Suppose I have a matrix $U: n \times 2$ where $n>2$ which is not all in $\mathbb{Q}$ and has no collinear rows. That is, there is at least one irrational member of $U$. In addition, every row of $U$ has unit norm. I would like to prove, or show a counterexample for the following statement: the is no vector $x: 2\times 1$ such that: $$ U\cdot x \in \mathbb{Z}^n $$ I cannot give a counterexample and I think the statement is true, but cannot prove it fully. I have been so far thinking along this path: suppose such vector $x$ exist. Then we have: $$ U\cdot x = \left|x\right|\begin{pmatrix}cos(a_1) \\ \ldots \\ cos(a_n)\end{pmatrix}, $$ where $a_i$ is the angle between the respective row of $U$ and $X$, and where then we must have that $\forall i,\ \left|x\right|cos(a_i) \in \mathbb{Z}$, or alternatively $\forall i,j,\ \frac{cos(a_i)}{cos(a_j)} \in \mathbb{Q}$. I got stuck at this point. Is this statement even true to begin with?
Edit: $U$ doesn't have to be rational. Consider any invertible matrix $R: 2 \times 2$ as a change of coordinates so that: $$U \cdot R \cdot R^{-1} x=U\cdot x.$$ So maybe a more well-defined version of the question: suppose $U$ cannot be rationalized by a change of coordinates. is there vector $x$ that brings it to the integers?