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Lecture

# Lecture 22.pdf

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University of Waterloo

Mathematics

MATH 136

Robert Sproule

Winter

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Monday, March 3 − Lecture 22 : Finding a basis for a vector space
Concepts:
1. Expanding a linear independent set to a basis.
2. Reducing a spanning family to a basis.
22.1 Lemma − The Removal Lemma. (Sometimes called the spanning theLet U = {v , 1 ,.2., v }k
and V = span{v , v1,..2, v } k span U. Suppose the vector v in U = {v1, v ,..., v1} i2 a k
linear combination of all the other vectors in that set. Then Span{v , v , ...,1v }2 k
= Span{ v , 2.., v }k
Proof:
- We are given that v is 1 linear combination of the other vectors in U. That is
v 1 β v 2 2 v +.3 3.+ β v . .k k
- Let w ∈ V = Span{v , v , 1..,2v }. Thkn, for some scalars α , α , ...,1α .2 k
w = α v 1 1 v + 2 2 + α v k k
= α (1 v2 2β v +3 3. + β v ) +k kv + α 2 2 ... 3 3 v . k k
= (α β1 2α )v 2 (2 β + α1 3 + ..3 + 3α β + α )v 1 k vectk2 ik Span{v , ..., v }. 2 k
- So V is contained in the Span{ v , ...2 v }. k
- Hence V = Span{v , …, 2 }. k
- Since V = Span{v , v 1 ...2 v } thkn Span{v , v , ..1, v 2 = Spak{v , ..., v }2 k
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 Removal 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 = {v , v ,..., v } is not linearly
1 2 k
independent does not mean that every vector in Uis redundant. For example, in the
set {e , e , 2e }, 2e is redundant but e is not redundant.
1 2 2 2 1
22.2 Theorem − The vectors B = {v , v , ..1, v2} be akspanning family of non-zero vectors
for the vector space V. Then a subset of B is a basis of V.
This follows from the Removal lemma. 22.2.1 Remark − Every finite spanning family of a vector space V can be reduced (or
"shrunk") to be a basis.
22.3 Proposition − Let U = {v , v ..1., 2,} be aklinearly independent set of vectors in the
vector space V such that span(v , v .1..,2, ) isknot all of V. Then, for any vector v in
V /span(U), the set {v , v1...2, v , vk forms a linearly independent set.
Proof (Outline):
- Let U = {v , 1 ..2,,v } ke linearly independent and v be a vector not in the span of U.
- Suppose α v +1 1v +...2 2 α v + αv =k k
- We claim α = 0:
Suppose α ≠ 0.
Then v = −(α /α) 1 + −(α1/α) v 2 , ...2 + −(α /α)v ∈ spkn(U )k a contradiction.
Thus α = 0.
- Also α v1 1α v +..2 2+ α v = 0 k klies α = 0 for allii. Why?
- Thus α = i for all i and α = 0.
- Thus {v 1, v2,...,vk, v} forms a linearly independent set.
22.3.1 Consequence of Proposition 22.3:
“Any linearly independent subset of a finite dimensional vector space V can be
completed to a basis for V.”
A Flow chart is given in class.
The loop in the chart can only run finitely many times since V is finite dimensional
and there cannot be more linearly independent vectors then dim(V).
22.4 Example − Find a basis for the subspace W = {(x, y, z) : 2x − y + 3z = 0} ( plane
containing the vector (0, 0, 0) ) and find dim(W).
- Note that if (x, y, z) is in W then x = (1/2)y

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