College Algebra Exam Review 156

College Algebra Exam Review 156 - n Definition 3.3.23 A...

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166 3. PRODUCTS OF GROUPS (b) B is a minimal spanning set for V . That is, B spans V and no proper subset of B spans V . (c) B is a maximal linearly independent subset of V . That is, B is linearly independent and no subset of V properly containing B is linearly independent. Proof. The implications (a) H) (b) and (a) H) (c) both follow from Lemma 3.3.20 . If B is a minimal spanning set, then B is linearly independent, by Lemma 3.3.21 , so B is a basis. Finally, if B is a maximal linearly independent set, and v 2 V n B , then f v g [ B is linearly dependent, so we have a linear relation ˇv C X i ˛ i v i D 0 with not all coefficients equal to zero and v i 2 B . Note that ˇ ¤ 0 , since otherwise we would have a nontrivial linear relation among elements of B . Solving, we obtain v D ± .1=ˇ/ X i ˛ i v i ; so v 2 span .B/ . It follows than span .B/ D V . Thus, a maximal linearly independent set is spanning, and therefore is a basis.
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Unformatted text preview: n Definition 3.3.23. A vector space is said to be finite–dimensional if it has a finite spanning set. Otherwise, V is said to be infinite–dimensional . Proposition 3.3.24. If V is finite dimensional, then V has a finite basis. In fact, any finite spanning set has a subset that is a basis. Proof. Suppose that V is finite dimensional and that S is a finite subset with span .S/ D V . Since S is finite, S has a subset B that is minimal spanning. By Proposition 3.3.22 , B is a basis of V . n Let V be a vector space over K . Represent elements of the vector space V n by 1 –by– n matrices (row “vectors”) with entries in V . For any n –by– s matrix C with entries in K , right multiplication by C gives an...
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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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