Leslie matrices
Master Leslie matrices for Year 12 Specialist Mathematics in Queensland (QCAA). A Leslie matrix models an age-structured population: the birth rates sit across the top row and the survival rates on the sub-diagonal, so multiplying by the population vector projects the population one cycle at a time.
You will learn to build the matrix from fecundity and survival rates, project the population over several cycles, find total and individual age-class numbers, and read the long-term growth rate from the dominant eigenvalue — a key application of matrices in the Further matrices topic and a favourite of the external exam.
Theory
A Leslie matrix models an age-structured population: in Year 12 Specialist Mathematics (QCAA, Queensland) the fecundity (birth) rates fill the top row and the survival rates sit on the sub-diagonal. Multiplying by the population vector, \(\mathbf{n}_{k+1}=L\mathbf{n}_k\), projects the population one cycle ahead, and the dominant eigenvalue gives the long-term growth rate.
A population is split into age classes (for example young, juvenile, adult), and the numbers in each class are listed in a population vector \(\mathbf{n}=\begin{pmatrix}n_1\\ n_2\\ n_3\end{pmatrix}\), youngest first.
The Leslie matrix \(L\) stores two kinds of rate. The fecundity rates \(f_1,f_2,f_3\) — the average offspring produced by a member of each class per cycle — fill the top row. The survival rates \(s_1,s_2\) — the fraction of each class that lives into the next class — sit on the sub-diagonal (just below the leading diagonal). Every other entry is \(0\).
Multiplying gives the population one cycle later: \(\mathbf{n}_{k+1}=L\mathbf{n}_k\). The first entry of \(L\mathbf{n}\) counts the newborns (from the fecundities), and each lower entry counts the survivors moving up an age class. Projecting \(k\) cycles ahead is \(\mathbf{n}_k=L^k\mathbf{n}_0\).
After many cycles the population settles into a fixed stable age distribution and the total then grows by a constant factor each cycle. That factor is the dominant eigenvalue \(\lambda\) of \(L\) — the long-term growth rate. If \(\lambda>1\) the population grows, \(\lambda=1\) it holds steady, and \(\lambda<1\) it declines toward extinction.
For three age classes the Leslie matrix and the projection one cycle ahead are:
Projecting \(k\) cycles from the start uses a matrix power:
In the long term the total grows by the dominant eigenvalue \(\lambda\), the largest real root of the characteristic equation:
How to work with a Leslie model
- Build \(L\). Put the fecundity (birth) rates across the top row and the survival fractions on the sub-diagonal, in order; fill every other entry with \(0\).
- Set up the population vector \(\mathbf{n}_0\), listing the age classes youngest first.
- Project. Multiply \(L\mathbf{n}\) one age class at a time — row 1 gives the newborns, the lower rows the survivors — and repeat (or use \(L^{k}\mathbf{n}_0\)) for several cycles.
- Analyse the long term. Find the dominant eigenvalue \(\lambda\); compare it with \(1\) to decide growth, steady state or decline, and read the stable age ratio from the eigenvector.
Put the fecundities (births) across the top row and the survival fractions on the sub-diagonal:
| \(\text{top row (births)}\) | \(=\) | \((0,\ 4,\ 6)\) |
| \(\text{class 1 survival}\) | \(=\) | \(0.5\) |
| \(\text{class 2 survival}\) | \(=\) | \(20\% = 0.2\) |
| \(L\) | \(=\) | \(\begin{pmatrix}0 & 4 & 6\\ 0.5 & 0 & 0\\ 0 & 0.2 & 0\end{pmatrix}\) |
\(L=\begin{pmatrix}0 & 4 & 6\\ 0.5 & 0 & 0\\ 0 & 0.2 & 0\end{pmatrix}\).
