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My goal is to prove or disprove the following claim:

For $2\times 2$ matrices $A$ and $B$, if $(AB)^2=0$, then $(BA)^2=0$.

My thoughts on the question:

  1. I know that $AB=O$ does not imply that $BA=O$, so my first impression was that it is false. I tried the counter-example I know but it leads to $(BA)^2=O$.
  2. (edited) As pointed out by Friedrich Philipp in the comments, $A$ or $B$ is not invertible. If one of them is invertible, then the question is easily shown to be true. I wish to avoid density arguments though, so I am still stuck with the case where $A$ and $B$ are both singular.
  3. The question is in a list of prove or disprove questions for square matrices of any order. This specific one precise that $A$ and $B$ are of order $2$, so maybe the property is true for these matrices. I seems to remember that if $M$ is a square matrix of order $n$ and is nilpotent, then the order of nilpotence (the smallest $p$ such that $M^p=O$) is at most $n$, but I don't know how to use it here.

Beside this, I am clueless. I am looking for ideas or hints on the problem.

Taladris
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    $AB=0$ implies $(BA)^2=BABA=B(AB)A=0$. – Emre Mar 15 '16 at 01:59
  • @Wilf-Fine So what? ;-) Taladris: "$A$ and $B$ are not invertible": I doubt that. – Friedrich Philipp Mar 15 '16 at 02:03
  • But I don't know if $AB=O$. Maybe my first comment is confusing: I meant that I initially thought the statement is false, and to find a counter-example, I can just take $A$ and $B$ such that $AB=O$ but $BA\neq O$. Then trivially $(AB)^2=O$ but hopefully we would have $(BA)^2\neq O$... which doesn't happen in the example I tried. My question is really about the statement in the title. – Taladris Mar 15 '16 at 02:04
  • @FriedrichPhilipp: you are right! One of them is not invertible, but maybe not both.I'll edit my question – Taladris Mar 15 '16 at 02:06
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    $(BA)^3=BABABA=B0A=0$ – Empy2 Mar 15 '16 at 02:15
  • @Michael That's good. So, the minimal polynomial of $BA$ divides $t^3$. Thus, it is either $t$ or $t^2$ so that $(BA)^2 = 0$. – Friedrich Philipp Mar 15 '16 at 02:22
  • If A invertable – Doug M Mar 15 '16 at 02:27

2 Answers2

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Use the fact that for a $2\times 2$ matrix $C$, $C^2 - tr(C) C + \det(C) I_2 = 0_2$. Conclude that if $C$ is a $2\times 2$ matrix and $C^2=0$, then both the trace and the determinant of $C$ are 0. Then use the fact that $AB$ and $BA$ have the same trace and determinant.

Catalin Zara
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In general, if $A$ is $m\times n$ and $B$ is $n\times m$ are two arbitrary matrices, the nonzero eigenvalues of $AB$ and $BA$ (counting multiplicity) must be identical . Therefore, if $AB$ is nilpotent, $BA$ must be nilpotent too and $(BA)^{\max\{m,n\}}=0$.

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user1551
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