Solutions for Chapter Solutions for exercises in section 0 a X b x, y 6, and z 0 a Neither b Sew symmetric c Symmetric d Neither The zero matrix trivially satisfies all conditions, and it is the only possible answer for part a The only possible answers for b are real symmetric matrices There are many nontrivial possibilities for c 4 A A T and B B T A + B T A T + B T A + B Yes the sew-symmetric matrices are also closed under matrix addition 5 a A A T a ij a ji If i j, then a jj a jj a jj 0 b A A a ij a ji If i j, then a jj a jj Write a jj x+iy to see that a jj a jj x +iy x +iy x 0 a jj is pure imaginary c B ia ia ia T ia T ia B 6 a Let S A+A T and K A A T Then S T A T +A T T A T +A S Liewise, K T A T A T T A T A K b A S + K is one such decomposition To see it is unique, suppose A X+ Y, where X X T and Y Y T Thus, A T X T +Y T X Y A+ A T X, so that X A+AT S A similar argument shows that Y A A T K 7 a [A + B ] ij [A + B] ji [A + B] ji [A] ji +[B] ji [A ] ij +[B ] ij [A + B ] ij b [αa ] ij [αa] ji [ᾱa] ji ᾱ[a] ji ᾱ[a ] ij 0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 Solutions for exercises in section Functions b and f are linear For example, to chec if b is linear, let a b A and B, and chec if fa + B fa +fb and a b
Solutions fαa αfa Do so by writing a + b fa + B f a + b a + b a + b a a + b b fa+fb, αa αa fαa f α αa αa Write fx n i ξ ix i For all points x all scalars α, it is true that x x x n a a αfa and y y y y n, and for fαx + y ξ i αx i + y i ξ i αx i + ξ i y i α i ξ i x i + i i i i ξ i y i αfx+fy There are many possibilities Two of the simplest and most common are Hooe s law for springs that says that F x see Example and Newton s second law that says that F ma ie, force mass acceleration 4 They are all linear To see that rotation is linear, use trigonometry to deduce x u that if p, then fp u, where x u u cos θx sin θx u sin θx + cos θx f is linear because this is a special case of Example To see that reflection x x is linear, write p and fp erification of linearity is x x straightforward For the projection function, use the Pythagorean theorem to x conclude that if p, then fp x+x x Linearity is now easily verified
Solutions Solutions for exercises in section 4 4 Refer to the solution for Exercise 4 If Q, R, and P denote the matrices associated with the rotation, reflection, and projection, respectively, then cos θ sin θ Q sin θ cos θ, R 0, and P 0 4 Refer to the solution for Exercise 4 and write 0 cos θ sin θ cos θ sin θ RQ 0 sin θ cos θ sin θ cos θ If Qx is the rotation function and Rx is the reflection function, then the composition is cos θx sin θx R Qx sin θx cos θx 4 Refer to the solution for Exercise 4 and write a x PQR + a x cos θ sin θ 0 a x + a x sin θ cos θ 0 cos θ + sin θ sin θ cos θ cos θ + sin θ sin θ cos θ Therefore, the composition of the three functions in the order ased for is P Q Rx cos θ + sin θx + sin θ cos θx cos θ + sin θx + sin θ cos θx Solutions for exercises in section 5 0 5 5 a AB 8 b BA does not exist c CB does not exist 8 5 9 d C T B 0 e A 6 f B does not exist 6 7 64 g C T C 4 h CC T 4 6 i BB T 5 8 7 8 6 8 6 9 7 8 58 0 j B T B C 69 T AC 76
4 Solutions 5 a A 4 0, x x x, b 0 b s 0 x c b A A +A A 5 a EA A A + A b AE A +A A A 54 a A j b A i c a ij 55 Ax Bx x Ae j Be j e j A j B j j A B The symbol is mathematical shorthand for the phrase for all 56 The limit is the zero matrix 57 If A is m p and B is p n, write the product as B B AB A A A p A B + A B + + A p B p B p p A B 58 a [AB] ij A i B j 0 0 a ii b j a in b jj 0 is 0 when i>j 0 b When i j, the only nonzero term in the product A i B i is a ii b ii c Yes 59 Use [AB] ij a ib j along with the rules of differentiation to write d[ab] ij d a ib j da i b j dai b db j j + a i da i b j + [ ] [ da B + A db ] [ ] da B + AdB ij ij ij a i db j 50 a [Ce] i the total number of paths leaving node i b [e T C] i the total number of paths entering node i
Solutions 5 5 At time t, the concentration of salt in tan i is xit /gal For tan, dx sec coming in sec r x t sec going out 0 sec r gal sec x t gal For tan, dx sec coming in sec going out r x t sec r gal r x t sec r x t sec r x t x t, sec x t gal and for tan, dx sec coming in sec going out r x t sec r gal r x t sec r x t sec r x t x t sec x t gal This is a system of three linear first-order differential equations dx dx dx r x t r x t x t r x t x t that can be written as a single matrix differential equation dx / dx / r 0 0 x t 0 x t dx / 0 x t
