204Lecture92006

# 204Lecture92006 - Economics 204 Lecture 9Thursday Revised...

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Economics 204 Lecture 9–Thursday, August 10, 2006 Revised 8/10/06, Revisions indicated by ** Section 3.3, Isomorphisms Definition 1 Two vector spaces X, Y over a field F are iso- morphic if there is an invertible (recall this means one-to-one and onto) T L ( X, Y ). T is called an isomorphism . Isomorphic vector spaces are essentially indistinguishable as vector spaces. Theorem 2 (3.3) Two vector spaces X, Y over the same field are isomorphic if and only if dim X = dim Y . Proof: Suppose X, Y are isomorphic, via the isomorphism T . Let U = { u λ : λ Λ } be a basis of X , and let v λ = T ( u λ ) , V = { v λ : λ Λ } Since T is one-to-one, U and V are numerically equivalent. If y Y , then there exists x X such that y = T ( x ) = T n i =1 α λ i u λ i = n i =1 α λ i T ( u λ i ) = n i =1 α λ i v λ i 1

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which shows that V spans Y . To see that V is linearly indepen- dent, note that if 0 = m i =1 β i v λ i = m i =1 β i T ( u λ i ) = T m i =1 β i u λ i Since T is one-to-one, ker T = { 0 } , so m i =1 β i u λ i = 0 Since U is a basis, we have β 1 = · · · = β m = 0, so V is lin- early independent. Thus, V is a basis of Y ; since U and V are numerically equivalent, dim X = dim Y . Now suppose dim X = dim Y . Let U = { u λ : λ Λ } and V = { v λ : λ Λ } be bases of X and Y ; note we can use the same index set Λ for both because dim X = dim Y . By Theorem 3.2, there is a unique T L ( X, Y ) such that T ( u λ ) = v λ for all λ Λ. If T ( x ) = 0, then 0 = T ( x ) = T n i =1 α i u λ i = n ı=1 α i T ( u λ i ) = n ı=1 α i v λ i α 1 = · · · = α n = 0 since V is a basis x = 0 2
ker T = { 0 } T is one-to-one If y Y , write y = m i =1 β i v λ i Let x = m i =1 β i u λ i Then T ( x ) = T m i =1 β i u λ i = m i =1 β i T ( u λ i ) = m i =1 β i v λ i = y so T is onto, so T is an isomorphism and X, Y are isomorphic. Section 3.3 Supplement, Quotient Vector Spaces (not in de la Fuente): Definition 3 Given a vector space X and a vector subspace W of X , define an equivalence relation by x y x y W Form a new vector space X/W : the set of vectors is { [ x ] : x X } where [ x ] denotes the equivalence class of x with respect to . Note that the vectors are sets ; this is a little weird at first, but ... . Define [ x ] + [ y ] = [ x + y ] α [ x ] = [ αx ] 3

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You should check on your own that is an equivalence relation and that vector addition and scalar multiplication are well-defined, i.e.
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