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

Question

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?

Answer 1

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 $A(x+ty)=Ax\in \mathbb Z^m$.