Perron-Frobenius and Markov chains on countable state space 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 $\lambda=1$?
 A: 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).
