Proof. First, note that \(v_1,\ldots,v_k\) being linearly-independent over \(\mathbb{Z}\) is equivalent to \(v_1,\ldots, v_k\) being linearly-independent over \(\mathbb{Q}\). In particular, this is because if \(\sum_{i\in [k]} \frac{a_i}{b_i} v_i = 0\) for integers \(a_i,b_i\) then \[\sum_{i\in [k]} v_i a_i \prod_{j\neq i} b_j = 0\] is an integer linear combination which is \(0\), and if \(\sum a_i v_i = 0\) for integers \(a_i\) then because \(\mathbb{Z}\subset\mathbb{Q}\subset \mathbb{R}\) this is also a rational linear combination, or even a real linear combination.

Thus, it remains to show that \(v_1,\ldots, v_k\) being linearly-independent over \(\mathbb{Q}\) suffices to guarantee that they are linearly-independent over \(\mathbb{R}\). Assume that there are \(c_i\in \mathbb{R}\) such that \(\sum_{i\in [k]} c_i v_i = 0\); we aim to show that this is only possible if \(c_i=0\) for all \(i\). Consider the set \[C = \left\{ q_{i} c_{i}\; : \;q_{i}\in \mathbb{Q} \right\},\] i.e. the \(\mathbb{Q}\)-span of the \(c_i\)’s. If we view \(\mathbb{R}\) as a \(\mathbb{Q}\) vector space we see that there must be a basis \(b_1,\ldots, b_\ell \in \mathbb{R}\) for \(C\) with \(\ell\leq k\). In particular, because this is a basis we can write each \(c_i\) as a \(\mathbb{Q}\) linear combination of the basis elements, say \[c_i =\sum_{j\in [\ell]} q_{i,j}b_j.\] Now, substituting into the equation dictating that \(\sum_{i\in [k]}c_iv_i = 0\) gives \[\sum_i \sum_j q_{i,j}b_j v_i = \sum_j \left( \sum_i q_{i,j} v_i \right) b_j = 0.\] Now considering this for each component of the vector gives \[\sum_j \left( \sum_i q_{i,j} v_{i,k} \right) b_j = 0.\] But this is a \(\mathbb{Q}\) linear combination of the basis elements \(b_j\), which are linearly-independent over \(\mathbb{Q}\). Thus, \(\sum_i q_{i,j}v_{i,k}=0\) for all \(j\). Then we also have \[\sum_i q_{i,j}v_{i}=0\] for all \(j\). But this is a \(\mathbb{Q}\) linear combination of \(v_i\)’s, which are linearly-independent over \(\mathbb{Q}\), hence \(q_{i,j}=0\) for all \(i,j\). But then \(c_i = 0\) for all \(i\), as desired.