In Exercises 3–6, assume that any initial vector \({x_0}\) has an eigenvector decomposition such that the coefficient \({c_1}\) in equation (1) of this section is positive.

5. In old-growth forests of Douglas fir, the spotted owl dines mainly on flying squirrels. Suppose the predator–prey matrix for these two populations is \(A = \left( {\begin{aligned}{}{.4}&{}&{.3}\\{ - p}&{}&{1.2}\end{aligned}} \right)\).Show that if the predation parameter p is .325, both populations grow. Estimate the long-term growth rate and the eventual ratio of owls to flying squirrels.

Given the value is \(A = \left( {\begin{aligned}{}{.4}&{}&{.3}\\{ - .325}&{}&{1.2}\end{aligned}} \right)\).

For finding eigenvalue,

\(\det \left( {A - \lambda I} \right) = \left( {\begin{aligned}{}{0.4 - \lambda }&{0.3}\\{ - 0.125}&{1.2 - \lambda }\end{aligned}} \right)\)

So, the characteristics equation is,

\(\)

\(\begin{aligned}{}0 &= \left( {0.4 - \lambda } \right)\left( {1.2 - \lambda } \right) - \left( {0.3} \right)\left( { - 0.125} \right)\\0 &= {\lambda ^2} - 1.6\lambda + .5775\end{aligned}\)

Solve the roots.

\(\begin{aligned}{}\lambda &= \frac{{1.6 \pm \sqrt {{{1.6}^2} - 4\left( {.5775} \right)} }}{2}\\\lambda &= \frac{{1.6 \pm \sqrt {.25} }}{2}\\\lambda &= 1.05,0.55\end{aligned}\)

Because one of the eigenvalues is greater than one, both populations expand. The entries in the eigenvector correspond to \(1.05\) define their respective sizes at the end.

So, for \(\lambda = 1.05\), find eigenvector as:

\(\begin{aligned}{}\left( {A - 1.05I} \right) &= 0\\{E_1} &= \left( {\begin{aligned}{}{ - 0.65}&{}&{0.3}&{}&0\\{ - 0.325}&{}&{.15}&{}&0\end{aligned}} \right)\\{E_1} &= \left( {\begin{aligned}{}{ - 13}&{}&6&{}&0\\0&{}&0&{}&0\end{aligned}} \right)\end{aligned}\)

An eigenvector is \({\rm{v}} = \left( {\begin{aligned}{}6\\{13}\end{aligned}} \right)\).

There will be around 6 spotted owls for every 13 (thousand) flying squirrels in the future.