How complicated is infinite-dimensional "undergraduate linear algebra"? The name "undergraduate linear algebra" in the title is a bit of a joke, and so I don't know how widely spread it is.  To wit:


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*High school linear algebra is the theory of a finite-dimensional vector space — i.e. it consists of the finite-dimensional representation theory of a field $\mathbb F$, or, better, for representation theory over $\mathbb F$ of the (trivial) monoid of size $1$ (the free monoid on zero generators).  The complete structure theory is well-known: every finite-dimensional vector space is (noncanonically isomorphic to) a direct sum of copies of the one-dimensional (free) vector space: $V \cong \mathbb F^n$ for some integer $n$.  In fact, assuming the axiom of choice, one can extend this result to arbitrary vector spaces: if $\mathbb F$ is a field and $V$ is a $\mathbb F$-module, then $V \cong \mathbb F^\kappa = \coprod^\kappa \mathbb F$ for some cardinal $\kappa$.

*Undergraduate linear algebra is the theory of a finite-dimensional vector space with a choice of matrix, which is to say it is the finite-dimensional representation theory of the polynomial ring $\mathbb F[x]$, or equivalently the finite-dimensional representation theory of the free monoid $\mathbb N$ on one generator.  When $\mathbb F$ is algebraically closed, the complete structure theory is given by the Jordan Canonical Form theorem.  The category of finite-dimensional representations of $\mathbb N$ is not semisimple, but the indecomposables are parameterized by pairs $(\lambda,n)$ where $\lambda \in \mathbb F$ and $n\in \mathbb Z_{>0}$ representing "Jordan blocks", and the irreducibles are precisely the indecomposables with $n = 1$, so that the irreducibles are parameterized by $\lambda \in \mathbb F$.  If $\lambda \neq \mu$, then the irreducibles corresponding to $\lambda,\mu$ have no nontrivial extensions; there is a unique nontrivial extension of the irreducible corresponding to $\lambda$ by itself.


I recently realized, however, that I never learned much of anything about the infinite-dimensional representation theory of the monoid $\mathbb N$, even assuming that my field $\mathbb F$ is algebraically closed (and, if you like, characteristic zero), and even assuming the axiom of choice.  There are plenty of vector spaces with endomorphisms that have no eigenvectors — a standard example with $\mathbb F = \mathbb C$ is $V = $ the vector space of smooth $\mathbb C$-valued functions on $\mathbb R$ that vanish outside the interval $[0,1]$, and the endormorphism is differentiation.
With some restrictions (words like "compact operator" and "spectral theorem" come up), there are lots of results that try to reproduce the Jordan theorem.  But since I hate all types of analysis, my question is:

Assuming the axiom of choice, to what extent is it possible to describe the full (infinite-dimensional) $\mathbb C$-representation theory of the monoid $\mathbb N$?  (Equivalently, the category of $\mathbb C[x]$-modules.)

