Finding the characteristic polynomial of a linear transformation Let $T: M_n(\Bbb{R}) \to M_n(\Bbb{R})$ be a linear transformation defined by $T(A)=A^t+A$. What is the characteristic polynomial of $T$?
If I use the basis $E_{ij}$, I get $T(E_{ij})=E_{ij}+E_{ji}$, but don't know how to compute the  characteristic polynomial.
 A: Let $E_{ij}$ be the matrix with a $1$ at the $(i,j)$-th entry and zero elsewhere. Then all the $E_{ij}$s form a basis of $M_n$. Now, order this basis as follows: $E_{11},E_{22},\ldots,E_{nn}$ go first, followed by every pair of $\{E_{ij},E_{ji}\}$ with $i\ne j$. The order of the pairs are unimportant. What is crucial is that each $E_{ij}$ is adjacent to $E_{ji}$ in the ordered basis.
The matrix of $T$ with respect to this basis is then $\operatorname{diag}(2I_n, J, J, \ldots, J)$, where there are $\frac{n(n-1)}{2}$ copies of $J=\pmatrix{1&1\\ 1&1}$. You can calculate the characteristic polynomial of this block-diagonal matrix easily.
A: $M_n(\mathbb{R})$ is an $n^2$ dimensional space.  So you are looking for $n^2$ eigenvalues, counting repetition.  One of the eigenvalues is 2.  For what sort of matrices does $T(A)=2A$?  What is the dimension of that subspace of $M_n(\mathbb{R})$?
A: It is clear that $f:A\mapsto A^t$ is an involution (it satisfies $f^2=1$), so it is diagonalisable (since $X^2-1=(X-1)(X+1)$ is split with simple roots) and has characteristic polynomial $\chi_f=(X-1)^s(X+1)^a$ where $s$ is the dimension of the eigenspace for$~1$ (the symmetric matrices) and $a$ is the dimension of the eigenspace for$~-1$ (the anti-symmetric matrices). Now your $T=f+I_{M_n(\Bbb R)}$, so its characteristic polynomial is obtained from $\chi_f$ by substituting $X-1$ for $X$ (which adds$~1$ to all eigenvalues). Filling in the remaining details should be easy.
A: This is a simpler way to do it. 
Suppose $\lambda \in \mathbb{R}$ is an eigenvalue of T. Then
$T(A)=A^{t}+A=\lambda A$  $\iff$  $A^t=(\lambda-1)A$
Now, this means that if we write $A=(a_{ij})$, 
$a_{ij} = (\lambda-1)a_{ji} = (\lambda-1)^{2}a_{ij}$
Hence, $A = (\lambda -1)^{2} A$; that is, $(\lambda-1)^2 = 1$. 
We claim that $f(x) = x^{2}-2x$ is the characteristic polynomial of $T$. This is easy to check:
$(T^{2}-2T)(A) = T(A^{t}+A)-2(A^{t}+A) = 2(A^{t}+A)-2(A^{t}+A)=0$
for all $A \in M_{n\times n }(\mathbb{R})$. 
