Proving full rank of a special type of a (0,1,2)-integer matrices

60 Views Asked by At

My question arise at the consideration of Newton polytopes. In that context

I consider integer matrices $ A =(a_{ij})\in \mathbb{Z}^{(n+1) \times N} $ with $n+1 \leq N$ having the following special properties

  • $a_{ij}\in \{0;1;2\}$
  • first row is fixed: $a_{1j} = 1$ for all $j$
  • the column-sums have only two possible values $\sum_{i=1}^{n+1} a_{ij} \in \{K;K+1\}$ with $2\leq K \leq n+1$
  • columns which satisfy $\sum_{i=1}^{n+1} a_{ij} = K$ can only contain $a_{ij}\in\{0;1\}$
  • all columns are distinct: for $j\neq j^\prime$ there is an $i$ with $a_{ij} \neq a_{ij^\prime}$

and I want to check if/prove that those matrices have the full rank: $\operatorname{rank} (A) = n+1$.

Also a proof for the special case $a_{ij}\in\{0;1\}$ would help me.

The following equivalent statements I have found:

  • the column vectors without the first entry are affinely independent
  • the convex hull of $A$ without the first row has the full dimension $n$

And I tried (currently without success) to interprete $A$ as an incidence matrix of a hypergraph with $K$- and $(K+1)$-regular part.

Thanks a lot

1

There are 1 best solutions below

1
On BEST ANSWER

No. $$ \pmatrix{ 1&1&1&1&1\\ 0&1&1&1&0\\ 1&0&1&1&1\\ 1&1&0&0&1\\ 1&1&1&0&0}\pmatrix{1\\ 0\\ -1\\ 1\\ -1}=0. $$