A good answer might consist of a minimal "generating" set, i.e. a set $\mathcal S$ of $\mathbb N$-representations to that every $\mathbb N$-rep is a colimit of a diagram whose objects are all in $\mathcal S$.  (E.g. for infinite-dimensional high school algebra, $\mathbb S$ consists of the one-dimensional vector space.)  I could also imagine an answer consisting of a "no-go theorem" that the representation theory of $\mathbb N$ is hard; maybe that's why I never learned it.
 A: In the structure theory of an operator acting on an infinite-dimensional vector space over a field $F$, there is some good news and a lot of bad news.  The best news (as Matthew Emerton mentions) is the classification of finitely generated modules over a PID.  This beautiful theorem immediately generalizes the Jordan canonical form theorem and the classification of finite or finitely generated abelian groups.  It says that every f.g. module over a PID $R$ is a direct sum of cyclic modules $R/d$, and the summands are unique in either the primary decomposition (which comes from the proof by isotypic submodules and Jordan blocks) or the invariant factor decomposition (which comes from the Smith normal form proof).  You can take $R = \mathbb{Z}$, or $R = F[x]$ for a field $F$, or of course there are other choices.
I suspect that the bad news is likely to be uniformly bad for countably generated modules over any PID.  I found a reference for the case of torsion-free abelian groups, so I will concentrate on that case first.  Consider the abelian groups $A$ that lie between $\mathbb{Z}^2$ and $\mathbb{Z}[\frac1p]^2$ for a prime $p$.  Those choices for $A$ that are free abelian are just lattices in the plane (whose coordinates have denominators that are powers of $p$).  There is an important theorem from the theory of affine buildings that you can make an infinite tree of valence $p+1$ out of these lattices.  Two lattices are equivalent if they are the same lattice up to homothetic expansion (by a power of $p$).  Two lattices are connected by an edge if they are nested and their quotient is $\mathbb{Z}/p$.  So, the theorem says that this graph whose vertices are lattices is acyclic and has degree $p+1$ everywhere; the graph is an affine Bruhat-Tits building of type $A_1$.
Given a general $\mathbb{Z}^2 \subseteq A \subseteq \mathbb{Z}[\frac1p]^2$, you can let $A_n = (\frac1{p^n}\mathbb{Z}^2) \cap A$.  This is the type of direct limit that Matthew Emerton describes, which looks good, but bad news is coming.  For a while, $A_n/A_{n-1}$ can be $(\mathbb{Z}/p)^2$, then it can be $\mathbb{Z}/p$, and then $A_n$ might stop growing.  If it stops growing, then $A$ is free with two generators.  If it always grows by $(\mathbb{Z}/p)^2$, then $A = \mathbb{Z}[\frac1p]^2$.  But in the middle case, if it grows by $\mathbb{Z}/p$, then $A$ is described by a path in the affine building from the origin to infinity.  There are uncountably many ($2^{\aleph_0}$) such paths, and they have a $p$-adic topology.
So when are two such modules $\mathbb{Z}^2 \subseteq A \subseteq \mathbb{Z}[\frac1p]^2$ and $\mathbb{Z}^2 \subseteq B \subseteq \mathbb{Z}[\frac1p]^2$ isomorphic?  The outer module is obtained by tensoring either $A$ or $B$ with $\mathbb{Z}[\frac1p]$, so the question is when $A$ and $B$ are equivalent under the action of $\text{GL}(2,\mathbb{Z}[\frac1p])$, which is a countable group.  After localizing at all of the other primes, this question was studied by Hjorth, Thomas, and other logicians and algebraists.  As Brian Conrad suggests, they have several theorems that indicate that the answer is wild, similar in spirit to the classification of infinite graphs up to isomorphism.
Now suppose that $F$ is a field, most conveniently a countable or finite field, and suppose that $V$ is an $F[x]$-module lying between $F[x]^2$ and $F[x,x^{-1}]^2$.  Then you get the same geometric picture as before:  The $F[x]$-free choices for $V$, up to homothety by powers of $x$, form an affine building which is an infinite tree, such that the neighbors of each vertex are a projective line $FP^1$.  The complicated submodules are the ones that correspond to paths in this tree from the origin to infinity.  If $F$ is finite or countable, then $\text{GL}(2,F[x,x^{-1}])$ is also countable, but the set of these paths is always uncountable and the group action is always far from transitive.  I don't know if Hjorth's and Thomas' results extend to this case (because I am not qualified to read their papers), but my guess is that they would be comparably pessimistic.
There are other cases in which you get good news, basically by making infinite-dimensional operators look like finite-dimensional operators.  The module $V$ in the previous paragraph has no $x$-eigenvectors even after passing to the algebraic closure $\bar{F}$.  This is the same statement as that $V$ is torsion-free over $F[x]$.  I think that a torsion module over a PID $R$ behaves much better, if it is the direct sum of its isotypic summands.   For instance, if $x$ is locally nilpotent on $V$, then I think that it is a direct sum of Jordan blocks, possibly including infinite Jordan blocks (like the differentiation operator $x = \partial/\partial t$ on $\mathbb{Q}[t]$).
