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 Defnition 1 Let X be a vector space over a feld F .A linear combination oF x 1 ,...,x n is a vector oF the Form y = n X 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 ,sp an V denotes the set oF all linear combinations oF V . Aset V X is linearly dependent iF there exist v 1 ,...,v n X and α 1 n F not all zero such that n X i =1 α i v i =0 V X is linearly independent iF it is not linearly dependent. 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 .I f( x, y )=(0 , 0), α = 0+0 2 =0 = 0 0 2 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 Fnite number of elements of V . ( Aside: the unique representation of 0 is 0 = i ∈∅ α i b i .) Proof: Let x X .S in c e V spans X ,wecanwr ite x = X s S 1 α s v s where S 1 is Fnite, α s F , α s 6 , v s V for s S 1 . Now, suppose x = X s S 1 α s v s = X s S 2 β s v s 2
where S 2 is fnite, β s F , β s 6 =0 ,and v s V For s S 2 . Let S = S 1 S 2 , and defne α s =0 For s S 2 \ S 1 β s s S 1 \ S 2 Then 0= x x = X s S 1 α s v s X s S 2 β s v s = X s S α s v s X s S β s v s = X s S ( α s β s ) v s Since V is linearly independent, we must have α s β s ,so α s = β s , For all s S . s S 1 α s 6 β s 6 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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## 204Lecture82009web - Economics 204 Lecture 8Wednesday,...

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