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?



Does a non-singular matrix have a large minor with disjoint rows and columns and full rank?
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 $i\_1<\ldots<i\_n$ and $j\_1<\ldots<j\_n$.


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


