
#1
Dec1512, 03:12 PM

P: 116

So, obviously one can diagonalize any selfadjoint 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
Dec1512, 03:21 PM

Mentor
P: 16,554

What you want to look at is diagonalization of compact selfadjoint 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 AscoliArzela 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 selfadjoint operator can be diagonalized if you interpret diagonalization suitably. 



#3
Dec1512, 11:55 PM

P: 116





#4
Dec1612, 03:00 PM

P: 116

Diagonalization of Operators
Anyone?



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