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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.