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T(f) = f + f', show that T is not diagonalizable

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    T: k[x]n -> k[x]n

    T(f) = f + f'

    show that T is not diagonalizable for n >= 1


    2. Relevant equations



    3. The attempt at a solution

    I would usually start by getting a characteristic polynomial but I don't know how to do that here?
     
  2. jcsd
  3. Oct 30, 2012 #2
    I also thought I could make it into an ODE and solve for the eigenvalue but i couldn't get that to work
     
  4. Oct 30, 2012 #3

    Dick

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    k[x] are polynomials? If so look at the ODE an eigenvector has to satisfy and think about degrees of polynomials.
     
  5. Oct 31, 2012 #4
    It doesn't say whether k[x] are polynomials.

    So would the ODE be λf = f + f'

    And then I solve for λ?
     
  6. Oct 31, 2012 #5

    Dick

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    The definition of k[x] is pretty important to how you solve this problem. You need to find out what it is.
     
  7. Oct 31, 2012 #6
    Looking at my notes I think that k[x] are polynomials
     
  8. Oct 31, 2012 #7

    Dick

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    Write your equation as (λ-1)f = f'. If λ isn't 1, what can you say about the degrees of the polynomials on each side?
     
  9. Oct 31, 2012 #8
    That the degree of the polynomial on the right is one less than on the left?
     
  10. Oct 31, 2012 #9

    Dick

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    What do you conclude from that?
     
  11. Oct 31, 2012 #10
    That there aren't enough eigenvalues for it to be diagonalizable?

    I'm not really sure!
     
  12. Oct 31, 2012 #11

    Dick

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    You going to try to show there aren't enough eigenvectors or functions. To do that you have to figure out all the possible ways (λ-1)f = f' can be solved. Go back to your comment about the λ≠1 case. Can two polynomials of different degree be equal?
     
  13. Oct 31, 2012 #12
    No I don't think so?
     
  14. Oct 31, 2012 #13
    I solved it to get f(x) = ke^((λ-1)x)

    Should i rearrange for λ?
     
  15. Oct 31, 2012 #14

    Dick

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    That doesn't work because the exponential isn't generally a polynomial. You only want polynomial solutions. You've ruled out λ≠1. Are there polynomial solutions if λ=1? What do they look like?
     
  16. Oct 31, 2012 #15
    Thanks so much for all of your help.

    If λ=1, f'(x)=0

    and so f(x) = constant?
     
  17. Oct 31, 2012 #16

    Dick

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    Very welcome! Ok, so your only possible eigenvalue is 1 and all of the eigenfunctions are constants. If T were diagonalizable in k[x]_n how many linearly independent eigenfunctions would you need? What's the dimension of k[x]_n?
     
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