If $n$ is even, every skew-symmetric $n\times n$ matrix $A$ can be factored as $A=SBS^T$ where $S$ is a invertible matrix and $B$ has the form $B = \left( \begin{array}{ccc} 0 & a_1 & 0 & 0 &0&0\\ -a_1 & 0 &0&0 &0&0\\ 0&0&0&a_{2}&0&0\\ 0 & 0&-a_{2}&0&0&0\\ 0&0&0&0&0&a_{n/2}\\ 0&0&0&0&-a_{n/2}&0 \end{array} \right)$
The $a_{n/2}$ is supposed to represent that $B$ can be any even $n\times n$ size (the size of $A$).
So my book shows this result without proving it, and I was wondering why it's true. It doesn't seem clear to me why this specific factorization holds. How can you prove that it's true? What would $S$ or $S^T$ look like?
Some useful results:
If $E$ is an elementary matrix obtained from $I_n$ by carrying out one elementary row operation on $I_n$, then $EA$ is a matrix obtained by carrying out a single elementary row operation on $A$, and $AE$ is a matrix obtained by carrying out a single elementary column operation on $A$.
$A$ is invertible iff it can be factored as a product of elementary matrices: $A = E_1...E_m$.