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 $$v_2,....v_j$$?

You've a mistake in the line after "Let $$\;j\;$$ be the first ..." . It must be;

$$a_jv_j=-a_1v_1-\ldots-a_{j-1}v_{j-1}\implies v_j=-\frac{a_1}{a_j}v_1-\ldots-\frac{a_{j-1}}{a_j}v_{j-1}$$

and thus

$$v_j\in\text{Span}\,\{v_1,...,v_{j-1}\}$$

I have misunderstood the proof. It should be :

Let $$j$$ be the largest element of {1,...,m} such that $$a_j \neq 0$$. It means that $$a_{j+1} = 0, a_{j+2} = 0$$, and so on.

Therefore, $$v_j \in span(v1,...,v_{j-1})$$ actually means if the list is linearly independent, one of the vectors is in the span of the PREVIOUS ones.

So, in the case of $$j=1$$, $$v_1 \in span()$$

More reference here