Name of Jordan Canonical Form in infinite dimensions? I tend to think of Jordan canonical form as the generalized spectrum theorem.  I read it as saying, every matrix cannot be diagonalized, but they can be "jordanized".  In functional, I've seen the spectral theorem again.  The situation there however is much more complex.  What does the Jordan canonical form go by in infinite dimensions, or does it really not hold there well?  Maybe just Hilbert Spaces would be easiest to consider so I can get a feeling for what is out there.
Thanks!
 A: Spectral Theory for a closed densely-defined linear operator $A : \mathcal{D}(A) \subseteq X\rightarrow X$ can be viewed in terms of properties of the resolvent operator $R(\lambda)=(A-\lambda I)^{-1}$. In fact, spectrum is defined in terms of the resolvent. For a selfadjoint operator $A$ on a Hilbert space $X$, the resolvent exists for all $\lambda\in\mathbb{C}$ except for all or part of the real axis, and the resolvent can be represented in a limiting form as
$$
           R(\lambda)x = \int_{-\infty}^{\infty}\frac{1}{t-\lambda}dE(t)x
$$
This can be viewed as a limiting form of the holomorphic calculus where the contour is squashed down to the real axis. Selfadjoint operators have resolvents that cannot have worse than a first order pole anywhere on the real axis because $\|R(\lambda)\| \le 1/|\Im\lambda|$ for $\lambda\notin\mathbb{R}$. This is quite easy to show based on the symmetry of $A$. For such $A$, the resolvent has a pole at some $\lambda_{0}\in\mathbb{R}$ iff it is a first order pole; and the residue of such a pole is the orthogonal projection onto the eigenspace associated with $\lambda_{0}$.
For a matrix $A$, the resolvent $R(\lambda)=(A-\lambda I)^{-1}$ has only poles at the zeros of the minimal polynomial $m$ of $A$. The negative of the residue of the resolvent at such a zero $\lambda_{j}$ is a projection $P_{j}$ onto the space associated with all of the Jordan blocks associated with $\lambda_{j}$. And, because $R(\lambda)$ has a finite pole at $\lambda_{j}$, it can be shown that
$$
                   (A-\lambda_j I)^{n_j}P_{j} = 0,\;\;\; (A-\lambda I)^{n_j-1}P_{j} \ne 0,
$$
where $n_j$ is the order of the pole of $R(\lambda)$ at $\lambda_j$. So, if the pole is of order $1$, then $P_j$ is a projection onto an eigenspace. The largest Jordan block associated with $\lambda_j$ has size $n_j$.
For any operator $A$, if $\lambda_0$ is an isolated singularity of the resvolent $R(\lambda)=(A-\lambda I)^{-1}$, then the negative of the residue of $R(\lambda)$ at $\lambda_0$ is a projection $P_0$, and $(A-\lambda I)^{n}P_0=0$, just as for matrices. You do get cyclic subspaces if the singularity of the resolvent at $\lambda_0$ is a pole, but these subspaces may not be finite-dimensional. You get a chain of subspaces
$$
            \{ 0 \} = (A-\lambda_0 I)^{n}P_0 X \subsetneq (A-\lambda_0 I)^{n-1}P_0 X \subsetneq \cdots \subsetneq (A-\lambda_0)P_0 X \subsetneq P_0 X.
$$
And from this you can extract cyclic subspaces as you do for Jordan form.
A general operator can have an essential singularity for its resolvent. A simple example is the quasinilpotent operator
$$
               Vf = \int_{0}^{x}f(t)dt.
$$
on a space such as $C[0,1]$. The resolvent for $V$ has only one singularity at $\lambda=0$, which is an essential singularity. Resolvent analysis doesn't give much information about such operators. Resolvents of arbitrary bounded operators can have arbitrary closed and bounded sets of singularities, unlike the resolvent of a matrix which has only a finite number of singularities that are all poles. So you can see why a general Jordan form does not exist for arbitrary bounded operators.
A: You might be looking for the classification of finitely generated modules over a PID. It's the natural generalization of Jordan canonical forms and smith normal forms for operators on nice infinite dimensional spaces. This and related topics were discussed recently in an MO thread.
Edit: And by recently I mean five years ago. Time flies.
A: You might be interested in holomorphic functional calculus.  We can extend the notion of Jordan canonical form to compact operators on Banach spaces (which includes Hilbert spaces), but this generalization fails in general for bounded linear operators.
