# Symmetric Matrix over a finite field of Characteristic 2

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

• 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. – Jyrki Lahtonen Jun 2 at 20:05

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.