Let $L \in End(V)$ and $V=A\oplus B$ be it's Fitting Decomposition. This means $L$ is nilpotent and invertible when restricted to $A$ and $B$ respectively. $A$ and $B$ are constructed by taking successive images and nullspaces of powers of $L$.
Let $M$ be another endomorphism of $V$ such that for some $s \in \mathbb{N}$, $ad(L)^s(M)=0$.
I want to show $A$ and $B$ are invariant under $M$.
The formula $ad(L)^s(M)=\sum_{0}^{s} {s\choose p}L^pML^{s-p}$ will probably be useful but I'm not sure how to apply it.
Can someone please show me how to proceed?
Edit: I'm reading Varadarajan's $Lie groups$, $Lie$ $Algebras$ $and$ $Representations$. There are a bunch of theorems like this which he states. Another similar theorem is that the Jordan factors of $L$ are invariant under $M$ if and only if the above condition holds.
Let $f=ad(L)$ and $spectrum(L)=(\lambda_i)_i$. Then $ker(f^s)\subset \ker(f^{n^2})$ and we may assume that $M\in \ker(f^{n^2})$. Note that $f=L\otimes I-I\otimes L^T$ (when we stack the matrix row by row) and $spectrum(f)=\{\lambda_i-\lambda_j|1\leq i,j\leq n\}$; remark that $0$ is a semi simple eigenvalue of $f^{n^2}$.
Let $L=D+N$ where $D$ is diagonalizable, $N$ nilpotent and $DN=ND$ (Jordan-Chevalley decomposition) and let $g=ad(D)$. Using the Jordan form of $L$, it is not difficult to see that $\ker(f^{n^2})=\ker(g^{n^2})$. Since $g$ is diagonalizable, $\ker(g^{n^2})=\ker(g)$ and finally $\ker(f^{n^2})=\ker(g)$, that is $M\in \ker(f^{n^2})$ iff $DM=MD$.
Conclusion: $M\in \ker(f^{n^2})$ iff the generalized eigenspaces of $L$ (the $\ker((L-\lambda I)^n)$ when $\lambda$ runs through $spectrum(L)$) are invariant under $M$.