A Leslie Matrix is given by $$L =\begin{pmatrix}0 & (3/2)a^2 & (3/2)a^3\\1/2 & 0 & 0\\ 0 & 1/3 & 0\end{pmatrix}\cdot$$
Find the Eigenvalues and determine the dominant eigenvalue and eigenvector. I am struggling with how to do this without a calculator, even when I use Wolfram alpha, the answers I am getting do not seem to make sense. Should I guess a root?
On
$\det(L-\lambda I)=-\lambda\begin{vmatrix}-\lambda & 0\\\frac{1}{3} &-\lambda\end{vmatrix}-\frac{1}{2}\begin{vmatrix}\frac{3}{2}a^2 & \frac{3}{2}a^3\\\frac{1}{3}&-\lambda\end{vmatrix}=-\lambda^3-\frac{1}{2}(\frac{3}{2}a^2)\begin{vmatrix}1&a\\\frac{1}{3}&-\lambda\end{vmatrix}=-\lambda^3-\frac{3}{4}a^2(-\lambda-\frac{1}{3}a)$
$=-\frac{1}{4}(4\lambda^3-3a^2\lambda-a^3)=-\frac{1}{4}(\lambda-a)(4\lambda^2+a\lambda+a^2)$, so $\lambda=a$ is the dominant eigenvalue.
Reducing $\begin{bmatrix}-a &\frac{3}{2}a^2&\frac{3}{2}a^3&0\\\frac{1}{2}&-a&0&0\\0&\frac{1}{3}&-a&0\end{bmatrix}$ gives $\begin{bmatrix}1&-2a &0&0\\0&1&-3a &0\\0&-\frac{1}{2}a &\frac{3}{2}a^2 &0\end{bmatrix}$ and then
$\hspace{.6 in}\begin{bmatrix}1&-2a&0&0\\0&1&-3a&0\\0&0&0&0\end{bmatrix}$ and $\begin{bmatrix}1&0&-6a^2&0\\0&1&-3a&0\\0&0&0&0\end{bmatrix}$.
Therefore $\begin{bmatrix}6a^2\\3a\\1\end{bmatrix}$ is an eigenvector for the dominant eigenvalue.
The Eigenvalues are those that satisfy: $Av= \lambda v$, $(A-\lambda I_n)v=0$ where $v \neq 0$. That implies that $(A-\lambda I_n)=0$, in other words $rank(A-\lambda I_n) < n$. We can then conclude $det\{A- \lambda I_n\}=0$. You following me?
So, just solve $det\{A- \lambda I_3\}=0$ in your case by applying a Laplace expansion along column 3.