6 Solutions Solutions for exercises in section 6 6 A A AB A B B A A A A B + A B + A B A B B + A B + A B 0 9 0 9 6 Use bloc multiplication to verify L I be careful not to commute any of the terms when forming the various products I C 6 Partition the matrix as A, where C 0 C and observe that C C Use this together with bloc multiplication to conclude that A I C+ C + C + + C I C 0 C 0 C 0 0 00 00 00 0 0 00 00 00 Therefore, A 00 0 0 00 00 00 0 0 0 / / / 0 0 0 / / / 0 0 0 / / / 64 A A A A A A and AA A A AA 65 AB T B T A T BA AB It is easy to construct a example to show that this need not be true when AB BA 66 [D + EF] ij D + E i F j [D + E] i [F] j [D] i +[E] i [F] j [D] i [F] j +[E] i [F] j [D] i [F] j + [E] i [F] j D i F j + E i F j [DF] ij +[EF] ij [DF + EF] ij 67 If a matrix X did indeed exist, then I AX XA trace I trace AX XA n trace AX trace XA 0,
Solutions 7 which is impossible 68 a y T A b T y T A T b T T A T y b This is an n m system of equations whose coefficient matrix is A T b They are the same 69 Draw a transition diagram similar to that in Example 6 with North and South replaced by ON and OFF, respectively Let x be the proportion of switches in the ON state, and let y be the proportion of switches in the OFF state after cloc cycles have elapsed According to the given information, x x + y y x 9 + y 7 so that p p P, where 9 p x y and P 7 Just as in Example 6, p p 0 P Compute a few powers of P to find P 80 70, P 44 756 40 760 5 748 P 4 5 749, P 5 50 750 50 750 50 750 /4 /4 and deduce that P lim P Thus /4 /4 p p 0 P 4 x 0 + y 0 4 x 0 + y 0 4 For practical purposes, the device can be considered to be in equilibrium after about 5 cloc cycles regardless of the initial proportions 60 4 6 5 6 a trace ABC trace A{BC} trace {BC}A trace BCA The 4 other equality is similar b Use almost any set of matrices to construct an example that shows equality need not hold c Use the fact that trace C T trace C for all square matrices to conclude that trace A T B trace A T B T trace B T A T T trace B T A trace AB T 6 a x T x 0 n x i 0 x i 0 for each i x 0 b trace A T A 0 [A T A] ii 0 A T i A i 0 i i [A T ] i [A] i 0 [A] i [A] i 0 i i [A] i 0 i [A] i 0 for each and i A 0
8 Solutions Solutions for exercises in section 7 7 a b Singular c 4 4 7 4 5 8 4 0 0 e 0 0 0 0 7 Write the equation as I AX B and compute d Singular X I A B 0 4 0 0 7 In each case, the given information implies that ran A < n see the solution for Exercise 74 a If D is diagonal, then D exists if and only if each d ii 0, in which case d 0 0 0 d 0 0 0 d nn /d 0 0 0 /d 0 0 0 /d nn b If T is triangular, then T exists if and only if each t ii 0 If T is upper lower triangular, then T is also upper lower triangular with [T ] ii /t ii 75 A T A T A 76 Start with AI A I AA and apply I A to both sides, first on one side and then on the other 77 Use the result of Example 65 that says that trace AB trace BA to write m trace I m trace AB trace BA trace I n n 78 Use the reverse order law for inversion to write [ AA + B B ] B A + BA B + A and [ BA + B A ] A A + BB B + A 79 a I Sx 0 x T I Sx 0 x T x x T Sx Taing transposes on both sides yields x T x x T Sx, so that x T x 0, and thus x 0
Solutions 9 recall Exercise 6 The conclusion follows from property 78 b First notice that Exercise 76 implies that A I + SI S I S I + S By using the reverse order laws, transposing both sides yields exactly the same thing as inverting both sides 70 Use bloc multiplication to verify that the product of the matrix with its inverse is the identity matrix 7 Use bloc multiplication to verify that the product of the matrix with its inverse is the identity matrix A B D T B 7 Let M and X T C D C T A T The hypothesis implies that MX I, and hence from the discussion in Example 7 it must also be true that XM I, from which the conclusion follows Note: This problem appeared on a past Putnam Exam a national mathematics competition for undergraduate students that is considered to be quite challenging This means that you can be proud of yourself if you solved it before looing at this solution Solutions for exercises in section 8 8 a B 0 4 b Let c 0 0 and d T 0 to obtain C 0 4 8 A j needs to be removed, and b needs to be inserted in its place This is accomplished by writing B A+b A j e T j Applying the Sherman Morrison formula with c b A j and d T e T j yields B A A b A j e T j A +e T j A b A j A A be T j A e j e T j A +e T j A b e T j e j A A b[a ] j e j [A ] j [A ] j b A A b e j [A ] j [A ] j b 8 Use the Sherman Morrison formula to write z A + cd T b A A cd T A +d T A b A b A cd T A b c +d T A c x ydt x +d T y 84 a For a nonsingular matrix A, the Sherman Morrison formula guarantees that A + αe i e T j is also nonsingular when + α [ A ] 0, and this certainly ji will be true if α is sufficiently small