Is there a solution to these matrix equations?

Question

Let $A$ and $B$ be matrices such that:

$$A\neq 0$$ $$B \neq 0$$ $$A^2=0$$ $$B^2=0$$ $$AB=A$$ $$BA=B$$

Is there a solution to this set of equations? I tried looking for a solution in $2\times 2$ matrices but couldn't find one. Is there a way to prove no such solution exists for $n \times n$ matrices?

(Or if it makes a difference the last two can be replaced with $AB=B$ and $BA=A$ .)

There doesn't seem to be a solution for any $n\times n$ matrices. If $B=BA$ and $A=AB$, then $A=ABA$. But $ABA=AA=A^2=0$, so $A=0$. Similarly for $B$. — doobdood, Nov 05 '21 at 02:35
I know. I guess I should've been more clear. We have $A=ABA$, and we also have $ABA=A^2$. Since $A^2=0$ by hypothesis, and since $A=A^2$ by transitivity, we get $A=0$. — doobdood, Nov 05 '21 at 02:41

Keeley Hoek · Accepted Answer · 2021-11-05T02:41:15.607

2

I like the version with $A B = B$ better. This means in particular that if $v \in \operatorname{im} B$ then $A v = v$. In turn this means that either $\operatorname{im} B = 0$ or no power of $A$ can ever be zero (including $A^2$). But your other hypotheses prohibit both of these things.

(The version with $A B = A$, etc. is converted to this version by taking the transpose.)

edited Nov 05 '21 at 02:41

answered Nov 05 '21 at 02:36

Keeley Hoek

2,929

Thanks! That's a nice proof. – zooby Nov 05 '21 at 02:40

Is there a solution to these matrix equations?

1 Answers1