Theorem

Note that is is the most important theorem of all of Chapter 2.

Suppose ${latex.inlinev_{1}, ..., v_{m}} is a linearly dependent list in V. Then there exists ${latex.inlinek \in {1, 2, ..., m}} such that ${latex.inlinev_{k} \in span(v_{1}, ..., v_{k-1})}. Furthermore, if k satisfies the condition above and the kth term is removed from the list, the span does not change.

Proof

Because the list ${latex.inlinev_{1}, ..., v_{m}} is linearly dependent, there exist numbers ${latex.inlinea_{1}, ..., a_{m}}, not all 0, such that ${latex.inlinea_{1}v_{1} + ... + a_{m}v_{m} = 0}. Let k be the largest element of {1, ..., m} such that ${latex.inlinea_{k} \neq 0}. Then:

1. We can subtract ${latex.inlinev_{k}} from both sides and get that: ${latex.inline-a_{k}v_{k} = v_{1}a_{1} + ... + a_{k-1}v_{k-1}}.
2. We can divide both sides by ${latex.inline-a_{k}} and get ${latex.inlinev_{k} = -(a_{1}/a_{k})v_{1} + ... + (-a_{k-1}/a_{k})v_{k-1}}
3. Realize that this means that ${latex.inlinev_{k}} is a linear combination of the terms before it and thus it is in the span of the terms before it.
4. Realize that the reason we needed linear dependence is to ensure that there is an ${latex.inlinea_{k}} that’s not 0 so that we can divide both sides by that and not get something that’s undefined.

Now we show that if the kth term is removed, the span does not change.

Suppose k is any element of {1, ..., m} such that ${latex.inlinev_{k} \in span(v_{1},...,v_{k-1})}. Let ${latex.inlineb_{1},...,b_{k-1} \in F} be such that ${latex.inlinev_{k} = b_{1}v_{1} + ... + b_{k-1}v_{k-1}}. Suppose ${latex.inlineu \in span(v_{1}, ..., v_{m})} . Then there exists ${latex.inlinec_{1}, ..., c_{m}} s.t ${latex.inlineu = c_{1}v_{1} + ... + c_{m}v_{m}}. But, because ${latex.inlinev_{k}} is in the span of the terms previous, we can replace it with ${latex.inlineb_{1}v_{1} + ... + b_{k-1}v_{k-1}} which shows u is in the span of the list obtained by removing the kth term. So removing the kth term does not change the span of the list.