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5 Cards in this Set
- Front
- Back
a linear transformation s from V->W is completely determined by what it does to a _____ of __
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basis of V
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range of T
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{u in W: u=Tv for some v in V}
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show that range of T is a subspace of W
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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 |
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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 |
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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. |