Suppose that $A, B$ are idempotent matrices ($A^2=A$), such that $A + B$ is idempotent, prove that $AB = BA = 0$

  • 1
    $\begingroup$ Are you sure you don't want to prove $AB+BA=0$ ? $\endgroup$
    – Belgi
    Jan 26, 2013 at 23:49
  • 1
    $\begingroup$ No it is definitely AB=BA=0 $\endgroup$
    – Chance
    Jan 26, 2013 at 23:51
  • 2
    $\begingroup$ Probably there is a typo on your book/list/etc... $\endgroup$
    – Sigur
    Jan 26, 2013 at 23:57
  • $\begingroup$ @Belgi Thanks.. $\endgroup$
    – Git Gud
    Jan 26, 2013 at 23:58
  • $\begingroup$ @GitGud - No problem, I was doing the same thing and I thought it would work $\endgroup$
    – Belgi
    Jan 27, 2013 at 0:00

2 Answers 2


This is true in every ring where you can divide by $2$. So it is true in particular for matrices.

We have $(A+B)^2=A^2+AB+BA+B^2=A+AB+BA+B$. Since $A+B$ is idempotent, it follows that $$AB+BA=0.$$ We call this equation (E).

Right-multiply (E) by $B$. We find $AB+BAB=0$, hence $AB=-BAB$.

Left-multiply (E) by $B$. This yields $BAB+BA=0$, hence $BA=-BAB$.

Equating the last two equations, we find $AB=BA$.

Now $AB+BA=0$ clearly yields $2AB=2BA=0$. Dividing by $2$, we get $$ AB=BA=0. $$

  • $\begingroup$ Very nice! I thought in this direction when I saw I coldn't get the result, but I figured it was a typo and I tried to find a trivial example so I tooked $B=A$ and then I saw that in $F_2$ I get such an example. Did you know this before, or just proved it ? (+1) $\endgroup$
    – Belgi
    Jan 27, 2013 at 0:17
  • $\begingroup$ Could you give me such example? I'm not able of find one $\endgroup$ Apr 24, 2018 at 19:25

This is incorrect.

For example take $F=\mathbb{F}_{2}$ the field with two elements and $A=B=I$ over $F$.

$A,B$ are clearly idempotent and $$A+B=I+I=2I=0$$ hence $$(A+B)^{2}=0^{2}=0=A+B$$ is also idempotent.

But $$AB=BA=I^{2}=I\neq0$$

From your assumption you can only get $AB+BA=0$.

  • $\begingroup$ Yes I can only seem to get $AB+BA=0$ from this assumption. Could possibly be a typo. $\endgroup$
    – Chance
    Jan 27, 2013 at 0:02
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    $\begingroup$ @Chance: It is true for square matrices over a field in characteristic distinct from $2$. See my answer. $\endgroup$
    – Julien
    Jan 27, 2013 at 0:13
  • $\begingroup$ Can the downvoter please explain ? maybe I can improve my answer. $\endgroup$
    – Belgi
    Jan 27, 2013 at 0:17
  • $\begingroup$ If you combine your answer with Julien's, you can arrive at $AB=BA=0$ $\endgroup$
    – Chance
    Jan 27, 2013 at 0:21
  • $\begingroup$ @Chance - Not in general, You have to assume $1+1\neq 0$ $\endgroup$
    – Belgi
    Jan 27, 2013 at 0:22

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