Given this matrix equation:
$$\begin{bmatrix} c_{k+1} \\ t_{k+1} \\ a_{k+1} \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0.33 \\ 0.18 & 0 & 0 \\ 0 & 0.71 & 0.94 \\ \end{bmatrix} \begin{bmatrix} c_k \\ t_k \\ a_k \\ \end{bmatrix}$$
where $c_k$ represents the female child population, $t_k$ represents the female teen population, and $a_k$ represents the female adult population of a species of animal. There is no given starting population.
From this matrix equation, what can be said about the long term survival of the species?
I know it will obviously die out from doing some basic algebra. I just don't know how to prove it given that 2 eigenvalues and eigenvectors for A are imaginary and I don't know what to do with imaginary eigen-pairs.
Can you suggest an increase in the survival rate of teens which would result in an increase of the species population?
This I managed to estimate to be around 25.5%, but that isn't the exact answer and it was using a roundabout way. I'm sure there is a much easier way to find this out, just not sure how.
Any direction on how to just approach the problems would be appreciated
If the matrix (in general, let's say $n \times n$) is $A$ and the initial population vector ${\bf v}$, after $k$ generations the population vector is $A^k {\bf v}$. If there are $n$ linearly independent eigenvectors ${\bf u}_j$ and $\lambda_j$, we have ${\bf v} = \sum_j c_j {\bf u}_j$, and $A^k {\bf v} = \sum_j c_j \lambda^k {\bf u}_j$. If all eigenvalues have absolute value $< 1$, the population will certainly die out as $k \to \infty$. If there is at least one eigenvalue with absolute value $\ge 1$, it can survive. This is still true in the "defective" case with fewer than $n$ linearly independent eigenvectors, but slightly more complicated to prove.
By the Perron-Frobenius theorem, for a matrix with nonnegative entries the eigenvalue with largest absolute value is positive. So basically you're interested in positive roots of the characteristic polynomial.
If you replace the $0.18$ in your matrix with $r$, the characteristic polynomial becomes $\lambda^3 - 0.94 \lambda^2 - 0.2343 r$. $\lambda = 1$ would satisfy this for $r = .06/.2343 \approx 0.2560819462$.