Finding the minimal polynomial, the eigenvalues, and the characteristic polynomial of $T(A)=A+A^t$ Let $M_n$ denote the vector space of all $n\times n$ real valued matrices and for any matrix $A\in M_n$ let $A^t$ denote its transpose. Define the linear map $T:M_n\rightarrow M_n$ as
$$T(A)=A+A^t.$$
What is the minimal polynomial of $T$? What are the eigenvalues and the corresponding eigenspaces of $T$? What is the characteristic polynomial of $T$?
I learned about minimal polynomials of operators from $\textit{Linear Algebra Done Right}$ by Axler. In his description he explains how in order to find the minimal polynomial you must find the smallest positive integer $m$ such that $T^m$ is a linear combination of $I, T, T^2, \ldots, T^{m-1}$. That is, the smallest integer $m$ such that
$$T^m = a_0I+a_1T+a_2T^2+ \cdots +a_{m-1}T^{m-1}.$$
As for the eigenvalues...since we are working with real-valued matrices, then $A$ and $A^t$ have the same eigenvalues, so if $\lambda \in \mathbb{R}$ is an eigenvalue of $A$ (and hence of $A^t$) wouldn't $2\lambda$ be an eigenvalue of $A+A^t$? I apologize if I am saying utter nonsense...
 A: For the minimal polynomial we note that
$$T^2(A)=T(T(A))=T(A+A^t)=(A+A^t)+(A+A^t)^t=2(A+A^t)=2T(A)\ ;$$
thus $T^2-2T=\bf0$ and the minimal polynomial is $z^2-2z$ (it's easy to check that no polynomial of smaller degree works).
For eigenvalues: we need
$$A+A^t=\lambda A\ .$$
Taking the transpose, $A^t+A=\lambda A^t$ and so
$$\lambda A=\lambda A^t\ .$$
Hence
$$\lambda=0\ ,\quad A^t=-A$$
or
$$A^t=A\ ,\quad \lambda=2\ .$$
The eigenvalues and eigenspaces are
$$\lambda=0\ ,\ \{\hbox{skew-symmetric matrices}\}\quad\hbox{and}\quad
  \lambda=2\ ,\ \{\hbox{symmetric matrices}\}\ .$$
The geometric multiplicities are
$$\frac{n(n-1)}2\ ,\quad \frac{n(n+1)}2$$
respectively, and since they add up to $n^2$, the dimension of $M_n$, they must also be the algebraic multiplicities.  So the characteristic polynomial is
$$z^{n(n-1)/2}(z-2)^{n(n+1)/2}\ .$$

Your last paragraph is wrong but I think Gerry has explained it clearly in his comment so I won't say anything more.
A: Though not an exact duplicate (as far as I can tell), this question is very similar to the following existing questions on this site, about the minimal polynomials of $A\mapsto A^t-A$, of $A\mapsto A-2A^t$, and of $A\mapsto-A^t$. In each case the question is about a linear combination of the transposition map (from square matrices to themselves) and the identity map $I$ (on square matrices). Since for eigenvalue problems, linear combinations with the identity map are easily handled (and more generally linear combinations of commuting operators) it is a good idea to start with the transposition map $\tau:A\mapsto A^t$.
It is clear that $\tau$ is an involution (that is, $\tau^2=I$), and it follows that the minimal polynomial divides $X^2-1=(X+1)(X-1)$. As the latter two factors are relatively prime (the real numbers do not have characteristic$~2$), $\tau$ is diagonalisable, and its eigenvalues are contained in the set $\{-1,1\}$ Indeed the eigenspace for $\lambda=-1$ is the set of antisymmetric matrices, which has positive dimension when $n>1$, and  the eigenspace for $\lambda=1$ is the set of symmetric matrices, which has positive dimension when $n>0$. So the minimal polynomial of $\tau$ is $1$ when $n=0$, it is $X-1$ when $n=1$, and it is $(X+1)(X-1)=X^2-1$ as soon as $n\geq2$. For the characteristic polynomial you must raise each factor $X-\lambda$ to a power equal to the dimension of the eigenspace for$~\lambda$, which is easily computed.
Now for a linear combination $T=aI+b\tau$ with $b\neq0$, the eigenspaces do not change; the only thing that is needed is to replace the eigenvalue $\lambda$ by $a+b\lambda$ (because the term $aI$ acts by a scalar $a$, and the term $b\tau$ by a factor $b\lambda$, on the eigenspace of$~\tau$ for $\lambda$). In the polynomials one must similarly replace the factors $X-\lambda$ by $X-(a+b\lambda)$ as only change (the condition $b\neq0$ ensures the two factors remain relatively prime). In your case ($a=1=b$) you get eigenvalues $0$ from $\lambda=-1$ (antisymmetric matrices) and $2$ from $\lambda=1$ (symmetric matrices). The other linked cases are entirely similar.
