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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).)

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    $\begingroup$ 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 $i$. $\endgroup$ – Federico Poloni Jul 16 at 11:12
  • $\begingroup$ 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? $\endgroup$ – H A Helfgott Jul 16 at 11:17
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    $\begingroup$ @HAHelfgott yes there's one in the matrix you give (for which $r=2$), namely the whole matrix. $\endgroup$ – YCor Jul 16 at 11:19
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    $\begingroup$ Aha: mathoverflow.net/questions/22180/… $\endgroup$ – H A Helfgott Jul 16 at 11:22
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    $\begingroup$ Good find! I think this settles the question then. $\endgroup$ – Federico Poloni Jul 16 at 12:15
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The question has been solved by combining ideas found in the comments. I'm writing down this (community wiki) answer to prevent the question from showing up among the unsolved ones.

  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 $r \geq \varepsilon n$).

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