Finding eigenvalues of a $2\times 2$ matrix that is a linear transformation 
We have a linear transformation. We need to find the eigenvalues and eigenvectors of it.
I tried the following:
$$\begin{bmatrix} 
0 & 1\\
1 & 0
\end{bmatrix} \begin{bmatrix} 
a & b\\
c & d
\end{bmatrix}-
\begin{bmatrix} 
a & b\\
c & d
\end{bmatrix}
\begin{bmatrix} 
1 & 0\\
1 & 0
\end{bmatrix} = 
\begin{bmatrix} 
(c-a-b) & d\\
(a-c-d) & b
\end{bmatrix}$$
However if I want to find the eigenvaules for this matrix I get a very complicated characteristic polynomial:
$$\det\begin{bmatrix} 
(c-a-b)-\lambda & d\\
(a-c-d) & b-\lambda
\end{bmatrix} = ((c-a-b)-\lambda)\cdot (b-\lambda) - d(a-c-d) = \\ =\lambda^2 + \lambda(a-c) + (cb-ab-b^2-da+dc+d^2) = 0$$
There must be something wrong since it gets so complicated.
 A: It seems to me that you are attempting to answer the wrong question.  $\mathcal F$ is a linear transformation from $\mathbb R^{2\times2}$ to itself, and we are asked for the eigenvalues and eigenvectors of $\mathcal F$.  Now, $\mathbb R^{2\times2}$ can be identified with $\mathbb R^4$, so you should be looking at a $4\times 4$ matrix.  The eigenvector will be $2\times2$ matrices.
We can write a standard basis for $\mathbb R^{2x2}$ as
$$\begin{align}
e_1&=\begin{bmatrix}1&0\\0&0\end{bmatrix}\\
e_2&=\begin{bmatrix}0&1\\0&0\end{bmatrix}\\
e_3&=\begin{bmatrix}0&0\\1&0\end{bmatrix}\\
e_4&=\begin{bmatrix}0&0\\0&1\end{bmatrix}\\
\end{align}$$
Now to get the matrix of $\mathcal F$, we jsut need to find its action of the basis vectors.  For example:
$$\mathcal F e_1 = \begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}-\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}1&0\\1&0\end{bmatrix}=\begin{bmatrix}-1&0\\1&0\end{bmatrix}=-e_1+e_3$$  Therefor the first column of the matrix of $\mathcal F$ with respect to the basis $<e_1,e_2,e_3,e_4>$ is $$\begin{bmatrix}-1\\0\\1\\0\end{bmatrix}$$
Proceed in this manner to find the matrix of $\mathcal F$ and the compute its eigenvalues and eigenvectors as usual.  You'll get the eigenvectors as $4\times1$ vectors, but you should translate them back to $2\times2$ matrices by reversing the process used above.
