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Consider matrices with entries in a field $F$ of characteristic $2$. Let $\Omega$ denote the $2n\times2n$ matrix $\left[\begin{array}{ll}0&1_n\\1_n&0\end{array}\right]$. Then $X^t\Omega X$ is symmetric with $0$ diagonal, for each $2n\times2n$-matrix $X$.

Question: can we express each symmetric matrix with zero diagonal in the form $X^t\Omega X$, for some $X$?

Note: This is an simpler version of a question I asked yesterday.

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In fact, the symmetric matrix with zero diagonal over $F$ with $\mathop{\rm char} F=2$ is skew symmetric. It is a standard fact that every skew symmetric (bilinear) form in some basis has matrix $\Omega$ surrounded by zeroes. Each such matrix can be easily obtained from $\Omega$ by an appropriate $X$.

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  • $\begingroup$ I think I understand. More precisely: for $1\leq m\leq n$ the $2n\times2n$-block matrix $\left[\begin{array}{lll}0&1_m&0\\1_m&0&0\\0&0&0\end{array}\right]$ is $X^t\Omega X$ where $X$ is a simple $0,1$-matrix. $\endgroup$ – John Murray Sep 12 '14 at 15:31
  • $\begingroup$ Thanks very much, although I prefer the terminology 'symplectic form' (a symmetric bilinear form which is zero on the diagonal) to `skew symmetric' in characteristic $2$. $\endgroup$ – John Murray Sep 12 '14 at 15:42
  • $\begingroup$ I just removed the requirement that $F$ is quadratically closed, which is redundant. $\endgroup$ – John Murray Sep 12 '14 at 19:19

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