Smart way to find eigenvectors of $L\colon \mathbb R^{2\times 2}\to\mathbb R^{2\times 2}$ Consider the following linear map:
$$
L\colon\mathbb R^{2\times2}\to\mathbb R^{2\times2}\colon X\mapsto
\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}
X-X
\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}.
$$
I have to calculate the eigenvalues and eigenvectors of $L$.
If I made no mistake, the images of the standard basis vectors are:
$$
L\begin{bmatrix}1&0\\0&0\end{bmatrix}=\begin{bmatrix}0&-1\\1&0\end{bmatrix},\\
L\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}-1&0\\0&1\end{bmatrix},\\
L\begin{bmatrix}0&0\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\\
L\begin{bmatrix}0&0\\0&1\end{bmatrix}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}.
$$
Using the standard basis $\{e_{11},e_{12},e_{21},e_{22}\}$, we get the following matrix representation:
$$
\begin{bmatrix}
0&-1&1&0\\
-1&0&0&1\\
1&0&0&-1\\
0&1&-1&0
\end{bmatrix}
$$
I tried finding the characteristic polynomial, so I calculated the determinant of:
$$
\begin{bmatrix}
-\lambda&-1&1&0\\
-1&-\lambda&0&1\\
1&0&-\lambda&-1\\
0&1&-1&-\lambda
\end{bmatrix}
$$
Calculating the characteristic polynomial would take up some work (the eigenvalues are $0,2,-2$) and I would have to come up with 4 eigenvectors (multiplicity of 0 is 2).
Now this was a subquestion on an exam, not worth many points. So I'm guessing there must be a quicker, "smarter", way of doing this. I see that $L$ basically switched the rows of $X$ and then subtracts it from a $X'$, that resulted from switching the columns of $X$. I see that symmetric matrices are mapped to zero. However, I can't see how to get the full solution in a smart way.
Could someone give me a hint?
 A: Your observation about symmetric matrices is a useful starting point, if not quite correct: we can use the basis
$$ E_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I, \\
E_2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\
E_3 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\
E_4 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$
The first three are symmetric, the last is antisymmetric. It's easy to check this really is a basis. Note also that $E_3$ is in fact the matrix we have in $L$. Since $E_1=I$ commutes with anything, $E_1E_3-E_3E_1=0$, so there's one eigenvector immediately. $E_3$ also commutes with itself, so there's your other eigenvector with eigenvalue zero. This leaves only two eigenvectors to find, which you can do by sticking the general linear combination of $E_2$ and $E_4$ in: you find that the answers are $E_2 \pm E_4$.
This is a useful basis to remember, since it also immediately tells you there are three LI symmetric matrices and only one antisymmetric one: generalisations to higher dimensions are also useful.
