ long to the pattern. The pattern must contain an en­9. True 11. The angle is 8 try in the ith row to the right of au as well as an 13. The two norms are equal, by Theorem 5.5.6. entry in the ith column below au. 15. If b = c and b2 < d 17. Ifker T = [O} 53. Only one pattern has a nonzero product, and that product is I. Since there are n2 inversions in that 19. The matrices A = l~ ~ J such that b = c, a > 0, pattern, we have det A = (_l)n2 = (_1)n. and b2 < ad. 55. Yes, since the determinants of all principal subma­ trices are nonzero. See Exercise 2.4.93. 21. Yes, (v, It') = 2(v . w) 57. Only one pattern has a nonzero product, and that 23. 1, 2t - 1 product is 1. Thus, del A = I or del A = -1. 25. VI + ~ + ~ + ... = ~ 59. a. Yes b. No c. No 1 . 1 0] 27. ao = viz' Ck = 0 tor all k 61. F,;I.<lo be ,"'emati",..,;"" F [~ () 1 = I but o 0 2 if k is odd bk = Okn { F [~ ~ ~] = O.if k is even 1 () () 21 n 29. L -- ­ k odd k2 - 8 6.2 Answers to more theoretical questions are omitted. 1. 6 3. -24 5. -24 33. b. IIfl12 = (f, f) = i~ w(t)dt = 1, so that 7. 1 9. 24 11. -72IIIII = 1 If 13. 8 15. 8 17. 8 35. a. IItl132 = ~ ill r2dt = and IItl134 = 19. -1 21. 1 23. (_1)n(I1-I)/2. This is 1 if either nor (n - I) is di­ )~jl ~t2dt=V~'~=~ visible by 4, and -1 otherwise. V n -J Jr 8 2 25. 16 27. a2 + b2 29. det(P)) = 1 and det(Pn) = det(Pn_I), by expan­b. For f(t) = ~ we have Ilfll32 = VI and sion down the first column, so det( Pn ) = 1 for all n. IIfl134 = If 31. a. det [ 1 1 J = aI - ao ao al CHAPTER 6 b. Use Laplace expansion down the last column to 6.1 Answers to more theoretical questions are omitted. see that f (t) is a polynomial of degree "S: n. The 1. 0 3. -2 5. 110 coefficient k of t n is IT (ai - a)). Now n-I~i» 7. 0 9. -36 11. k =F 3/2 det A = f(an) = k(an - ao)(an - al) ... (an ­ an-I) = IT (ai - a) ). as claimed. 13. Lie () 15. k =F 1/2 n,?:-i>j n17. [f k is neither I nor -1 33. IT ai' IT (ai -a)) (use linearity in the columns and 19. [f k is neither 0 nor I i=1 i» Exercise 31) 21. If k is neither 1 nor -2 123. If A is 1 or 4 25. If Ais 2 or 8 A A35. [ ~21] [aa21] and [~21] [hb2] are solutions. The 27. If A is 2, 3, or 4 29. If Ais 3 or 8 equation is of the form pXl + qX2 + b = 0; that is, 31. 24 33. 99 35. 18 it defines a line. 37. 55 39. 120 41. 24 37. ±l 2 43. det(-A) = (-l)ndetA 39. det(A T A) = (detA) > 0 ANSWERS TO ODD-NUMBERED EXERCISES 487 41. detA = det(A T ) = det(-A) = (-l)ndetA b. V(VI, V2, V3, VI x V2 x V3) -det A, so det A = 0 =ldet[vl xv2 XV3 VI V2 V3]! 2 = Ilvl x V2 x v311 2, by definition of the cross 43. AT A = [IIVI1 v· w] product v·iL, Ilw11 2 ' c. V(VI, V2. V3) = Ilvl x V2 x v311, by parts (a) so det(A T A) = IIvI1 2 11wll 2 - (v·w)2 ~ 0, by the and (b) Cauchy-Schwarz inequality. Equality holds only if 19. det [VI V2 V3] = VI . (V2 x V3) is positive ifvand ware parallel. (and only if) VI and V2 x V3 enclose an acute angle. 