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I was wondering if the product of two nonsymmetric matrices can ever be a symmetric matrix. Honestly I would not know how to tackle this problem.

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    $\begingroup$ For any matrix $A$, $A^TA$ is symmetric (and positive semi-definite) $\endgroup$ – eepperly16 Dec 21 '16 at 22:31
  • $\begingroup$ Shortest answer(?): LU-decomposition of a symmetric matrix A $\endgroup$ – Gottfried Helms Jan 30 '17 at 11:33
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Try something simple first:

$$\begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&1\end{bmatrix}\;.$$

More generally, if $A$ is any square real matrix, $AA^T$ is symmetric: the $(i,j)$-entry is the dot product of the $i$-th row of $A$ and the $j$-th column of $A^T$, and the $j$-th column of $A^T$ is the $j$-th row of $A$, so the $(i,j)$-th entry of $AA^T$ is the dot product of the $i$-th and $j$-th rows of $A$. The $(j,i)$-th entry of $AA^T$ is then the dot product of the $j$-th and $i$-th rows of $A$, which is of course the same.

This is not the only kind of example, however:

$$\begin{bmatrix}0&0&0\\1&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$

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    $\begingroup$ Good observations. Anyway I should have picked a different solution than $A$ and its transpose, just not to end up in this generalization, if that solution even exists, I should think a little bit. $\endgroup$ – nbro Dec 21 '16 at 22:38
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    $\begingroup$ @nbro: Thanks; you just reminded me to add an example that doesn’t arise in that way. $\endgroup$ – Brian M. Scott Dec 21 '16 at 22:42
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As an alternative counterexample that isn't $AA^T$ or $A^TA$: Let $A$ be any non-symmetric, invertible matrix. Then the inverse matrix $A^{-1}$ will not be symmetric either (*). However, $A \cdot A^{-1} = A^{-1} \cdot A = I$, the identity matrix, which is of course symmetric.

(*) Proof: Suppose $A^{-1} = (A^{-1})^T$. Then $$A^{-1} = (A^{-1})^T = (A^T)^{-1},$$ hence $A = A^T$. But we assumed that $A$ was non-symmetric.

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Here's a very simple example.

Take for example the following matrix $A = \begin{pmatrix}1 & 3 \\ 2 & 2 \end{pmatrix}$ and its transpose $A' = \begin{pmatrix}1 & 2 \\ 3 & 2 \end{pmatrix}$, the product of $$A*A' = \begin{pmatrix}10 & 8 \\ 8 & 8 \end{pmatrix}$$ which is symmetric. So this holds quite trivially.

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    $\begingroup$ You don't need to have a special $A$. As long as $A$ is non symmetric, $AA^T$, you have a counterexample. $\endgroup$ – Jean Marie Dec 21 '16 at 22:32
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Note that if $A$ is invertible, then $(A^{-1})^T=(A^T)^{-1}$ so $A$ is symmetric iff $A^{-1}$ is symmetric. So if $A$ is any nonsymmetric invertible matrix, $A$ and $A^{-1}$ are both nonsymmetric but their product $AA^{-1}=I$ is symmetric.

(To prove that $(A^{-1})^T=(A^T)^{-1}$, just note that $A^T(A^{-1})^T=(A^{-1}A)^T=I^T=I$.)

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There are also infinitely many solutions possible without the given examples of pairs of transposed or inverse matrices. You can assume an arbitrary symmetric matrix $A$ , use a rotation on the columns, by a rotation matrix $R$ and get $B = A \cdot R^{-1}$. Then $B$ is (at least very likely) not symmetric, not a transpose or inverse of $R$ and still $A = B \cdot R$

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