Is $(B - A)(A + B)$ symmetric if $A = A^T$ and $B = B^T$? I have a problem where I have to say if a matrix is symmetric or not, if $A = A^T$ and $B = B^T$.
According to what I know, a matrix $A$ is symmetric if $A = A^T$.

The specific matrix I am a bit confused about is:
$$(B - A)(A + B)$$
One of the hints of the problem is to use the following properties of matrix transposition:
$$(A + B)^T = A^T + B^T \\
(AB)^T = B^TA^T$$

So this is my attempt:
I have to show that: $$(B - A)(A + B) = ((B - A)(A + B))^T$$
I know that: 
$$((B - A)(A + B))^T = (A + B)^T(B - A)^T \\
(A + B)^T(B - A)^T = (A^T + B^T)(B^T - A^T) \\
(A^T + B^T)(B^T - A^T) = (A + B)(B - A) 
$$
In what cases $(A + B)(B - A) = (B - A)(A + B)$ ? It would be like $AB = BA$? Well, if either $A$ or $B$ are the $0$ matrix or $A$ is the inverse of $B$, right? Are there other situations?
So what can I conclude? 
 A: Your idea is correct, i.e. $(B-A)(A+B)$ is symmetric if $AB=BA$, and it's easy to continue your proof from what you've done already. Just expand your terms, simplify, and you will end up with $AB=BA$. 
There are, however, counterexamples for the converse. For instance, take
\begin{align*}
A=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix},\qquad B=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}.
\end{align*}
Then 
\begin{align*}
(B-A)(A+b)=\begin{pmatrix}-1 & -1 \\ 1 & 0\end{pmatrix},
\end{align*}
which is not symmetric. Note that in this case $AB\neq BA$.
A: $((B-A)(A+B))^T = (A^T+B^T)(B^T-A^T) = A^TB^T - (A^T)^2 + (B^T)^2 +B^TA^T = /A^T = A, B^T = B/ = AB - A^2 + B^2 - BA$
As you say you get BA - AB instead in the lhs. So you need AB = BA. In other terms A and B must commute. In general there are quite strict rules for which matrices commute.
A: As you say, you need $AB=BA$, that is, $A$ and $B$ must commute.
While you don't explicitly say so, I assume this is about real matrices (that is, matrices whose entries are real numbers).
Since A and B are, by assumption, symmetric, they commute exactly if their eigenvectors are the same.
