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### 5 Cards in this Set

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 a linear transformation s from V->W is completely determined by what it does to a _____ of __ basis of V range of T {u in W: u=Tv for some v in V} show that range of T is a subspace of W Let w1, w2 be a range of T Let v1, v2 be in V: Tv1=w1, Tv2=w2 since w1, w2 is in range T T(v1+v2)=Tv1+Tv2 Therefore, w1 + w2 is in range T If T is 1-1, and Show {v1...vn} is linearly independent c1v1+c2v2+...+cnvn=0 T(c1v1+c2v2+...+cnvn)=T(0) Therefore, c1v1+c2v2+...+cnvn is in the kernel of T. Since T is 1-1, its kernel just 0 vector. c1v1+c2v2+...+cnvn=0 Since {v1,...Vn} is linearly independent where c1,c2,...=0 Given If T:V->W is a linear transformation such that Tv1...Tvn is linear independent, then must {v1,...,vn} be linearly independent? c1v1+c2v2+...+cnvn=0 T(c1v1+c2v2+...+cnvn)=T(0) c1Tv1 + c2Tv2+...+cnTvn=0 Then, {Tv1...Tvn} are vectors which are linearly independent. c1v1+c2v2+...+cnvn=0 because its in the kernel and the v in V: Tv=0. Therefore, c1, c2, cn=0.