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If $A$ is a symmetric integral matrix with zero diagonal, then I want to prove $2-rank(A)$ (i.e. the dimension of $C_A$ ) is even?

$2-rank(A)$ means dimension A on field $\mathbb{F}_2$.

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    **Hint:** Reduce your matrix modulo 2, and find the determinant (modulo 2, of course). What can you say if the determinant is 1? What can you say if it is zero?2012-12-05

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When considering elements of $\mathbb{F}_2$ a symmetric matrix with zero diagonal values is actually a skew symmetric matrix since $1 \equiv -1 \mod(2)$. It is easier to prove that a (real) skew symmetric matrix of odd dimension has zero determinant.

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    A $ 1 \times 1$ matrix is symmetric, and (in $\mathbb{F}_2$) skew symmetric with determinant of $0$ or $1$, but if it has a zero diagonal, it must be the $1 \times 1$ matrix $0$, and it has determinant of $0$.2012-12-05