204Lecture82009 (1)

204Lecture82009 (1) - Economics 204 Lecture 8Wednesday,...

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Economics 204 Lecture 8–Wednesday, August 5, 2009 Revised 8/5/09, Revisions indicated by ** and Sticky Notes 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 , span V denotes the set oF all linear combinations oF V . Aset V X is linearly dependent iF there exist v 1 ,...,v n V ∗∗ and α 1 ,...,α n F not all zero such that n X i =1 α i v i =0 Aset V X is linearly independent iF it is not linearly depen- dent. Aset 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 1
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α = x + y 2 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 =0 so the coefficients are all zero, so { (1 , 1) , ( 1 , 1) } is linearly in- dependent. Since it is linearly independent and spans R 2 ,itisa 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 ince V spans X , we can write x = X s S 1 α s v s
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where S 1 is fnite, α s F , α s 6 =0 , v s V For s S 1 .N o w , suppose x = X s S 1 α s v s = X s S 2 β s v s 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 =0 For 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 =0 ,so α s = β
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This note was uploaded on 10/04/2010 for the course ECON 204 taught by Professor Anderson during the Summer '08 term at Berkeley.

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204Lecture82009 (1) - Economics 204 Lecture 8Wednesday,...

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