Multiply \(L\mathbf{n}\) one age class at a time: row 1 is the newborns, rows 2 and 3 the survivors:
| \(L\mathbf{n}\) | \(=\) | \(\begin{pmatrix}2(200)+3(100)\\ 0.5(600)\\ 0.5(200)\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}400+300\\ 300\\ 100\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}700\\ 300\\ 100\end{pmatrix}\) |
Add the entries for the new total:
| \(\text{total}\) | \(=\) | \(700+300+100\) |
| \(=\) | \(1100\) |
After one cycle \(\begin{pmatrix}700\\ 300\\ 100\end{pmatrix}\), a total of \(1100\).
Project once to get \(L\mathbf{n}_0\):
| \(L\mathbf{n}_0\) | \(=\) | \(\begin{pmatrix}3(200)+8(80)\\ 0.5(400)\\ 0.25(200)\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}600+640\\ 200\\ 50\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}1240\\ 200\\ 50\end{pmatrix}\) |
Apply \(L\) again to that vector to reach \(L^2\mathbf{n}_0\):
| \(L^2\mathbf{n}_0\) | \(=\) | \(\begin{pmatrix}3(200)+8(50)\\ 0.5(1240)\\ 0.25(200)\end{pmatrix}\) |
| \(=\) | \(\begin{pmatrix}600+400\\ 620\\ 50\end{pmatrix}\) | |
| \(=\) | \(\begin{pmatrix}1000\\ 620\\ 50\end{pmatrix}\) |
\(L^2\mathbf{n}_0=\begin{pmatrix}1000\\ 620\\ 50\end{pmatrix}\), a total of \(1670\).
The growth rate is the dominant eigenvalue \(\lambda\), the largest root of \(\det(L-\lambda I)=0\):
| \(\det(L-\lambda I)\) | \(=\) | \(-\lambda^3+0.75\lambda+0.25\) |
| \(-\lambda^3+0.75\lambda+0.25\) | \(=\) | \(0\) |
| \(\lambda\) | \(=\) | \(1 \ \text{(largest root)}\) |
For the stable ratio solve \(L\mathbf{v}=\lambda\mathbf{v}\) with \(\lambda=1\), using \(v_2=\dfrac{0.5\,v_1}{\lambda}\), \(v_3=\dfrac{0.5\,v_2}{\lambda}\):
| \(\text{take } v_1\) | \(=\) | \(4\) |
| \(v_2\) | \(=\) | \(\dfrac{0.5\times4}{1}=2\) |
| \(v_3\) | \(=\) | \(\dfrac{0.5\times2}{1}=1\) |
| \(\text{ratio}\) | \(=\) | \(4:2:1\) |
\(\lambda=1.00\): the total settles to a constant size, in the stable ratio \(4:2:1\).
Common pitfalls
Frequently asked questions
What is a Leslie matrix?
A Leslie matrix \(L\) models an age-structured population. Multiplying it by the population vector, \(\mathbf{n}_{k+1}=L\mathbf{n}_k\), projects the numbers in each age class one cycle into the future.
What do the entries of a Leslie matrix mean?
The top row holds the fecundity (birth) rates \(f_1,f_2,f_3\) — average offspring per member of each class — and the sub-diagonal holds the survival rates \(s_1,s_2\), the fraction of each class that lives into the next. All other entries are \(0\).
How do you project a population several time steps ahead?
Apply \(L\) once per cycle: \(\mathbf{n}_1=L\mathbf{n}_0\), \(\mathbf{n}_2=L\mathbf{n}_1\), and so on, which is \(\mathbf{n}_k=L^k\mathbf{n}_0\). Each product is worked out one age class at a time.
What is the long-term growth rate of a Leslie model?
It is the dominant eigenvalue \(\lambda\) of \(L\) — the largest real root of \(\det(L-\lambda I)=0\). Once the population settles to its stable ratio, the total is multiplied by \(\lambda\) each cycle.
What does the stable age distribution mean?
After many cycles the age classes settle into a fixed ratio (the dominant eigenvector). From then on every class grows by the same factor \(\lambda\), so the shape of the population no longer changes.
What happens when the growth rate is less than 1?
If \(\lambda<1\) the total is multiplied by a number below \(1\) each cycle, so the population declines toward extinction; \(\lambda=1\) holds it steady and \(\lambda>1\) makes it grow without bound.