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Function that is an isomorphism

  1. Jan 1, 2014 #1
    1. The problem statement, all variables and given/known data

    So my text states the proposition:
    If V and W are finite dimensional vector spaces, then there is an isomorphism T:V→W ⇔ dim(V)=dim(W).

    So, in an example the text give the transformation T:P[itex]_{3}[/itex](R)→P[itex]_{3}[/itex](R)
    defined by T(p(x)) = x dp(x)/dx.

    Now I understand T is not an isomorphism since ker(T) = span(1) , the set of all constant polynomial functions. But by the above theorem since the dim(V) = dim(W), T would an isomorphism.
    So I'm a bit confused.
     
  2. jcsd
  3. Jan 1, 2014 #2
    The theorem says that there is some function that is an isomorphism. It does not say that any function ##T## is an isomorphism.

    In your example, the theorem is satisfied since ##S(x) = x## is an isomorphism. But the theorem does not say that any arbitrary ##T## is an isomorphism. So you can not deduce that your ##T## is an isomorphism from the theorem.
     
  4. Jan 1, 2014 #3

    LCKurtz

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    No. The theorem says that if dim(V) = dim(W) then there is an isomorphism. It doesn't say any old map you choose is an isomorphism. Your map isn't onto.
     
  5. Jan 1, 2014 #4
    Ok thanks alot. Its makes sense now.
     
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