45. Expand down the first column: j(x) = 21. a. Reverses b. Preserves c. Reverses -xdet(A41) + constant, so j'(x) = -det(A41) = -24. det [~ -~] 7 47. T is linear in the rows and columns. 23. XI = 17' [ 51 det -6 -~] 49. A = 2 ; ], [0< ",mp[e. Start with a 1 14 det [_~ triangular [: matrix with determinant 13, such as ~] 6 X2 = ­ [~ 1 0 17 det [_~ -~] I :] . 'od add the r.", row to the ,~ood 0 13 .[] and to the third to make all entries nonzero. 25. adj A ~ [ ~ -1 o ; 51. det A = (_I)n -2 0 1 53. a. Note that det(A) det(A -I) = 1, and both factors 1 A-I = ---adj A = -adj A are integers. det A b. Use the formula for the inverse of a 2 x 2 matrix -1 0 I](Theorem 2.4.9b). o 1 0[ 59. No 2 0 -1 a -b61. Take the determinant of both sides of 27. X = -2--2 > 0; Y = -2--2 < 0; X decreases a +b a +bIn 0] [A B] [A B] [ -C A C D - 0 AD - CB ' as b increases. and divide by det A. 29. dXI = -D- I R2(1 - RIHI - a)2de2 dYI = D- I (1-a)R2(RI(1-a) +a)de2 > 0 65. a. dn = dn-I + dn-2, a Fibonacci sequence dp = D- I RI R2de2 > 0 b. dl = I, d2 = 2, d3 = 3, d4 = 5, .... dlO = 89 c. Invertible for all positive integers n [ ·6 12 0 0 [24 0 0 31. -~ 5 33. ~ 0 8-~] ~j-56.3 Answers to more theoretical questions are omitted. 0 0 I. 50 3. 13 7. 110 35. det(adj A) = (detA)/-1 11. Idet A I = 12, the expansion factor of T on the par­ 37. adj(A- I ) = (adj A)-I = (detA)-lA allelogram defined by v1 and V2 39. Yes. Use Exercises 38: If AS = S B, then 13. ,j2O (adj S)(adj A) = (adj B)(adj S). 15. We need to show that if VI, .. , , vm are linearly 43. A(adj A) = (adj A)A = (detA)In = 0dependent, then (a) V(VI .... ,Vm ) = 0 and (b) det(A T A) = O. 45. [~ -~] a. One of the Vi is redundant. so that vf = 0and V (v 1, ... , Vm ) = 0, by Definition 6.3.5. b. ker A i- (OJ and ker A ~ ker(A T A), so CHAPTER 7 that ker(A T A) i- (OJ. Thus, AT A fails to be 7.1 Answers to more theoretical questions are omitted. invertible. I. Yes; the eigenvalue is A. 3. 17. a. V(VI. V2. V3, VI x V2 x V3) 3. Yes; the eigenvalue is A. + 2. = V(VI,V2,v3)11vI x V2 x v311 because VI x V2 x v3 is orthogonal to v I , V2, and V3 5. Yes 490 ANSWERS TO ODD·NUMBERED EXERCISES 7. Eigenbasis: eI, e2, e" with eigenvalues 1. 2, 3 0 0 ~]9. I o and B = 0 Is ~ [~ -I] ['0 1 0 0 II. Eigenbasis: [:],Hl [ _~], with ,lg,,,,,I,,, 3,0,0 I 0 b. Al = [AIel Al ih] approaches ~ 3 [;_ 13. S ~ [~ -3 Ii] '''d R ~ [~ J]0 I 0part a. I -+--1 [bc. A b +c 1; c ~] 15. Elg,,,,,clo~ m, [ i], willi ,Ig'"vm", 0 29. Ae = e, so that eis an eigenvector with associated no eigenbasis eigenvalue 1. 