A Note on the Eigenvalues and Eigenvectors of Leslie matrices. Ralph Howard Department of Mathematics University of South Carolina

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1 A Note on the Eigenvalues and Eigenvectors of Leslie matrices Ralph Howard Department of Mathematics University of South Carolina Vectors and Matrices A size n vector, v, is a list of n numbers put in a column: v v v := 2 v n When for the values n = 2 and n = 3 this looks like v v =, v = v 2 v v 2 2 where v,v 2,v 3 are numbers (often called scalars when also talking about vectors) Examples of size 2, 3 and 4 vectors are [ 3, 2] 4 52, For use a matrix, A, is an n n array of numbers Thus 2 2 and 3 3 matrices look like a a 2, a a 2 a 3 a a 2 a 2 a 22 a a 3 a 32 a 33 where the entries a i j are scalars The formula for multiplying a matrix A with a vector v in the cases n = 2 and n = 3 is [a a 2 a 2 a 22 ][ v v 2 ] = v 3 a v + a 2 v 2 a 2 v + a 22 v 2 The general definition of a matrix is an m n array, as we will only be working with the case of square matrices it seems pointless to complicate things with the more general rectangular matrices

2 2 a a 2 a 3 a 2 a 22 a 23 v v 2 = a v + a 2 v 2 + a 3 v 3 a 2 v + a 22 v 2 + a 23 v 3 a 3 a 32 a 33 v 3 a 3 v + a 32 v 2 + a 33 v 3 Thus a matrix times a vector yields a vector We can also multiply two matrices together If A and B are 2 2 matrices then let b b B = 2 = B b 2 b,b 2 22 b where B = b2 and B b 2 = are the columns of B Note these 2 b 22 columns are vectors and thus we can multiple them by the matrix A to get AB and AB 2 Then the product AB is AB = [ AB,AB 2 ] That is AB is the matrix whose columns are the result of multiplying the columns of B by A In full detail this is a a AB = 2 b b 2 a b = + a 2 a 2 a b 2 + a 2 a 22 a 2 a 22 b 2 b 2 2 a 2 b + a 22 a 2 a 2 b 2 + a 22 a 22 In the 3 3 case this is in terms of the columns of B: AB = A [ B,B 2,B 3 ] = [ AB,AB 2,AB 3 ] The full blown, and fully hideous, formula is 2 a a a 3 b b 2 3 b 3 AB = 4a 2 a 2 2 a b 2 b 2 2 b a 3 a 3 2 a 3 3 b 3 b 3 2 b a b + a 2 b 2 + a 3 b 3 a b 2 + a 2 b a 3 b a b 3 + a 2 b a 3 b 3 3 = 4a 2 b + a 2 2 b 2 + a 2 3 b 3 a 2 b 2 + a 2 2 b a 2 3 b 3 2 a 2 b 3 + a 2 2 b a 2 3 b a 3 b + a 3 2 b 2 + a 3 3 b 3 a 3 b 2 + a 3 2 b a 3 3 b 3 2 a 3 b 3 + a 3 2 b a 3 3 b 3 3 We can also define powers A n of a matrix So A 2 = AA, A 3 = AAA, A 4 = AAAA etc Fortunately we can have the calculator multiply and take powers of a matrices 2 Eigenvectors and Eigenvalues of Matrices Let A be a square matrix (that is A has the same number of rows and columns) Let v be a vector and λ a number Then v and λ number is an eigenvector of A with eigenvalue λ iff For a 2 2 matrix Av = λv A = a b c d

3 the eigenvalues are the roots of the characteristic equation x a b det (xi A) = det c x d = (x a)(x d) cd = x 2 (a + d)x + (ad bc) = 0 (If you don t know what det and I are in the above, don t worry, in the cases we consider these will not be important) 3 Eigenvalues and Eigenvectors of Leslie matrices Assume we have a population of organisms where we will count their numbers of each age once during progressive time periods of the same length (which to be concrete we assume to be a year) Let m be the maximum reproductive age of the organism For each x with x m, let n x (t) = number of females that have age x during the census year t Let n(t) be the vector which keeps track of all the ages in the year t, that is n (t) n n(t) = 2 (t) n m (t) For example of the m = 4 (so that only females of age at most 4 are reproductively active) we would have n (t) n(t) = n 2 (t) n 3 (t) n 4 (t) To be a little more specific, n (t) is the number of females that were born in the year before the census year t and survived until the time of the census (which is different from the number of births), n 2 (t) is the number of two year olds at the time of the year t census, and in general n x (t) the number of x-year olds at the time of the year t census For x m let p x be the proportion of the age x organisms from the year t census that survive until the year t + census This means that n x+ (t + ) = p x n x (t) for x m If b x is the net fecundity (which can also be thought of as the per capita birth rate) of females of age x, that is the number average number of 3

4 4 females offsprings of an age x female that survive until the next census, then n (t + ) = p n (t) + p 2 n 2 (t) + + p m n m (t) (In most realistic cases p = 0, but there is no reason to rule it out mathematically) In the case that m = 4 and b = 0 all this can summarized by saying the given the vector n (t) n(t) = n 2 (t) n 3 (t) n 4 (t) which records the number of females of age, 2, 3 and 4 in the census year t that we find ages in the next year by use of the loop diagram: b 4 b 3 b 2 n (t) s n 2(t) n 3(t) n 4(t) p 2 p 3 Figure For most of the rest of these notes we will simplify notation and assume that m = 4 Then from either the loop diagram or the equations preceding it () n (t + ) = b n (t) + b 2 n 2 (t) + b 3 n 3 (t) + b 4 n 4 (t) n 2 (t + ) = p n (t) n 3 (t + ) = p 2 n 2 (t) n 4 (t + ) = p 3 n 3 (t) We rewrite this as a matrix equation Let n (t) b b 2 b 3 b 4 n t := n 2 (t) n 3 (t), and A := s s n 4 (t) 0 0 s 3 0

