Assume $A^T + B^T = A - B$, prove $A$ is symmetric and $B$ is antisymmetric 
Let $A, B$ be square matrices ($n \times n$). Assume $A^T + B^T = A - B$ prove $A$ is symmetric and $B$ is antisymmetric.


Having trouble with this question. Thought about using  $A^T + B^T = (A + B)^T$ but got stuck.
 A: You may rearrange the equation to get $B+B^T=A-A^T$. The LHS is symmetric, while the RHS is anti-symmetric. The only matrix that is both symmetric and anti-symmetric is zero. Hence $B+B^T=0=A-A^T$.
A: $A^T + B^T = (A+B)^T$
Therefore, $(A-B)^T = A+B$
Hence, $A = (A - B)^T - B$ and $B = (A - B)^T - A$
Now, transpose $A$ and $B$ to find what you want.
A: $$A^T+B^T=A-B$$
Transpose the whole equation, using that transposition is linear and involutive:
$$A+B=A^T-B^T$$
Which is the same as
$$A^T-B^T=A+B$$
Now, add this equation to the original one:
$$2 A^T=2A$$
Or substract it from the original one:
$$2B^T=-2B$$
A: All answers are necessarily similar, so the point is just arranging the computations in an appealing way. I like this:
$$
\begin{aligned}
A^\top+B^\top&=A-B & & \\
A+B&=A^\top-B^\top & \text{Transpose}  \\
A^\top-B^\top &=A+B& \text{Rearrange}  \\
2A^\top&=2A & \text{Subtract } (1)-(3) \\
2B^\top&=-2B & \text{Add } (1)+(3) \\
\end{aligned}
$$
A: A good way to look at it is to contruct A and B as sums of $A + B$ an $A - B$.
You can then calculate $A^T$ and $B^T$ easily.
A: We have $a_{ij}+b_{ij}=a_{ji}-b_{ji}$ and also $a_{ji}+b_{ji}=a_{ij}-b_{ij}$.
If $i\neq j$ then these equalities together imply that $b_{ij}+b_{ji}=0$ and $a_{ij}=a_{ji}$.
A: You use that $A = A^T - B^T - B$ and then $A^T = (A^T - B^T - B)^T = A - B - B^T = A$. In similar way you get $B^T = (A-A^T-B^T)^T = A^T - A - B = -B$. 
You also use that transposing twice get you back, ie that $A^{TT} = A$ and $B^{TT} = B$.
