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I've been given a class assignment to try and prove or disprove the following:

$A^2=B^2=0$ where $A,B \in M_n(\mathbb R)$ for $n≥2$ . If $AB=BA$, then is $(A+B)^2=0$ ?

I've been trying to prove this by saying that given the above conditions, $(A+B)^2 = 0$ is always true, $0=0$. My attempt:

$$(A+B)^2 = A^2 + AB + BA + B^2 = AB + BA = 2AB$$

Now because: $(A+B)^2=0$ then: $2AB=0 \implies AB=0 $ but because $A^2=B^2=0$ then if we multiply by A on the left side we get: $AAB = A0 \implies A^2B = 0 \implies 0 = 0$.

Did I prove this correctly? or am I not allowed to multiply by $A$ when there is $0$ on one side? I tried disproving it and just could not find any counter-examples that worked.

Dor Weid
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    It seems that you use $(A+B)^2=0$, the statement you want to prove. Or, do I misunderstand what you want to do? – mickep Dec 30 '16 at 08:45
  • The statement I want to prove is $(A+B)^2 = 0$, yes. So if given the conditions, I can prove that $(A+B)^2 = 0$ is actually $0=0$ then the statement is correct. – Dor Weid Dec 30 '16 at 08:48
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    The type of reasoning you apply seems entirely backward. Assuming the desired conclusion and concluding the trivially true $0=0$ buys you nothing, on a double count: you started from something that might not be true, and you concluded nothing you did not know already. – Marc van Leeuwen Dec 30 '16 at 08:48
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    Basically untill the point where you get $(A+B)^2=2AB$ your reasoning is correct. But the last line proves nothing. So it seems you need to also enforce $AB=0$ for your initial statement to be true. – Jesús Ros Dec 30 '16 at 08:58
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    @DorWeid the problem with your proof is that you have used $(A+B)^2=0$ as as assumption, although that is what you are trying to prove\disprove. Essentially you have to think that you're trying to prove $(A+B)^2=0$ and you have $A^2=B^2=0$ and $AB=BA$ at your disposal as assumptions. Therefore, your proof isn't valid. –  Dec 30 '16 at 09:03
  • @DorWeid: Since $A$ is not invertible, the equality $AB = 0$ is not equivalent to the equality $A^2 B = 0$. In a ring, multiplication by a non-invertible element introduces new information that wasn't previously there. For instance, in the ring $\Bbb Z / (6)$ the equation $\hat 3 x = \hat 3$ has the solutions $\hat 1, \hat 3, \hat 5$, but if you multiply it by $\hat 2$ you get the equation $\hat 0 x = \hat 0$ which has more solutions. – Alex M. Dec 30 '16 at 09:29

1 Answers1

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When $A,B$ commute you can apply the binomial theorem, as you did, which gives that $(A+B)^2=A^2+2AB+B^2$. Since you have the hypothesis $A^2=0$ and $B^2=0$, this means that $(A+B)^2=2AB$, and the question is whether this is forced to be $0$ under the given hypotheses.

If you have played a bit with the possibilities of (commuting) nilpotent matrices, you may know that they can often be modelled by partial maps from a finite set to itself, each element of the set representing a basis vector, and the map describing basis vectors being mapped to other basis vectors, with the elements where the map is undefined corresponding to basis vectors mapped to$~0$. Here I'll take a finite set$~S$ of points in$~\Bbb Z^2$, the maps being a shift in the horizontal and vertical direction (which ensures commutation) whenever that lands inside$~S$. The conditions $A^2=0$ and $B^2=0$ mean the $S$ can have no more than $2$ successive points horizontally and vertically, and $AB$ is a shift by $(1,1)$. It seems clear that the hypotheses do not imply that such a shift always takes a point of $S$ outside the set$~S$. The simplest counterexample is with $S$ being the four corners of a unit square. This give rise to matrices $$ A=\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\\} \quad\hbox{and}\quad B=\pmatrix{0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\\} $$ which indeed verify the hypotheses, but fail the conclusion: $(A+B)^2=2AB\neq0$.

Marc van Leeuwen
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