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lecture_23 - MA 265 LECTURE NOTES MONDAY MARCH 3 Bases and...

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MA 265 LECTURE NOTES: MONDAY, MARCH 3 Bases and Dimension Bases and Span. Let ( V, + , · ) be a real vector space. Let W = span S be the span of some collection S = { v 1 , v 2 , . . . , v k } of vectors from V . Then W has a basis T S . We sketch the proof via an algorithm: #1. Start with j = 1 and let S 1 = { v 1 } . #2. Set j + 1 j . We assume that is has already been shown that S j - 1 is linearly independent. #3. Choose v j S , and consider the following equation: a 1 v 1 + · · · + a j - 1 v j - 1 + a j v j = 0 . If there is a solution with a j 6 = 0, then we can write v j = - a 1 a j v 1 + · · · + - a j - 1 a j v j - 1 = v j span S j - 1 . In this case choose S j = S j - 1 . Otherwise, choose S j = S j - 1 ∪ { v j } . In either case the set S j is linearly independent. #4. If j < k , return back to Step #2. Otherwise, give the result T = S k . Example. Let V = R 3 , and consider W = span S where S = { v 1 , v 2 , v 3 , v 4 , v 5 } in terms of the vectors v 1 = 1 0 1 , v 2 = 0 1 1 , v 3 = 1 1 2 , v 4 = 1 2 1 , and v 5 = - 1 1 - 2 . We perform the following: We find a basis for W i.e. we compute a spanning set T S that is linearly independent.
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