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.