The rank of an antisymmetric matrix

Question

Let $A$ be an antisymmetric matrix of order $(2n+1)\times(2n+1)$, whose non-zero elements are either $1$ or $-1$, and that has no zero rows. I know that

rank$(A)<2n+1$ since it is not invertible.
rank$(A)$ must be an even number.

I have a feeling (or rather a wish since it will help me a lot in a different problem I'm working on) that in this particular case it has rank $2n$ but I don't know if it's true \ if it is, how can I prove it.

I suggest showing that the determinant of the leading $2n\times 2n$ principal submatrix is odd (hence non-zero); mechanically it is the same procedure that I did here: https://math.stackexchange.com/questions/3867140/how-can-you-calculate-the-rank-of-an-nxn-matrix-with-the-given-conditions/3867230#3867230 starting at "Since all components of are integers..." — user8675309, Apr 20 '22 at 16:23

score 1 · Accepted Answer · answered Apr 20 '22 at 14:14

It is indeed the case that we must have $\operatorname{rank}(A) = 2n$.

As you have noted, $A$ cannot be invertible, so $\operatorname{rank}(A) \leq 2n$. To see that $\operatorname{rank}(A) \geq 2n$, this is the case, it suffices to note that the upper-left $(2n)\times(2n)$ submatrix is a square matrix of even size whose diagonal entries are non-zero and whose non-diagonal entries are either $1$ or $-1$. From this, it follows that this submatrix has a non-zero determinant.

I use a similar technique on my answer here. – Ben Grossmann Apr 20 '22 at 14:16 — Ben Grossmann, Apr 20 '22 at 14:16

The rank of an antisymmetric matrix

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