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Lecture 21.pdf

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University of Waterloo
MATH 136
Robert Sproule

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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