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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.

Invincible
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  • This answer may help. – Chrystomath Dec 17 '20 at 17:45
  • I agree the ideas behind the linked question and answer will be very helpful, but one has to be quite careful because I think power notation there means with respect to the Lie bracket, whereas here OP is interested in powers of $A, B, A+B$ w.r.t. the associative multiplication of matrices. – Torsten Schoeneberg Dec 17 '20 at 20:39

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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.)

user1551
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  • I had a feeling when dabbling with the problem that there is some tedious but more elementary proof just using the relation $[A, [A, B]]=0$ written out as $ABA = \frac12 (AAB + BAA)$ (and likewise with $A,B$ switched) to "move $A$'s and $B$'s past each other until there's many enough to make zero", which would work for any associative $F$-algebra provided $char(F) \neq 2$. Admittedly I did not look into it further, but do you have any counterexample for $char(F) \ge 3$? – Torsten Schoeneberg Dec 21 '20 at 05:55
  • @TorstenSchoeneberg I think the statement is true when the characteristic of $F$ is not $2$. I have a sketch of proof in mind that uses a strengthened version of Jacobson's lemma, but that still relies on finite dimensionality and it's probably too tedious to write out. I agree with you that statements of this sort can likely be proved using only properties of the Lie bracket. The key is to find a magic expression that automatises the proof, but I have no idea what this right expression is. – user1551 Dec 21 '20 at 08:56