My problem: Find the biggest linearly independent set of the symmetric matrices $S_n(\Bbb F)$ and prove that each of its superset must be linearly dependent.
My solution: I think that this biggest linearly independent set would be the basis of $S_n(\Bbb F)$. I will consider just matrices $3 \times 3$, because it is easier to write it down here.
The set: $$ \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix},\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix} \right\} $$
With the elements in that set above we can construct any symmetric matrix, like the identical matrix, etc.
For the check of the linear independece I will use an isomorphism.
$$ a_1\left(\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{matrix} \right) + a_2\left(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{matrix} \right) +a_3\left(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ \end{matrix} \right) + a_4\left(\begin{matrix} 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{matrix} \right)+a_5\left(\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ \end{matrix} \right) + a_6\left(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ \end{matrix} \right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{matrix} \right) $$
$a_1=a_2=...a_6 =0$
Then we proved that our set is linearly independent.
Dimension of the symmetric matrices is equal to: $\frac{n(n+1)}2$, in our case it equals to 6 and that is the number of elements in our set.So it spans whole $S_3(\Bbb F)$, then it's a basis. Basis is by the definition the smallest set that can generate given vector space, or in other words, the biggest linearly independent set. If we put another vector to our set, we can write as the linear combination of the others, so the set wouldn't be linearly independent. We would have a nontrivial linear combination which would give us zero.