TITLE: Dimensions of symmetric and skew-symmetric matrices
QUESTION [35 upvotes]: Let $\textbf A$ denote the space of symmetric $(n\times n)$ matrices over the field $\mathbb K$, and $\textbf B$ the space of skew-symmetric $(n\times n)$ matrices over the field $\mathbb K$. Then $\dim (\textbf A)=n(n+1)/2$ and $\dim (\textbf B)=n(n-1)/2$.

Short question: is there any short explanation (maybe with combinatorics) why this statement is true?
EDIT: $\dim$ refers to linear spaces.

REPLY [3 votes]: There have been many very good answers, so this is going to be another perspective of counting. We tackle the case of skew-symmetric first, which yields the condition that for any matrix $A$ with real entries, $a_{ij} = -a_{ij}$, so this means one side of the matrix is completely determined by the other as shown.
$$\begin{matrix}
\begin{pmatrix}
0 &*' \\
*& 0 \\
\end{pmatrix}  &  *
\end{matrix}$$
$$\begin{matrix}
\begin{pmatrix}
0 &*' & *' \\
*& 0 &  *' \\
* & * & 0 \\
\end{pmatrix} & \begin{matrix}
*&   \\ 
* & *
\end{matrix}
\end{matrix}$$
$$\begin{matrix}
\begin{pmatrix}
0 &*' & *' &*'\\
*& 0 &  *' &*'\\
* & * & 0  &*'\\
* & * &* & 0 
\end{pmatrix} &  \begin{matrix}
* &  & \\ 
 *&*  & \\ 
* & * &* 
\end{matrix}
\end{matrix}$$
$$\begin{matrix}
\begin{pmatrix}
0&*'&*'&*'&\cdots \\
*&0&*'&*'& \\
*&*&0&*'& \\
*&*&*&0& \\
\vdots&&&&\ddots
\end{pmatrix} & \begin{matrix}
*& \\
*&*& \\
*&*&*& \\
\vdots&&&&\ddots
\end{matrix}
\end{matrix}$$
Notice that the triangle keeps getting larger and larger and at each stage and is incremented by $+1$. Since we have $n-1$ entries to consider we add up the number of entries, that is $$1+ 2 + \dots + (n-1).$$
So by Gauss, we have $$\frac{(n-1)(n-1+1)}{2} = \frac{(n-1)n}{2}$$ degree of freedom and that is the dimension we seek. 
For the symmetric case we need to add back the $n$ objects in diagonal zeroed out, so $$\frac{(n-1)n}{2} + n = \frac{n^2 + n}{2}$$