This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ﬁeld. 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 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...
View
Full Document
 Spring '10
 MOORE
 Linear Algebra, Algebra, Vector Space, bn wn, Professor J. Douglas Moore

Click to edit the document details