How to come up with a Leslie matrix with convenient eigenvalues?

Question

A three by three Leslie matrix looks like $$ \begin{bmatrix} f_0 & f_1 & f_2 \\ s_0 & 0 & 0 \\ 0 & s_1 & 0 \end{bmatrix}, $$ where $f_0 \ge 0$ and everything else is strictly positive.

The application is in population dynamics; $f_0$ tells how many children a zero-year old creature has on average, $f_1$ how many a first year old has, and $f_2$ how many a second year old has. $s_0$ tells the proportion of zero year olds that survive to become one year old, $s_1$ that of the one year olds that survive.

I would like to come up with nice and easy numbers for $f_j$ and $s_k$ so that:

• $f_0 = 0$ or a small fraction
• $f_1 > 1$, maybe around 5, and an integer or a friendly fraction.
• Ideally $1 \le f_2 < f_1$ and $f_2$ is also an integer or a friendly fraction.
• $0 < s_0, s_1 < 1$ and they should also be friendly fractions. They should not be microscopically small, and probably not very close to 1, either.

By a "friendly fraction" I mean a fraction which looks friendly to a student, rather being a horrible mess or intimidating. The other restrictions and wishes are to make the model correspond to the application.

How do I choose the parameters so that the eigenvalues are nice and friendly, or at least so that the highest eigenvalue is friendly? Preferably, the highest eigenvalue would be greater than one, as the population is roughly $\lambda_3^t$, where $\lambda_3$ is the greatest eigenvalue and $t$ is the time. Given the model, this should be achievable with the parameters at roughly the desired values, though whether the values can be friendly is a different question.

The eigenvalues are, modulo miscalculations, the zeroes of the polynomial $$ -\lambda^3 +f_0 \lambda^2 +f_1 s_0 \lambda + f_2 s_0 s_1. $$ The second order term vanishes if one takes $f_0 = 0$, which one can do if one wishes to.

There exists a closed formula for the solutions, but I do not immediately see how to proceed from there to choosing nice parameters. See the Cardano formula at Wikipedia: https://en.wikipedia.org/wiki/Cubic_function#Cardano_formula

For some intuition on the model, suppose $f_0 = f_2 = s_1 = 0$. Then $s_0$ tells the proportion of animals that survive the winter, while $f_1$ tells how much each procreates (on average); then all of them die. Hence, the yearly growth should be multiplication by $\sqrt{f_1 s_0}$ and the eigenvalues tell the same story.

Answer 1

If I use your simplification that $f_0 = 0$, then I suggest just choosing a real eigenvalue $\lambda$ and writing out the relation for the other parameters:

$$-\lambda^3+f_1s_0\lambda + f_2s_0s_1 = 0$$ Now isolate a parameter, say $s_1$:

$$s_1 = \frac{\lambda^3-\lambda f_1 s_0}{f_2 s_0}$$

Then just find values that satisfy your requirements. It should be clear that choosing nice values of these parameters (especially $\lambda$) will help to make $s_1$ "friendly". Also, because of real-world constraints, you'll need $\lambda^3 > \lambda f_1 s_0$ (so $s_1$ will be positive) and for $f_2 s_0 > \lambda^3 - \lambda f_1 s_0$ (so $s_1$ will be a proper fraction). Since you want $f_1 > f_2$, we will need the numerator to be fairly small.

Messing around, here are a few examples:

• Choose $\lambda = 2$, $f_1 = 5$ and $s_0 = \frac{3}{4}$.

This makes the numerator $\frac{1}{2}$. Now, $f_2$ can be any integer greater than zero, so I choose $f_2 = 2$, and we get $s_1 = \frac{1}{3}$, and our matrix becomes: $$\left[\begin{matrix} 0 & 5 & 2\\ \frac{3}{4} & 0 & 0\\ 0 & \frac{1}{3} & 0 \end{matrix}\right]$$

And we can confirm that $$\left[\begin{matrix} 16\\ 6\\ 1 \end{matrix}\right]$$ is the eigenvector for $\lambda = 2$.

• Just because I tried this one out and it didn't quite work:

Let $\lambda = 3$, $f_1 = 5$ and $s_0 = \frac{6}{10}$. This makes the numerator $18$, so $f_2$ must be larger than $30$ in order to have a positive value of $s_1$. Therefore, you'll have to be thoughtful concerning the sizes of the parameters you choose.

Answer 2

1. Choose three numbers $\lambda_1,\lambda_2,\lambda_3$.
2. Compute the coefficients $a,b,c$ of the polynomial $(x-\lambda_1)(x-\lambda_2)(x-\lambda_3) = x^3-ax^2-bx-c$.
3. Choose $s_i$ and $f_i$ such that $f_0=a$, $f_1s_0=b$, $f_2s_0s_1=c$. It shouldn't be hard to play around with the coefficients so that those inequalities are satisfied, since you have so many degrees of freedom.

If you want to have $f_0=0$, you'll need to choose the numbers so that $\lambda_1+\lambda_2+\lambda_3=0$.