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Criteria for isomorphism.

  1. Mar 21, 2008 #1
    Can anyone tell me clearly what the criteria for isomorphism in linear algebra is?

    For instance, my book gives the following reason:

    Transformation T is not isomoprhic because T((t-1)(t-3)) = T(t^2 - 4t +3) = zero vector.

    I don't get why this means T is not an isomorphism. Can anyone explain?

    PS. T is a transformation from P_2 to R^(3).

    the actual T is: T(f(t)) =

    f(1)
    f'(2)
    f(3)

    Thanks.
     
  2. jcsd
  3. Mar 21, 2008 #2
    Let V and W be vector spaces. A linear transformation T:V->W is an isomorphism if it's bijective. That's basically the definition.

    It's easy to show a linear transformation is injective iff kerT = {0}.

    So in your example, T(polynomial) = 0 ,but polynomial != 0, so T is not injective, so it certainly can't be an isomorphism.
     
    Last edited: Mar 21, 2008
  4. Mar 21, 2008 #3
    so, it is not injective if the polynomial is not equal to 0?
    How does the statement
    "T((t-1)(t-3)) = T(t^2 - 4t +3) = zero vector."
    show this?
     
  5. Mar 21, 2008 #4
    because that implies t^2 - 4t + 3 is in kerT, so kerT != {0}, so T is not injective.
     
  6. Mar 21, 2008 #5
    Oh, I see; you mean the only solution for the T has to be 0 and in this case there are more?
     
    Last edited: Mar 21, 2008
  7. Mar 21, 2008 #6
    i'll show all the steps once more,

    We know T(t^2 - 4t + 3) = 0, so t^2 - 4t + 3 is in kerT, and since t^2 - 4t + 3 is not the zero polynomial we know kerT contains a nonzero element, so kerT != {0}, so T is not injective. (this last part is because T is injective iff kerT = {0})

    Where exactly did I lose you?
     
  8. Mar 21, 2008 #7
    I was jsut asking for confirmation. So, the isomorphism exists only if ker = 0 and if you can write a formula for the inverse, correct? thanks!
     
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