Find the eigenvalues of this matrix I was wondering if anyone wanted to give a shot at finding the eigenvalues of this $4\times 4$ system. I have tried without success, mainly because I end up having to solve a very nasty cubic polynomial that wolfram seems to be having trouble with as well. There may be some method to solve this that I didn't try, so I wanted to post this here just in case someone could help me out,
$$J(\epsilon)=\begin{bmatrix} -\mu &&0&&0 && -\beta b/\mu\\ 0&& -\sigma && 0&& \beta b/\mu\\0&& \sigma&& -(\phi+d)&& 0\\ 0 &&0&&\alpha&&-(\phi+\epsilon)\end {bmatrix}$$
Thanks!
 A: After removing the obvious eigenvalue $-\mu$, you are indeed left to solve the cubic polynomial
$$ {\lambda}^{3}\mu+ \left( d\mu+\epsilon\,\mu+2\,\mu\,\phi+\mu\,\sigma
 \right) {\lambda}^{2}+ \left( d\epsilon\,\mu+d\mu\,\phi+d\mu\,\sigma+
\epsilon\,\mu\,\phi+\epsilon\,\mu\,\sigma+\mu\,{\phi}^{2}+2\,\mu\,\phi
\,\sigma \right) \lambda-\alpha\,\sigma\,\beta\,b+d\epsilon\,\mu\,
\sigma+d\mu\,\phi\,\sigma+\epsilon\,\mu\,\phi\,\sigma+\mu\,{\phi}^{2}
\sigma
=0$$
which seems to have no special algebraic features: in fact, for suitable values of the parameters I believe it can be any cubic polynomial.  A "closed-form" expression in radicals expressed in terms of all
these parameters will be a nightmare.  In practice, for particular 
numerical values of the parameters you may be best off using a 
numerical solver, but you can also use the general formula.
A: You can do row reduction  of the matrix $$J(\epsilon)=\begin{bmatrix} -\mu &&0&&0 && -\beta b/\mu\\ 0&& -\sigma && 0&& \beta b/\mu\\0&& \sigma&& -(\phi+d)&& 0\\ 0 &&0&&\alpha&&-(\phi+\epsilon)\end {bmatrix}$$
add row 2 and row 3
$$J(\epsilon)=\begin{bmatrix} -\mu &&0&&0 && -\beta b/\mu\\
 0&& -\sigma && 0&& \beta b/\mu\\
 0&& 0&& -(\phi+d)&& \beta b/\mu\\
 0 &&0&&\alpha&&-(\phi+\epsilon)\end {bmatrix}$$
row 3 $\times \alpha /(\phi+d)+$row 4
$$J(\epsilon)=\begin{bmatrix} -\mu &&0&&0 && -\beta b/\mu\\
 0&& -\sigma && 0&& \beta b/\mu\\
 0&& 0&& -(\phi+d)&& \beta b/\mu\\
 0 &&0&&0&&-(\phi+\epsilon)+(\beta b/\mu )(\alpha /(\phi+d))\end {bmatrix}$$
now determinant will be the multiplication of the diagonal elements 
$$|J-\lambda I|=(-\mu-\lambda)(-\sigma-\lambda)(-(\phi+d)-\lambda)(-(\beta b/\mu )(\alpha /(\phi+d))-\lambda)=0$$
