MATH136 Lecture Notes - Lecture 22: Linear Combination, Spanx
14 May 2018
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Monday, June 19
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Lecture 22 : Finding a basis for a vector space (Refers to 4.2)
Concepts:
1. Expanding a linear independent set to a basis.
2. Reducing a spanning family to a basis.
22.1 Lemma
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(Sometimes called the spanning theorem) Let U = {v1, v2,..., vk}. If the
vector v1 in U is a linear combination of all the other vectors in that set then
Span{v1, v2, ..., vk} = Span{ v2, ..., vk}.
Proof:
- We are given that v1 is a linear combination of the other vectors in U. That is
v1 = β2v2 + β3v3 +.....+ βkvk. .
- Let w ∈ V = Span{v1, v2, ..., vk}. Then, for some scalars α1, α2, ..., αk.
w = α1v1 + α2v2 + ... + αkvk
= α1(β2v2 + β3v3 + ... + βkvk) + α2v2 + α3v3 + ... + αkvk.
= (α1β2 + α2)v2 + (α1β3 + α3)v3 + ... + (α1βk + αk2)vk, a vector in Span{v2, ..., vk}.
- So V is contained in the Span{ v2, ..., vk}.
- Hence V = Span{v2, …, vk}.
- Since V = Span{v1, v2, ..., vk} then Span{v1, v2, ..., vk} = Span{v2, ..., vk}.
22.1.1 Note – A spanning family that is not linearly independent always contains a
vector which is a linear combination of the others. We call a vector in a spanning
family that is a linear combination of the other a redundant vector.
The lemma says that we can eliminate redundant vectors from a spanning family
without affecting it’s span.
A word of caution: The fact that a spanning family U = {v1, v2,..., vk} is not linearly
independent does not mean that every vector in U is redundant. For example, in the
set {e1, e2, 2e2 }, both e2 and 2e2 is redundant but e1 is not redundant.
22.2 Theorem − The vectors B = {v1, v2, ..., vk} be a spanning family of non-zero vectors
for the vector space V. Then B contains a basis of V.
This follows from the lemma.
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