Sum of nilpotent matrices with certain property is nilpotent I want to prove that if $A$ and $B$ are nilpotent matrices such that $[A, [A,B]] = [B, [A,B]] = 0$ then $A + B$ is a nilpotent matrix as well.
I know that it must follow from some general theorem from Lie algebras theory. I am sure, though, that in this particular case it can be done by  simple methods.
My attempt uses rather the language of the Lie theory than its results  themselves. I noticed that such a matrices (if $dim \langle A, B, [A,B] \rangle = 3$) is a (faithful) representation if the Heisenberg algebra.  And since it is true for this algebra it must be true in any faithful representation. I think. I could't prove it.
I am looking for hints for proving it my way and for the solution via elementary methods.
 A: The statement is not true in general. E.g. consider
$$
A=\pmatrix{1&1\\ 1&1}
\text{ and } B=\pmatrix{0&1\\ 0&0}
$$
over $GF(2)$. We have $A^2=B^2=0,\,[A,B]=I_2$ and $[A,[A,B]]=[B,[A,B]]=0$, but $A+B$ is nonsingular.
Yet, the statement is true if $[A,B]$ is also nilpotent. (This occurs, for instance, when $A,B\in M_n(F)$ where $\operatorname{char}(F)=0$ or $\operatorname{char}(F)>n$, by Jacobson's lemma.) Since $A$ commutes with $[A,B]$, they share a common eigenvector $x$. Therefore $Ax=[A,B]x=0$. Hence we can prove by mathematical induction that $AB^kx=0$ for every nonnegative integer $k$, where in the inductive step we have
\begin{aligned}
AB^{k+1}x
&=ABB^kx\\
&=\left([A,B]+BA\right)B^kx\\
&=[A,B]B^kx\\
&=B^k[A,B]x\\
&=0.
\end{aligned}
However, since $B$ is nilpotent, there exists a nonnegative integer $k$ such that $B^kx\ne0=B^{k+1}x$. Therefore $B^kx\in\ker(A)\cap\ker(B)$, i.e. $A$ and $B$ share a common eigenvector. Proceed recursively, $A$ and $B$ can be simultaneously triangularised. Hence $A+B$ is nilpotent.
(Note that we don't need the condition $[B,[A,B]]=0$ in the above proof.)
