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Homework Help: How to prove an isomorphism

  1. Jul 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I have D(p(x)) = (the second derivative of p with respect to x ) - (2 derivative of p with respect to x) + p

    Proof that D(p(x)) is not an isomorphism

    2. Relevant equations

    3. The attempt at a solution

    Just by watching the problem it seems I can assign one unique value to D(p(x)) and receive an unique image. It seems by inspection that it is one to one. It also seems onto by inspection.
  2. jcsd
  3. Jul 16, 2012 #2
    What are the domain and codomain??
    Isomorphism of what?? Vector spaces, groups, rings, modules???
  4. Jul 16, 2012 #3
    Sorry, It is a linear Operator D from P3 to P3 where P3 is the polynomial space of degree equal or less than 3.
  5. Jul 16, 2012 #4
    Did you check whether it is injective or surjective??
  6. Jul 16, 2012 #5
    can I say that it is not onto because it is not defined for degrees greater than 3?
  7. Jul 16, 2012 #6
    No, degrees greater than 3 don't matter because they are not in the domain or codomain.
  8. Jul 16, 2012 #7
    How can I start doing that?
  9. Jul 16, 2012 #8
    Do you know about the matrix form of a linear operator?? That might help you.
  10. Jul 16, 2012 #9
    No I haven't, how can I do that, and where can I find some more information about it? that might come useful for nexts week test. I would appreciate it
  11. Jul 16, 2012 #10
    Oh well, if you haven't seen it yet, then you probably aren't allowed to use it. You'll see it when the time comes!!

    Anyway, can you calculate the kernel of the linear map?? Find out which p are such that D(p)=0??
  12. Jul 16, 2012 #11
    well I guess that would be just P = 0 because there is a + p , thats the reason why I thought it would be injective.
  13. Jul 16, 2012 #12
    Hmm, yes, it does turn out to be an isomorphism... Weird...
  14. Jul 16, 2012 #13
    Are you sure? because the problem states that it is not :P
  15. Jul 16, 2012 #14
    how did you find out it was onto just by inspection or is there a another way?
  16. Jul 17, 2012 #15


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    Something is an isomorphism if there exists a linear bijective transformation T such that :

    T(T-1) = Id

    Where Id is the identity transformation ( The do nothing transformation ).

    So your question is abit vague, but you have a transformation :

    D(p(x)) = p''(x) - 2p'(x) + p ?

    You phrased it terribly so im guessing that's what you meant? Also is p a real number?
  17. Jul 17, 2012 #16
    Yes it is like that before that I'm asked to find D with respect to the base B = { p1=1, p2=x, p3=x^2 and p^4 =x^3 }

    I already did that, then I'm asked to prove that D is not an isomorphism I guess p can be a real number in P3 which is the space of polynomials less or equal to 3.
  18. Jul 17, 2012 #17


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    What was your answer to this question?
  19. Jul 17, 2012 #18


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    I believe p is an element of P3, i.e. it is a polynomial of degree <= 3. This means p could be a real number (which is a polynomial of degree 0), but it doesn't have to be.
  20. Jul 17, 2012 #19


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    I get that it's an isomorphism too.

    What did you mean by this? It sounds like you found the matrix representing D relative to B, but you told micromass you didn't know how to find the matrix representation of a linear operator.
  21. Jul 17, 2012 #20
    Well I'm a bit confused because I'm taking this subject for the first time but I know how to find a linear transformation with respect to a given base.

    For D with respect to the basis B = { p1=1, p2=x, p3=x^2 and p^4 =x^3 }

    the result was D relative to B =
    1 -2 2 0
    0 1 -4 6
    0 0 1 6
    0 0 0 1
  22. Jul 17, 2012 #21


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    Okay, that matrix is what micromass was referring to earlier. If its determinant isn't 0, it's invertible, which tells you D is bijective.

    Could just be a typo, but I got a -6 in the third row.
  23. Jul 17, 2012 #22
    Yes it is -6 thanks a lot for your time.
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