## 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 $$v_2,….v_j$$?