# Does an antisymmetric matrix with high rank have a minor with disjoint rows and columns and high rank?

This is a generalized version of Does a non-singular matrix have a large minor with disjoint rows and columns and full rank?

Let $$A$$ be an $$n$$-by-$$n$$ antisymmetric matrix of rank $$r\geq \epsilon n$$. Is there a minor of $$A$$ with disjoint row and column indices $$I,J\subset \{1,2,\dotsc,n\}$$ and rank $$k\geq \lfloor r/1000\rfloor$$?

(Some argument involving a Pfaffian might work, but the matter does not seem self-evident (to me).)

• Doesn't it work if you take a full-rank principal submatrix and apply the other result to it? A Hermitian matrix with signature $(n_+,n_0,n_-)$ has rank $r = n_+ + n_-$ and one can find a principal submatrix of rank $\max(n_+, n_-) \geq r/2$; I presume the same result applies to antisymmetric matrices after multiplying it by $i$. – Federico Poloni Jul 16 '19 at 11:12
• Why would it be true that, for $A$ antisymmetric, there is an $r$-by-$r$ principal submatrix of rank $r$? It isn't true for the matrix $\left(\begin{matrix} 0 & 1\\ -1 & 0\end{matrix}\right)$. Perhaps there's a weaker version of the statement for antisymmetric matrices? – H A Helfgott Jul 16 '19 at 11:17
• @HAHelfgott yes there's one in the matrix you give (for which $r=2$), namely the whole matrix. – YCor Jul 16 '19 at 11:19
• – H A Helfgott Jul 16 '19 at 11:22
• Good find! I think this settles the question then. – Federico Poloni Jul 16 '19 at 12:15

1. Take a full-rank principal submatrix $$A(I,I)$$ of $$A$$ (with $$|I|=r=\operatorname{rk} A$$): it is proved here that one always exists.
2. Apply the result in the linked question to $$A(I,I)$$: it shows that $$A(I,I)$$ has a non-principal submatrix $$A(J,K)$$ with $$|J|=|K|=|I|/2=r/2$$, and $$J \sqcup K = I$$.
Hence the result is proved with $$k=r/2$$ (and without the need for the hypothesis $$r \geq \varepsilon n$$).