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

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, $T: W \rightarrow W$ $$rank(T) + null(T) = dim(W)$$


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.

  • $\begingroup$ Why are principal minors enough? $\endgroup$ Apr 8 '16 at 14:18
  • $\begingroup$ @darijgrinberg Checking them you can find out what the matrix's rank is. Observe that the biggest pricipal minor is the whole matrix's determinant. $\endgroup$
    – DonAntonio
    Apr 8 '16 at 14:20
  • $\begingroup$ Judging by the principal minors alone, the matrix $\left(\begin{matrix}0&1&0\\ -1&0&0\\ 0&0&0\end{matrix}\right)$ would have rank $0$. $\endgroup$ Apr 8 '16 at 14:22
  • $\begingroup$ @darijgrinberg How come? The principal minor $\;2\times 2\;$ is $\;-1\;$ ... $\endgroup$
    – DonAntonio
    Apr 8 '16 at 14:23
  • 1
    $\begingroup$ I am still curious about how you get your criterion with principal minors. I can confirm it in the case when the base field is $\mathbb{R}$, but the general proof is elusive. $\endgroup$ Apr 8 '16 at 15:48

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