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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?

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  • $\begingroup$ $Ax$ is a vector, not an integer... $\endgroup$ – Don Thousand Nov 7 '19 at 7:57
  • $\begingroup$ An integer vector I mean $\endgroup$ – Luna Nov 7 '19 at 7:59
  • $\begingroup$ Then this is not true... Every matrix will satisfy this. $\endgroup$ – Don Thousand Nov 7 '19 at 8:01
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    $\begingroup$ 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. $\endgroup$ – Luna Nov 7 '19 at 8:06
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    $\begingroup$ 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. $\endgroup$ – Eike Schulte Nov 7 '19 at 8:50
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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 $A(x+ty)=Ax\in \mathbb Z^m$.

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