108APracticeQuizF

108APracticeQuizF - T ), w = T ( v ) = b 1 T ( w 1 ) + + b...

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Name: Mathematics 108A: Practice Quiz F April 28, 2010 Professor J. Douglas Moore One of the Main Theorems from Chapter 2 of the text is: Theorem. If V is a finite dimensional vector space and T : V W is a linear map into a vector space W , then dim V = dim N ( T ) + dim R ( T ) . The proof of this theorem is based upon the theory of linear independence and bases presented in Chapter 1. We start by choosing a basis ( u 1 ,..., u m ) for N ( T ). The Extension Theorem states that we can extend this to a basis ( u 1 ,..., u m , w 1 ,..., w n ) of V . If we can show that ( T ( w 1 ) ,...,T ( w n )) is a basis for R ( T ), then dim R ( T ) = n . It will then follow that dim V = m + n = dim N ( T ) + dim R ( T ) , and the theorem will be proven. Thus we need only carry out the following two steps: 1. Prove that the list ( T ( w 1 ) ,...,T ( w n )) spans R ( T ). Solution: Suppose that w R ( T ). Then there exists v V such that T ( v ) = w . We can write v = a 1 u 1 + ··· + a m u m + b 1 w 1 + ··· + b n w n , and then T ( v ) = a 1 T ( u 1 ) + ··· + a m T ( u m ) + b 1 T ( w 1 ) + ··· + b n T ( w n ) , and since u 1 ,..., u m lie in null(
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Unformatted text preview: T ), w = T ( v ) = b 1 T ( w 1 ) + + b n T ( w n ) . Thus w lies in the span of ( T ( w 1 ) ,...,T ( w n )). 2. Prove that the list ( T ( w 1 ) ,...,T ( w n )) is linearly independent. Solution: Suppose that b 1 T ( w 1 ) + + b n T ( w n ) = , 1 for some elements b 1 ,...,b n of the eld. Then T ( b 1 w 1 + + b n w n ) = , so b 1 w 1 + + b n w n N ( T ) . But then b 1 w 1 + + b n w n = a 1 u 1 + + a m u m , for some choice of scalars a 1 ,...,a m . Thus a 1 u 1 + + a m u m-b 1 w 1--b n w n = . Since the list ( u 1 ,..., u m , w 1 ,..., w n ) is linearly independent, a 1 = = a m = b 1 = b n = 0 . From this we conclude that ( T ( w 1 ) ,...,T ( w n )) is linearly independent. 2...
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108APracticeQuizF - T ), w = T ( v ) = b 1 T ( w 1 ) + + b...

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