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Unformatted text preview: MA 265 LECTURE NOTES: MONDAY, MARCH 3 Bases and Dimension Bases and Span. Let ( V, + , Â· ) 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 âŠ† S . We sketch the proof via an algorithm: #1. Start with j = 1 and let S 1 = { v 1 } . #2. Set j + 1 â†’ j . We assume that is has already been shown that S j 1 is linearly independent. #3. Choose v j âˆˆ S , and consider the following equation: a 1 v 1 + Â·Â·Â· + a j 1 v j 1 + a j v j = . If there is a solution with a j 6 = 0, then we can write v j = a 1 a j v 1 + Â·Â·Â· + a j 1 a j v j 1 = â‡’ v j âˆˆ span S j 1 . In this case choose S j = S j 1 . Otherwise, choose S j = S j 1 âˆª { 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 1 , v 2 = 1 1 , v 3 = 1 1 2 , v 4 = 1 2 1 , and v 5 =  1 1 2 ....
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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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