Linear Dependence Lemma 2 I'm trying to interpret the very last sentence. He says nothing about $a_{1}$, so in the case of $a_{1}=0$ all the other coefficients can not be zero, which would imply linearly independence. The case where $a_{1}\neq 0$ all the other coefficients can not be zero because, if they were, we would have $a_{1}v_{1}=0$ which is never true since $v_{1}\neq 0$.
Is this correct? Thank you in advance.
I have seen this post post about lin.dep. lemma., but I think my question is unanswered in that post.

 A: What you've said is perfectly correct. A slightly quicker/neater/different way to see it (without splitting into cases) might be the following:
Suppose instead that all of $a_2,\dots,a_m$ are equal to zero. Then $a_1v_1=0$. But $v_1\not=0$ so it must be that $a_1=0$, contradicting our original assumption that not all of the $a_i$ are $0$.
A: It seems to combine two cases together:


*

*If $a_0=0$, then since $\{a_1,a_2,\ldots,a_m\}$ contains a non-zero element and $a_0=0$, there must be a non-zero member of $\{a_2,a_3,\ldots,a_m\}$.

*If $a_0 \neq 0$, then $a_1 v_1 \neq 0$ since $v_1 \neq 0$.  Hence, if $a_1 v_1 + x=0$ where $x=a_2v_2+a_3v_3+\cdots+a_mv_n$, then $x = -a_1v_1 \neq 0$ and hence $\{a_2,a_3,\ldots,a_m\}$ cannot all be zero.
A: If you note that $span{} = \{0\}$, then it is clear that $v_1$ can also be equal to $0$, because if we have an empty span generated by an empty set of preceding vectors, then if $v_1 \in span\{0\}$, $v_1 = 0$.
The empty set is independent by itself but when an extra $v_1 = 0$ gets added, then the set of vectors becomes linearly dependent.
Stating that $v_1 \neq 0$ is kind of a confusing way to prove this theorem IMO.
