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Homework Help: Eigenvalues and eigenfunction

  1. Jan 20, 2010 #1
    1. The problem statement, all variables and given/known data
    T(f(x)) = 5 f(x)
    T is defined on C. Find all real eigenvalues and real eigenfunction. V:R -> R

    2. Relevant equations
    Not sure.

    3. The attempt at a solution
    No, clue. I can find eigenvalues for matrices, that's not a problem. I'm having problem that its a T(function) = something function, how do I solve a problem like this in general?

    Any hints?

  2. jcsd
  3. Jan 20, 2010 #2


    Staff: Mentor

    Isn't 5 the eigenvalue?
  4. Jan 20, 2010 #3
    :redface: Bad example. So the eigenfunction is any function?

    Uh how to do theses in general? Can you make up something more complicated and explain how to do it?

    Or a link?
  5. Jan 20, 2010 #4


    User Avatar
    Science Advisor

    How about using the definition of "eigenvalue": If T is a linear transformation that maps functions into functions, then [itex]\alpha[/itex] is an "eigenvalue" and a non-zero function, f(x), is an eigenvector if and only if [itex]Tf(x)= \alpha f(x)[/itex]. If you are told that Tf(x)= 5f(x) for all f, then, yes, 5 is the only eigenvalue and every function in the space is an eigenvector. T just "multiplies by 5" and is exactly the same as a diagonal matrix having all "5" on its diagonal.
  6. Jan 20, 2010 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    If all T does is multiply a function by 5, then yes.

    Usually, you solve a differential equation. Equations like the Legendre's differential equation, the Bessel's differential equation, and the Schrodinger equation are all of this form.

    The differential equation [itex]y'-\lambda y=0[/itex] is a simple example. You could write it as

    [tex]D(y) = \lambda y[/tex]

    where D is the derivative operator. The solution to this equation [itex]y=e^{\lambda x}[/itex] is an eigenfunction of D.

    Did you have a specific type of problem in mind?
  7. Jan 20, 2010 #6
    Ok cool. :)

    How about T(f(x)) = 4f(-x) + f'(x) + 6f(6)? (Might not be do able) Or something like that?

  8. Jan 21, 2010 #7


    Staff: Mentor

    Give us an actual problem, not something you just made up.
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