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Prove one to one using rank-nullity theorem

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Let V and W be finite dimensional vector spaces.Given dim(V)≤ dim(W)
    , prove that there exists a one-to-one linear transformation T : V -> W .
    2. Relevant equations



    3. The attempt at a solution
    What I want to prove here is to show that nullity=0
    dim(V) ≤ dim (W)
    dim(N(T)) + dim(R(T)) ≤ dim(W)
     
  2. jcsd
  3. Feb 26, 2012 #2

    HallsofIvy

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    You can't prove that the nullity of T until after you have T! How are you defining T?
     
  4. Feb 26, 2012 #3
    I really have no idea no how to prove this question . Do I need to prove it for general T or I just need to find an example which satisfies all the properties above ??
     
  5. Feb 26, 2012 #4

    HallsofIvy

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    I suggest you go back and reread the problem! "prove that there exists"... First define an obvious linear transformation from V to W, then show that it is one-to-one.

    Here's a simple special case: suppose dim(V)= 2, dim(W)= 3. Then, choosing some basis for both V and W, you can write any vector in V as <x, y> and any vector in W as <a, b, c>. Okay, what can you say about T(<x, y>)= <x, y, 0>?
     
  6. Feb 26, 2012 #5

    Deveno

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    hint 2: if dim(V) = n, we have a basis {v1,v2,...,vn}.

    since dim(W) ≥ dim(V), what can you say about any basis of W?
     
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