5 The vector n t gives the age distribution of females at the year t census, and A is the Leslie matrix Then the system () of four scalar equations can be written as the single matrix equation: (2) n t+ = An t Our next goal is to find eigenvectors for A That is vector v for some scalar λ Av = λv If we have such an eigenvector, then n t = λ t v is a solution to the matrix equation (2) To see this note if n t = λ t v n t+ = λ t+ v = λ t λv = λ t Av = A(λ t v) = An t, where we have used that Av = λv Also note that if v is an eigenvector and c is a scalar, then cv is also an eigenvector (Exercise: Show this) Therefore given an eigenvector with first element v we can multiple by the scalar c = v and get a new eigenvector cv where the first entry is That is we assume we have an eigenvector of the form v = v 2 Then computing Av and λv and setting them equal we get b + b 2 v 2 + b 3 v 3 + b 4 v 4 λ (3) Av = p p 2 v 2 = λv = λv 2 λv 3 p 3 v 3 λv 4 Comparing the last three entries of these vectors gives the equations v 3 v 4 p = λv 2, p 2 v 2 = λv 3, p 3 v 3 = λv 4 We can solve successively for v 2, v 3, and v 4 to get v 2 = λ p, v 3 = λ p 2 v 2 = λ 2 p p 2, v 4 = λ p 3 v 3 = λ 3 p p 2 p 3 Using these values in (3) gives (4) b + b 2 λ p + b 3 λ 2 p p 2 + b 4 λ 3 p p 2 p 3 Av = p λ p 2 λ 2 p p 2 p 3 = λv = λ p λ p p 2 λ 2 p p 2 p 3 5

6 6 So for v to be an eigenvector the only condition left is make the first entries agree That is (5) b + b 2 λ p + b 3 λ 2 p p 2 + b 4 λ 3 p p 2 p 3 = λ For x from to m let l = and for 2 x m Let l x be the product of p, p 2, up to p x : x l x = p p x, that is l x = p j In our case of m = 4 we have j= l =, l 2 = p, l 3 = p p 2, l 4 = p p 2 p 3 Then l x the proportion of one year olds that survive to the beginning of the x-th year Using this notation we can rewrite (5) as (6) b l + b 2 l 2 λ + b 3 l 3 λ 2 + b 4 l 4 λ 3 = λ Now divide this by λ to get (7) b l λ + b 2 l 2 λ 2 + b 3 l 3 λ 3 + b 4 l 4 λ 4 = This is the Lotka-Euler equation Note if we write it in summation notation it becomes m (8) b x l x λ x = x= Just to be specific about the dependence of the Lotka-Euler equation on the survival rates p x we note it can be written as (9) λ b + b 2 λ 2 p + b 3 λ 3 p p 2 + b 4 λ 4 p p 2 p 3 =, which in the general case looks like m b x p p 2 p x λ x = x= If we multiple (5) by λ move all the terms of the result to one side of the equation we get (0) λ 4 b λ 3 b 2 p λ 2 b 3 p p 2 λ b 4 p p 2 p 3 = 0 which is the characteristic equation (that is the equation det(λi A) = 0 which is the equation for λ to be an eigenvalue of the matrix A, see any text on linear algebra) of the Leslie matrix A This can be rewritten in terms of the l x s as () λ 4 b l λ 3 b 2 l 2 λ 2 b 3 l 3 λ b 4 l 4 = 0

7 (And this can also be derived by multiplying (7) by λ 4 and rearranging a bit) Note that as the characteristic equation () results from the Lotka- Euler equation by just multiplying by λ 4 the two equation have the same collection of non-zero roots As both equations only have one positive root (this is not really quite elementary, but can be shown without too much trouble) we have: Proposition The Lotka-Euler equation has exactly one positive root We call it the dominate eigenvalue of Leslie matrix Finally we note that when λ is a solution to the Lotka-Euler equation then (4) becomes Av = λv where v = λ p λ 2 p p 2 λ 3 p p 2 p 3 Therefore (2) v = λ 3 p p 2 p 3 λ p λ 2 p p 2 gives the stable age distribution normalized so that n (t) = Written in terms of the l x s this is (3) v = λ l 2 λ 2 l 3 λ 3 l 4 To be explicit about the general case (that is for general values of m, not just m = 4) the Leslie matrix is b b 2 b 3 b m b m p p (4) A = p m 0 The characteristic equation is λ m b l λ m b 2 l 2 λ m 2 b m l m λ b m l m = 0 7

8 8 which in summation notation is m λ m b m k l m k λ k = 0 k=0 (Dividing by λ m and rearranging gives the Lotka-Euler equation (8)) It has exactly one positive root (the others are negative or complex) and this positive root is the dominate eigenvalue of A The stable age distribution, normalized so that n (t) =, is given by the column vector λ l 2 λ 2 l 3 v = λ (m 2) l m λ (m ) l m