17. Eigenbasis: C2, e4, el, e3 - e2, with eigenvalues I,31. A and AT have the same eigenvalues, by Exer­ 1,0,0cise 22. Since the row sums of AT are I, we can use the results of Exercises 29 and 30: I is an eigenvalue of A; if A is an eigenvalue of A, then 19. Eg,,,,,,,to" [ j],[ ~] with ,Igm,'ue, 0, 1. -1 < A ~ 1. eneed not be an eigenvector of A; Matrix A fails to be diagonalizable.. [0.9 0.9] consIder A = 0.1 0.1 . 21. We want A [~] = [;] and A m=2 [~] [:]; 33. a. fA (A) = _A3 + CA2 + bA + a that is, A [~ ~] = [~ : ] . The unique solution b. M = [~ ~ ~l IT -5 17 is A = [~ =~J. -1 0 o 0 23. The only eigenvalue of A is l, with EI = span(el). 35. There is no eigenbasis. A represents a horizontalo 0 -~] shear.o I 25. The geometric multiplicity is always I.37. We can write fA (A) = (A - AO)2g (A). By the prod­ uct rule, f~ (A) = 2(A - AO)g(A) + (A - Ao)2l (A), 27. fA(A) = A2 - 5A + 6 = (A - 2)(A - 3), so that the so that f~ (AO) = O. eigenvalues are 2. 3. 39. It's a straightforward computation. 29. Both multiplicities are tl - r. 41. tr(S-I(AS)) = tr((AS)S-t) = trA 3 I. They are the same. 43. No. since tr(AB - BA) = 0 and tr(ln) = 11 33. If B = S-I AS, then B - M n = S-1 (A - Aln)S. 45. For k = 3 35. No (consider the eigenvalues) 47. If we write M = [v w], then it is required that 37. a. AD· II) = (AiJ)T w = iJTATw = iJTAw = Ai! = 2i! and Aw = 3w. Thus, a nonzero M with iJ· Au) the given property exists if 2 or 3 is an eigenvalue b. Suppose Ai! = AV and AU'} = Jlw. Then of A. Av· w= i-,(v· w) and v· AUI = (L(V' UI). By part a, A(V' U}) = {i(v, UI), so that (),. -/l)(V' w) = O.49. If 2, 3, or 4 is an eigenvalue of A Since A i {i, it follows that v . w = 0, as claimed.7.3 Answers to more theoretical questions are omitted. 39. a. E, = V and Eo = V-L, so that the geometric 1. Eigenbasis: [~J, [~], with eigenvalues 7, 9 multiplicity of ] is 111 and that of 0 is 11 -111. The algebraic multiplicities are the same. See Exer­ cise 31. 3. S = [_~ qand B = [~ ~] b. £) = VandE_J = V-L,sothatthemuItiplicity 5. No real eigenvalues. of lis 111 and that of -1 is 11 - m. ANSWERS TO ODD-NUMBERED EXERCISES 491 41. Diagonalizable for all (/ 21. hm. A, = -1[I'") 2I] t--+oo 3­43. Diagonalizable for positive a 45. Diagonalizable for all a, b, c 7 7 7 ] 23. lim At = ~ 10 10 10 47. Diagonalizable only if a = b = c = 0 '--+00 22 [ 5 5 5 49. Never diagonalizable ,~ 1[10]25. lim (A xo) = ­51. fA(A)=-A 3 +d2 +bA+a t--+oo 17 7 0 0 a 7 I 0 b * * * 27. lim (Ar.'ro) = ~ [10]B2 ] t--+oo 2253. a. B = 10 I c * :1 [ ~I B3 5 0 0 0 w x 0 0 0 y " 29. lim (Alxo) = ~ [~~]b. Note that A is similar to B. Thus, ,--+ 00 84 fA (A) = fB (A) = fB3 (A) fBI (A) 31 = h(A)( _A3 + d 2 + bA + a), where h(A) = ,h,,(A).SeeExercise51. 