Prove that the direct sum of a symmetric and skew symmetric matrix belongs to $M_n(K)$ using $A_{ij}$ and $A_{ji}$ notation. Basically 
Let $M_n(K)$ be an $n\times n$ matrix of a $K$ vector space.
$U =\{A\in M_n(K)\;|\;A_{ij}=A_{ji}\}$
$W =\{A\in M_n(K)\;|\;A_{ij}=−A_{ji}\}$
So I don't understand my mark scheme. It says 
let $A\in U$ and $B\in W$ and $X\in M_n(K)$
So we want to show that $X = A + B$. 
So $X_{ij} = A_{ij} + B_{ij}$ and likewise
$X_{ji} = A_{ji}+ B_{ji} = A_{ji} - B_{ij}$
I don't understand what comes after. How do they find that $A_{ij} = 0.5(X_{ij} + X_{ji})$ and  $B_{ij} = 0.5(X_{ij} - X_{ji})$ ? And adding them together, how does that show that it is the direct sum? Sorry for such a simple question but i feel like that brings me back to square one again. I know the mark scheme is right i just don't get how that shows that it is a direct sum. I know how to prove that their intersection is 0 but this part i don't. Thanks!
 A: Perhaps it is easier to understand this without the $ij$ notation. 
To show that $M_n(K)$ is the direct sum of $U$ and $W$, we need to show that we can write any matrix $X \in M_n(K)$ as the sum of a matrix in $U$ plus a matrix in $W$ (this is the definition of a direct sum).
Note that for any $n \times n$ matrix $X$, we have $X = \dfrac{1}{2}(X+X^T) + \dfrac{1}{2}(X-X^T)$. 
You can check that $\dfrac{1}{2}(X+X^T)$ is a symmetric matrix, so $\dfrac{1}{2}(X+X^T) \in U$.
Also, you can check that $\dfrac{1}{2}(X-X^T)$ is a skew-symmetric matrix, so $\dfrac{1}{2}(X-X^T) \in W$.
This shows that we can write any $X \in M_n(K)$ as the sum of a matrix in $U$ plus a matrix in $W$, which is precisely what it means for $M_n(K)$ to be the direct sum of $U$ and $W$. 
The proof you were reading is doing the same thing, except instead of psychicly knowing to write $X$ as $\dfrac{1}{2}(X+X^T) + \dfrac{1}{2}(X-X^T)$, they wrote $X$ as $A+B$ for some $A \in U$ and some $B \in W$, and then they figured out what $A$ and $B$ needed to be to make the proof work out. 
