How to formally define eigenvectors of T$:M_{n\times n}(\mathbb{R})\to M_{n\times n}(\mathbb{R}) \quad TA=A^t $ I found out that the eigenvalues of this transformation is $\pm 1$. So for $\lambda =1$ the eigenvector is any symmetric matrix and for $\lambda = -1$ the eigenvector is any Skew-symmetric matrix. So what I don't know is how to formally define this eigenvectors can I say:
$v_1={A\in M_{n\times n}:A=A^t}$
$v_2={A\in M_{n\times n}:-A=A^t}$
 A: Let $A\in M_n(\mathbb{R})$ be a matrix so if we put $S=\frac{1}{2}(A+A^t)$ and $S'=\frac{1}{2}(A-A^t)$ then we have :


*

*$S^t=S$

*$S'^t=S'$

*$S+S'=A$


we can deduce easily that if we put $\mathcal{S}$ (resp $\mathcal{S}'$) the space of symmetric (resp Skew-symmetric) matrix :
$$
M_n(\mathbb{R})=\mathcal{S}\oplus\mathcal{S}'
$$
So let now the operator 
$$
\begin{array}{c}
T: & M_n(\mathbb{R})& \to & M_n(\mathbb{R})\\
& A & \mapsto & A^t
\end{array}
$$
then $T$ is a linear operator and we need to see the eigenvalues of $T$ :
$$
\lambda \in \sigma(T) \iff \exists A \in( M_n(\mathbb{R})-\{0\})\;  \textrm { such that } T(A)=\lambda A
$$
so it mean that :
$$
A^t=\lambda A 
$$
two solution are obvious $\lambda=1,-1$, but using the fact that the sum of eigenspaces is a direct sum and the decomposition of our space we can see that $1$ and $-1$ are the only eigenvalues of $T$.
A: An eigenvector for the eigenvalue $\lambda$ is any nonzero vector $v$ satisfying $Tv = \lambda v$. Equivalently, it is a non-zero element of the eigenspace $\ker\{ T- \lambda I\} = \{ v \mid Tv = \lambda v \}$.
Adapted to your situation, this means that $A$ is an eigenvector for the eigenvalue $1$ iff it belongs to the set $\{A \in M_n \setminus \{ 0\}\mid A^t = A\}$ (and similarly for the other eigenvalue).
A: The subspaces (check they are subspaces) of $M_{n\times n}(\Bbb{R})$ defined as follows
$$\begin{align}S_n=&\{A\in M_{n\times n}(\Bbb{R}),\,A=A^T\}\\
A_n=&\{A\in M_{n\times n}(\Bbb{R}),\,A=-A^T\}\end{align}$$
have the zero matrix as intersection and have ${n(n+1)\over 2}$ and ${n(n-1)\over 2}$ respectively as dimensions and so they are supplementary. They correspond to the $\pm 1$ only eigenvalues.
Without using any dimension argument one can see that any matrix can be decomposed into a symmetric and an anti symmetric part
$$A={A+A^T\over 2}+{A-A^T\over 2}$$