1[0 1 I] 33. a. A = - 1 0 I c. tA(A)v = h(A)(-A3 + cA2 + bA + als)v = 2 1 1 0 h(A)(-A 3 v+cA2 v+bAv+av)=O' J o 55. We want A - 7h to be noninvertib]e. For example, c. xU) (I + ~~) \. v m+ HY [J + 8 1 1] A = 1 8 1 . , [-1][ (-D [;0 -~ 1 ] 8 7.4 Answers to more theoretical questions are omitted. t Carl wins if he chooses Co < 1. t _ 3 ­[I I]1. A - , 0.1 0.2] ~ [1]o 3 35. a. A = [ 0.4 0.3 ' b = 2 , ][5'+2(-1)' 5 _ (-I)' ] 3. A ="3 2. 5t - 2(-1)' 2·5'+(_1)' b. B = [~ ~] 5. At = 7'-1 [I 2] c. The eigenvalues of A are 0.5 and -0.1, those of3 6 Bare 0.5, -0.], ]. If vis an eigenvector of A, t _ ~ [I + 2(0.25)t 1 - (0.25)'] 7. A - 3 2 - 2(0.2W 2 + (0.2W then [~] is an eigenvector of B. Furthermore, o 0] (I - A)-lb] [2] A' = is an eigenvector of B 2 1+ (-I)t ] - (-1)t. 2 9. = ~ [ -2(~l)t 2(-1)t 0 [ 2 I ~ with eigenvalue 1. 4 - 2' 2 - 2' t 4 - 3· 2 ]I I A' =- -4 -2 -4 d. Will approach (h - A)-lb = [~], for any ini­I . 2 [ 2' 2' 3·2t tial value ret)] 13. Atxo= []+2.3'] 37. Let'r(t) = pet) . Then xU + 1) = Ax(t) where 2·3t [ wU) 15. Atxo = ~ [1 + 2(0.25)' ] 3 2 - 2(0.25)' 1 1 01 2: 4 I 1 t 4 - 2 ] A ~ ~ ~ 1 Rig'o""i"o' A [;] l~l 17. Atxo = ;: [ r t I + 2' + 3 . 6 ] [-~lwith ,igeov",",", I, !, 0, 19. Ar.'ro = _2' +5 ·6t [ -1 + 2· 6t 492 ANSWERS TO ODD·NUMBERED EXERCISES Xo ~ '1 ~ ±m+ ~ U] + ~ H] ,"X(I) ~ ~ + Gf' U] [0<1 > 0m The proportion in the long run is I :2: I. 39. All real numbers A are eigenvalues, with corre­ sponding eigenfunctions Ce(H I)t. 41. The symmetric matrices are eigenmatrices with eigenvalue 2, and the skew-symmetric matrices have eigenvalue O. Yes, L is diagonalizable, since the sum of the dimensions of the eigenspaces is 4. 43. I and i are "eigenvectors" with eigenvalues I and -I, respectively. Yes, T is diagonalizable; I, i is an eigenbasis. 45. No eigensequences 47. The nonzero polynomials of the form a + ex 2 are eigenfunctions with eigenvalue I, and bx (with b =I­ 0) has eigenvalue -1. Yes, Tis diagonalizable, with eigenbasis I, x, x 2 . 49. I, 2x - I, and (2x - 1)2 are eigenfunctions with eigenvalues I, 3, and 9, respectively. These func­ tions form an eigenbasis, so that T is indeed diago­ nalizable. 51. The only eigenfunctions are the nonzero constant functions, with eigenvalue O. ~ ] , for example 55. A = [~ 59. Exercise 58[i;Pl~es t3]at A and B are both similar to the matrix 0 0 0 , so that A is similar to B. 000 65. A basis of V is [~ ~], [~ _:J.and dim V = 2. 67. The dimension is 32 + 22 = 13. 71. The eigenvalues are I and 2, and (A - 13)(A - 2/j) = O. Thus A is diagonalizable. 73. If AI, .... Am are the distinct eigenvalues of A, then fA(A) = (A - A1)'" (A - Am)h(A) for some polynomial h(A), so that fA(A) (A-A]I,,) .. ·(A-Amln)h(A) = 0, by Exer­ \. .; o cise 70. 7.5 Answers to more theoretical questions are omitted. 