Axler's "linear dependence lemma" The theorem in Axler I'm looking at is:

Suppose $v_1, \ldots, v_m$ is a linearly dependent list in $V$. Then there exists $j \in \{1, 2 \ldots, m\}$ such that the following hold:
(a) $v_j \in \mathrm{span}(v_1, \ldots, v_{j-1})$;
(b) if the $j$th term is removed from $v_1, \ldots, v_m$, the span of the remaining list equals $\mathrm{span}(v_1, \ldots, v_m)$.

The proof begins by asserting that by linear dependence, there are coefficients $a_1, \ldots, a_m$, not all of which are zero, such that $\sum\limits_{i=1}^m a_i v_i = 0$, and defines $j$ to be the largest element of $\{1, \ldots, m\}$ such that $a_j \neq 0$.
The proof makes sense, as does this step, but my only question is: are these coefficients unique? If I picked a different set of coefficients, the value of $j$ would be different. Does that not matter? If the coefficients weren't unique, then I would probably need there to be two vectors in the span of the others which could be removed, and using different sets of coefficients would only impact the order of the steps.
 A: The only "problem" of this theorem is the numbering of the vectors. You could just use the fact that if the family is linearly dependent, one vector $v_i$ can be expressed (in one way or another, it doesn't matter) as a linear combination of the others, and show that ${\rm span}(v_1,\dots,v_n) = {\rm span}(v_i,\dots,v_{i-1},v_{i+1},\dots,v_n)$.
A: No, the coefficients are not unique. That's not important anyway, since the goal is to prove that there is some $j\in\{1,2,\ldots,n\}$ with a certain property, rather than proving that there is one and only one such $j$.
However, in general it is not true that you can remove two vectors. Consider, for instance, these vectors from $\Bbb R^2$: $v_1=(1,0)$, $v_2=(0,1)$, and $v_3=(1,1)$. You have $1\times v_1+1\times v_2+(-1)\times v_3=0$. So, in this case, every coefficient is different from $0$. Therefore, if you pick any $i\in\{1,2,3\}$ the span of $\{v_1,v_2,v_3\}$ is equal to the span of $\{v_1,v_2,v_3\}\setminus\{v_i\}$. But the span of $\{v_1,v_2,v_3\}$ is $2$-dimensional, and therefore, if you remove two vectors from $\{v_1,v_2,v_3\}$, you will get a set whose span is not the span of $\{v_1,v_2,v_3\}$.
