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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: $A = E_1...E_m$.

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    $\begingroup$ Hint: The eigenvalues of $A$ are all imaginary, and come in conjugate pairs. $\endgroup$ Nov 25, 2014 at 7:12
  • $\begingroup$ Is $S$ supposed to be orthogonal? $\endgroup$ Nov 25, 2014 at 7:25
  • $\begingroup$ @darijgrinberg it says $S$ is invertible, but does not specify it has to be orthogonal. $\endgroup$
    – Jacob
    Nov 25, 2014 at 7:28
  • $\begingroup$ I cannot really comment on the general case, but Travis's comment seems to refer to the case when $A$ has real entries -- I suspect that the book you are talking about is tacitly assuming so. See also math.stackexchange.com/questions/370272/… . $\endgroup$ Nov 25, 2014 at 7:42
  • $\begingroup$ Oh, I see. Nice problem, and holds over every field of characteristic $\neq 2$. In basis-free terms, it asks you to show that, given a skew-symmetric form $B$ on an even-dimensional vector space $V$, you can find a basis $\left(e_1,f_1,e_2,f_2,...,e_n,f_n\right)$ of $V$ such that $B\left(e_i,f_j\right) = \delta_{i,j}$, $B\left(f_i,e_j\right) = -\delta_{i,j}$, $B\left(e_i,e_j\right) = 0$, and $B\left(f_i,f_j\right) = 0$ for all $i$ and $j$. (Here the form is $\left(x,y\right) \mapsto x^TAy$, and the basis will make the columns of $\left(S^T\right)^{-1}$.) This can be done by induction. $\endgroup$ Nov 25, 2014 at 7:54

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