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.
 A: This is not true in general. Consider for example the family of real skew-symmetric matrices i.e. $A=-A^T$. Clearly skew-symmetric matrices satisfy $AA^T=A^TA$.
The given matrix $B$ for this family is:
$$B=2AA^T(A^T-A)=-2A^2(-2A)=4A^3.$$
While the matrix $B^T$ for this family is:
$$B^T=2(A-A^T)AA^T=2(2A)(-A^2)=-4A^3.$$
A: Compute the transpose of $B$:
$$ B^T = (2AA^T(A^T-A))^T = (A^T-A)^T(2AA^T)^T = (A-A^T)2A^TA = 2(AA^TA - A^TA^T A) = 2(AA^TA - AA^TA^T) = 2AA^T(A-A^T).$$
Then
$$B-B^T = 4AA^T(A^T-A).$$
So the claim is true if and only if $A^T=A$, i.e. $A$ is symmetric.
A: Continuing from where you left off: 
$B^t = AAA^t2 - A^tAA^t2$
$B^t = 2AAA^t - 2A^tAA^t$ ($2$ is a scalar, so it commutes with any matrix)
$B^t = 2AA^tA - 2AA^tA^t$ (Use the fact that $AA^t = A^tA$)
$B^t = 2AA^t(A - A^t)$ (Factor the expression)
$B^t = -2AA^t(A^t - A) = -B$
So we get $B^t = -B$. Thus, the claim $B^t = B$ is true if and only if $B = 2AA^t(A^t-A) = 0$. 
This holds true in some cases, such as if $A$ is symmetric, but it does not hold in general. 
For instance, $A = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$ satisfies $AA^t = A^tA$ but $B = 2AA^t(A^t-A) = \begin{bmatrix}0&4\\-4&0\end{bmatrix}$. 
A: 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: $B^t=B$
HINT: 
$$\begin{align}
B &=2AA^t(A^t-A) \\
B^t &=((2AA^t)(A^t-A))^t
\\
B^t &= (A^t - A)^t(2AA^t)^t
\\
B^t &= 2(A - A^t)A^tA
\\
B^t &= 2(A - A^t)AA^t
\\
B^t + B &= 2((AAA^t - A^tAA^t - AA^tA^t-AA^tA) = 0
\\ \end{align}
$$
A: $$
\begin{eqnarray*}
B^T&{}={}&2\left(A-A^T\right)AA^T\newline
&{}={}&2\left(AAA^T-A^TAA^T\right)\newline
&{}={}&2A\left(AA^T-A^TA^T\right)\newline
&{}={}&2AA^T\left(A-A^T\right)\newline
&{}\neq{}& B\,.
\end{eqnarray*}
$$
Almost, but not quite, in general.
A: Let's write it out fully:
$B=2AA^tA^t-2AA^tA$.
At this point use $AA^t=A^tA$ to get:
$B=2A(A^t)^2-2(A^2)A^t$.
So, $B^t=2A^2A^t-2A(A^t)^2=-B$.
A: Hint. Consider any $2\times2$ rotation matrix $A=\pmatrix{\cos t&-\sin t\\ \sin t&\cos t}$ for an angle $t$ such that $\sin t\ne0$.
