Any $n \times n$ matrix $A$ can be written as $A = B + C$ with $B$ is symmetric and $C$ skew-symmetric. How can I proof the following statement?

Any $n \times n$ matrix $A$ can be written as a sum
  $$
  A = B + C
$$
  where $B$ is symmetric and $C$ is skew-symmetric.

I tried to work out the properties of a matrix to be symmetric or skew-symmetric, but I could not prove this.
Does someone know a way to prove it?
Thank you.
PS: The question Prove: Square Matrix Can Be Written As A Sum Of A Symmetric And Skew-Symmetric Matrices may be similiar, in fact gives a hint to a solution, but if someone does not mind in expose another way, our a track to reach to what is mentioned in the question of the aforementioned link.  
 A: Suppose
$$A=B+C$$
If $$B^T=B, $$
$$C^T=-C,$$ then according to the known property of transposition of sum of matrices
$$A^T=(B+C)^T=B^T+C^T=B+(-C)=B-C$$
Now we have $$A=B+C \tag 1\\ $$
$$A^T=B-C\tag 2\\$$
Adding $(1)$ to $(2)$ gives $$B={(A+A^T)\over 2}\\ $$
Subtracting $(2)$ from $(1)$ gives $$C={(A-A^T)\over 2}\\ $$
A: Given a matrix $A$, let $B=\frac{A+A^T}{2}$ and $C=\frac{A-A^T}{2}$.  Observe that $B^T=B$, so $B$ is symmetric. Also, $C^T = -C$, so $C$ is skew-symmetric.
A: Let $A$ be our matrix and
$$\begin{align}B&={A^T+A\over 2}\\C&={A-A^T\over 2}\end{align}$$
We have $B^T=B$ and $C^T=-C$ and $A=B+C$
A: For a matrix $A$, 
$A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$, where the first one is symmetric, and second one is skew-symmetric
A: Let $A$ be any square matrix. We can write
$A=\frac12(A+A^\top)+(A-A^\top)=P+Q$, say,
where $P=\frac12(A+A^\top)$ And $Q=\frac12(A-A^\top)$
we have $$P^\top=(\frac12(A+A^\top))^\top=\frac12(A+A^\top)^\top=
\frac12{A^\top+(A^\top)^\top}=\frac12(A+A^\top)=P.$$
Therefore $P$ is a symmetric matrix.
Now $$Q^\top=(\frac12(A-A^\top))^\top=\frac12(A-A^\top)^\top\\
                   =\frac12(A^\top-(A^\top)^\top)
                   =\frac12(A^\top-A)=-\frac12(A-A^\top)=-Q.$$
Therefore $Q$ is a skew-symmetric matrix
A: "if someone does not mind to expose another way"...
Though nothing is mentioned in the statement about this, I will assume the matrices considered have entries in a field$~F$ that is not of characteristic$~2$ (in other words $1$ and $-1$ are distinct elements in $F$), since the statement is false for matrices with entries in a field of characteristic$~2$.
The set of $n\times n$ matrices with entries in $F$ is (with usual addition and scalar multiplication) a vector space $V$ over$~F$, and the operation of transposition defines a linear map $T:V\to V$. Since $T\circ T=I$, the polynomial $X^2-1=(X-1)(X+1)\in F[X]$ is an annihilating polynomial of $T$. Since this polynomial, which is split as indicated, has distinct roots $1$ and $-1$ in $F$, the operator $T$ is diagonalisable with eigenvalues in $\{1,-1\}$. This means that the sum of the two eigenspaces $V_\lambda$ for $\lambda=1$ and $\lambda=-1$, which is a direct sum, fills the whole space: $V=V_1\oplus V_{-1}$. But $V_1$ is the subspace of matrices $M$ satisfying $T(M)=M$, in other words of symmetric matrices, and similarly $V_{-1}$ is the subspace of anti-symmetric matrices.
The fact that $V=V_1+V_{-1}$ says that every matrix in $V$ can be written as sum of a symmetric and an anti-symmetric matrix (that was the question); moreover the fact that the sum is direct says that they can be so written in a unique manner.
