$T:M_{4\times 4}^{\mathbb{R}}\to M_{4\times 4}^{\mathbb{R}}\:$ defined by $T(M)=-2M^t + M$. Find the minimal polynomial of $T$ For start, I've noticed that if $M$ is symmetric, we get:
$T(M)=-M$ , therefore the space of symmetric matrices in $M_{4\times 4}^{\mathbb{R}}$ is an eigenspace of the eigenvalue $-1$
Also , if $M$ anti-symmetric then :
$T(M)=3M$,  therefore the space of anti-symmetric matrices in $M_{4\times 4}^{\mathbb{R}}$ is an eigenspace of the eigenvalue $3$
I know that $M_{4\times 4}^{\mathbb{R}} = V_{\lambda =3}\:\bigoplus \:V_{\lambda =-1}$ 
So those eigenvalues of $T$ must be the only one. 
How can I now find the minimal polynomial of $T$?
 A: I shall generalize and work with $\text{Mat}_{n\times n}(\mathbb{K})$ for any integer $n>0$ instead.  Here, $\mathbb{K}$ is an arbitrary field.
Note that 
$$T^2(M)=T\big(-2\,M^\top+M\big)=-2\,\big(-2\,M^\top+M\big)^\top+\big(-2\,M^\top+M\big)$$
for all $M\in\text{Mat}_{n\times n}(\mathbb{K})$.
Thus,
$$T^2(M)=-4\,M^\top+5\,M=2\,T(M)+3\,M\text{ for all }M\in\text{Mat}_{n\times n}(\mathbb{K})\,.$$
Consequently,
$$T^2-2\,T-3\,\boldsymbol{1}=0\,,$$
where $\boldsymbol{1}:\text{Mat}_{n\times n}(\mathbb{K})\to\text{Mat}_{n\times n}(\mathbb{K})$ is the identity map.  Hence, the minimal polynomial $\mu(x)\in\mathbb{K}[x]$ of $T$ divides $$x^2-2x-3=(x-3)(x+1)\,.$$

  For $n=1$, it is obvious that $\mu(x)=x+1$.  From now on, suppose that $n>1$.  First, if $\mathbb{K}$ is of characteristic $2$, then we see that $T=\boldsymbol{1}$, whence $\mu(x)=x+1$.  If $\mathbb{K}$ is a field of characteristic not equal to $2$, then $\text{Mat}_{n\times n}(\mathbb{K})=S_n(\mathbb{K})\oplus A_n(\mathbb{K})$, where $S_n(\mathbb{K})$ is the set of $n$-by-$n$ symmetric matrices over $\mathbb{K}$, whilst $A_n(\mathbb{K})$ denotes the set of $n$-by-$n$ antisymmetric matrices over $\mathbb{K}$.  Observe that $S_n(\mathbb{K})$ is the eigenspace of $T$ associated to the eigenvalue $-1$, and $A_n(\mathbb{K})$ is the eigenspace of $T$ associated to the eigenvalue $3$.


Even more generally, suppose that $a,b\in\mathbb{K}$ and $T:\text{Mat}_{n\times n}(\mathbb{K})\to\text{Mat}_{n\times n}(\mathbb{K})$ is defined by
$$T(M):=a\,M+b\,M^\top\text{ for all }M\in\text{Mat}_{n\times n}(\mathbb{K})\,.$$
Then, the minimal polynomial $\mu(x)\in\mathbb{K}[x]$ of $T$ is
$$\mu(x)=\left\{
\begin{array}{ll}
x-(a+b)&\text{if }n=1\text{ or }b=0\,,
\\
\big(x-(a-b)\big)\big(x-(a+b)\big)&\text{if }n>1\text{ and }b\neq0\,.
\end{array}
\right.$$
If $\mathbb{K}$ is of characteristic not equal to $2$, then $S_n(\mathbb{K})$ is the eigenspace associated to the eigenvalue $a+b$, whereas $A_n(\mathbb{K})$ is the eigenspace associated to the eigenvalue $a-b$.  If $\mathbb{K}$ is a field of characteristic $0$, then there is only one eigenvalue $a+b$, with the eigenspace $S_n(\mathbb{K})$.  However, the generalized eigenspace associated to the eigenvalue $a+b$ is the whole $\text{Mat}_{n\times n}(\mathbb{K})$.
