Go to CCP Homepage Go to Materials Page Go to Linear Algebra Materials Go to Table of Contents
Go Back One Page Go Forward One Page

Leslie Growth Models

Part 2: Properties of Leslie matrices

In Part 1 we found the general form of a Leslie matrix and of its characteristic polynomial. In this part we list some properties of such matrices (without proof), and we explore the effect of iterating the transition many times, that is, of allowing the population to pass through many reproductive cycles.

Theorems about Leslie Matrices

1. A Leslie matrix L has a unique positive eigenvalue lambda1. This eigenvalue has multiplicity 1, and it has an eigenvector x1 whose entries are all positive.
2. If lambda1 is the unique positive eigenvalue of L, and lambdai is any other eigenvalue (real or complex), then |lambdai| < lambda1. That is, lambda1 is a dominant eigenvalue.
3. If any two successive entries aj and aj+1 of the first row of L are both positive, then |lambdai| < lambda1 for every other eigenvalue. That is, if the females in two successive age classes are fertile (almost always the case in any realistic population) then lambda1 is a strictly dominant eigenvalue.
4. Let x(k) denote the state vector Lkx(0) after k growth periods. If lambda1 is a strictly dominant eigenvalue, then for large values of k, x(k+1) is approximately lambda1x(k), no matter what the starting state x(0). That is, as k becomes large, successive state vectors become more and more like an eigenvector for lambda1.

We saw in Part 1 that the New Zealand sheep population illustrates the first three of these theorems. In particular, we saw that the unique positive eigenvalue is 1.175..., that it has multiplicity 1, and that is has an eigenvector with entries all of the same sign. (If you have not yet seen an eigenvector, one will appear soon.) If all the entries of the eigenvector are negative, multiplication by -1 produces an eigenvector with all entries positive. In this case there is only one other real eigenvalue, -0.658..., definitely smaller in magnitude. We did not check the magnitudes of the complex eigenvalues, but they are all smaller than 1.175 also. (In fact, none of them has magnitude as large as 0.8.)

Theorem 4 needs careful interpretation. It does not say that the sequence of states converges -- in particular, if the dominant eigenvalue is > 1, the sequence does not converge at all. On the other hand, if we "normalize" the state vector at each step -- say, by making its entries sum to 1 -- the sequence of modified state vectors does converge to an eigenvector. Normalized or not, the sequence shows us an equilibrium age distribution of the female population, which is approached over time.

  1. For an arbitrary starting vector x(0), compute x(k) = Lkx(0) for a modest value of k. Then compute Lx(k) and lambda1x(k) and explain the meaning of each of these vectors. How do Lx(k) and lambda1x(k) compare?
  2. Vary k in the preceding step to find an eigenvector for lambda1. Modify your eigenvector (if necessary) to make it a state vector. (Hint: Your state vector will be most meaningful if its entries sum to 1.) Describe what your state vector tells you about a distribution into age classes. If you completed Step 8 in Part 1, how do your two state vectors compare?

Go to CCP Homepage Go to Materials Page Go to Linear Algebra Materials Go to Table of Contents
Go Back One Page Go Forward One Page


modules at math.duke.edu