Dimension of the space of symmetric matrices

Question

So dimension of the space of symmetric matrices would be

$\frac{n(n+1)}{2}$ for sure. And since $A=A^T$ this would imply that $A$ can be formed by $n$ different rank one projection matrices. In first case n is being formed by $\frac{n(n+1)}{2}$ matrices but in the second it is being formed by $n$ matrices.

Since the dimension $\frac{n(n+1)}{2}$ exceeds $n$, how is this difference explained?

Answer 1

Rank of a matrix is the dimension of the column space of the matrix. Which happens to be provably equal to the dimension of the row space of the matrix. It is not the dimension of the matrix space itself. These are different vector spaces.