Computing maximum dimension of a vector subspace given that it's every element is a symmetric matrix and is closed under matrix multiplication 
Q. Let $S$ be a subspace of the vector space of all $11 \times 11$ real matrices such that (i) every matrix in $S$ is symmetric and (ii) $S$ is closed under matrix multiplication. What is the maximum possible dimension of $S$?

Attempt : Subspace of symmetric matrices has dimension $\frac {n(n+1)}2$ for the vector space of $n \times n$ matrices.
Here in this case, it is $\frac {11\times 12}2=66$.
Next, using (ii) we have that $(AB)^T=AB$. But $(AB)^T=B^T A^T=BA. (\because A^T=A, B^T=B)$. Hence $AB=BA$. So we conclude that elements in $S$ commute with each other.
In case of vector spaces of $2\times 2$ and $3 \times 3$ symmetric matrices, I found that only diagonal matrices commute with each other.
For e.g. in $2 \times 2$ case, I just multiplied the basis elements of the subspace of symmetric matrices $$
        \begin{pmatrix}
        1 & 0  \\
        0 & 0  \\
        \end{pmatrix}
\cdot
        \begin{pmatrix}
        0 & 0  \\
        0 & 1  \\
        \end{pmatrix}
= \begin{pmatrix}
        0 & 0  \\
        0 & 1  \\
        \end{pmatrix} \cdot  \begin{pmatrix}
        1 & 0  \\
        0 & 0  \\
        \end{pmatrix}$$ but $$ \begin{pmatrix}
        1 & 0  \\
        0 & 0  \\
        \end{pmatrix} \cdot  \begin{pmatrix}
        0 & 1  \\
        1 & 0  \\
        \end{pmatrix} \neq \begin{pmatrix}
        0 & 1  \\
        1 & 0  \\
        \end{pmatrix} \cdot \begin{pmatrix}
        1 & 0  \\
        0 & 0  \\
        \end{pmatrix}$$
Similarly checking the commutativity of basis elements  of space of $3 \times 3$ symmetric matrices also gives only the space of diagonal matrices as a candidate having commuting elements.
Like, $$
        \begin{pmatrix}
        0 & 1 & 0 \\
        1 & 0 & 0 \\
        0 & 0 & 0 \\
        \end{pmatrix}
\cdot \begin{pmatrix}
        0 & 0 & 0 \\
        0 & 0 & 1 \\
        0 & 1 & 0 \\
        \end{pmatrix} \neq \begin{pmatrix}
        0 & 0 & 0 \\
        0 & 0 & 1 \\
        0 & 1 & 0 \\
        \end{pmatrix} \cdot \begin{pmatrix}
        0 & 1 & 0 \\
        1 & 0 & 0 \\
        0 & 0 & 0 \\
        \end{pmatrix}$$
Hence the vector space of all diagonal matrices satisfies (i) & (ii). Thus minimum possible 'maximum dimension' of $S$ is $11$. So we have that $11 \le \max \dim S \lt 66$.
Am I in right direction so far? What can be my next step?

If $A$ and $B$ are symmetric, then $AB$ symmetric means $AB=(AB)^T=B^TA^T=BA$. All the matrices therefore commute. A symmetric
real matrix is diagonalisable, and pairwise commuting real matrices
are simultaneously diagonalisable. So the dimension of such a space is
at most $11$.
