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
 A: Seems good. I'd just be more careful with the word "assume" - you're actually proving these statements. I think this is more about grammar than math, but anyway (I'll understand english is not your first language, it's not mine either). 
For example, you don't assume that $S$ and $A$ are subspaces of $M_3(\Bbb R)$, they actually are subspaces (even more fundamentally, they're subsets of $M_3(\Bbb R)$. You have proven that too. I'd have skipped that, being too lazy).
Also, you don't assume that $\dim S + \dim A = 9$. This is true, and you have shown us a proof.
Lastly, just to write less, I'd call $E_{ij}$ the matrix which has $1$ in position $(i,j)$ and zero everywhere else just to say that $B_s = \{E_{ij}+E_{ji} \mid 1  \leq i\leq j \leq 3\}$ and $B_a = \{ E_{ij}-E_{ji} \mid 1 \leq i < j \leq 3\}$.
A: So, you've proven that $\dim M_3(\Bbb R) = \dim S + \dim A$. But you should also show that $\dim(S \cap A) = 0$. You assert this at the bottom of your post, and it's pretty easy to show.

EDIT
Like you said above, $\dim(S + A) = \dim S + \dim A - \dim(S \cap A)$. So you have $\dim(S + A) = 9 - \dim(S \cap A)$. But unless $\dim(S \cap A) = 0$, you haven't shown what you need to.
Nothing forces $S \cap A = \{ 0 \}$, and so this condition must actually be checked. For example, let $T$ be a $3$-dimensional subspace of $S$ (for example, the diagonal matrices). Since $S + T = S$, we certainly don't have $S + T = M_3(\Bbb R)$.
A: 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*}
A: Here is a marginally different viewpoint (really an addendum to Ivo's answer
above):
When dealing with subspaces, it is usually fruitful to look for operators
such that the kernel or range yields those subspaces.
(Also, using the Frobenius norm for matric operators often simplifies calculations.)
Let $L(X) = {1 \over 2} (X-X^T)$, then it is easy to check that $S=\ker L$ and $A={\cal R}L$.
If we use the inner product $\langle A, B \rangle = \operatorname{tr} (A^T B)$
we can check that $L=L^T$ (in the sense that $\langle A, L(B) \rangle
= \langle L(A), B \rangle$) and so ${\cal R}L = {\cal R}L^T$ and hence
$M_{3\times 3}(\Bbb R) = \ker L \oplus (\ker L)^\bot = \ker L \oplus {\cal R}L^T = \ker L \oplus {\cal R}L$.
If we let $E_{ij}$ be the matrix of zeroes except for a $1$ at the $i,j$ position, then we can quickly check that the collection
$\{E_{ij} -E_{ji} \}_{i <j} \cup \{ E_{ij} + E_{ij} \}_{i \le j}$
is a basis for $M_{3\times 3}(\Bbb R) $.
Since $L(E_{ij} -E_{ji}) = E_{ij} -E_{ji}$ and
$L(E_{ij} + E_{ij}) = 0$, we see that $\{ E_{ij} + E_{ij} \}_{i \le j}$ is a basis for $S=\ker L$ and hence $\{E_{ij} -E_{ji} \}_{i <j} $ is a basis for
$A={\cal R}L$.
In particular, $L$ is the orthogonal projection onto $A$.
The operator $M(X) = {1 \over 2} (X+X^T)$ would work equally well, and
result in the orthogonal projection onto $S$ (note that $M=I-L$).
