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Eigenvalue of Polynomial Transformation

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Let T:P2→P2 be defined by
    T(a0+a1x+a2x2)=(2a0-a1+3a2)+(4a0-5a1)x + (a1+2a2)x2
    1) Find the eigenvalues of T
    2) Find the bases for the eigenspaces of T.

    I believe the 'a' values are constants.
    2. Relevant equations
    None.

    3. The attempt at a solution
    The problem I am having is actually pulling out the matrix for T. I know how to find eigenvalues (by solving det(λI-A) - A being the matrix) and from that finding the bases of the eigenspaces comes from substituting the eigenvalues into (λI-A) and performing elementary row operations to find the eigenvectors which form the bases.

    What I have tried to do is separate the basis (1,x,x2) from the rest and come up with the matrix :
    2a0 -a1 3a2
    4a0 -5a1 0a2
    0a0 a1 2a2
    Am I on the right track here or am I barking up the wrong tree? If so what method should i follow?
     
  2. jcsd
  3. Mar 29, 2012 #2
    Well, sort of. You need to figure out which elements the polynomials [itex]1,x,x^2[/itex] get sent to by [itex]T[/itex]. Once you do that, put them into a matrix and proceed as you have described. So, if you put the polynomial [itex]p_1(x) = x[/itex] into [itex]T[/itex] what would come out "the other side"?
     
  4. Mar 29, 2012 #3

    tiny-tim

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    Hi PandaGunship! Welcome to PF! :smile:
    leave all the a's out! :wink:
     
  5. Mar 29, 2012 #4
    Just to check, my matrix is correct?
    As per your direction, a0 goes to 2a0 + 4a0x +0x2 which I make into a column like the matrix I put it already?
    Thanks Tim :) I thought of leaving the a's out, but what of their importance? Are they just to show what each constant of the basis is mapped into?
     
  6. Mar 29, 2012 #5

    tiny-tim

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    No, your vectors will have a's. :smile:

    (The 1 x x2 are just a basis like x y z …

    you wouldn't expect x y z to appear in a vector or matrix, would you? :wink:)​
     
  7. Mar 29, 2012 #6
    Well, I said to find out where [itex]p_1(x) = x[/itex] goes, and what you did was find where [itex]p(x) = a_0 [/itex] goes. I only mentioned to do it as an example, so its OK that you did it for [itex]p(x) = a_0 [/itex].

    So, the [itex]a_0, a_1, a_2 [/itex] are there because you need to know what [itex]T[/itex] does to a general quadratic polynomial, and that is the form of a general quadratic polynomial.

    Now, the three basis polynomials we are using are: [itex] p_0(x) = 1, p_1(x) = x, p_2(x) = x^2[/itex]. Now, for each of these basis polynomials, what is [itex]a_0, a_1, a_2[/itex]? As an example, for [itex]p_0[/itex] we have that [itex]a_0 = 1, a_1 = 0, a_2 = 0[/itex]. So, [itex]T(p_0(x)) = 2a_0 + 4a_0x = a + 4x[/itex]. So, you got it correct, but as Tiny Tim said, you need to get rid of the a's because for the basis polynomials they are all 1 or 0.
     
  8. Mar 29, 2012 #7
    I suppose not :P Thanks :)
    Ok I understand properly now. In the beginning when I was going over this problem I thought that I must remove the a's but I didnt want to remove them 'just because'. I wanted to know that what I was doing was actually correct, even if my reasoning was a bit off the mark.
    Thank you both for your help:biggrin:
     
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