Diagonalization of Operators

1. Dec 15, 2012

the_kid

So, obviously one can diagonalize any self-adjoint transformation on a finite dimensional vector space. This is pretty simple to prove. What I'm curious about is integral operators. How does this proof need to be adapted to handle integral operators? What goes wrong? What do we need to account for? Are there any corresponding theorems I might want to look at?

2. Dec 15, 2012

micromass

Staff Emeritus
What you want to look at is diagonalization of compact self-adjoint operators. See: http://en.wikipedia.org/wiki/Spectral_theorem#Compact_self-adjoint_operators
This theorem is a direct generalization of the theorem in finite dimensions.

So in order to apply this theorem, you will want to prove that your integral operator is compact. This will use some form of the Ascoli-Arzela theorem. But you want to look at Fredholm operators: http://en.wikipedia.org/wiki/Fredholm_integral_operator

If your operators are not compact, then there are other diagonalization theorems that may apply. But those theorems are more abstract and are not (immediately) of the form as the theorem in finite dimensions. But in any case, it is nice to know that any self-adjoint operator can be diagonalized if you interpret diagonalization suitably.

3. Dec 15, 2012

the_kid

Thanks so much for the help, micromass. The proof for the finite dimensional case relies on the fact that an eigenvalue of the operator can be found (at least the proof I'm familiar with). The operator Tf(x)=xf(x) on L^2[(0,1)] is obviously self-adjoint but has no eigenvalues. So, trying to apply the finite dimensional proof would break down here. Thus, this leads us to demanding that our operator T be compact--one can prove that it them must have an eigenvalue. Is this all correct? My question is where do integral operators fit into this discuss? What does it mean for an integral operator to be compact? Thanks for your help!

4. Dec 16, 2012

Anyone?