Eigenvalues in Applications

Transcription

1 Eigenvalues in Applications Abstract We look at the role of eigenvalues and eigenvectors in various applications. Specifically, we consider differential equations, Markov chains, population growth, and consumption. 1 Differential equations We consider linear differential equations of the form = Au. (1) The complex matrix A is n n and u 1 (t) u 2 (t) u = u(t) =.. (2) u n (t) The equations are linear, since if v and w are two solutions, then so is u = αv + βw, as shown by 1.1 Scalar case (n = 1) = d (αv + βw) = α dv + β dw = αav + βaw = A(αv + βw) = Au. In the scalar case, i.e., n = 1, the differential equation (1) reces to the scalar differential equation = λu, (3) where λ C. Its general solution is u = e λt c where c is an arbitrary constant. 1

2 By specifying an initial condition u(0) = u 0 we can determine c from u 0 = u(0) = e 0 c = c. Thus, the general solution given an initial condition is u(t) = e λt u 0. (4) The real part of λ determines growth or decay according to Re(λ) < 0: exponential decay, Re(λ) > 0: exponential growth, Re(λ) = 0: neither growth nor decay. A nonzero imaginary part adds oscillation in the form of a sine wave. 1.2 General case (n > 1) In the general case, i.e., n > 1, we can solve (1) similar to the scalar case if the matrix A is diagonalizable. Using the spectral decomposition A = SΛS 1 and multiplying both sides of (1) from the left by S 1, we get d (S 1 u) = Λ(S 1 u). (5) After changing variables from u to y = S 1 u we have which is nothing more than the set dy = Λy, (6) dy 1 = λ 1y 1, dy 2 = λ 1y 2,. =. (7) dy n = λ 1 y n, of n decoupled scalar differential equations. The solutions to (7) are y i (t) = e λit c i for i = 1, 2,..., n. The parameters c i are some arbitrary constants. Using linear algebra notation, we have the solutions y(t) = e λ1t e λ2t... c. (8) e λnt 2

3 The constants c are determined by the n initial conditions u(0) = u 0 to the original differential equation (1) by u(0) = Sy(0) = Sc = u 0 c = S 1 u 0. (9) We put the pieces back together and find the general solution u(t) = Sy(t) = S e λ1t e λ2t... S 1 u 0 (10) e λnt to the differential equation (1) with initial conditions u Matrix exponential We would like to express the general solution of (1) as u(t) = e At u 0. (11) But how do we generalize the exponential function to matrices? Recall that the scalar exponential function e x is defined by the infinite series e x = 1 + x + (1/2)x 2 + (1/6)x 3 +. (12) Let us define the matrix exponential e A by simply replacing x with A in (12), as in e A = I + A + (1/2)A 2 + (1/6)A 3 +. (13) The series (13) is defined for square matrices and it always converges. But is u(t) = e At u 0 (14) actually a solution of (1) if we use the definition (13)? The answer is yes since the derivative of e At u 0 with respect to t is d eat u 0 = (A + A 2 t + (1/2)A 3 t 2 + )u 0 = Ae At u 0. If A has the spectral decomposition A = SΛS 1, then from e At = Se Λt S 1 we see immediate connections with the previous section. Note that the matrix exponential solves the differential equation even when A is not diagonalizable. 3

4 2 Markov A Markov chain is a random process that can be in one of a finite number of states at any given time and the next state depends only on the current state. We are given n 2 transition probabilities p ij [0, 1]. The number p ij gives the probability that the next state is state i if the current state is state j. By putting the transition probabilities in a square matrix A such that a ij = p ij we obtain a Markov matrix. Definition 1 (Markov matrix). A square matrix A is a Markov matrix if it satisfies the following two properties. First, every entry in A is nonnegative. Second, every column of A adds to 1. Two facts about Markov matrices follow directly from the definition. First, multiplying a Markov matrix A with a nonnegative vector u 0 proces a nonnegative vector u 1 = Au 0. Second, if the components of u 0 add to 1, then so does the components of u 1 = Au 0. The first fact is trivial, and the second fact can be shown as follows. Let e = ( 1 1 ) T be a vector of all 1s. Then e T u 1 = e T (Au 0 ) = (e T A)u 0 = e T u 0 = 1. After k steps in a Markov chain, the initial probability distribution u 0 changes to u k = A k u 0. For many Markov matrices, the limit of u k as k exists and is unqiue. We say that the Markov chain approaches the steady state u. The existence of a steady state also shows that 1 is an eigenvalue of A and the corresponding eigenvector is a steady state. A famous theorem e to Perron and Frobenius shows that a Markov matrix with strictly positive entries has 1 as its largest eigenvalue. That eigenvalue is also simple (multiplicity equal to 1) and the corresponding eigenvector can be scaled so that it has positive entries. Let us show that 1 is indeed an eigenvalue of any Markov matrix. The rows of A I sum to 1 1 = 0, which means that A I is singular. Hence, λ = 1 is an eigenvalue of A. Suppose that A is diagonalizable, i.e., A = SΛS 1. We have u k = A k u 0 = SΛ k S 1 u 0. (15) Thus, if λ 1 = 1 and λ i < 1 for all i 1, then u k approaches a steady state in the direction of the dominant eigenvector s 1. 3 Population A Leslie model models the long-term age distribution and growth rate of a population. It is popular in, e.g., ecology, and it works as follows. Partition the population into n disjoint age groups. The population in the age groups are represented as a vector p with n components. After one time step, the population in each age group is given by Ap, where A is an n n Leslie matrix. 4

5 The long-term growth rate and age distribution come from the largest eigenvalue and its corresponding eigenvector. The Leslie matrix is constructed as follows. The members of the youngest age group is a proct only of reproctive activity. Formally, we write p (k+1) 1 = n i=1 f i p (k) i, (16) where the coefficients f i reflect the rate of reproction in age group i. If f i = 0, then no reproction occurs in that age group and if f i = 2, then each indivial in the age group proces two (unique) offsprings. In this model, all the members of the oldest age group die off in one time step. The members of the other age groups have a chance of surviving and thereby transition to the next age group. Formally, we write this as p (k+1) i+1 = s i p (k) i, (17) where the coefficients s i reflect the chance of surviving from age group i and transition to age group i + 1. For example, consider n = 4 and combine the formulas (16) and (17) to obtain, in matrix form, the Leslie model f 1 f 2 f 3 f 4 p (k+1) = Ap (k) = s s p(k). 0 0 s Consumption Let A be a consumption matrix, p the proction levels, and y the demand. The proction levels are given by p = (I A) 1 y. (18) The question is: When does (I A) 1 exist and when is it a nonnegative matrix? Intuitively, if A is small then (I A) 1 exists and the economy can meet any demand, but if A is large then the proction consumes too much of the procts and as a consequence the economy can not meet the demand. What we consider small and large depends entirely on the eigenvalues of A. We show that when the series B = I+A+A 2 + converges, then B(I A) = I and thus B = (I A) 1. Suppose that A is diagonalizable and write B as B = S(I + Λ + Λ 2 + )S 1. (19) The infinite series within the parenthesis is nothing but n independent scalar geometric series. The n scalar series converge if and only if λ i < 1. The matrix B is obviously nonnegative since it is the sum of nonnegative matrices. The inverse of (I A) exists and is nonnegative when all the eigenvalues satisfy λ i < 1. 5