TITLE: Show that $A$ is symmetric, with $A \in M_n(\mathbb R)$
QUESTION [10 upvotes]: Let $A \in M_n(\mathbb R)$. 
Show that if $A(\,{}^t\!A)A$  is symmetric, then $A$ is also symmetric.

My attempt:
If $A \in Gl_n(\mathbb{R})$,
We have : ${}^t(A^{-1})=(\,{}^t\!A)^{-1}$
Then :
$${}^t\!AA\,{}^t\!A= A{}^tAA
A\,{}^t\!A=(\,{}^\!tA)^{-1}A\,{}^t\!AA$$
This does not allow me to continue ...
I started with $A$ invertible for use after density. However, it may be a bad idea, what do you think? 
Thank you
Edit : ${}^tA$ is the transpose of A.

REPLY [2 votes]: By an orthogonal change of basis, we may assume that the singular value decomposition of $A$ is $US$, where $U$ is real orthogonal and $S$ is nonnegative and diagonal. Then the given condition implies that $US^3$ is symmetric. Hence $(US^3)(S^3U^T) = (S^3U^T)(US^3)$ (the credit of considering $(AA^TA)(A^TAA^T)=(A^TAA^T)(AA^TA)$ here goes to Hu Zhengtang), or $US^6U^T = S^6$. Since positive semidefinite matrices have unique positive semidefinite $n$-th roots, it follows that $USU^T = S$. Therefore both $U$ and $U^T$ commute with $S$, and in turn also with $S^+$, the Moore-Penrose pseudoinverse of $S$. Consequently,
$$
A=US=US^3(S^+)^2=(US^3)^T(S^+)^2=S^3U^T(S^+)^2=S^3(S^+)^2U^T=SU^T=A^T.
$$