Transition Matrices We want to apply this method of computing A to the analysis of a certain type of physical system that can be described by means of the following kind of mathematical model. Suppose that the sequence xo,Xi,x2, ...,xk,... (4) of «-vectors is defined by its initial vector xo and an n x n transition matrix A in the following manner: Xjt+i = Axt for each k > 0. (5) We envision a physical system—such as a population with n specified subpopula-tions—that evolve through a sequence of successive states described by the vectors in (4). Then our goal is to calculate the kůl state vector xk. But using (5) repeatedly, we find that xi = Axo, X2 = Axj = A2xo, x3 = Ax2 = A3xo, and in general that Xk = A*x0. (6) Thus our task amounts to calculating the k power A* of the transition matrix A. Consider a metropolitan area with a constant total population of 1 million individuals. This area consists of a city and its suburbs, and we want to analyze the changing urban and suburban populations. Let Q denote the city population and 5* the suburban population after k years. Suppose that each year 15% of the people in the city move to the suburbs, whereas 10% of the people in the suburbs move to the city. Then it follows that C*+i=0.85Q + 0.10S* (7) Sk+i=0.l5Ck + 0.90Sk 6.3 Applications Involving Powers of Matrices CHAPTER 6 .375 for each k > 0. (For instance, next year's city population Ck+i will equal 85% of this year's city population C* plus 10% of this year's suburban population £*.) Thus the metropolitan area's population vector Xk = [ Ck Sk] satisfies Xjfc+i = Axt and hence x* = A*xo (8) (for each k > 0) with transition matrix [0.85 0.10] [0.15 0.90J- The characteristic equation of the transition matrix A is GH(*-*)-(s)(š)- (17 - 20X)(9 - 10A) -3 = 0 200A.2 - 350A + 150 = 0 4A.2 - 1\ + 3 = 0 (A. - 1)(4A. - 3) = 0. Thus the eigenvalues of A are A i = 1 and X% = 0.75. For k\ = 1, the system (A - Al)v = 0 is [-0.15 0.10] [*] _ [0] L 0.15 -0.10j[jJ-[0j' so an associated eigenvector is vi = [ 2 3 ] . For Xz = 0.75, the system (A — AI)v = 0 is [0.10 0.10][x]_[0] L0.15 0.15 J L^ J _ L°J ' so an associated eigenvector is V2 = [ — 1 1 ] . It follows that A = PDP-1, where P = and 2 "M and D = 1 3 i o *--i[4 1] Now suppose that our goal is to determine the. long-term distribution of population between the city and its suburbs. Note first that (f )* is "negligible" when k is sufficiently large; for instance, (|)40 « 0.00001. It follows that if k > 40, then the formula A* = PD^P"1 yields Ak = [l ~l][J (|°)*]ö)[-3 2j ^H3 i\i° °\i-3 2J ^^L3 °JL—3 2J==5L3 3J (9) 376 CHAPTER 6 Eigenvalues and Eigenvectors Hence it follows that, if k is sufficiently large, then x* = A*x0 « i = (Q + So) 2 2 3 3 0.4 0.6 Co So 0.4 0.6 because Co+So = 1 (million), the constant total population of the metropolitan area. Thus, our analysis shows that, irrespective of the initial distribution of population between the city and its suburbs, the long-term distribution consists of 40% in the city and 60% in the suburbs. ■ Remark The result in Example 2—that the long-term situation is independent of the initial situation—is characteristic of a general class of common problems. Note that the transition matrix A in (8) has the property that the sum of the elements in each column is I. Ann x n matrix with nonnegative entries having this