Recent content by L.Wil

  1. L

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

    Thanks so much for all of your help. If λ=1, f'(x)=0 and so f(x) = constant?
  2. L

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

    I solved it to get f(x) = ke^((λ-1)x) Should i rearrange for λ?
  3. L

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

    No I don't think so?
  4. L

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

    That there aren't enough eigenvalues for it to be diagonalizable? I'm not really sure!
  5. L

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

    That the degree of the polynomial on the right is one less than on the left?
  6. L

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

    Looking at my notes I think that k[x] are polynomials
  7. L

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

    It doesn't say whether k[x] are polynomials. So would the ODE be λf = f + f' And then I solve for λ?
  8. L

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

    I also thought I could make it into an ODE and solve for the eigenvalue but i couldn't get that to work
  9. L

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

    Homework Statement T: k[x]n -> k[x]n T(f) = f + f' show that T is not diagonalizable for n >= 1 Homework Equations The Attempt at a Solution I would usually start by getting a characteristic polynomial but I don't know how to do that here?
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