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The following question naturally arises in the theory of Markov chains with countable state space to which I would be curious to know the answer:

Let $A:\ell^1 \rightarrow \ell^1$ be a contraction, i.e. $\Vert A \Vert \le1 ,$ such that $A$ is positivity preserving. Moreover, let $A$ have the property that it preserves probabilities, i.e. let $x=(x_i) \in \ell^1$ such that $x_i \ge 0$ and $\sum_i x_i =1$, then also $\sum_i (Ax)_i=1.$

If we then know that the spectrum of $A$ on the unit circle consists of isolated point spectrum and let $\lambda \in \sigma(A)$ be one of them, i.e. $\vert \lambda \vert=1$. We can then study the spectral projection $\text{Proj}_{\lambda}$ associated with $\lambda$. Does it follow that $$(A-\lambda) \text{Proj}_{\lambda}=0?$$-Maybe at least for $\lambda=1$?

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  • $\begingroup$ Are you only assuming that $\lambda$ is isolated in the spectrum, or also that it is a pole of the resolvent of $A$? (The latter is, for instance, automatically satisfied if $A$ is compact or, more generally, quasi-compact). $\endgroup$
    – Jochen Glueck
    Apr 5, 2021 at 11:35
  • $\begingroup$ @JochenGlueck sure, let's add that it is a pole of the resolvent, does that help? $\endgroup$
    – Landauer
    Apr 5, 2021 at 11:44
  • $\begingroup$ Under this assumption the answer is yes, due to the power-boundedness of $A$. I don't have much time right now, but I'll try to write an answer in a few hours. $\endgroup$
    – Jochen Glueck
    Apr 5, 2021 at 12:40
  • $\begingroup$ @JochenGlueck thanks, I am curious to learn more. $\endgroup$
    – Landauer
    Apr 5, 2021 at 12:48
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    $\begingroup$ @BenoîtKloeckner sorry yes, typo. In general the spectral projection associated with $\lambda$ is not onto $\ker(A-\lambda),$ but may also contain a nilpotent part. $\endgroup$
    – Landauer
    Apr 5, 2021 at 15:58

1 Answer 1

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What you are looking for is actually true for every power-bounded operator, without any appeal to positivity:

Theorem. Let $E$ be a Banach space and let $A: E \to E$ be a bounded linear operator such that $\sup_{n \in \mathbb{N}_0} \|A^n\| < \infty$.

If $\lambda$ is an isolated spectral value of $A$ and a pole of the resolvent, and has modulus $1$, then the corresponding pole order is $1$ and hence, $\lambda$ is a semi-simple eigenvalue. In particular the range of the corresponding spectral projection $P$ coincides with the eigenspace $\ker(\lambda-A)$.

Sketch of proof. This is all classical spectral theory. The essence of the proof is as follows: if the eigenvalue was not semi-simple, then we could find a generalized eigenvector $x$ of rank $2$. For the non-zero vector $y := (A-\lambda)x$ we would then obtain $$ A^nx = n\lambda^{n-1}y + \lambda^n x \qquad \text{for each integer } n \ge 0, $$ which contradicts the power-boundedness of $A$. $\square$

I summed up several such results in Appendix A here: DOI: 10.18725/OPARU-4238 (without many proofs, bit with detailed references to the literature).

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  • $\begingroup$ thanks a lot. Btw. do you know if an isolated point in the spectrum on the boundary has to be point spectrum? $\endgroup$
    – Landauer
    Apr 5, 2021 at 17:28
  • $\begingroup$ I also wonder why I should be able to find this rank $2$ generalized eigenvector actually. Is there a theorem that ensure this?-I am asking because in your thesis you state a lot of things for matrices which have a nice finite Jordan decomposition, but here with the operator? $\endgroup$
    – Landauer
    Apr 5, 2021 at 18:20
  • $\begingroup$ @Martinique: Let my first answer your second comment: The results in Appendix A of the thesis are all formulated for operators on infinite-dimensional spaces. Matrices do not occur in Appendix A (except for Remark A.2.2). The existence of the rank-$2$ generalized eigenvector is the content of Proposition A.2.3(b). The argument from the post above is then given in Proposition A.2.4(a). The relation to poles of the resolvent and to the associated spectral projection is given in Proposition A.3.2(d). $\endgroup$
    – Jochen Glueck
    Apr 5, 2021 at 19:06
  • $\begingroup$ @Martinique: Now my answer to your first comment: Poles of the resolvent are always eigenvalues (see for instance Proposition A.3.2(a)). (Or did I misunderstand your first comment, and you wanted to know something else?) $\endgroup$
    – Jochen Glueck
    Apr 5, 2021 at 19:10
  • $\begingroup$ Thanks, I will have a closer look at the appendix. Regarding my first comment: not assuming point spectrum or a pole of the resolvent, can an operator as described in the question have isolated spectrum on the unit circle that is no point spectrum? $\endgroup$
    – Landauer
    Apr 5, 2021 at 19:15

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