# lecture_22 - MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 29...

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MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 29 Linear Independence (cont’d) Linear Independence and Span. Let ( V, + , · ) be a real vector space. Let S = { v 1 , v 2 , ..., v k } be a collection of vectors from V . Recall that W = span S is a subspace of V , and S is linearly dependent if the equation a 1 v 1 + a 2 v 2 + ··· + a k v k = 0 has a solution with not all of the a i equal to zero. There is an inductive way to determine whether S is linearly dependent. Consider the following subsets of S : S 1 = { v 1 } S 2 = { v 1 , v 2 } . . . S j = { v 1 , v 2 , ..., v j } ⊆ S, j = 1 , 2 , ..., k. Note that S j - 1 S j , so that span S j - 1 span S j . Recall that if some S j is linearly dependent, then so is S ; and if S is linearly independent, then so are all S j . We use this to create an algorithm to “weed out” those linearly dependent vectors. Consider the following steps: #1. Initialize j = 1, and consider S 1 = { v 1 } . This is linearly dependent if and only if v 1 = 0 . Hence we may as well assume that v 1 6 = 0 is a nonzero vector. Otherwise, replace v 2 v 1 and repeat step. #2. Increase j + 1 j . We assume that we’ve already veriﬁed that S j - 1 is linearly independent. #3. Consider S j . If it is linearly independent, then return to Step #2. Otherwise, assume that it is linearly dependent. Consider an equation in the form a 1 v 1 + a 2 v 2 + ··· + a j - 1 v j - 1 + a j v j = 0 . We know that there exists a solution with not all of the

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## This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.

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lecture_22 - MA 265 LECTURE NOTES: FRIDAY, FEBRUARY 29...

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