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Math Help - prove linear transformation is not surjective

  1. #1
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    prove linear transformation is not surjective

    I know how to prove that it is one to one, but i don't know how to prove that it's not onto, can some1 help me ?

    ps: P(R) is a polynomial
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The endomorphism is:

    T:P(\mathbb{R})\rightarrow P(\mathbb{R})\,,\quad T[p(x)]=\int_0^x p(t)\;dt

    Suppose there exists p(x)\in P(\mathbb{R}) such that T[p(x)]=1 , use the Fundamental Theorem of Calculus and you will find a contradiction.


    Fernando Revilla
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  3. #3
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    Quote Originally Posted by mystic View Post
    I know how to prove that it is one to one, but i don't know how to prove that it's not onto, can some1 help me ?

    ps: P(R) is a polynomial

    Hint: Check that for any polynomial F(x)\,,\,\,F(x)-F(0) is a polynomial with zero free coefficient...

    Tonio
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  4. #4
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    well isn't p(x) = 0 a polynomial ? so even we let T (p(x))= 1, p(x) = 0 still works right ?
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    Quote Originally Posted by mystic View Post
    well isn't p(x) = 0 a polynomial ? so even we let T (p(x))= 1, p(x) = 0 still works right ?


    No, p(x)=0 is not a polynomial, of course. It looks like an equation, though.

    Tonio
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  6. #6
    Senior Member roninpro's Avatar
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    Quote Originally Posted by tonio View Post
    No, p(x)=0 is not a polynomial, of course. It looks like an equation, though.

    Tonio
    If we don't treat the zero function as a polynomial, then P(\mathbb{R}) fails to be a vector space. Could you clarify what you meant?
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    Quote Originally Posted by roninpro View Post
    If we don't treat the zero function as a polynomial, then P(\mathbb{R}) fails to be a vector space. Could you clarify what you meant?


    I meant the whole expression "p(x)=0" is not a pol. but an equation, of course.

    And in this case it is the zero polynomial, not the zero function. Had he written "p(x):=0" things would

    have been clearer (for me, at least).

    Tonio

    Ps. By the way, I think my first message shows easily and clearly why the given transf. isn't surjective, but

    there hasn't been any discussion on it...
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