MHB Do Commuting Linear Integral Operators Share Eigenfunctions and Eigenvalues?

  • Thread starter Thread starter sarrah1
  • Start date Start date
  • Tags Tags
    Operators
Click For Summary
Commuting linear integral operators do not necessarily share the same eigenfunctions or eigenvalues, despite their kernels commuting. An example illustrates that two operators can commute while having distinct eigenfunctions and eigenvalues. Specifically, operators defined with kernels like $k(x,s) = s^2$ and $l(x,s) = x^2$ demonstrate this phenomenon. The relationship between the eigenvalues of the inverse of one operator and the other is also not guaranteed. Therefore, the properties of eigenfunctions and eigenvalues for commuting operators differ from those of matrices.
sarrah1
Messages
55
Reaction score
0
I have two linear integral operators

$Ky=\int_{a}^{b} \,k(x,s)y(s)ds$

$Ly=\int_{a}^{b} \,l(x,s)y(s)ds$

their kernels commute

Do they have same eigenfunctions like matrices and for instance in this case their product is the product of their eigenvalues. I am poorly read in operator theory but well read in matrices. Does the inverse of one has inverse eigenvalues
thanks
Sarrah
 
Physics news on Phys.org
A:The answer to your question is no.Let $K$ and $L$ be the operators you describe in your question, and let $\lambda$ be an eigenvalue of $K$ with eigenvector $v$. Then, by definition, $Kv = \lambda v$. However, it does not follow that $Lv = \lambda v$; in particular, it is not necessarily true that $Lv$ will have the same eigenvalue as $Kv$. It is possible that $Lv$ has the same eigenvalue as $Kv$, but there is no general guarantee that this is the case.In fact, in many cases it is possible to construct explicit examples of two linear integral operators $K$ and $L$ for which their kernels commute, but for which $K$ and $L$ do not have the same eigenfunctions and/or eigenvalues. For instance, let $k(x,s) = s^2$ and $l(x,s) = x^2$. It is straightforward to check that the kernels of $K$ and $L$ commute, but $K$ and $L$ do not have the same eigenfunctions (nor the same eigenvalues).
 

Similar threads

Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K