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Is it true that for any square matrix of real numbers $A$, there exists a square matrix $B$, such that $AB$ is a symmetric matrix? This is obviously true if $A$ is invertible, but how about if $A$ is not invertible?

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    $\begingroup$ Yes, it is true. Do you know about transposes of matrices? $\endgroup$ – hardmath Jan 1 '13 at 15:58
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    $\begingroup$ @hardmath oh, you are right, I forgot about transposes. I think A times A transpose will always be symmetric. $\endgroup$ – Sunny88 Jan 1 '13 at 16:02
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    $\begingroup$ @Sunny88 Please write that as an answer and accept it. The question will be removed from the unanswered list. $\endgroup$ – WimC Jan 1 '13 at 16:05
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Yes, $AB$ will be symmetric if we let $B=A^{T}$.

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    $\begingroup$ Similarly, $AB$ will be symmetric if $B=SA^T$ for some symmetric matrix $S$. $\endgroup$ – user1551 Jan 1 '13 at 16:34
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You can simply let $B=0$. Then $AB=0$ is symmetric.

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