Let $X, Y \in M_n(\mathbb{C})$ such that $Y^2=YX-XY$ and $\text{rank}(X+Y)=1$.
Prove that $Y^3=YXY=0$
The problem is from a university contest for first and second year math majors and it is not supposed to use results from Lie Algebras such as Jacobson's Lemma (if $A$ and $AB-BA$ commute then $AB-BA$ is nilpotent) or results such as if $A(AB-BA)=(AB-BA)B$ then $A$ and $B$ are simultaneously triangularizable.
As a student who is preparing for a contest I have no ideea how to approach a problem like this. I knew that there were some results about commutators of matrices (such as the ones above) but even if I take them for granted I still cannot figure out how to solve this problem.
I will denote $[X, Y] = XY-YX$
My ideea was to (somehow) prove that $[Y, X]^2=0$ (this has to be true because $[Y, X]^2=Y^4$ which has to be $0$ otherwise the problem would be false). So $\text{rank}(YX-XY)=1$ which implies that $X$ and $Y$ are simultaneously triangularizable by this. Therefore there is an invertible matrix $U$ such that $X=UAU^*$ and $Y=UBU^*$, where $A$ and $B$ are upper triangular (simultaneous triangularization on $\mathbb{C}$ is equivalent to simultaneous unitary triangularization because we can use $QR$ decomposition on the transition matrix).
By the hypothesis $\text{rank}(X+Y)=\text{rank}(A+B)=1$ which means that either $(A+B)^2=0$ or $\text{Trace}(A+B) \neq 0$ and $A+B$ is diagonalizable (don't really know if this helps). This tells us that the eigenvalues of $X$ are $-1$ times the eigenvalues of $Y$, except possibly one eigenvalue.
Now $Y^2=YX-XY \iff B^2=BA-AB$ and the diagonal entries of $A$ and $B$ are their eigenvalues. Now assume $\text{Trace}(A+B)=0 \iff \lambda_A=-\lambda_b \iff \lambda_b^2=-\lambda_B^2+\lambda_b^2 \iff \lambda_B=0$. If $\text{Trace}(A+B) \neq 0$ then there is at most one nonzero eigenvalue of $B$ (however that is not possible since $B^4=0$, therefore all the eigenvalues of $B$ are $0$)
Therefore all the eigenvalues of $B$ are $0$ then so are the eigenvalues of $A$.
How should I continue/approach a problem like this?
Note that all my proof is invalid since I haven't been able to prove that $[Y, X]^2=0$ (it has to be true if the problem is true, and it would be an intuitive guess because [Y, X] is nilpotent by Jacobson's Lemma), but this seems like the correct way to go.
Also assuming that I actually got this far (and that I proved $[Y, X]^2=0$) I should somehow get that $B^3=BAB=0$. I don't know how to get to this. One way that pops up in mind is studying the Jordan chains of $A$ and $B$, but that seems really tedious.
Any hints or help with this would be gladly appreciated.