1. ~ (cos ( - ~) + i sin ( - ~ ) ) 3. cos C:k) + i sin C:k). for k= 0, ... , n - 1 5. If z = r (cos ¢ + i sin ¢ ), then 27Tk )),w =:yr (cos (¢ +n2nk ) + i sin (¢ +n for k = 0, ... , n - I. 7. Clockwise rotation through an angle of i followed by a scaling by a factor of .Ji 9. Spirals outward since Izi > I 11. f(A) = (A - I)(A - I - 2i)(A - I + 2i) 13. s= [~ ~],forexamPle 15. s= [~. ~],forexamPle 17. s= [_~ ~],forexamPle 19. a. tr A = m, detA = 0 b. tr B = 2m - n, det B (_1),,-m. Compare with Exercise 7.3.39 1./3 21.2±3i 23.L-2"±Ti 25. ±I, ±i 27. -I, -1,3 29. trA = Al + A2 + A3 = 0 and detA = AIA2A3 = bed> O. Therefore, there are one positive and two negative eigenvalues; the positive one is largest in absolute value. 31. b. Eigenvalues A] = I, A2,3 ~ -0.2 ± 0.136i, A4,5 ~ 0.134 ± 0.132i. Note that IA)I < I for j = 2,3,4,5. Having five distinct eigenvalues, matrix A is diagonalizable. c. xequ = [0.2 0.2 0.2 0.2 0.2]T d. The proof is analogous to the proof of Theo­ rem 7.4.1. Note that lim (At),) = 0 for j = t--> 00 2, 3,4,5 since IA) I < 1. See Example 5. e. lim At is the 5 x 5 matrix whose entries are t-->oo all 0.2. 33. c. Hint: Let AI, A2, ... , A5 be the eigenvalues, with Al > IA)I, for j = 2, ... ,5. Let v] ,V2, ... , V5 be corresponding eigenvectors. Write ei = CIVI + .. ,+ C5V5. Then ith col­ umn of At = Atei = C[A~ Vi + ... +C5A~V5 is nearly parallel to V] for large t. 45. If a is nonzero. 47. If a is nonzero. 49. If a is neither I nor 2. 51. Q is a field. 53. The binary digits form a field. 55. H is not a field (multiplication is noncommuta­ tive). ANSWERS TO ODD-NUMBERED EXERCISES 495 43. im T = span(xT), rank T = I, ker T = span(X]X2, xi), nullity T = 2 45. im T = P2, rank T = 3, ker T = span(x5 - xi, XlX3 - XlX2, X2 X3 - xi), nullity T = 3 47. The determinant of the mth principal submatrix is positive if m is even, and negative if m is odd. au aij ] 255. Note that det = aiia jj - aij > 0, so [ a ji a jj that aii > aij or a jj > aij. 57. q(x) = AICT + A2C~ + A3cj = I, with positive Ai, defines an ellipsoid. 59. q (x) = AIcT = 1, with positive AI, defines a pair of I parallel planes, CI = ± ~. vA] _1 2" 1 'h61. q(x) = AICI + A2C2 + A3C3 = ,WIt Al > 0, A2 > 0, A3 < 0 defines a hyperboloid of one sheet. 63. q(cI WI + ... + cnwn) = (CJ WI + ... + cnwn) . ( - - 2 II - 2CJAIW] + ... + CnAnWn) = CIAI Will + ... + 2 -22 -21. 1 cnAnllwnl1 = c l + ... + c~ smce IIwill = -, by • AI constructlOn. 