Generalization of the Jordan form for infinite matrices Under what conditions is it the case that for a matrix $M$ whose rows and columns are indexed by a countably infinite set $S$ one has a Hamel basis consisting of generalized eigenvectors (i.e. $v \in \ker(M - \lambda I)^n$) of $M$? Must $M$ be a compact operator (I have a norm)?
The matrix I am working with has non-negative entries, row sums not exceeding $1$ (substochastic), is irreducible and aperiodic. However, I suspect this question may be of general interest to others, so any solution not employing these properties would be all the more useful.
EDIT
Here is some more information: the matrix $M$ which I am working with is $R$-positive. This means that none of the sequences $\{ M^n_{ij}\}_{n \in \mathbb{N}}$, $i,j \in S$, converge to $0$, where 
$$
   R^{-1}
:= \lim_{n \to \infty} (M_{ij}^n)^{1/n}.
$$
In such a case, it is known that $R^{-1}$ is the spectral radius of $M$, and moreover that $R^{-1}$ is an eigenvalue for $M$ for which there are unique left and right eigenvectors $\alpha,\beta$ which are strictly positive and satisfy 
$$
  \sum_{k \in S} \alpha(k) \beta(k)
< \infty.
$$
In particular the set of eigenvalues for $M$ cannot be empty.
Thanks!

Let $H$ be a countably-dimensional Hilbert space with basis $\{e_n\}$. Let $T$ be the weighted shift given by 
$$
Te_n=\frac1n\,e_{n+1}.
$$
This operator is compact (actually, Hilbert-Schmidt). 
If $Tv=\lambda v$, with $v=\sum_n\alpha_n e_n$, then
$$
\sum_{n=1}^\infty\alpha_n\lambda e_n=\sum_{n=1}^\infty\alpha_n Te_n=\sum_{n=1}^\infty\alpha_n\,\frac1n\,e_{n+1}=\sum_{n=2}^\infty\frac{\alpha_{n-1}}{n-1}\,e_n.
$$
If $\lambda=0$, we deduce that $\alpha_n=0$ for all $n$, so $v=0$. If $\lambda\ne0$, then $\alpha_1=0$, and $\alpha_{n+1}=\alpha_n/n$, implying again that $\alpha_n=0$ for all $n$. So $v=0$. This shows that $T-\lambda I$ has trivial kernel for all $\lambda$. 
So $T$ has no nonzero generalized eigenvectors, and it cannot have a Jordan form, at least in the obvious sense. 
