Finding the characteristic polynomial of the $T: \mathcal{M}_{n} (\mathbb{R}) \to \mathcal{M}_{n} (\mathbb{R})$ given by $T(M)=M^{\text{tr}}$ Here's a problem from Larry Smith's Linear Algebra textbook:

Let $\mathcal{M}_{n} (\mathbb{R})$ be the set of real matrices of
  order $n \times n$. Let $T: \mathcal{M}_{n} (\mathbb{R}) \to \mathcal{M}_{n} (\mathbb{R})$ be the linear transformation given by
  $T(M)=M^{\text{tr}}$.
  
  
*
  
*Find the characteristic polynomial of $T$
  
*Find the eigenvalues and corresponding eigenspaces
  
*Is $T$ diagonalizable?
  

I have tried desperately to solve this problem and still have not been able to. I am stuck on the first part. I first tried to  bruteforce the problem for $n=1,2,3,4$ and I conjectured that the characteristic polynomial is 
$$\Delta (x) = (x-1)^{\frac{n(n+1)}{2}} (x+1)^{\frac{n(n-1)}{2}}$$
(I wrote a code which generates the matrix of $T$ provided $n$ is given: https://ideone.com/Aycf3N )
I know I should be using induction to prove this but I don't see how. I'll appreciate it if hints are provided. Nevertheless, is there any other way of attacking this problem?
 A: $T^2=Id$ implies the  minimal polynomial is $X^2-1=0$ since $T$ is different of the $M\rightarrow -M$ map and the identity, its minimal polynomial cannot be $X$ or $X-1$.
The eigenspace associated to $1$ is the space of symmetric matrices and the eigenspace associated to $-1$ is the space of antisymmetric matrices.
The characteristic polynomial is $(X-1){{n(n+1)}\over 2}(X+1)^{{n(n-1)}\over 2}$ since the space of symmetric matrices has dimension ${{n(n+1)}\over 2}$ and the space of antisymmetric matrices has dimension ${{n(n-1)}\over 2}$ and you can compute the characteristic polynomial of $T$ by decomposing $M(n,\mathbb{R})=Sym(n,\mathbb{R})\oplus Antisym(n,\mathbb{R})$.
A: $\newcommand{\tr}[0]{\text{tr}}$The answer by Tsemo Aristide is crystal clear. Let me add a couple of comments. Assume $n > 1$.
Given any $M \in \mathcal{M}_{n} (\mathbb{R})$, write
$$
M = \frac{1}{2} (M + M^{\tr}) + \frac{1}{2} (M - M^{\tr}).
$$
Here the first summand is a symmetric matrix ($N = N^{\tr}$) and the second summand is anti-symmetric ($N = -N^{\tr}$), so that
$$
\begin{cases}
T(\frac{1}{2} (M + M^{\tr})) = \frac{1}{2} (M + M^{\tr})\\
T(\frac{1}{2} (M - M^{\tr})) = - \frac{1}{2} (M - M^{\tr})\\
\end{cases}
$$
We thus have that $\mathcal{M}_{n} (\mathbb{R})$ is the direct sum of the space $\mathfrak{S}$ of symmetric matrices, on which $T$ acts as the identity matrix, and the space $\mathfrak{A}$ of anti-symmetric matrices, on which $T$ act acts as minus the identity. Hence the minimal polynomial of $T$ is $x^{2} - x$, the eigenspaces are $\mathfrak{S}$ for the eigenvalue $1$ and $\mathfrak{A}$ for the eigenvalue $-1$, and $T$ is clearly diagonalizable.
As to the characteristic polynomial, note that $\mathfrak{S}$ has dimension 
$$
s = \dbinom{n+1}{2},
$$
and $\mathfrak{A}$ has dimension 
$$
a = \dbinom{n}{2},
$$
so that the characteristic polynomial is
$$
(x - 1)^{s} (x + 1)^{a}.
$$