65. Adapt the method outlined in Exercise 63. Consider an orthonormal eigenbasis vI, V2 for A with associ­ ated eigenvalues Al > 0 and A2 < O. Now let WI = vIlv'Al and W2 = V2/J-A2, so that IIwII1 2 = I/A[ and IIw2112 = -1/A2. Thenq(cI WI +C2W2) = - - - - 1-2(CIWI +C2W2)'(AICI WI +A2C2W2) = Alqllw) II + 2 - 2 2A2c211W211 = CI - ci· 1 67. Adapt the method outlined in Exercises 63 and 65. Consider an orthonormal eigenbasis v I, ... , vp, ... , Vr , ... , vn for A such thatthe asso­ ciated eigenvalues Aj are positive for j = 1, ... , p, negative for j = p + I, ... , r, and zero for j = r + 1, ... , n. Let Wj = vj / y1IjT for j = 1, ... , r and Wj = vj for j = r + I, ... , n. 69. Note thaLr T R T ARx = (Ri)T A(Rx) ~ 0 for all x in lR m . Thus R T AR is positive semidefinite.fT AR is positive definite if (and only if) ker R = {O}. 71. Anything can happen: The matrix R T AR may be positive definite, positive semidefinite, negative def­ inite, negative semidefinite, or indefinite. 8.3 Answers to more theoretical questions are omitted. 1. (JI = 2, (J2 = I 3. All singular values are 1 (since AT A = In). 5. (J\ = (J2 = vip2 + q2 7. [_~ ~] [~ ~] [~ ~] 1 [I -2] [5 0] 1 [ 1 9. vis 2 1 0 o vis -2 ~] II. 0[! 1 ~W !] [: ~] 0 13.2I [3v1s 0] 1 [ 2 0 vis vis -1 ~] 15. Singular values of A -I are the reciprocals of those of A. 21. [ 0.8 0.6] [ 9 -~]-0.6 0.8 -2 23 A T-. _ {(J?Ui fori = I, ... ,r .. A u/ - - . o for I = r + 1, ... , 11 The nonzero eigenvalues of A T A and AA T are the same. 25. Choose vectors VI and V2 as in Theorem 8.3.3. Write 17 = CI VI + C2V2· Note that ~ 2 2 2 II ull =CI +C2 = 1. Now Au = CJAvl + C2Av2, so that 2 IIAul1 = = 1:s (Cl+C2)(J1 = (JI2 . We conclude that II Au II :s (JI. The proof of (J2 :s II Au II is analogous. 27. Apply Exercise 26 to a unit eigenvector vwith as­ sociated eigenvalue A. 33. No; consider A = [~ ~]. I _ _ - Vi for i = I, ... , m 35. (AT A)-J ATUi = (Ji { 6 for i = m + 1, ... , n CHAPTER 9 9.1 Answers to more theoretical questions are omitted. I. xU) = 7e 51 3. pet) = 7eO.031 5. yet) = -0.8eO.8f 7. x (t) = I~1 ' has a vertical asymptote at t = 1. 9. xU) = ((I - k)t + 1) 1/(1-k) 11. x(t) = tan t 13. a. About 104 billion dollars b. About 150 billion dollars 15. The solution of the equation ekT /100 = 2 is 100In(2) 69 T= ~- k k 2 cTIIAvII1 + c~IIAv2112 2 2 2 2ci (JI + C2(J2 2 2 496 ANSWERS TO ODD·NUMBERED EXERCISES 17. 37. £1.1 = span [~] and £16 = span [~]. Looks roughly like the phase portrait in Figure 10. 39. £1 = span [_~] and £IA = span [_~]. Looks roughly like the phase portrait in Exercise 35. 41. 19. [I ~21. ~ t] 23. x (t) is a solution. x(t) = 0 I Xo 43. a. Competition ~ 2 -6t [ 3] -t C]27. xU) = O. e -2 + OAe 1 b. y St29. x(t) = 2 [_~] +e [~] 31. i(n ~,' HJ x S . 1" "'f yeO) 233. c. peCies 'WInS I -- < . x(O) 45. b. y x . 1'" "'f Y (0) 1c. SpeCies WInS' I -- < -. x(O) 2 a. Symbiosis h. The eigenvalues are ~ (-5 ± )9 + 4k 2). There are two negative eigenvalues if k < 2; if k > 2, there is a negative and a positive eigenvalue. 