Basis for symmetric and skew symmetric $n\times n$ matrices 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?
 A: 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 $n(n-1)/2+n = n(n+1)/2$.
A: For symmetric matrices entries above the diagonal are same as entries below the diagonal. So you just need to find how many entries are there above the diagonal including diagonal entries. For that subtract the number of entries below the diagonal from $n^2$(total number of entries) to get your answer.
Number of entries below the first diagonal entry = $n-1$
Number of entries below the second diagonal entry = $n-2$
.
.
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Number of entries below the $n^{th}$(last) diagonal entry = $0$
Thus, number of entries below diagonal= $(n-1)+(n-2)+\cdots+1+0=\frac{n(n-1)}{2}$
Now, $n^2-\frac{n(n-1)}{2}=\frac{n(n+1)}{2}$
which is the required answer.
For Skew- symmetric case diagonal entries are all zero and entries below diagonal are just the negative of entries above diagonal. So only entries below the diagonal are enough to get the whole matrices, which is determined by $\frac{n(n-1)}{2}$(calculated above ).
