Krein–Rutman theorem is a generalization of Perron–Frobenius theorem, I know that things could be more subtle in infinite dimension, yet there's an important result in Perron–Frobenius that's missing in Krein-Rutman and I don't quite understand.

In Perron–Frobenius theorem, we know that for a irreducible non-negative matrix, its positive eigenvector is unique(up to scaling), corresponding to its largest eigenvalue. the analog for positive eigenfunction is not stated in Krein–Rutman theorem. So is it possible that we have a positive operator that has two positive eigenfunctions corresponding to two distinct eigenvalues?

If it helps to narrow thing down, I'm interested in integral operators in $L_2(R)$ space.

Thanks in advance!

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    $\begingroup$ FYI, symmetry is not needed in Perron-Frobenius. Which makes more unexpected the fact that the dominating eigenvalue is real. $\endgroup$ Commented Jun 17, 2014 at 9:50
  • $\begingroup$ Yes, I've always found that striking too! $\endgroup$ Commented Jun 17, 2014 at 11:00
  • $\begingroup$ @FedericoPoloni thank you for pointing out this, I've edited it. $\endgroup$
    – Sylvan
    Commented Jun 17, 2014 at 19:22

1 Answer 1


Beware of Wikipedia! It is true that the infinite dimensional setting makes things slightly more delicate, but actually not so much.

Assuming that the positive cone $C\subset X$ under consideration is solid (i-e has non empty interior) and that your operator $T:X\to X$ is compact and strongly positive (i-e maps the positive cone $C$ into its interior $\overset{\circ}{C}$), then the following stronger conclusion holds: the spectral radius is a simple eigenvalue associated with a strictly positive eigenvector $v\in \overset{\circ}{C}$, and there is no other eigenvalue associated with (non necessarily strictly) positive eigenvectors.

You can find an elementary proof here (theorem 1.2)

  • $\begingroup$ Welcome. But my reference/link showed up as answer #2 after googling "Krein Rutman", so I really deserve no credit here... $\endgroup$ Commented Jun 16, 2014 at 22:41
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    $\begingroup$ Showing that the cone of positive functions has non-empty interior can be very tricky sometimes (or not even true) but you don't need it. It is enough to assume that $\overline{K-K}=X$ (and that holds for positive functions..), see here. $\endgroup$
    – username
    Commented Jun 17, 2014 at 9:11
  • $\begingroup$ Yes, I should have mentioned that. Thanks for pointing that out! $\endgroup$ Commented Jun 17, 2014 at 9:45
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    $\begingroup$ The link to the elementary proof is broke. Could you please write out the full reference which will be independent of the validity of any link and supply a new link? Thank you. $\endgroup$
    – Hans
    Commented Jul 12, 2018 at 8:02
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    $\begingroup$ I apologize for the resurrection; but given that the link provided by @username does not seem to work anymore, and that the cone in most functions spaces that occur in applications actually has empty interior, I thought an additional reference might be useful: on the class of Banach lattices, one case use the - rather weak - notion of irreducibility to derive simplicity of the leading eigenvalue, no matter whether the cone has empty interior or not. See for instance [H. H. Schaefer: Banach lattices and positive operators (Springer, 1974), Theorem V.5.2 and its corollary]. $\endgroup$ Commented Mar 25, 2020 at 8:58

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