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lecture_24 - MA 265 LECTURE NOTES WEDNESDAY MARCH 5 Bases...

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MA 265 LECTURE NOTES: WEDNESDAY, MARCH 5 Bases and Dimension Dimension. Let ( V, + , · ) be a real vector space. Let W = span S be a subspace of V in terms of a linearly independent set S = { v 1 , v 2 , . . . , v n } . If T = { w 1 , w 2 , . . . , w m } is any basis for W , then m = n . Last time be showed that m n as follows: We saw that W is spanned by S 1 = { w 1 , v 2 , . . . , v n } . Inductively, say that we have shown that W is spanned by S j - 1 = { w 1 , . . . , w j - 1 , v j , . . . , v n } . We want to show that W is spanned by S j = { w 1 , . . . , w j , v j +1 , . . . , v n } . Since w j is a vector contained in W = span S j - 1 , we see that there are scalars a i , b i R such that w j = ( b 1 w 1 + · · · + b j - 1 w j - 1 ) + ( a j v j + · · · + a n v n ) . Note that the subset T j = { w 1 , . . . , w j } of T must be linearly independent since T is linearly independent. If all of the a i = 0 then T j would be linearly dependent, so without loss of generality say that a j 6 = 0. Then we have the expression v j = - b 1 a j w 1 + · · · + - b j - 1 a j w j - 1 + 1 a j w j + - a j +1 a j v j + · · · + - a n a 1 v n .

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