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T:P2 to R2, find rank or nullity of T

  1. Jul 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Find either the rank or nullity of T.
    T:P2--> R2 defined by T(p(x)) = [p(0)
    p(1)]

    2. Relevant equations
    Null(T)={x:T(x)=0}
    I think its usually easier to to find Nullity as opposed to Rank.


    3. The attempt at a solution
    I have only done these questions within the same vector spaces, I don't have any idea where to begin. Hints or clues would be very much appreciated.
     
  2. jcsd
  3. Jul 4, 2010 #2

    Hurkyl

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    P2 is the space of quadratic polynomials?


    Then transform the problem to one where both vector spaces are the same, solve, then transfer back.

    BTW, why does it matter to you whether or not they're the same?
     
  4. Jul 4, 2010 #3
    I'm still new to linear transformstions and am not comfortable with the concept yet. And yes P2 is the space of quadratic polynomials.
     
  5. Jul 4, 2010 #4

    HallsofIvy

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    If P2 is the space of quadratic polynomials, the any p in P2 can be written as [itex]ax^2+ bx+ c[/itex]. p(0)= c and p(1)= a+ b+ c. The nullity of T consists of all quadratic polynomials, [itex]ax^2+ bx+ c[/itex] such that p(0)= c= 0 and p(1)= a+ b+ c= 0. What polynomials are those?

    Actually, I would think it very easy to find the rank of T. If (x, y)= (c, a+ b+ c) where a, b, and c can be any real numbers, the what can x and y be?
     
  6. Jul 4, 2010 #5
    If I stick with looking for nullity, I get nullity=1.
     
  7. Jul 4, 2010 #6

    Dick

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    Yes, the dimension of the null space is 1. Can you write down a polynomial that spans it?
     
  8. Jul 4, 2010 #7
    1+2x
     
  9. Jul 4, 2010 #8

    Dick

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    Uh, look at your original definition. If p(x)=1+2x, then T(p(x))=[p(0),p(1)]=[1,3]. It doesn't look like 1+2x is in the null space to me.
     
  10. Jul 4, 2010 #9
    x^2-x+0

    If this is not right then I am definitly confused
     
    Last edited: Jul 4, 2010
  11. Jul 5, 2010 #10

    Dick

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    Now that's right. So the elements of the null space are c*(x^2-x) for any constant c. Hence one dimensional. Right?
     
  12. Jul 5, 2010 #11
    Yes, nullity(T)=1, then with the rank theorem, rank(T)=2,
    Thanks for the help, I think I have to just keep doing problems until I can grasp this better.
     
  13. Jul 5, 2010 #12

    HallsofIvy

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    To find the rank directly (which I have to do because I said "Actually, I would think it very easy to find the rank of T.":smile:), note that T(ax^2+ bx+ c)= (c, a+ b+ c). Since a, b, and c can be any numbers, so can c and a+ b+ c. That is, the image of T is all or [itex]R^2[/itex] and so has dimension 2.

    Of course, once you have the nullity, using the rank theorem is simpler.
     
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