Is it possible to form a $\mathbb{R}^{n \times n}$ basis only with symmetric matrices? It is a simple question but I don’t know if it is possible.
Can only symmetric matrices form a $\mathbb{R}^{n \times n}$ basis?
 A: No. The set $V$ of symmetric matrices forms a proper subspace of $\mathbb{R}^{n \times n}$, so the span of any set of symmetric matrices must be a subspace of $V$.
A: No, since the set of symmetric matrices has dimension $\frac{n(n+1)}{2}<n^2$ whatever set of $\big(\frac{n(n+1)}{2}+1\big)$ you take will be linearly dependent and by definition won't be a basis for $\mathbb{R}^{n\times n}$
A: If $n=1$, then yes. Otherwise, no.
Any linear combination $A$ of symmetric matrices $A_k=A^T_k$ is itself symmetric
$$
A^T = \left(\sum_k a_k A_k\right)^T = \sum_k a_k A_k^T = \sum_k a_k A_k = A
$$
but if $n>1$ then $\mathbb{R}^{n\times n}$ includes matrices which are not symmetric. Therefore, there is no basis of $\mathbb{R}^{n\times n}$ which consists of symmetric matrices only.
A: A symmetric matrix is determined by the entries on and above  (or below) the diagonal.   There are $1+2+\dots+n=n (n+1)/2$ such entries.   It is easy to see that they form a subspace  whose dimension is $n (n+1)/2$.  Thus the answer in general is no.
