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Math Help - linear independence

  1. #1
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    linear independence

    how do you prove that if {f(v1), f(v2),...f(vn) } is linearly independent where f:V to W then f is injective?


    by def of LI, a linear combination of f(v1), f(v2),...f(vn) = 0 iff the coefficients are 0. so can i say that for all w in W, there is a unique linear combi of {f(v1), f(v2),...f(vn) } which will give w.
    this means that w in W is unique...

    and then how do i continue?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Prove that {v_1,...,v_n} is linearly independent also and use the following theorem:


    Let V and U be vector space over field K. Let {v_1,...,v_n} a basis of V and {u_1,...,u_n} a basis of U. Then exist a unique linear transformation F:V-->U which:

    F(v_1)=u_1, ... , F(v_n)=u_n.
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  3. #3
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    Why are we able to use the second theorem? I thought by saying that there is a unique linear transformation, we are assuming that it is injective.. Which is something we need to prove.

    Thanks!
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  4. #4
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    Quote Originally Posted by alexandrabel90 View Post
    how do you prove that if {f(v1), f(v2),...f(vn) } is linearly independent where f:V to W then f is injective?
    You are leaving out a lot here. In order for this to be true, {v1, v2, ..., vn} has to be a basis for V.


    by def of LI, a linear combination of f(v1), f(v2),...f(vn) = 0 iff the coefficients are 0. so can i say that for all w in W, there is a unique linear combi of {f(v1), f(v2),...f(vn) } which will give w.
    this means that w in W is unique...

    and then how do i continue?
    Let u and w be such that f(u)= f(v). Then, since {v1, v2,..., vn} is a basis for V, there exist a1, a2, ..., an such that a1v1+ a2v2+ ...+ anvn= u and b1, b2, ..., bn such that b1v1+ b2v2+ ...+ bnvn= v. Then, since f(u)+ f(v), a1f(v1)+ a2f(v2)kl+ ...+ anf(vn)= b1f(v1)+ a2f(v2)+ ...+ bnf(vn). Rewrite that so that you have a linear combination of f(v1), f(v2), ..., f(vn) equal to 0 and use the fact that they are independent to show that a1= b1, a2= b2, ..., an= bn and so u= v.
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  5. #5
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    Opps ya. v1 to vn is a basis for V. Sorry!
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