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MATH 136 (145)
Lecture

# Lecture 22.pdf

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School
University of Waterloo
Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter

Description
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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