TITLE: Existence of $A^2B - BA^2 = 2A \textrm{ and } AB^2 - B^2A = 2B$. in $\mathcal{M}_n({\mathbb{C}}) $
QUESTION [8 upvotes]: This question arose in this classical exercise : Do there exist two matrices such that $AB-BA=I_n$ in $\mathcal{M}_n({\mathbb{C}}) $.
Wich is impossible (by using trace to prove this) 
But if $AB-BA=I_n$ then $A^2B - BA^2 = 2A \textrm{ and } AB^2 - B^2A = 2B$.

So my question is :


Do there exist  $A$ and $B$ (non-zero) in $\mathcal{M}_n({\mathbb{C}})$ such that : 
  $$
\textrm{(1) } A^2B - BA^2 = 2A 
$$
  $$
\textrm{(2) } AB^2 - B^2A = 2B
$$

For $n=2$ it's impossible (by putting $A^2$ for example).
By induction, we could conclude, if all matrices of rank one switch with matrices with zero trace.
Unfortunately, this is not the case.
Any ideas to prove it's impossible ?
Thank you.

REPLY [5 votes]: When $n\ge 3$, such $A$ and $B$ exist. The following is an example for $n=3$, where $x\in\Bbb C$ is arbitrary.
$$A=\begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},\quad B=\begin{pmatrix}0 & x & 0\\ -2 & 0 & x \\ 0 & 2 & 0 \end{pmatrix}.$$

Remark: From $A^2B-BA^2=2A$ we can easily see that ${\rm tr} A^k=0$ for every $k\ge 1$, so $A$ is nilponent. As a result, when $n=2$, $A=0$. Similar argument can be applied to $B$.