Let $M \in \{0,1\}^{n{\times}m}$ be a $n{\times}m$-matrix $(m \leq n)$ with entries in $\{0,1\}$ whose column vectors are linearly independent over $\mathbb R$. This means that, for any $x \in \mathbb R^m$, $Mx=0$ implies $x=0$.
My question is, does under these conditions also $Mx \in \mathbb Z^n$ imply $x \in \mathbb Z^m$? Proof/counterexample? More generally, if instead $Mx$ is "close" to $\mathbb Z^n$ in some suitable metric, does it follow that $x$ is "close" to $Z^m$?
Edit: Clive has produced a simple counterexample showing that the answer to my question above is "no". Let me therefore slightly modify the question. Assume that the entries of $M$ are all in $\{-1,1\}$ instead; does $\frac12 Mx \in \mathbb Z^n$ now imply $x \in \mathbb Z^m$?
The answer to your first question is 'no', for example $$\begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1/2 \\ 1/2 \\ -1/2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$ As for your second question, I suspect the answer is also 'no', but it's not really clear what you mean. You can cook up metrics to make things arbitrarily close to other things, but that doesn't necessarily give you useful information. I think $(\frac{1}{2},\frac{1}{2},{-\frac{1}{2}})$ is about as far from $\mathbb{Z}^3$ as you can reasonably expect to be, though!