Matrix $A$

\[A - \lambda I_2 = \begin{pmatrix} \var{a11}-\lambda & \var{a12}\\ \var{a21} & \var{a22}-\lambda \end{pmatrix}\]
Hence the characteristic polynomial $p(\lambda)$ is: \[\begin{eqnarray*} \mathrm{det}\left(A-\lambda I_2 \right)&=&\simplify[zeroTerm]{({a11}-lambda)({a22}-lambda)-{a12}*{a21}}\\ &=& \simplify[std]{lambda^2-{trA}*lambda+{dA}}\\ &=&\simplify[std]{(lambda-{a})(lambda-{b})} \end{eqnarray*} \]
We see that on solving $p(\lambda)=0$ we get the eigenvalues:
\[\lambda_1=\var{mnA},\;\;\;\lambda_2=\var{mxA}\]
Note: We could have found the characteristic polynomial by noting that for a 2 × 2 matrix $A$ then the characteristic polynomial is
\[\lambda^2-\mathrm{trace}(A)+\mathrm{det}(A)\]
where $\mathrm{trace}(A) = \var{trA},\;\;\;\mathrm{det}(A)=\var{dA}$

Finding the eigenvectors:

1. $\lambda_1=\var{mnA}$

We have the eigenspace is given by all $v=(x,y)^\mathrm{T}$ such that $(\simplify{A-{mnA}}I_2)v=(0,0)^\mathrm{T}$ i.e.

\[\begin{pmatrix} \var{a11-mnA}&\var{a12}\\ \var{a21}&\var{a22-mnA} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \end{pmatrix}\]

This gives the two equations:

\[ \begin{eqnarray*} \simplify[std]{{a11-mnA}x + {a12}y}&=&0\\ \simplify[std]{{a21}x + {a22-mnA}y}&=&0 \end{eqnarray*} \]
There is only one equation here as we see that the equations are the same (one is a multiple of the other).

So putting $x=1$ in the first equation we get $y_1=\var{-s*(a11-mnA)}$

Hence the eigenvector we want is \[\begin{pmatrix} 1 \\ \var{-s*(a11-mnA)} \end{pmatrix}\]

2. $\lambda_2=\var{mxA}$

In this case we have the equations:

\[ \begin{eqnarray*} \simplify[std]{{a11-mxA}x + {a12}y}&=&0\\ \simplify[std]{{a21}x + {a22-mxA}y}&=&0 \end{eqnarray*} \]

Once again there is only one equation, so putting $x=1$ in the first equation we get $y_2=\var{-s*(a11-mxA)}$

Hence the eigenvector we want is \[\begin{pmatrix} 1 \\ \var{-s*(a11-mxA)} \end{pmatrix}\]