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If $a, b$ are two numbers such that $(a+b)^2 = a^2 + b^2$, then $a.b = 0$.

Is there a similar statement for square matrices.

"If $A, B$ are square matrices such that $(A+B)^2 = A^2 + B^2$, then $A.B = 0$."

Note that if $(A+B)^2 = A^2 + B^2$, then $AB = -BA$, hence $tr(AB) = 0$

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closed as off topic by Todd Trimble, Andreas Blass, Noah Snyder, Denis Serre, Ryan Budney Sep 19 '11 at 0:25

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If looks like you found the statement -- it's $AB+BA=0$. –  Ryan Budney Sep 18 '11 at 17:33
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Unfortunately, this is not a good question for MO. It would better belong at math.stackexchange.com. I have voted to close. –  Todd Trimble Sep 18 '11 at 18:57
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1 Answer 1

up vote 7 down vote accepted

If $$ A=\begin{pmatrix} 0&M\\ M&0\end{pmatrix},\qquad B =\begin{pmatrix}-I&0\\ 0&I\end{pmatrix} $$ then $AB+BA=0$, so $A^2+B^2=(A+B)^2$ and neither $A$ nor $B$ is zero.

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Thanks Chris Godsil. I forgot Quaternion group en.wikipedia.org/wiki/Quaternion_group $ij = -ji$ –  Pham Hung Quy Sep 18 '11 at 17:32
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