# Number of $x \in [0,1]^n$ such that Ax integer

If $$A$$ is an integer matrix in $$\mathbb{Z}^{m \times n}$$, why is there only a finite number of vectors $$x \in [0,1]^n$$ such that $$Ax$$ is a vector of integers?

• $Ax$ is a vector, not an integer... – Don Thousand Nov 7 '19 at 7:57
• An integer vector I mean – Luna Nov 7 '19 at 7:59
• Then this is not true... Every matrix will satisfy this. – Don Thousand Nov 7 '19 at 8:01
• No, by [0,1] I mean the interval. If A is a 2x2 matrix of all ones and x = [1/3, 1/3], then Ax is not a vector of integers. – Luna Nov 7 '19 at 8:06
• This is not true in general: If $A$ is the zero matrix, $Ax$ is an integer vector for every vector $x$. There has to be some extra condition. – Eike Schulte Nov 7 '19 at 8:50

The claim is true if and only if $$\ker A=\{0\}$$.
Let $$\ker A=\{0\}$$. Then necessarily $$m\ge n$$ and there is $$B\in \mathbb R^{n,m}$$ such that $$BA=I_n$$.
Since multiplication with $$A$$ is a linear and bounded map, the image $$A([0,1]^n)$$ of $$[0,1]^n$$ is bounded, so $$A([0,1]^n) \cap \mathbb Z^m$$ is a finite set. Then the set of vectors $$x\in [0,1]^n$$ with $$Ax\in \mathbb Z^m$$ is precisely $$B \left( A([0,1]^n) \cap \mathbb Z^m \right),$$ which is finite as well.
Let now $$y \in \ker A$$, $$y\ne 0$$. Since $$A$$ is an integer matrix, it holds $$Ax\in \mathbb Z^m$$ for all $$x\in \{0,1\}^n$$.
Now define $$x$$ as $$x_i = \begin{cases} 1 & \text{ if } y_i<0,\\ 0 & \text{ if } y_i\ge0. \end{cases}$$ Then for all $$t\ge0$$ small enough, $$x+ty\in [0,1]^n$$ and $$A(x+ty)=Ax\in \mathbb Z^m$$.