I'm reading Axler's Linear Algebra Done Right and am hung up on one part of Axler's Replacement Theorem proof.
His proof states the following:
"Suppose $u_1,...,u_m$ is linearly independent in V. Suppose also that $w_1,...,w_n$ spans V.
...the list $$u_1,w_1,...,w_n$$ 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 $V$."
So if we follow what Axler says to do then our set $B=\{u_1,w_1,...,w_n\}$ spans V. And when we add $u_2$ we get $B_1=\{u_1,u_2,w_1,...,w_n\}$ which is also linearly dependent so we can remove a vector and the set will still span V. My question is how does the Linear Dependence Lemma allow us to always remove a $w_i$ rather than (in this case) $u_1$ or $u_2$?
(note: I understand that for the set B we don't remove $u_1$ because it is not equal to zero, but I am concerned with after the initial step when there are multiple u's)
Ok...so I actually answered my own question with the help of Sheldon Axler (who commented). We know $B_1=\{u_1,u_2,w_1,...,w_n\}$ is linearly dependent because $u_2$ can be written as a linear combination of the vectors in $B$. This would like like so: $$u_2=c_0u_1+c_1w_1+...+c_nw_n$$ It follows that there are vectos $w_i\in B$ with $c_i\neq 0$ and at least one of these vectors is not $u_1$ because that would mean $u_2$ was a multiple of $u_1$ which Sheldon Axler pointed out is not possible. So if we fix a $w_j$ with $c_j\neq 0$ we can write: $$w_j=(-u_2+c_0u_1+c_1w_1+...+c_nw_n)*-c_j^{-1}$$ And now we can remove $w_j$ from $B_1$ and the resulting set will still span $V$.