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Let $M$ be a $n$ by $n$ symmetric matrix over a finite field of Characteristic 2. Suppose that the entries in the diagonal of $M$ are all zero, and $n$ is an odd number. I found that the rank of $M$ is at most $n-1$. Is my observation true? How do we prove it? Thanks

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    $\begingroup$ I believe this has something to do with symplectic bilinear forms in characteristic two having an even rank. See for example Keith Conrad's notes. $\endgroup$ – Jyrki Lahtonen Jun 2 at 20:05
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Yes, this is true. In general, over any field, if $M$ is a skew-symmetric matrix with a zero diagonal (i.e. if it represents an alternating bilinear form), the rank of $M$ must be even.

Suppose $M\ne0$. By a simultaneous permutation of the rows and columns of $M$, we may assume that $c_1:=m_{21}=m_{12}\ne0$. So, we may write $$ M=\pmatrix{R&-Y^T\\ Y&Z},\ \text{ where }\ R=c_1\pmatrix{0&-1\\ 1&0} $$ and $Z$ is a symmetric matrix with a zero diagonal. Thus $M$ is congruent to $R\oplus S$, where $S=Z+YR^{-1}Y^T$ is the Schur complement of $R$ in $M$. Since $Z$ and $R^{-1}$ represent alternating bilinear forms, so must $S$. Therefore, we may proceed recursively and $M$ will eventually be congruent to a matrix of the form $c_1R\oplus c_2R\oplus\cdots\oplus c_kR\oplus0$. Hence its rank is even.

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