On the rank of antisymmetric matrices

Question

Let $\Bbb F$ be a field, and $\mbox{char}(F) \neq 2$. Let $A$ be an antisymmetric matrix ($A^{T} = -A$). Show that $\mbox{rank}(A)$ is even.

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I think the theorem bellow might help, but I don't know how to use it:

$T$ is a translation and $W$ is a vector space, $T: W \to W$ $$\mbox{rank}(T) + \mbox{null}(T) = \dim(W)$$

Answer 1

Observe that

$$\det A=\det A^T=\det(-A)=(-1)^n\det A$$

and this much is true for any principal minor of the matrix, so if we have an odd numbered minor and $\;\text{char}\,\Bbb F\neq2\;$ , the determinant vanishes.