Prove that a matrix equals to its transpose Let $A$ be a $(n\times n)$ matrix that satisfies: $AA^t=A^tA$
Let $B$ be a matrix such that: $B=2AA^t(A^t-A)$
Prove/disprove that: $B^t=B$
I started with:
$$\begin{align}
B &=2AA^t(A^t-A) \\
B^t &=(2AA^t(A^t-A))^t
\\
B^t &=((A^t-A)^t(A^t)^tA^t2)
\\
B^t &=((A-A^t)AA^t2)
\\
B^t &=(AAA^t2-A^tAA^t2) \end{align}
$$
At this point I see no clue how to turn it into this form: $2AA^t(A^t-A)$, I thought it might be not true got no idea how to disprove that either.
Suggestions?
Thanks.

$$
\begin{eqnarray*}
B^T&{}={}&2\left(A-A^T\right)AA^T\newline
&{}={}&2\left(AAA^T-A^TAA^T\right)\newline
&{}={}&2A\left(AA^T-A^TA^T\right)\newline
&{}={}&2AA^T\left(A-A^T\right)\newline
&{}\neq{}& B\,.
\end{eqnarray*}
$$
Almost, but not quite, in general.
