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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 mb 1 w 1b 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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 Spring '10
 MOORE
 Linear Algebra, Algebra, Vector Space

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