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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.

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  • $\begingroup$ What vector is "1"? $\endgroup$
    – abx
    Mar 6, 2021 at 6:39
  • $\begingroup$ Oh it is just a number. I made a typo of "over $\mathbb{Q}^n$". $\endgroup$
    – HAORAN ZHU
    Mar 6, 2021 at 6:40

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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 $m[v_j]=\sum_{i\in I} n_{j,i}[v_i]$.

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  • $\begingroup$ Thanks a lot for your great explanation! $\endgroup$
    – HAORAN ZHU
    Mar 6, 2021 at 12:39

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