When do symmetric matrices commute? Let $S$ be subspace of the vector space of $10*10$ real matrices so that (1) every matrix in $S$ is symmetric and (2) $S$ is closed under matrix multiplication. what is the maximal possible dimension of $S$.
Here is my work.
Say $A\ \&\ B$ be elements of $S$. Then $A^\intercal = A \ \&\ B^\intercal = B$. Hence $(AB)^\intercal = B^\intercal A^\intercal= BA $ but if $ AB \in S$ then $(AB)^\intercal = AB$ hence $AB=BA$. If $AB=BA$ and $A^\intercal = A \ \&\ B^\intercal = B$, then $AB \in S$.
Hence we need to find dimension of symmetric abelian group of 10*10 real matrices. Any help now.
 A: Hint: If $A$ is a diagonal matrix, 
$$(AB - BA)_{ij} = (a_{ii} - a_{jj}) b_{ij} $$
Further hint: Every real symmetric matrix can be diagonalized.
A: EDIT. The item (1) and (2) imply that the matrices of $S$ pairwise commute. Then we assume that the matrices of $S$ are real symmetric and pairwise commute.
Clearly, the OP did not work more than one minute on his question. Moreover, the first version of the question was so poorly written that the first readers did not understand what it was. The last but not tje least,, the OP did not even  have the effort to study the case $n=2$ as proposed by Bubaya. He would have seen that we can reduce ourselves to diagonal matrices; this fact can be generalized for every $n$. More generally, one has
$\textbf{Proposition}$. Let $K$ be a field and $S\subset M_n(K)$ be a subvector space st any matrix of $S$ is diagonalizable over $K$ and any two matrices of $S$ commute.
Then the maximum of $dim(S)$ is $n$.
$\textbf{Proof}$. Any basis of $S$ is constituted by matrices that are simultaneously diagonalizable over $K$. cf, for example with google
"Simultaneous commutativity of operators" , Keith Conrad.
and we are done.  $\square$