property is called a stochastic matrix. It can be proved that, if A is a stochastic matrix having only positive entries, then Ai = 1 is one eigenvalue of A and |A; | < 1 for the others. (See Problems 39 and 40.) Moreover, as k -> oo, the matrix A* approaches the constant matrix [vi Vi ••• Vi], each of whose identical columns is the eigenvector of A associated with X\ = 1 that has the sum of its elements equal to 1. The 2x2 stochastic matrix A of Example 1 illustrates this general result, with X\ = 1, A.2 = |, and Vj = (|, |). ■ Predator-Prey Models Next, we consider a predator-prey population consisting of the foxes and rabbits living in a certain forest. Initially, there are Fo foxes and i?o rabbits; after k months, there are Fk foxes and Rk rabbits. We assume that the transition from each month to the next is described by the equations F/fc+i = 0AFk + 0.3Rk Rk+l = -rFk + l.2Rk, (10) where the constant r > 0 is the "capture rate" representing the average number of rabbits consumed monthly by each fox. Thus Xfc+i = Axjt and hence x* = A xo, where *-[S] - -[-? I (11) (12) 6.3 Applications Involving Powers of Matrices CHAPTER 6 377 The term OAFk in the first equation in (10) indicates that, without rabbits to eat, only 40% of the foxes would survive each month, while the term 0.3 Rk represents the growth in the fox population due to the available food supply of rabbits. The term 1.2Rk in the second equation indicates that, in the absence of any foxes, the rabbit population would increase by 20% each month. We want to investigate the long-term behavior of the fox and rabbit populations for different values of the capture rate r of rabbits by foxes. The characteristic equation of the transition matrix A in (12) is (0.4-A)(1.2-A) + (0.3)r = 0; (4 - 10X)(12 - 10A.) + 30r = 0; 100A2 - 160A + (48 + 30r) = 0. The quadratic formula then yields the equation X = 2ÖÖ [16° ± ^(160)2 ~ (400)(48 + 30r)] = i- (8 ± V16 - 30r) , (13) which gives the eigenvalues of A in terms of the capture rate r. Examples 3, 4, and 5 illustrate three possibilities (for different values of r) for what may happen to the fox and rabbit populations as k increases: • Fk and Rk may approach constant nonzero values. This is the case of stable limiting populations that coexist in equilibrium with one another. • Fk and Rk may both approach zero. This is the case of mutual extinction of the two species. • Fk and Rk may both increase without bound. This is the case of a population explosion. If r — 0.4, then Equation (13) gives the eigenvalues A.i = 1 and A.2 = 0.6. For Ai = 1, the system (A — AJ)v = 0 is -0.6 0.3] [x] _ ["0" -0.4 0.2j[vJ_[oJ' so an associated eigenvector is Vj = [ 1 2 ] . For Xi = 0.6, the system (A — k\)\ = 0 is " -0.2 0.3 -0.4 0.6 mn- so an associated eigenvector is V2 = [ 3 2 ] . It follows that A = PDP ', where P = 1 3 2 2 D = 1 0 0 0.6 and P"1 "*[ 2 -3 -2 1 378 CHAPTER 6 Eigenvalues and Eigenvectors We are now ready to calculate A*. If k is sufficiently large that (0.6)* « 0 (for instance, (0.6)25 « 0.000003), then the formula A* = PD^p-1 yields ~ [ 2 2 J [ 0 (0.6)* J ( 4) [ -2 1 J 2 -3 -2 i] ^"5[2 2J [o 0J L _ _i f 1 Oil" 2 -3] ~ 4[2 0J L-2 lj _ ir-2 31 ~ 4 [ -4 6 J • Hence it follows that if k is sufficiently large, then x,_Ax0-4_4 6j|_/?oJ"46/?