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204Lecture82009web

204Lecture82009web - Economics 204 Lecture 8Wednesday...

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Economics 204 Lecture 8–Wednesday, August 5, 2009 Chapter 3, Linear Algebra Section 3.1, Bases Definition 1 Let X be a vector space over a field F . A linear combination of x 1 , . . . , x n is a vector of the form y = n i =1 α i x i where α 1 , . . . , α n F α i is the coefficient of x i in the linear combination. If V X , span V denotes the set of all linear combinations of V . A set V X is linearly dependent if there exist v 1 , . . . , v n X and α 1 , . . . , α n F not all zero such that n i =1 α i v i = 0 A set V X is linearly independent if it is not linearly dependent. A set V X spans X if span V = X . A Hamel basis (often just called a basis ) of a vector space X is a linearly independent set of vectors in X that spans X . Example: { (1 , 0) , (0 , 1) } is a basis for R 2 . { (1 , 1) , ( 1 , 1) } is another basis for R 2 : ( x, y ) = α (1 , 1) + β ( 1 , 1) x = α β y = α + β x + y = 2 α α = x + y 2 1
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y x = 2 β β = y x 2 ( x, y ) = x + y 2 (1 , 1) + y x 2 ( 1 , 1) Since ( x, y ) is an arbitrary element of R 2 , { (1 , 1) , ( 1 , 1) } spans R 2 . If ( x, y ) = (0 , 0), α = 0 + 0 2 = 0 , β = 0 0 2 = 0 so the coefficients are all zero, so { (1 , 1) , ( 1 , 1) } is linearly independent. Since it is linearly independent and spans R 2 , it is a basis. Example: { (1 , 0 , 0) , (0 , 1 , 0) } is not a basis of R 3 , because it does not span. Example: { (1 , 0) , (0 , 1) , (1 , 1) } is not a basis for R 2 . 1(1 , 0) + 1(0 , 1) + ( 1)(1 , 1) = (0 , 0) so the set is not linearly independent. Theorem 2 (1.2’, see Corrections handout) Let V be a Hamel basis for X . Then every vector x X has a unique representation as a linear combination (with all coefficients nonzero) of a finite number of elements of V . ( Aside: the unique representation of 0 is 0 = i ∈∅ α i b i .) Proof: Let x X . Since V spans X , we can write x = s S 1 α s v s where S 1 is finite, α s F , α s = 0, v s V for s S 1 . Now, suppose x = s S 1 α s v s = s S 2 β s v s 2
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where S 2 is finite, β s F , β s = 0, and v s V for s S 2 . Let S = S 1 S 2 , and define α s = 0 for s S 2 \ S 1 β s = 0 for s S 1 \ S 2 Then 0 = x x = s S 1 α s v s s S 2 β s v s = s S α s v s s S β s v s = s S ( α s β s ) v s Since V is linearly independent, we must have α s β s = 0, so α s = β s , for all s S . s S 1 α s = 0 β s = 0 s S 2 so S 1 = S 2 and α s = β s for s S 1 = S 2 , so the representation is unique.
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