# Does a non-singular matrix have a large minor with disjoint rows and columns and full rank?

Say I have an $$n$$-by-$$n$$ non-singular matrix $$A$$ all of whose diagonal entries are $$0$$. We call an $$m$$-by-$$m$$ minor of $$A$$ good if its set $$I$$ of row indices and its set $$J$$ of column indices ($$I,J\subset \{1,2,\dotsc,n\}$$) are disjoint. Can one give a good lower bound on the size $$m$$ of the largest non-singular good minor of $$A$$?

(Perhaps $$m = \lfloor n/2\rfloor$$?)

EDIT: All right, so obviously there aren't enough conditions - the answer is too easy. What if $$A$$ is antisymmetric?

• If I understand correctly, $m = \lfloor n/2 \rfloor$ is not possible (at least when $n$ is even), as evidenced by the matrix with $0$ on the diagonal and $1$ everywhere else. – Nathaniel Johnston May 15 at 12:13
• is it intended that in the question both $I$ and $J$ are sets of column indices? I suppose $I$ is the set of row indices. – Manfred Weis May 15 at 17:35
• Fixed. I must have been half-asleep when I posted the question. – H A Helfgott May 15 at 17:38

## 2 Answers

I am assuming the question is for antisymmetric matrix. Then $$n$$ is even. The claim follows from the properties of Pfaffian (see wikipedia):

If $$M$$ is $$2n$$ by $$2n$$ anti-symmetric matrix, then $$\det(M)=Pf(M)^2$$, where

$$Pf(M) = 2^{-n} \sum_{I\sqcup J=[1,2n]} \pm \det(M_{I,J})$$,

where $$I, J$$ specify partition of the set $$\{1,\dots,2n\}$$ into two subsets of size $$n$$. For each such partition we take the corresponding minor. The sign is the sign of the permutation $$(i_1,j_1,i_2,j_2,\dots,i_n,j_n)$$ where $$I=\{i_1,\dots,i_n\}$$ and $$J=\{j_1,\dots,j_n\}$$ so that $$i_1<\ldots and $$j_1<\ldots.

If all the minors were zero, then the Pfaffian would be zero.

The matrix with zeroes on the diagonal and ones everywhere else is nonsingular, but all its "good" minors of size bigger than 1 are singular, since they have all entries equal to 1.

• Ah, thanks, that was silly of me. Let me try to see whether I can improve the question. – H A Helfgott May 15 at 12:15
• @HAHelfgott Maybe your "true" question is answered by this or this? – Federico Poloni May 15 at 12:21
• @FedericoPoloni Thanks but not really. See above. – H A Helfgott May 15 at 12:33
• OK, updated the question. – H A Helfgott May 15 at 12:43
• If $A$ is antisymmetric and nonsingular, $n$ must be even. In the case $n=4$ I can confirm that there must be a good minor of size $2$. – Robert Israel May 15 at 12:59