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Kernal, range and linear transformations

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data

    T: P2 --> P2 be a linear transformation defined by T(p(x)) = xp'(x)

    where ' is the derivative
    Describe the kernal and range of T and are any of the following polynomials in the range and or in the kernal of T?
    2
    x2
    1 - x

    2. Relevant equations

    power rule (for derivatives)

    3. The attempt at a solution
    I took an arbitrary a+bx+cx2 and turned it into bx + 2cx2 after applying T on it. I understand that the kernal means some vector that will turn it into the zero vector (like finding its null). so does that b = c = 0 in order for this to work? Then will it imply that the kernal of T is the set of all constant polynomials? If this is right, then from this I think x2 and 2 and 1 - x should be in ker(T).

    As for the range of T, I think it is all of P2, so that includes the 2, x2 and 1 - x.

    Is all of what I've just done correct? I like checking my work here when I get stuck, you guys are so nice and great at helping people! :)
     
  2. jcsd
  3. Mar 21, 2009 #2

    Mark44

    Staff: Mentor

    Or more mathematically, T(a+bx+cx2) = bx + 2cx2.
    To find the kernel (no such word as kernal) of T, what values of a, b, and c give you output polynomials that are zero for any value of x? IOW, for what values of a, b, and c is bx + 2cx2 identically zero?
    Constant polynomials are in the kernel, but the other two functions aren't.
    No, there are some polynomials in P2 that aren't in the range of T. Look at this equation again--T(a+bx+cx2) = bx + 2cx2--and notice that there are some polynomials that aren't in the range.
     
  4. Mar 21, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Okay

    Are you saying that you think x2 and 1- x are constant polynomials?

    You said above that T(a+ bx+ cx2)= bx+ 2cx2. How do you get 1- x from that?

    I'm not! I'm mean and grumpy!
     
  5. Mar 21, 2009 #4
    oups sorry for misspelling kernel.

    I looked up constant polynomial again and realized I made another mistake... only 2 is a constant polynomial in this case. So only 2 should be in ker(T). My bad...

    And so, I think the range should be something like the span of these basis: x and x2 ? I realized I couldn't get 1 - x in there, so x2 should be in the range of T.
    Thanks for pointing that out.

    HallsOfIvy, I see you as a nice person too, not grumpy and not mean, you're still helping me just like 2 months ago or so (can't remember exactly) but I can see why you're upset about me originally thinking about what constant polynomial was... :)

    Did I get it right this time?
     
    Last edited: Mar 21, 2009
  6. Mar 21, 2009 #5

    Mark44

    Staff: Mentor

    You're not saying that 2 is the only constant polynomial in the kernel of T, are you?
     
  7. Mar 21, 2009 #6
    no, 2 is not the only constant, but the question also asked me if 2 was included in there. Ker(T) is all constant values.

    *edit* I think I got it. Thanks Mark, and Ivy.
     
    Last edited: Mar 21, 2009
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