How to prove $A+ A^T$ symmetric, $A-A^T$ skew-symmetric. Prove that if $A$ is a square matrix, then:
a) $A+ A^T$ is symmetric.
b) $A-A^T$ is skew-symmetric.
c) Use part (a) and (b) to show $A$ can be written as the sum of a symmetric matrix $B$ and a
skew-symmetric matrix $C$, with $A = B + C$.
 A: Hint: use this property: If $M$ and $N$ are square matrices then $(M+N)^T=M^T+N^T$ (can you see why?)
Now add the equations on $a$ and $b$, and do something else to obtain $A$. 
Added: For a matrix to be symmetric it means that if you take the transpose you get the same back. You start with $A+A^T$. Take transposes, and use some properties. Do you get what you started with?
A: Hint$$(A+A^T)^T   =  A+A^T$$
and
$$(A-A^T)^T  =  -(A-A^T)$$
A: Think about it, for example, let $A$ be a $2\times 2$ square matrix, then
$A=\begin{pmatrix}
a&b\\c&d
\end{pmatrix}$, therefore $A^T=\begin{pmatrix}a&c\\b&d\end{pmatrix}$
$\therefore A+A^T=\begin{pmatrix}2a&b+c\\b+c&2d\end{pmatrix}$ and $A-A^T=\begin{pmatrix}&b-c\\c-b&\end{pmatrix}$
You can apply the same rule to any $n\times n$ square matrices.
A: $(a) \ $  Let $B = A + A^T$.  $ \ \ B_{ij} = A_{ij} + A_{ji} = B_{ij}^T$.

$(b) \ $  Let $B = A - A^T$.  $ \ \ B_{ij} = A_{ij} - A_{ji} = -A_{ji} - (-A_{ij}) = - (A_{ji} - A_{ij}) = -B_{ij}^T$
A: Assume $(A^T)_{ij}=A_{ji}$.
Then the $ij-th$ entries of $A+A^T$ are $ij+ji$ of $A$, and the $ji-th$ entries of $A+A^T$ are $ji+ij$ of $A$. 
By commutative prop of $\mathbb{R}$, $ij+ji=ji+ij$. So, since the $ij-th$ entries and $ji-th$ entries of $A+A^T$ are equal, $A+A^T$ is symmetric. 
