Given a vector $v = (v\_1, \ldots, v\_n) \in \mathbb{R}^n$, we can associate a rational linear subspace with this vector: assume $\{1, v\_i \text{ for }i \in I\}$ is a linear basis of $\{1, v\_1, \ldots, v\_n\}$ over $\mathbb{Q}$: there exist positive integer $m$ and integer $n\_{j,i}, i \in \{0\} \cup I$, such that for any $j \notin I, m v\_j = n\_{j,0} + \sum\_{i \in I} n\_{j,i} v\_i$. Then we can define rational linear subspace $V: = \{x \in \mathbb{R}^n \mid mx\_j = \sum\_{i \in I}n\_{j,i} x\_i, j \notin I\}$.


I'm sure that $V$ is actually independent of the actual choice of the basis $\{1, v\_i \text{ for }i \in I\}$, in other words, for different choice of the basis $I$, it will yield the same linear space $V$. My question is, is there any easy way to prove this point? Any reference or guidance will be much appreciated.



Rational linear subspace corresponding to an irrational vector
Invariant description (which yields your basis-independence claim) is the following: *$V$ is the space of sequences $(x\_1,\ldots,x\_n)$ such that $\sum m\_ix\_i=0$ whenever $\sum m\_iv\_i$ is rational.*


To prove it, note that $\sum m\_i v\_i$ is rational if and only if $\sum m\_i [v\_i]=0$, where $[v\_i]$ is a class of $v\_i$ in the factor-space $\mathbb{R}/\mathbb{Q}$ (over field $\mathbb{Q}$); the basis of the span of $\{[v\_1],\ldots,[v\_n]\}$ in this space is $\{[v\_i]:i\in I\}$, and all linear dependencies of $[v\_1],\ldots,[v\_n]$ are linear combinations of the basic linear dependencies $m[v\_j]=\sum\_{i\in I} n\_{j,i}[v\_i]$.


