This has already been asked but I can't seem to understand why its $\frac{1}{2}n(n+1)$ elements for the basis of symmetric $n\times n$ matrices and $\frac{1}{2}n(n-1)$ for skew symmetric $n\times n$ matrices. Any mechanical explanation that I can easily see?
[Math] Basis for symmetric and skew symmetric $n\times n$ matrices
linear algebramatrices
Best Answer
Basically a symmetric matrix is determined by its coefficients on the lower triangle (including the diagonal) because then the others are determined by symmetry. So you have $n(n+1)/2$ parameters.
For an anti-symmetric matrix, it's almost the same except that everything on the diagonal must be $0$. So you only have $n(n+1)/2-n=n(n-1)/2$ parameters.
To see these numbers, divide your square in $3$ parts : the strict lower triangle, the strict upper triangle and the diagonal. Obviously, the diagonal has $n$ spots. Now the rest ($n^2-n$ spots) is equally divided in the strict lower and strict upper triangles, so each have $(n^2-n)/2$ spots, ie $n(n-1)/2$. So the large lower triangle is $n(n-1)/2+n = n(n+1)/2$.