How prove this matrix $\sigma$ Characteristic polynomial $f(x)=(x+1)^n(x-5)^{\frac{n(n-1)}{2}}(x+1)^{\frac{n(n-1)}{2}}?$ Question:
let $$V=\{A|A\in F^{n\times n}\}$$,define linear transformation
$$\sigma:A\mapsto 2A-3A^T$$
show that
the matrix $\sigma$ Characteristic polynomial $$f(x)=(x+1)^n(x-5)^{\frac{n(n-1)}{2}}(x+1)^{\frac{n(n-1)}{2}}?$$
show that $\sigma$ can diagonalization
My try:this Characteristic polynomial reslut is my frend tell me,
so I can't  ensure this is true.
Thank you  very much!
 A: Let $\cal S$ ($\cal A$) denote the subspace of symmetric (antisymmetric
matrices). 
  Denote by $E_{ij}$ the matrix all of whose coefficients are zero except in
the entry at the intersection of the $i$-th line and the $j$-th column, equal
to $1$.  Thus $(E_{ij})$ is the canonical basis of $V$.
For any $i < j$, we have ${\sf span}(E_{ij},E_{ji})={\sf span}(S_{ij},A_{ij})$
    where $S_{ij}=\frac{E_{ij}+E_{ji}}{2}$ and $A_{ij}=\frac{E_{ij}-E_{ji}}{2}$.
So the family $\lbrace E_{ii} | 1 \leq i\leq n \rbrace \cup 
  \lbrace S_{ij} ; A_{ij} | 1 \leq i < j \leq n \rbrace $  is another basis
  of $A$.
The transpose map $\tau : A \mapsto A^{T}$ is easily seen to be diagonal
  in that basis : we have $\tau E_{ii}=E_{ii}, \tau S_{ij}=S_{ij}, \tau A_{ij}=-A_{ij}$.
Note that $\cal S$ is the eigenspace of $\tau$ corresponding to the eigenvalue $1$. 
So $\lbrace E_{ii} | 1 \leq i\leq n \rbrace \cup 
  \lbrace S_{ij}  | 1 \leq i < j \leq n \rbrace  $ is a basis of $\cal S$, and we deduce
  ${\sf dim}(\cal S)=\frac{n(n+1)}{2}$.
Similarly  $\lbrace S_{ij}  | 1 \leq i < j \leq n \rbrace  $ is a basis of $\cal A$, and we deduce ${\sf dim}(\cal A)=\frac{n(n-1)}{2}$.
For $A\in \cal S$ we have $A^{T}=A$ so $\sigma(A)=-A$.
For $A\in \cal A$ we have $A^{T}=-A$ so $\sigma(A)=5A$.
So the characteristic polynomial
  $\chi_A$ of $A$ is
$$
  \chi_A=(X+1)^{{\sf dim}({\cal S})} (X-5)^{{\sf dim}({\cal A})}=
  (X+1)^{\frac{n(n+1)}{2}} (X-5)^{\frac{n(n+1)}{2}}
  $$
A: We have
$$\sigma(\sigma(A))=\sigma(2A-3A^T)=2(2A-3A^T)-3(2A^T-3A)\\=13A-12A^T=13A+4(\sigma(A)-2A)=5A+4\sigma(A)$$
hence the polynomial $x^2-4x-5=(x+1)(x-5)$ with simple roots $-1$ and $5$ annihilates $\sigma$ so $\sigma$ is diagonalized and since $\sigma\ne k\mathrm {id}$ hence $-1$ and $5$ are two eigenvalues of $\sigma$. Now to find the multiplicity of the both eigenvalues we solve the equality
$$\sigma(A)=kA\quad k=-1,5$$
for example for $k=-1$ we find
$$2A-3A^T=-A\iff A=A^T\iff A\in S_n$$
and we know that
$$\dim S_n=\frac{n(n+1)}{2}$$
and we find also that the multiplicity of $5$ is $\dim A_n=\frac{n(n-1)}{2}$ and we conclude.
