Some context first: Consider $S=M_n(\mathbb{C})$ as an algebra over $\mathbb{C}$. For every $A \in S$, it's easy to check that $ad_A(M):=AM-MA$ is a derivation ($C$-homomorphism of $S$ that satisfies the Leibniz rule: $f(ab)=f(a)b+af(b)$). Its also easy to check that $A$ nilpotent $\implies ad_A$ nilpotent.
My question is:
Can you describe all matrices $A$ for which $ad_A$ is nilpotent?
If I'm not mistaken, induction gives, for all $M \in S$: $ad_A(M)^{(k)}=\sum\limits_{j =0}^{k} A^{k-j}MA^{j} {k \choose j} (-1)^{j}$. Hence, we want to classify all matrices $A \in S$ such that, for all $M \in S$, we have $M=\sum\limits_{j =0}^{k} A^{k-j}MA^{j} {k \choose j} (-1)^{j}$ for some $k \in \mathbb{N}$.
Any help? Thanks.
Since $ad_A$ is a linear operator on a finite dimensional space we have to see for which $A$ the spectrum of $ad_A$ is trivial, i.e., $\sigma(ad_A)=\{ 0\}$. We will show that this happens if and only if $A$ has only one eigenvalue. The proof in one direction is easy: if $\sigma(A)=\{ \lambda \}$, then $A-\lambda I$ is nilpotent. Since $ad_A=ad_{A-\lambda I}$ we conclude that $ad_A$ is nilpotent.
For the opposite direction, assume that $A$ has at least two eigenvalues, say $\lambda, \mu \in \sigma(A)$, $\lambda \ne \mu$. Let $u$ be an eigenvector of $A$ at $\lambda$ and let $v$ be an eigenvector of $A^*$ at $\overline{\mu}$. Denote bu $u\otimes v$ the operator which maps a vector $x$ to vector $\langle x,v\rangle u$. This is a nonzero operator because $u\ne 0$ and $v\ne 0$. It follows from $$ [A (u\otimes v)-(u\otimes v) A]x=\langle x,v\rangle Au-\langle Ax,v\rangle u=(\lambda -\mu)(u\otimes v)x $$ that $u\otimes v$ is an eigenvector of $ad_A$ at eigenvalue $\lambda -\mu\ne 0$. Hence $ad_A$ is not nilpotent.