TITLE: If the product of an invertible symmetric matrix and some other matrix is symmetric, is that other matrix also symmetric?
QUESTION [7 upvotes]: The thought came from the following problem:
Let $V$ be a Euclidean space.  Let $T$ be an inner product on $V$.  Let $f$ be a linear transformation $f:V \to V$ such that $T(x,f(y))=T(f(x),y)$ for $x,y\in V$.  Let $v_1,\dots,v_n$ be an orthonormal basis, and let $A=(a_{ij})$ be the matrix of $f$ with respect to this basis.
The goal here is to prove that the $A$ is symmetric.  I can prove this easily enough by saying:
Since $T$ is an inner product, $T(v_i,v_j)=\delta_{ij}$.
\begin{align*}
T(A v_j,v_i)&=T(\sum_{k=1}^n a_{kj} v_k,v_i)\\
&=T(a_{1j} v_1,v_i) + \dots + T(a_{nj} v_n,v_i)\\
&=a_{1j} T(v_1,v_i) + \dots + a_{nj} T(v_n,v_i)\tag{bilinearity}\\
&=a_{ij}\tag{$T(v_i,v_j)=\delta_{ij}$}\\
\end{align*}
By the same logic,
\begin{align*}
T(A v_j,v_i)&=T(v_j,A v_i)\\
&=T(v_j,\sum_{k=1}^n a_{ki} v_k)\\
&=T(v_j,a_{1i} v_1)+\dots+T(v_j,a_{ni} v_n)\\
&=a_{1i} T(v_j,v_1)+\dots+a_{ni} T(v_j,v_n)\\
&= a_{ji}\\
\end{align*}
By hypothesis, $T(A v_j,v_i)=T(v_j,A v_i)$, therefore $a_{ij}=T(A v_j,v_i)=T(v_j,T v_i)=a_{ji}$.
I had this other idea though, that since $T$ is an inner product, its matrix is positive definite.
$T(x,f(y))=T(f(x),y)$ in matrix notation is $x^T T A y=(A x)^T T y$
\begin{align*}
x^T T A y &= (A x)^T T y\\
&=x^T A^T T y\\
TA &= A^T T\\
(TA)^T &= (A^T T)^T\\
A^T T^T &= T^T A\\
TA &= A^T T^T\tag{T is symmetric}\\
&= (TA)^T\tag{transpose of matrix product}\\
\end{align*}
This is where I got stuck.  We know that $T$ and $TA$ are both symmetric matrices.  Clearly $T^{-1}$ is symmetric.  If it can be shown that $T^{-1}$ and $AT$ commute, that would show it.

REPLY [2 votes]: I did some numerical search for higher dimensions:
$n=3:$
$\left(
\begin{array}{ccc}
 1 & 1 & 0 \\
 1 & 1 & 1 \\
 0 & 1 & 1
\end{array}
\right).\left(
\begin{array}{ccc}
 -383 & 13 & -13 \\
 -36 & -445 & -36 \\
 -13 & 13 & -383
\end{array}
\right)=\left(
\begin{array}{ccc}
 -419 & -432 & -49 \\
 -432 & -419 & -432 \\
 -49 & -432 & -419
\end{array}
\right)$
$n=4:$
$\left(
\begin{array}{cccc}
 1 & 1 & 0 & 0 \\
 1 & 1 & 1 & 0 \\
 0 & 1 & 1 & 1 \\
 0 & 0 & 1 & 1
\end{array}
\right).\left(
\begin{array}{cccc}
 -383 & 13 & -36 & -23 \\
 85 & -360 & 49 & -49 \\
 -49 & 49 & -360 & 85 \\
 -23 & -36 & 13 & -383
\end{array}
\right)=\left(
\begin{array}{cccc}
 -298 & -347 & 13 & -72 \\
 -347 & -298 & -347 & 13 \\
 13 & -347 & -298 & -347 \\
 -72 & 13 & -347 & -298
\end{array}
\right)$
$n=5:$
$\left(
\begin{array}{ccccc}
 1 & 1 & 0 & 0 & 0 \\
 1 & 2 & 1 & 0 & 0 \\
 0 & 1 & 3 & 1 & 0 \\
 0 & 0 & 1 & 4 & 1 \\
 0 & 0 & 0 & 1 & 5
\end{array}
\right).\left(
\begin{array}{ccccc}
 -7526 & 3158 & -9379 & 7340 & -8405 \\
 5216 & -3477 & 6079 & -6570 & 6359 \\
 -3225 & 1486 & -3098 & 2500 & -3543 \\
 1159 & -1300 & 905 & -1249 & 970 \\
 -641 & 414 & -841 & 186 & -656
\end{array}
\right)=$
$=\left(
\begin{array}{ccccc}
 -2310 & -319 & -3300 & 770 & -2046 \\
 -319 & -2310 & -319 & -3300 & 770 \\
 -3300 & -319 & -2310 & -319 & -3300 \\
 770 & -3300 & -319 & -2310 & -319 \\
 -2046 & 770 & -3300 & -319 & -2310
\end{array}
\right)$
All matrices have full rank. However, for high dimensions they are quite ugly.