Proof of $j=1$ where $v_j \in span(v_1, …,v_{j-1})$

In Linear Algebra Done Right, it presents the Linear Dependence Lemma which states that :

Suppose $ v_1,…,v_m$ is a linearly dependent list in $ V$ . Then there exists $ j \in {1,2,…,m}$ such that the following hold: $ v_j \in span(v_1,…,v_{j-1})$ . I follow the proof, but get confused on a special case where $ j=1$ . The book said choosing $ j=1$ means that $ v_1=0$ , because if $ j=1$ then the condition above is interpreted to mean that $ v_1 \in span()$ .

I tried to follow the example in the proof:

Because the list $ v_1,…,v_m$ is linearly dependent, there exist numbers $ a_1,…,a_m \in \mathbb{F}$ , not all $ 0$ such that $ $ a_1v_1+…+a_mv_m = 0$ $

Let $ j$ be the first element of {1,…,m}, such that $ a_j \neq 0$ . Then $ $ a_1v_1 = -a_2v_2 -…-a_jv_j$ $ $ $ v_1 = \frac{-a_2}{a_1}v_2-…-\frac{a_j}{a_1}v_j$ $

Then does it mean $ v_1$ is the span of $ v_2,….v_j$ ?