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Polynomial Proof

  1. Oct 24, 2005 #1
    "Let Px be union of all polynomials.
    Choose a an element of R, and define ta : Px --> R by ta(f) = f(a)
    Let T=ker(Ta). Prove that the map
    p(x) |--> (x − a)p(x)
    is a linear, one-to-one, and onto transformation Px --> T ."

    Is the assertation in the problem correct?
    If so, how do you prove it?
  2. jcsd
  3. Oct 25, 2005 #2


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    Choose [tex]a\in\mathbb{R}[/tex] and define [tex]t_{a}:P_{x}\rightarrow\mathbb{R} \mbox{ by }t_{a}(f)=f(a)[/tex].
    Let [tex]T=\mbox{ker}(t_{a})[/tex]. Prove that if [tex]\Lambda\mbox{ maps } p(x) \mapsto (x-a)p(x)[/tex], then [tex]\Lambda[/tex] is a linear, one-to-one transformation of [tex]P_{x}\mbox{ onto }T[/tex].
    Proof: Let [tex]f,g\in P_{x}[/tex], and let [tex]b,c\in\mathbb{R}[/tex].
    Then [tex]\Lambda(bf(x)+cg(x))=(x-a)(bf(x)+cg(x))=b(x-a)f(x)+c(x-a)g(x)=b\Lambda(f(x))+c\Lambda(g(x))[/tex], so [tex]\Lambda[/tex] is linear.
    Since [tex]\Lambda(f(x))=\Lambda(g(x))\Rightarrow (x-a)f(x)=(x-a)g(x)\Rightarrow f(x)=g(x)[/tex], [tex]\Lambda[/tex] is one-to-one.
    To see that [tex]\Lambda[/tex] maps [tex]P_{x}\mbox{ onto }T[/tex], consider that
    [tex]\forall f\in P_{x}[/tex], [tex]f\in\mbox{ker}(t_{a})\Leftrightarrow f(a)=0\Leftrightarrow \exists q\in P_{x} \mbox{ such that } f(x)=(x-a)q(x)\Leftrightarrow \exists q\in P_{x} \mbox{ such that } f(x)=\Lambda(q(x))[/tex] by the definition of ker and by the fundamental theorem of algebra.
    Last edited: Oct 25, 2005
  4. Oct 25, 2005 #3


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    In less formal terms:

    Ta maps each polynomial p(x) into the number p(a).
    The "kernel" of any linear transformation is the set of vectors it maps into 0 so the kernel of Ta is the set of all p(x) that have value 0 at x=a: p(a)= 0.

    Let [itex]\Lambda[/itex] be the transformation that takes p(x) into (x-a)p(x).

    1. Linear: [itex]\Lambda[/itex](mp(x)+ nq(x))= (x-a)(mp(x)+ nq(x)= m{(x-a)p(x)}+ n{(x-a)p(x)}= m [itex]\Lambda[/itex]p(x)+ n[itex]\Lambda[/itex]q(x).

    2. One-to-One. Suppose [itex]\Lambda[/itex]p(x)= [itex]\Lambda[/itex]q(x). That is (x-a)p(x)= (x-a)q(x). For x not equal to a, we can divide both sides by x-a to get p(x)= q(x). Since p and q are both polynomials, and so continuous, their values at x=a must also be the same: p(x)= q(x).

    3. Range of [itex]\Lambda[/itex] is a subset of kernel of T.
    If q(x) is in the range of [itex]\Lambda[/itex], q(x)= (x-a)p(x) for some polynomial p. Trivially, q(a)= (a-a)p(a)= 0. Therefore q(x) is in the kernel of T.

    4. [itex]\Lambda[/itex] is "onto".
    Let q(x) be in the kernel of T. Then q(x) is a polynomial such that q(a)= 0 and therefore has a factor of the form (x- a). q(x)= (x-a)p(x)= [itex]\Lambda[/itex]p(x).
    Last edited: Oct 25, 2005
  5. Oct 25, 2005 #4


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    Translating a problem can often make it easier to solve:

    We want to say:

    p(x) --> (x-a)p(x)

    is a one-to-one, onto transformation of Px -> T

    Breaking it down into pieces:

    (1) This is a map Px -> T
    is the same as
    For any p(x) in Px, (x-a)P(x) is in T
    which is the same as
    For any p(x) in Px, (a-a)P(a) = 0

    (2) This is a one-to-one map
    is the same as
    If (x-a)p(x) = (x-a)q(x) then p(x) = q(x).

    (3) This is an onto map
    is the same as
    If q(x) is in T, then there is a p(x) s.t. q(x) = (x-a)p(x).
    which is the same as
    If q(a) = 0, then there is a p(x) s.t. q(x) = (x-a)p(x)
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