For a fixed $A \in M_n (\mathbb{R})$, define $\mathscr{L}(X)=AX$ on the set $M_n (\mathbb{R})$. We have to find how characteristic polynomials of $\mathscr{L}$ and $A$ are related. I need to get my proof verified.
Here's what I have: I have proved(?) they both have the same set of eigenvalues. For if $\alpha$ is an eigenvalue of $A$ with corresponding eigenvector $v$, then the matrix $X$ with each of it's columns as $v$ will be an eigenvector for $\mathscr{L}$ with eigenvalue $\alpha$. If $\beta$ is an eigenvalue of $\mathscr{L}$ then $AX=\beta X$ for some $X \in M_n (\mathbb{R})$, or $(A-\beta I)X=0$. I think the first column of $X$ is an eigenvector for $A$ with eigenvalue $\beta$. If this much is true, their characteristic polynomials will have the same roots, only one is of degree $n$ and the other is of degree $n^2$. Can we say something more about them from here? Is it possible that the characteristic polynomial of $\mathscr{L}$ is just that of $A$ but raised to power $n$? How do we prove that?
Hint: Let $U_j$ denote the subspace of $M_{n}(\Bbb R)$ consisting of matrices whose entries outside of the $j$th column are all zero. Verify that $U_j$ is an invariant subspace of $\mathscr L$. What is the characteristic polynomial of $\mathscr L|_{U_j}$?