A set in a lattice is $\mathbb{R}$-linearly independent iff it is $\mathbb{Z}$-linearly independent

Question

In the course of PETE L. CLARK about geometry of numbers page 5, in Exercice 1.5; http://alpha.math.uga.edu/~pete/geometryofnumbers.pdf that says let $L \subset \mathbb{R}^n$ be a lattice and let $S\subset L$. Show that $S$ is $\mathbb{R}$-linearly independent $\iff$ it is $\mathbb{Z}$-linearly independent. It is clear the direct direction $\implies$. For the inverse direction $\Longleftarrow$: Suppose $S$ is $\mathbb{Z}$-linearly independent; and let $\sum_{finite}\lambda_i v_i=0$ with $(\lambda_i,v_i)\in \mathbb{R}\times S$, so how to prove that $\lambda_i=0$ $\forall i$?.

Answer 1

Since $L$ is a lattice, it is the $\mathbb{Z}$-span of some $u_1, \dots, u_n$ that are $\mathbb{R}$-linearly independent. Now $v_i = \sum a_{ij} u_j$ for some choice of $a_{ij} \in \mathbb{Z}$. So $\sum \lambda_i a_{ij} u_j = 0$ which means that $\sum \lambda_i a_{ij} = 0$ for each $j$. But the rank and therefore nullity matrix $(a_{ji})$ is the same over $\mathbb{Q}$ as over $\mathbb{R}$, and if there is a solution over $\mathbb{Q}$ then clearing denominator gives a solution over $\mathbb{Z}$. So if there is some non-zero $(\lambda_i)$ over $\mathbb{R}$ then there is some non-zero $(\lambda_i)$ over $\mathbb{Z}$.