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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? –  andybenji Dec 5 '12 at 6:57
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up vote 1 down vote accepted

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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I suppose that symmetric is the same as skew symmetric in $\mathbb{F}_2$ really, but for this problem focus on the skew symmetric attribute. –  adam W Dec 5 '12 at 21:44
    
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 $0$. –  adam W Dec 5 '12 at 22:00
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