I'm trying to understand the proof of Axler's Theorem 2.2.3.
In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.
He starts by taking $u_1, \ldots, u_m$, a set of linearly independent vectors, and $w_1, \ldots,w_n$ a spanning set. We need to prove $m \leq n$.
He lets $B$ be the list $w_1, \ldots, w_n$ and adds $u_1$ to it. As $u_1 \in \mathrm{span}(w_1, \ldots, w_n)$ since the $w_i$ span $V$, this list $u_1, w_1, \ldots, w_n$ is linearly dependent.
I understand everything at this point, but the next step confuses me. He seeks to use the "linear dependence lemma" he proved earlier, which states:
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 $\mathrm{span}(v_1, \ldots, v_m)$.
I didn't have any trouble understanding this proof, but Axler states in his proof of 2.2.3.:
In particular, the list
$$
u_1, w_1, \ldots, w_n
$$
is linearly dependent. Thus by the Linear Dependence Lemma (2.21), we can remove one of the $w$'s so that the new list $B$ (of length $n$) consisting of $u_1$ and the remaining $w$'s spans $V$.
How do I know that the $j$ referred to in the linear dependence lemma is in fact of the $w$'s? If I relabel the above vectors as
$$
v_1, v_2, \ldots, v_{n+1}
$$
where $u_1 = v_1$, $v_2 = w_1$, $v_{n+1} = w_n$, and so forth, I would need $j > 1$. How do I know this is the case?
My only intuition is that Axler's proof of the linear dependence lemma proceeds as in this manner (with my new relabeling). The $v_i$, labelled above, are linearly dependent, so there exist $a_i \in F$ such that
$$
\sum\limits_{i=1}^n a_i v_i = 0
$$
and not all $a_i$ are $0$. Then he takes $j$ to be the largest index such that $a_j \neq 0$.
I think the reason he's able to remove a $w$ is because $j$ is defined to be the largest index. If $a_1 = 0$, then we have $\sum\limits_{i=2}^n a_i v_i = 0$. If $1$ is in fact the largest index, can't I conclude immediately that the $w_i$ are linearly independent?
Best Answer
I haven't read Axler, but I doubt he would invoke some details of the proof of his lemma, while saying "by the lemma". I think he is merely relying on the statement of the lemma.
In this application of the lemma, we can see that $j=1$ cannot hold as follows.
Suppose $j=1$. Now look at what (a) of the lemma tells us: $u_1$ belongs to the span of the empty set. The smallest subspace including the empty set is the subspace consisting of only the zero vector; hence $u_1=0.$
But $u_1,\dots,u_m$ is a linearly independent list, so $u_1\not = 0.$