Show that $A$ is symmetric, with $A \in M_n(\mathbb R)$

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.

Solution 1:

Hint: Let me denote the transpose of $A$ by $A^t$. Since $AA^tA$ is symmetric,

$$AA^tA=A^tAA^t\Longrightarrow (AA^t)^3=(A^tA)^3 \Longrightarrow AA^t=A^tA.$$

That is to say, $A$ is normal. Then use the fact that $A$ unitarily similar to a diagonal matrix to show $A$ is symmetric.

Update: Let me extend the statement to the complex valued case and add some details to the proof. To begin with, let me denote the Hermitian conjugate of $A\in M_n(\Bbb C)$ by $A^*$.

Claim: Given $A\in M_n(\Bbb C)$, if $AA^*A$ is Hermitian, then $A$ is also Hermitian.

Proof: Denote $B:=AA^*A$. As shown in the hint, since $B=B^*$, $$(AA^*)^3=BB^*=B^*B=(A^*A)^3.$$ Since $AA^*$ and $A^*A$ are Hermitian, the equality above implies that $AA^*=A^*A$. This is due to a general fact: if $H$ is Hermitian and $f:\Bbb R\to\Bbb R$ is a function, then there is a natural way to uniquely define a Hermitian matrix $f(H)$. In our situation, $H=(AA^*)^3=(A^*A)^3$, $f(x)=x^{\frac{1}{3}}$.

We have seen that $A$ is a normal matrix, so $A=U\Lambda U^*$ for some unitary $U$ matrix and some diagonal matrix $\Lambda$. Then $$\Lambda\Lambda^*\Lambda=U^*BU=U^*B^*U=\Lambda^*\Lambda\Lambda^*.$$ Since $\Lambda$ is diagonal, the equality above implies that it is real valued, so $A$ is Hermitian. $\quad\square$

Solution 2:

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, $$ A=US=US^3(S^+)^2=(US^3)^T(S^+)^2=S^3U^T(S^+)^2=S^3(S^+)^2U^T=SU^T=A^T. $$