# How do I prove or disprove this matrix identity: $ABA=0$ and $B$ is invertible implies $A^2=0$?

Let us have two $n \times n$ matrices $A$ and $B$ with real entries. Either prove, or disprove by providing a counterexample, that if $ABA=0$ and $B$ is invertible, then $A^2=0$.

My attempt:

I could not find a counterexample so I assume the statement is true and then try to prove it.

I first assume that AB=0, then $ABB^{-1}=0$ so $A=0$, which then gives the required result. A similar result is given by assuming $BA=0$. Next I assume $AB\neq0$,i.e. $AB=C$ where $C$ is some $n\times n$ matrix such that $CA=0$. However, I am stuck at this point because I believe I can show that the matrices $A$ and $C$ are non-invertible, and I'm afraid I don't really know how to proceed. I would really appreciate it if someone could give me a hint on how to prove or disprove this statement, or point out a flaw in my attempt.

Here is a $2\times 2$ counterexample, easily extendable to $n\times n$: Let $A$ orthogonally project onto one axis, and let $B$ rotate the plane by $90^\circ$. The operation of $ABA$ is to collapse everything down to one axis, then turn that axis, then collapse that axis down to the origin. However, $A^2 = A\neq 0$.
Specifically, $A = \left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ and $B = \left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$.
The counterexample: $B = \left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$, where $a+b+c+d=0$ and $A = \left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)$.
Take $a=b=c=1$ and $d=-3$.
We see that $B$ is invertible and $A^2\neq\mathbb 0$.