Symmetric and antisymmetric matrices as subspaces of $M_{3 \times 3}(\Bbb R)$ Show that the set $S$ compound of $3 \times 3$ symmetric matrices and the set $A$ of the antisymmetric matrices, are subspaces of $M_{3\times 3}(\Bbb R)$. Determine basis for $S$ and $A$. Show, using a $\dim$ relation, that $S \oplus A = M_{3\times 3}(\Bbb R)$.
My solution:
Set $S$ can be described as any matrix in the follow pattern:
$$
\begin{pmatrix}
a & d & e \\
d & b & f \\
e & f & c
\end{pmatrix}
$$
Set $A$ follows this other pattern:
$$
\begin{pmatrix}
0 & d & e \\
-d & 0 & f \\
-e & -f & 0
\end{pmatrix}
$$
Proof for both sets being subspaces of $M_{3\times 3}(\Bbb R)$:
To summarize this post I'll affirm that set $S$ and $A$ are closed under multiplication, addition, and contains the $ \left\{ 0 \right\} $, which make them subspaces. It's easy to see that.
Vector space $M_{3 \times 3}(\Bbb R)$ follows this pattern:
$$
 \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix} 
$$
To see that the set $S$ is included there, take $b=d, c=g, h=f$.
To see that the set $A$ is included there, take $d=-b, h=-c, f=-h$ and $a=e=i=0$
Hence, because of that and the affirmation made previously, we can assume that $S$ and $A$ are subspaces of $M_{3 \times 3}(\Bbb R)$
Basis:
For $S$:
$$
B_{s}= \left\{
 \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
, \begin{pmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{pmatrix} 
,  \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{pmatrix} 
,  \begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} 
,  \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{pmatrix} ,
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix} 
\right\} \\
\dim(S) = 6
$$
For $A$:
$$
B_{a}= \left\{
 \begin{pmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
, \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
-1 & 0 & 0
\end{pmatrix} 
,  \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{pmatrix} 
\right\}\\
\dim(A) = 3
$$
Proof for $S \oplus A = M_{3x3}(R)$ using a dimension relation:
Using that $\dim$ relation:
$$
\dim(a+b) = \dim(a) + \dim(b) - \dim(a \cap b)
$$ We can assume that:
$$
\dim(S) + \dim(A) = 9
$$ Since $\dim(M_{3 \times 3}(\Bbb R)) = 9$ we get:
$$
\dim(M_{3 \times 3}(\Bbb R)) = \dim(S) + \dim(A)
$$ Hence:
$$
S \cap A =  \left\{ 0 \right\}\\
S \oplus A = M_{3 \times 3}(\Bbb R)
$$
Is my approach to the answer correct? Do I need to be more rigorous? 
Thanks

Any $n \times n$ matrix $B$ can be written uniquely as a sum of a symmetric matrix and a skew symmetric matrix: 
\begin{align*}
B = \frac{1}{2}(B+B^t) + \frac{1}{2}(B-B^t)
\end{align*} 
where $B^t$ denotes the transpose of $B$. Thus $M_{n\times n}(\mathbb{R}) = S + A$. If $B \in S \cap A$, then $B^t = B$ and $B^t = -B$ and hence $B$ is the zero matrix. It follows that $S \oplus A = M_{n \times n}(\Bbb R)$.
We haven't used dimension argument. However, it is not difficult to see that
\begin{align*}
\dim(S) = \frac{n(n+1)}{2}, \qquad \dim(A) = \frac{n(n-1)}{2}
\end{align*}
