Is $BA$ a symmetric matrix?

Question

$A,B$ are both n-by-n matrices, and $A^TB=B^TA$. I know how to prove that $B^TA$ is symmetric, but I don't know how to prove that $BA$ is symmetric (or not).

Answer 1

Counterexample: Let $$ A=\begin{bmatrix} 6 & 0 \\ 0 & 3\end{bmatrix},\;B=\begin{bmatrix}0 & 1 \\ 2 & 0\end{bmatrix}, $$ then $$ A^T B = \begin{bmatrix} 6 & 0 \\ 0 & 3\end{bmatrix}\begin{bmatrix}0 & 1 \\ 2 & 0\end{bmatrix}=\begin{bmatrix}0 & 6 \\ 6 & 0\end{bmatrix}=(A^T B)^T = B^T A, $$ but $$ BA=\begin{bmatrix}0 & 1 \\ 2 & 0\end{bmatrix}\begin{bmatrix} 6 & 0 \\ 0 & 3\end{bmatrix}=\begin{bmatrix}0& 3\\ 12 & 0\end{bmatrix}\ne (BA)^T. $$