8tget) = 45e-0. - 15e-OAt and h(t) = _45e- 0.8t + 45e-OAt - Ilt [COS(qt J] .. .x(t) = e . ( ) ,a spIral If Pi=- 0 and a Circle SIn qt if P = O. Approaches the origin if p is negative. Eigenvalues A1.2 = ! (-q ± ) q2 - 4p); hoth eigenvalues are negative. U' w(O) door slams if -­ < A2. 8(0) 9.2 Answers to more theoretical questions are omitted. 1. I 3. he37ri / 4 5. e-O . 1t (cos(2t) - i sin(2t)); spirals inward, in the clockwise direction [ 0 1] [c~s t] I 1 sm t with corresponding eigenvec­ ~ sinU) - t sin(2t) vet) = mg (I _ e-kt / m ) k mg lim vet) = ­ = terminal velocity t--+oo k a. CI e-r + C2e-2t b. 2e-t - e-­ 2r c. -e-r + 2e-2t d. In part c the oscillator goes through the equi­ librium state once; in part b it never reaches it. x (1) = te­3t 39. e­t (CI + C2t + qt2) A is an eigenvalue with dim(EA) = n, because E A is the kernel of the nth-order linear differential op­ erator T(x) - AX. fa cos t + fa sin t + C] e­2r + C2 e ­ 3t et [I =12t] tor [1: J .r(t) [ sin t ] All' . h 1 k' .. . n e Ipse WIt c oc wIse Ort­ sm t + cos t entation. 35. 43. 37. 41. 29. 31. 45. 33. Eigenvalue ANSWERS TO ODD·NUMBERED EXERCISES 497 r [0 I 0]39. The system .5.. = 0 0 1 chas the solutions dt 0 0 0 _ [k l + k2t + k3t2 /2] e(t) = k2 + k3t k3 where kl' k2. h are arbitrary constants. The solu­ tions of the given system are .r(t) = eAt c(t), by Ex­ ercise 9.1.24. The zero state is a stable equilibrium solution if (and only if) the real part of A is negative. I [COS(2t) + Sin(2t)] " . e­ . (~ ) (2) . SpIrals mward, ttl the sm.!ot - cos t counterclockwise direction. 9.3 Answers to more theoretical questions are omitted. l. Ce5t 3. ~e31 + Ce-2t (use Theorem 9.3.13) 5. -I ­ t + Cel 7. cje-4t + C2 e3t 9. C1 e 3t + C2e-3t 11. et (CJ cos t + c2 sin t) 13. e­r(Cl + C2t) (compare with Example 10). 15. CI + c2t 17. e-t(CJ+c2t)-!Cost 19. cost+qcos(ht)+C2sin(ht) 21. C] et + C2e-t + C3e-2t 23. 3e5t 25. e-2t+2 27. - sin(3t) ______ trajectory 1 EJ. - [trajectory 2 EJ. 2 - trajectory 3 9. Stable Eo .I' "".... L. e 0.07.1' 7. Not stable II. a. B = 2A d. The zero state is a stable equilibrium of the sys­dx tern - = grad(q) if (and only it) q is negative dt definite (then, the eigenvalues of A and B are all negative). 13. The eigenvalues of A -I are the reciprocals of the eigenvalues of A; the real parts have the same sign. 15. 17. Iflkl<1 19. False; consider A with eigenvalues 1,2, -4. db - = 0.5b + dt 21. a. I ds dt b. bet) = 50,000e007t - 49,000eO . 5t set) = 1,000eo. OJt ~ [cos(3t) - sin(3t)] [0] 27. x (t) = sin(3t) cos(3t) b' where 0, b are arbitrary constants 29. Eigenvalue 2 + 4i with corresponding eigenvector [ _:] . Use Theorem 9.2.6, with p = 2, q = 4, w=[~lv=[_~]. xU) = e2r [1 0] [COS(4t) - Sin(4t)] [0] o -1 sin(4t) cos(4t) b 31. Eigenvalue -I + 2i with corresponding eigenvec­ [ i] ~( -t [COS(2t) -- Sin(2t)] [ l]tor . x t) = e . = I sm(2t) cos(2t) -I