What are the eigenspaces and the minimal polynomial of the "transposing about the anti-diagonal" endomorphism? Given a field $K$ of characteristics zero, let the endomorphism $f:M_2(K)\to M_2(K)$ be defined by $f\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=\begin{bmatrix}d&b\\c&a\end{bmatrix}$ for every $\begin{bmatrix}a&b\\c&d\end{bmatrix}\in M_2(K).$
How can I find the eigenspaces and the minimal polynomial of $f$? 
And why is $\operatorname{char} K=0$ important?

What I noticed is that $f$ is invertible (hence its eigenvalues are non-zero) and $f^{-1}=f$. But I couldn't go further than this
 A: Let$$E_1=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ E_2=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ E_3=\begin{bmatrix}0&0\\1&0\end{bmatrix},\text{ and }E_4=\begin{bmatrix}0&0\\0&1\end{bmatrix}.$$Then $\{E_1,E_2,E_3,E_4\}$ is a basis of $M_2(\mathbb{R})$ and the matrix of $f$ with respect to this basis is$$\begin{bmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{bmatrix}.$$Can you take it from here?
A: Since as you noted $f^2=1$, and $f\neq\pm1$, it follows that $X^2-1$ is indeed the minimum polynomial of $f$. The eigenvalues are $+1$ an $-1$. We have $f(A)=1\cdot A$ for $A\in$Mat$_2(K)$ if and only if $$\begin{pmatrix}d & b \\ c & a\end{pmatrix}=\begin{pmatrix}a & b \\ c & d\end{pmatrix},$$
if and only if $a=d$, so the eigenspace for $\lambda=+1$ is 
$$E_1=\left\{\begin{pmatrix}a & b \\ c & a\end{pmatrix} \ | \ a,b,c\in K\right\}.$$
Similarly, the eigenspace for $\lambda=-1$ is found by studying $f(A)=-A$, which gives $d=-a$, $b=-b$, $c=-c$, hence $d=-a$, $b=0=c$ when char$(K)\neq2$:
$$E_{-1}=\left\{\begin{pmatrix}a & 0 \\ 0 & -a\end{pmatrix} \ | \ a\in K\right\}.$$
By the way, $f$ is the symplectic involution, and satisfies $f(AB)=f(B)f(A)$ in addition. Moreover $E_1$ is the set of symmetric elements with respect to the involution and $E_{-1}$ is the set of skew-symmetric elements. You can define a symplectic involution on $n\times n$ matrices whenever $n$ is even.
