A basis of a null space of a binary matrix

An $m \times n$ matrix $A$ is called a binary matrix if all entries of $A$ are either $0$ or $1$. Also, a vector $\mathbf{x}$ in $\mathbb{R}^n$ is called a trinary vector if all entries of $\mathbf{x}$ are either $-1$, $0$, or $1$.

I want to prove/disprove the following assertion.

Given an $m \times n$ binary matrix $A$, it is always possible to construct a basis for a null space of $A$ with trinary vectors only.

I played with matlab, and didn't find any counter example at least for comparatively small $m,\, n$. Any idean would be appreciated.

Solution 1:

Here is a counterexample:

\begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{bmatrix}

The lines of this $3\times 4$ matrix being linearly independent, the kernel is of dimension 1. It is generated by the following vector

\begin{bmatrix} -2 \\ -1 \\ 1 \\ 1 \end{bmatrix} .

Therefore there is no trinary vector generating the null space of that binary matrix.

The question of determining which conditions on $A$ imply the existence of a trinary basis seems a very interesting (and non-trivial) task. Any ideas in this direction are welcome...!