# Distinct eigenvalues for integral operator?

Is there some sufficient condition on the kernel $K$ of a (say, finite rank, to simplify) integral operator $$\mathcal K:f(x)\in L^2(\mathbb R)\mapsto \int K(x,y)f(y)dy$$ so that it has all its non-zero eigenvalues $\textbf{distinct}$ (i.e. with multiplicity one) ?

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I can refer you to this paper. –  Mhenni Benghorbal Aug 25 '12 at 12:34
A finite-rank operator on an infinite-dimensional Banach space has infinite-dimensional kernel. Are you talking about the nonzero eigenvalues? –  Qiaochu Yuan Aug 25 '12 at 23:17

The OP added to his question that "the multiplicity of eigenvalues is strictly one". The following answer was given before that addition which does not assure the multiplicity of the eigenvalues to be one.

Note that, if $A$ is a compact operator, then the spectrum of $A$, excluding $0$, consists of a countable family of isolated points with finite multiplicity $|\lambda_1| \geq |\lambda_2| \geq \dots$ and either the spectrum of $A$ is finite or $\lim_{n\rightarrow \infty} \lambda_n = 0$.

Now, you can use the fact that "integral operators with kernels of finite rank are compact operators".

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This shows that the nonzero eigenvalues have finite multiplicity. I think the OP wants to know when they have multiplicity $1$. –  Qiaochu Yuan Aug 25 '12 at 23:18
I don't follow. What do you mean by "distinct eigenvalues" if it isn't "the eigenvalues have multiplicity $1$"? –  Qiaochu Yuan Aug 26 '12 at 0:40
That isn't what I asked. What do you mean by "distinct eigenvalues"? –  Qiaochu Yuan Aug 26 '12 at 3:31
Okay, so what would it mean to not have distinct eigenvalues? Does this just mean "has at least $2$ eigenvalues?" I don't think this is how anyone uses the word "distinct." –  Qiaochu Yuan Aug 26 '12 at 4:01
Sorry but I fail to see how this is, even remotely, an answer to the question asked. –  Did Sep 5 '12 at 20:43