# Rank of a 0-1-matrix

Suppose $K$ is a field of characteristic $0$. Let $M \in K^{n \times m}$ be a matrix such that every entry of $M$ is either $0$ or $1$. About this matrix, I know further that each sum over a column and a row is a constant number, i.e.

$$\sum_{i=1}^n M_{ij} = c_{\text{column}} ~~ \forall j=1,...,m$$

and

$$\sum_{j=1}^m M_{ij} = c_{\text{row}} ~~ \forall i=1,...,n$$

for some $c_{\text{column}}, c_{\text{row}}$ both lying in $\{1,2,3,...\}$.

Question: Are there criteria that enables us to say something about the rank of $M$? I.e. are there theorems of the form

'' If some condition is satisfied, then $$\text{rank}(M) \geq f(n,m,c_{\text{column}}, c_{\text{row}})$$

for some (hopefully not too complicated) function $f$''?

I have found a paper of Odlyzko from '79 in which he shows that a $0$-$1$-matrix with constant row-sums is of full rank if the number of distinct row vectors exceeds a certain number. Unfortunately, in my case I do not have sufficiently many row-vectors but I have some additional information, for example, I know that the column-sum is also constant.

Regards,

FW

Even in the special case $K = \mathbb R$, $n=m$ and $c_{\rm row} = c_{\rm column}$, this is not an easy problem!
• Given that the field is to be of characteristic zero, it is actually enough to do this for $\mathbb Q$ (or $\mathbb R$, if one prefers) because the entries are integers. The special case is therefore all one needs to do :-) – Mariano Suárez-Álvarez Jun 29 '13 at 3:23