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.
- 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?
- 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?
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