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.
Best Answer
All square matrices of a given size $n$ constitute a linear space of dimension $n^2$, because to every matrix element corresponds a member of the canonical base, i.e. the set of matrices having a single $1$ and all other elements $0$.
The skew-symmetric matrices have arbitrary elements on one side with respect to the diagonal, and those elements determine the other triangle of the matrix. So they are in number of $(n^2-n)/2=n(n-1)/2$, ($-n$ to remove the diagonal).
For the symmetric matrices the reasoning is the same, but we have to add back the elements on the diagonal: $(n^2-n)/2+n=(n^2+n)/2=n(n+1)/2$.