Help with the proof of the characterization of linearly dependent sets In Lay - Linear Algebra, 3rd ed. ch 1.7, Theorem 7 states that

An indexed set $S = \{ \mathbf{v}_1, \dots , \mathbf{v}_p \}$ of two or more vectors is linearly dependent if and only if at least one of the vectors in $S$ is a linear combination of the others. In fact, if $S$ is linearly dependent and $\mathbf{v}_1 \neq \mathbf{0}$ , then some $\mathbf{v}_j$ (with $j > 1$) is a linear combination of the preceding vectors, $\mathbf{v}_1,\dots , \mathbf{v}_{j-1}$ .

The proof later on in the chapter proves the theorem from two "directions", starting with $\mathbf{v}_j$ in $S$ as a linear combination of the other vectors, and showing that if this is the case then $S$ must be linearly dependent. Then Lay proves the theorem from the other direction, namely supposing $S$ is linearly dependent first, and then showing that some $\mathbf{v}_j$ must be a linear combination of the other vectors.
The whole second part is shown here for context, but my problem is already in the second line (the rest is fine):

[...] suppose $S$ is linearly dependent. If $\mathbf{v}_1$ is zero, then it is a (trivial) linear combination of the other vectors in $S$. Otherwise, $\mathbf{v}_1 \neq \mathbf{0}$, and there exists weights $c_1 , \dots , c_p$ , not all zero, such that $$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_p \mathbf{v}_p = \mathbf{0}$$ Let $j$ be the largest subscript for which $c_j \neq 0$. If $j=1$ , then $c_1 \mathbf{v}_1 = \mathbf{0}$, which is impossible because $\mathbf{v}_1 \neq \mathbf{0}$. So $j>1$, and $$\begin{align} c_1 \mathbf{v}_1 + \cdots + c_j \mathbf{v}_j + 0 \mathbf{v}_{j+1} + \cdots + 0\mathbf{v}_p = \mathbf{0} \\ c_j \mathbf{v}_j = -c_1 \mathbf{v}_1 - \cdots -c_{j-1} \mathbf{v}_{j-1} \\ \mathbf{v}_j = ( - \frac{c_1}{c_j} \mathbf{v}_1 ) + \cdots + ( -\frac{c_{j-1}}{c_j} ) \mathbf{v}_{j-1} \end{align}$$ 

Now, my concern here is quite trivial really (pun sort of intended). In the second line of the proof above, if $\mathbf{v}_1$ is zero, why must it be a trivial linear combination of the other vectors in $S$? Since $S$ is here defined to be linearly dependent, can't we find weights $c_1, \dots , c_p$ not all zero so that $$c_1 \mathbf{v}_1 = c_2 \mathbf{v}_2 + \cdots c_p \mathbf{v}_p $$, where $\mathbf{v}_1 = \mathbf{0}$? Or is it that $\mathbf{v}_1$ is the defining factor that determines the linear dependence of $S$? There is something here I just can't see. Maybe it's too late.
Sorry for the long post! I wasn't sure how much context would be appropriate. If somebody knows how to clean up the multiline code in the proof then I'd appreciate that.

The point is not that $\mathbf v_1$ must be a trivial linear combination of the other vectors, as you rephrased it, but that it is. That's enough to show that at least one of the vectors is a linear combination of the others.
Also, we did find weights $c_1,\dotsc,c_p$, not all zero, so that $c_1 \mathbf{v}_1 = c_2 \mathbf{v}_2 + \cdots c_p \mathbf{v}_p$, namely $c_1=1$, $c_2=\dotso=c_p=0$.
