The set of $n \times n$ symmetric matrices is a vector space with dimension $\frac{n(n + 1)}{2}$

Question

I am told that the set of $n \times n$ symmetric matrices,

$$S^{n} = \{ X \in \mathbb{R}^{n \times n} \vert X = X^T \},$$

is a vector space with dimension $\dfrac{n(n + 1)}{2}$.

Naively, I would initially think that it should be a vector space with dimension $n^2$. However, since the matrix is symmetric, I then wonder if there would be "duplicates", for lack of a better term, which would give us something like $\dfrac{n^2}{2}$. But I'm not sure how one gets $\dfrac{n(n + 1)}{2}$. Could someone please explain this? Thank you.

Answer 1

Denote by $E_{ij}$ the $n \times n$ matrix with a $1$ in position $(i, j)$ and zeroes elsewhere. Then the set of matrices $$\{ E_{ij} + E_{ji} : i, j \in \{1, \ldots, n\}, i < j \} \cup \{ E_{ii} : i \in \{1, \ldots, n \} \}$$ is a basis of the space of symmetric matrices. Its cardinality is $$\binom{n}{2} + n = \frac{n(n-1)}{2} + n = \frac{n(n+1)}{2} \,.$$

Answer 2

Each of the $\binom{n}{2}$ choices of matrix entries above the leading diagonal provides an obvious basis element for this vector space, as does each of the leading diagonal's $n$ entries. But$$\binom{n}{2}+n=\frac{n(n-1)}{2}+n=\frac{n(n+1)}{2}.$$Equivalently, the number of degrees of freedom is reduced by the symmetry constraints:$$n^2-\binom{n}{2}=\frac{n(n+1)}{2}.$$