()-4FoJ —that is, [S]"[i]- where a = ±(3/ř0 - 2F0). (14) Assuming that the initial populations are such that a > 0 (that is, 3/?o > 2Fo), (14) implies that, as fc increases, the fox and rabbit populations approach a stable situation in which there are twice as many rabbits as foxes. For instance, if Fo = R0 = 100, then when k is sufficiently large, there will be 25 foxes and 50 rabbits. EXAMPLE 4 If r = 0.5, then Equation (13) gives the eigenvalues X\ = 0.9 and X.2 = 0.7. For Mutual Extinction Xi = 0.9, the system (A — Xl)v = 0 is [£ £][;]-[!]■ so an associated eigenvector is vi = [ 3 5 ] . For X2 = 0.7, the system (A — Xl)\ = 0 is [=" s][;]-[;]. so an associated eigenvector is V2 = [ 1 1 ] . It follows that A = PDP-1, with P = 1]. D-[«? £]. ana ,H-J[_> ">]. 6.3 Applications Involving Powers of Matrices CHAPTER 6 379 Now both (0.9)* and (0.7)* approach 0 as k increases without bound (k -> +oo). Hence if k is sufficiently large, then the formula A* = PD*P_1 yields A* = L5 l\l ° (0.7)* J(~2)[_5 3 J *-*[5 !][o o][-5 Í] = [o o]' so Fk Rk *[%Y\i i][ZY\i\ Thus Fk and Rk both approach zero as k -> +oo, so both the foxes and the rabbits die out—mutual extinction occurs. ■ EXAMPLE 5 If r = 0.325, then Equation (13) gives the eigenvalues Ai = 1.05 and A2 = 0.55. Population Explosion For ki = 1.05, the system (A - AI)v = 0 is f-0.650 0.30] [x] _ ["O"] [-0.325 0.15 J [y J ~ [OJ' Each equation is a multiple of — 13x + 6y = 0, so an associated eigenvector is V! = [ 6 13 f. For A2 = 0.55, the system (A - AI)v = 0 is [-0.150 0.30] [xl _ f0] [ -0.325 0.65 JL^J L°J' so an associated eigenvector is V2 = [ 2 1 ] . It follows that A = PDP-1 with p_r 6 2] D_ri.o5 0 ] p-i__j.r 1 -2i F-[l3 lj' D-[ 0 0.55 J' ^ F - 20 [-13 6J- Note that (0.55)* approaches zero but that (1.05)* increases without bound as k -> +00. It follows that if k is sufficiently large, then the formula A* = PD*P_1 yields *_r 6 2] r (Los)* o j ,f 1 -2] L13 1J L ° (°-55)* J 2 L ~13 6 J ~_±ľ 6 2] r(1.05)* oi r 1 -2] ~ 2o[l3 lJL 0 0j[-13 6 J ___Lr (6)0.05)* 0]f 1-2] ~ 20 L (13X1.05)* 0 J L —13 6j' 380 CHAPTER 6 Eigenvalues and Eigenvectors and therefore A* * -^(1.05)* Hence, if k is sufficiently large, then r 6 -ni L 13 -26 J (16) Xjt A V L(1.05)* 20 6 -12 13 -26 ][i] -^(1.05)' f 6F0-12/?o] L13Fo-26Ä6_ľ and so [5 ]«<»»*[»]• where y = 4j(2/?0 - Fo) (17) If)/ > 0 (that is, if 2/?o > F0), then the factor (1.05)* in (17) implies that the fox and rabbit populations both increase at a rate of 5% per month, and thus each increases without bound as k -*■ +oo. Moreover, when k is sufficiently large, the two populations maintain a constant ratio of 6 foxes for every 13 rabbits. It is also of interest to note that the monthly "population multiplier" is the larger eigenvalue Xi = 1.05 and that the limiting ratio of populations is determined by the associated eigenvector Vi = [ 6 13 ] . ■ In summary, let us compare the results in Examples 3, 4, and 5. The critical capture rate r = 0.4 of Example 3 represents a monthly consumption of 0.4 rabbits per fox, resulting in stable limiting populations of both species. But if the foxes are greedier and consume more than 0.4 rabbits per fox monthly, then the result is extinction of both species (as in Example 4). If the rabbits become more skilled at evading foxes, so that less than 0.4 rabbits per fox are consumed each month, then both populations grow without bound, as in Example 5.