Diagonalization of Operators


by the_kid
Tags: diagonalization, operators
the_kid
the_kid is offline
#1
Dec15-12, 03:12 PM
P: 116
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?
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
micromass
micromass is offline
#2
Dec15-12, 03:21 PM
Mentor
micromass's Avatar
P: 16,690
What you want to look at is diagonalization of compact self-adjoint operators. See: http://en.wikipedia.org/wiki/Spectra...oint_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.
the_kid
the_kid is offline
#3
Dec15-12, 11:55 PM
P: 116
Quote Quote by micromass View Post
What you want to look at is diagonalization of compact self-adjoint operators. See: http://en.wikipedia.org/wiki/Spectra...oint_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.
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!

the_kid
the_kid is offline
#4
Dec16-12, 03:00 PM
P: 116

Diagonalization of Operators


Anyone?


Register to reply

Related Discussions
A is n by n matrix. It is diagonalized by P. Find the matrix that Calculus & Beyond Homework 1
simultaneous diagonalization of two hermitian operators Quantum Physics 2
Diagonalization Calculus & Beyond Homework 3
Quantum Operators (or just operators in general) Advanced Physics Homework 22
Simultaneous diagonalization and replacement of operators with eigenvalues...? Quantum Physics 1