What are the eigenvalues and eigenfunctions for T(f(x)) = 5f(x) on C for V:R->R?

[glueball8]

Homework Statement

T(f(x)) = 5 f(x)
T is defined on C. Find all real eigenvalues and real eigenfunction. V:R -> R

Homework Equations

Not sure.

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?

Thanks

 

[Mark44]

Isn't 5 the eigenvalue?

 

[glueball8]

Isn't 5 the eigenvalue?

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?

 

[HallsofIvy]

How about using the definition of "eigenvalue": If T is a linear transformation that maps functions into functions, then \alpha is an "eigenvalue" and a non-zero function, f(x), is an eigenvector if and only if Tf(x)= \alpha f(x). 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.

 

[vela]

Bad example. So the eigenfunction is any function?

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

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

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 y'-\lambda y=0 is a simple example. You could write it as

D(y) = \lambda y

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

Did you have a specific type of problem in mind?

 

[glueball8]

Ok cool. :)

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

Thanks

 

[Mark44]

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