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?
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.
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.
