What is the dimension of the solution vector space for $M=-M^t$ ($M\in M_{n\times n}^{\mathbb{C}})$?

Question

What is the dimension of the solution vector space for $M=-M^t$ ($M\in M_{n\times n}^{\mathbb{C}})$?

I saw someone say that it's equal to $\frac{n^2-n}{2}$ but I don't know why.

I'm not sure how to approach this problem, any tips?

Answer 1

Note that if $M=(a_{ij})_{1\leqslant i,j\leqslant n}$, then each $a_{ii}$ is $0$ and, if $i\neq j$, then $a_{ji}=-a_{ij}$. And there are no more restrictions. So, a basis of your space are those matrices such, for one $i$ and one $j$ in $\{1,2,\ldots,n\}$, with $i<j$, $a_{ij}=1$, $a_{ji}=-1$ and all other entries are $0$. So, the dimension is$$\#\bigl\{(i,j)\in\{1,2,\ldots,n\}\,|\,i<j\bigr\}=\binom n2=\frac{n^2-n}2.$$