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Lecture

# Lecture 21.pdf

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University of Waterloo

Mathematics

MATH 136

Robert Sproule

Winter

Description

Friday, February 28 − Lecture 21 : Basis and dimension of a vector space
Concepts:
1. Define a basis.
2. Recognize that any two bases have the same number of elements.
3. Verify that a set is linearly independent. Verify that a set is a basis.
4. Define dimension of a vector space.
21.1 Definition − If {v , v1...2, v } ik both
1. A spanning family of V and
2. Linearly independent
then we say that {v , v ....,v } is a basis of V .
1 2, k
21.2 Theorem − Any n linearly independent vectors inthe vector space ℝ forms a basis n
n
of ℝ .
Proof:
- Suppose {v , v1...2, v } lnnearly independent .
n
- To show that {v , v 1...2,v } isna basis it suffices to show that this set spans ℝ .
- Let A = [v 1 v2....,v n, a square n by n matrix
n n
- To show that spans ℝ it suffices to show that Col(A) = ℝ .
n
- To do this it suffices to show that for any vector v in ℝ , Ax = v is consistent.
- Since {v , v ....,v } is linearly independent Ax = 0 only has the trivial solution.
1 2, n
- This implies that A is the identity matrix.
RREF
- Then the system [A | v] row reduces to [ I | w] and so has a unique solution.
- Thus Col(A) = ℝ . And so {v , v ....,v } spans ℝ . n
1 2, n
- So {v ,1v .2,.,v }nis a basis of ℝ n
21.2.1 Remark – The trivial subspace {0} of a vector space V is a special vector space
which does not have a basis.
21.3 Theorem − Suppose B = {v , v , …,1v }2is a basns of V and {w , w , … w } i1 lin2arly k
independent in V. Then k ≤ n. (Proof is left to read from the text on page 124). 21.3.1 Corollary − If A and B are two finite bases of a vector space V then A and B
contain the same number of elements.
Proof is given in class. This follows almost immediately from the theorem.
21.4 Definition − If a vector space V has a finite basis,the common number of elements
in all its bases is called the dimension of V. This number is denoted by dim(V).
21.5 Definition − A vector space V which has a finite spanning family is referred to as
being finite dimensional. That is, there exists in V a finite set {v , v 1...2,v } ok vectors
such that V = span(v , 1 ..2,, v )k
21.5.1 Remark − There is a theorem that confirms that:

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