College Algebra Exam Review 167

College Algebra Exam Review 167 - 3.4. THE DUAL OF A VECTOR...

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Unformatted text preview: 3.4. THE DUAL OF A VECTOR SPACE AND MATRICES 177 Lemma 3.4.4. Let V be a finite dimensional vector space over a field K . For each non-zero v 2 V , there is a linear functional f 2 V such that f .v/ ¤ 0. Proof. We know that any linearly independent subset of V is contained in a basis. If v is a non-zero vector in V , then fv g is linearly independent. Therefore, there is a basis B of V with v 2 B . Let f be any function from B into K with f .v/ ¤ 0. By Proposition 3.3.32, f extends to a linear functional on V . I Theorem 3.4.5. If V is a finite dimensional vector space, then à W V V is a linear isomorphism. ! Proof. We already know that à is linear. If v is a non-zero vector in V , then there is an f 2 V such that f .v/ ¤ 0, by Lemma 3.4.4. Thus Ã.v/.f / D f .v/ ¤ 0, and Ã.v/ ¤ 0. Thus à is injective. Applying Proposition 3.4.2(c) twice, we have dim.V / D dim.V / D dim.V /. Therefore any injective linear map from V to V is necessarily surjective, by Proposition 3.3.35. I Finite dimensionality is essential for this theorem. For an infinite dimensional vector space, à W V ! .V / is injective, but not surjective. Duals of subspaces and quotients Let V be a finite dimensional vector space over K . For any subset S  V , let S ı denote the set of f 2 V such that hv; f i D 0 for all v 2 S . Likewise, for A  V , let Aı denote the set of v 2 V such that hv; f i D 0 for all f 2 A. (We identify V with V .) S ı is called the annihilator of S in V . Lemma 3.4.6. Let S and T be subsets of V , and W a subspace of V . (a) S ı is a subspace of V . (b) If S  T , then T ı  S ı and S ıı  T ıı (c) T  T ıı . (d) W D W ıı . (e) S ı D span.S /ı and S ıı D span.S /. ...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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