6
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

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 $B = \left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$.

$\endgroup$
4
$\begingroup$

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 $A^2\neq\mathbb 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.