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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)

pomelozest
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    The Linear Dependence Lemma states that in a linearly dependent list, one of the vectors in the list is a linear combination of the previous vectors. Because the list $u_1, u_2$ is linearly independent, $u_2$ is not a multiple of $u_1$. – Sheldon Axler Dec 06 '15 at 06:15
  • @SheldonAxler so is it true that since $u_2$ makes the set $B_1$ linearly dependent, there is at least one other vector in the set besides $u_2$ that makes the set linearly dependent and we know that vector is not $u_1$ because ${u_1,u_2}$ is linearly independent? – pomelozest Dec 06 '15 at 16:54
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    Please read carefully the proof on page 35 (third edition of Linear Algebra Done Right) where this is explained. But here is a summary: The list $u_1, u_2, w_1, \dots, w_n$ is linearly dependent. The Linear Dependence Lemma states that one of the vectors in the list is a linear combination of previous vectors in the list; furthermore that vector can be removed from the list without changing the span. The vector in question must be one of the $w_j$'s because $u_1, u_2$ is linearly independent and thus neither $u_1$ nor $u_2$ is a linear combination of the previous vectors in the list. – Sheldon Axler Dec 07 '15 at 03:07
  • @SheldonAxler thanks for responding! The reason that the explanation on p.35 confused me was because I thought that the order of the list was arbitrary so $u_1$ and $u_2$ being at the front of the list did not matter. – pomelozest Dec 07 '15 at 03:22
  • @SheldonAxler, is it possible that multiple such vectors exist in the same list? That is, multiple vectors which can be written as a linear combo of the vectors before it? If there are multiple linear combinations that yield zero, (in your proof on page 34) we could have different $v_j$'s from different combinations with non-zero coefficients. – Vishal Subramanyam Jul 21 '20 at 20:31
  • @VishalSubramanyam Yes, more than one vector in the list might be a linear combination of the previous vectors. Just pick the first time this happens in the list. – Sheldon Axler Jul 22 '20 at 02:13

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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 $V$.

pomelozest
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I'd like to rephrase the answer by tying it back to the definition of linear dependence and the linear dependence lemma used to prove this result.

For Step 1:

Recall that $\text{span}() = \{0\}$. By the linear dependence lemma, picking $u_1$ requires $u_1$ to satisfy the condition $u_1 \in \{0\}$, and that is only possible when $u_1 = 0$. However, any tuple with the null vector is linearly dependent, as it's possible to write $a0 = 0$ with $a \neq 0$, where $a$ is some scalar in $\mathbb F$. This violates the premise that $u_1,\ldots,u_m$ is linearly independent, which leads to a contradiction. Hence some $w$ vector must be chosen here.

  • "But why is the one-tuple containing the null vector linearly dependent?" Think of the definition of linear independence: it's equivalent to the statement that the only way to represent $v_j$ is $v_j$ itself. However, $0$ can be represented by $1 \cdot 0$, $2 \cdot 0$, and so on.1

For all subsequent steps (labeled step $j$):

The linear dependence lemma states that for some $v_k$ in a tuple, $v_k$ must satisfy the property that $v_k \in \text{span}(v_1,\ldots,v_{k-1})$. Picking $u_i$ for any $1 < i \leq j$ violates the premise is that $u_1,\ldots,u_m$ is linearly independent, as that implies $u_i \in \text{span}(u_1,\ldots,u_{i-1})$. This leads to a contradiction, and hence some $w$ vector must be chosen here as well.

1 "But why is the definition chosen that way?" Definitions are defined the way they are because either later results require entities to obey a definition precisely the way they are defined, or, if the later results don't require it, the edge cases are too irrelevant to be worth adapting the proof over. This result itself is actually a good example of the latter case; the result trivially extends to the case $m = 1$ and $u_1 = 0$. However, it's just easier to omit such a set from consideration, since satisfying the edge case $\{0\}$ is typically of no practical value.

dlq
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