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…
Best Answer
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.