linear algebra over a division ring vs. over a field When I was studying linear algebra in the first year, from what I remember, vector spaces were always defined over a field, which was in every single concrete example equal to either $\mathbb{R}$ or $\mathbb{C}$.
In Associative Algebra course, we sometimes mentioned (when talking about $R$-modules) that if $R$ is a division ring, everything becomes trivial and known from linear algebra.
During the summer, I'm planning to revisit my notes of linear algebra, write them in tex, and try to prove as much as possible in a general setting. 
Are there any theorems in linear algebra, that hold for vector spaces over a field and not over a division ring? How much linear algebra can be done over a division ring?
Also, what are some examples of division rings, that aren't fields? $\mathbb{H}$ is the only one that comes to mind. I know there aren't any finite ones (Wedderburn). Of course, I'm looking for reasonably nice ones...
 A: In my experience, when working over a division ring $D$, the main thing you have
to be careful of is the distinction between $D$ and $D^{op}$.
E.g. if $F$ is a field, then $End_F(F) = F$ ($F$ is the ring of $F$-linear endomorphisms of itself, just via multiplication), and hence $End(F^n) = M_n(F)$;
and this latter isomorphism is what links matrices and the theory of linear transformations.
But, for a general division ring $D$, the action of $D$ by left multiplication on itself is not $D$-linear, if $D$ is not commutative.  Instead, the action of $D^{op}$ on $D$ via right multiplication is $D$-linear, and so we find that
$End_D(D) = D^{op}$, and hence that $End_D(D^n) = M_n(D^{op}).$

As for examples of division algebras, they come from fields with non-trivial Brauer groups, although this may not help particularly with concrete examples.  
A standard way to construct examples of central simple algebra over a field $F$ is via a crossed product.  (Unfortunately, there does not seem to be a wikipedia entry on this topic.)
What you do is you take an element $a\in F^{\times}/(F^{\times})^n$, and
a cyclic extension $K/F$, with Galois group generated by an element $\sigma$
of order $n$, and then define a degree $n^2$ central simple algebra $A$ over $F$
as follows:
$A$ is obtained from $K$ by adjoining a non-commuting, non-zero element $x$,
which satisfies the conditions


*

*$x k x^{-1} = \sigma(k)$ for all $k \in K$, and

*$x^n = a$.


This will sometimes produce division algebras.  
E.g. if we take $F = \mathbb R$, $K = \mathbb C$, $a = -1$, and $\sigma =$ complex conjugation, then $A$ will be $\mathbb H$, the Hamilton quaternions.
E.g. if we take $F = \mathbb Q_p$ (the $p$-adic numbers for some prime $p$),
we take $K =$ the unique unramified extension of $\mathbb Q_p$ of degree $n$,
take $\sigma$ to be the Frobenius automorphism of $K$,
and take $a = p^i$ for some $i \in \{1,\ldots,n-1\}$ coprime to $n$,
then we get a central simple division algebra over $\mathbb Q_p$, which is called the division algebra over $\mathbb Q_p$ of invariant $i/n$ (or perhaps $-i/n$, depending on your conventions).
E.g. if we take $F = \mathbb Q$, $K =$ the unique cubic subextension of $\mathbb Q$ contained in $\mathbb Q(\zeta_7)$, and $a = 2$, then we will get 
a central simple division algebra of degree $9$ over $\mathbb Q$.
(To see that it is really a division algebra, one can extend scalars to $\mathbb Q_2$, where it becomes a special case of the preceding construction.)
See Jyrki Lahtonen's answer to this question, as well as Jyrki's answer here, for some more detailed examples of this construction.  (Note that a key condition for getting a division algebra is that the element $a$ not be norm from the extension $K$.)

Added: As the OP remarks in a comment below, it doesn't seem to be so easy to find non-commutative division rings.  Firstly, perhaps this shouldn't be so surprising, since there was quite a gap (centuries!) between the discovery of complex numbers and Hamilton's discovery of quaternions, suggesting that the latter are not so easily found.
Secondly, one easy way to make interesting but tractable non-commutative rings is to form group rings of non-commutative finite groups, and if you do this over e.g. $\mathbb Q$, you can find interesting division rings inside them.  The one problem with this is that a group ring of a non-trivial group is never itself a division ring; you need to use Artin--Wedderburn theory to break it up into a product of matrix rings over division rings, and so the interesting division rings that arise in this way lie a little below the surface.  
A: Let me quote a relevant paragraph in the Wikipedia article on "Division ring":

Much of linear algebra may be
  formulated, and remains correct, for
  (left) modules over division rings
  instead of vector spaces over fields.
  Every module over a division ring has
  a basis; linear maps between
  finite-dimensional modules over a
  division ring can be described by
  matrices, and the Gaussian elimination
  algorithm remains applicable.
  Differences between linear algebra
  over fields and skew fields occur
  whenever the order of the factors in a
  product matters. For example, the
  proof that the column rank of a matrix
  over a field equals its row rank
  yields for matrices over division
  rings only that the left column rank
  equals its right row rank: it does not
  make sense to speak about the rank of
  a matrix over a division ring.

I hope this helps!
A: The others have outlined the basic differences. Since you asked for reasonably nice examples let me give you the following. Let $F=\mathbf{Q}(i)$, $E=F(\sqrt5)$ and $\sigma$ be the non-trivial element of $Gal(E/F)$ determined by $\sigma(i)=i, \sigma(\sqrt5)=-\sqrt5$. Then the set of matrices of the form
$$
A(x_0,x_1)=\left(\begin{array}{rr}
x_0&\sigma(x_1)\\
ix_1&\sigma(x_0)
\end{array}\right),
$$
where $x_0$ and $x_1$ range over $E$, is a division algebra. It is easy to see that it is closed under multiplication. Whenever $(x_0,x_1)\neq(0,0)$ the determinant of $A(x_0,x_1)\neq0$. This is easiest to see as follows: $\det A(x_0,x_1)=N(x_0)-iN(x_1)$, where $N$ is the relative norm map $N:E\rightarrow F, x\mapsto x\sigma(x)$. For the determinant to vanish, we must have $i=N(x_0/x_1)$ by the multiplicativity of the norm. But a 5-adic study of this equation shows that $i$ does not belong to the image of the norm map. From the above we also see that $\det A\in F$, so the usual formula for the inverse of a 2x2 matrix shows that $A(x_0,x_1)$ has an inverse of the same form.
In the language of Matt E's answer this division algebra represents an element of the Brauer group $Br(F)$. It is of order two. To the initiated I will tell that its non-trivial Hasse invariants lie above the primes $2+i$ and $2-i$.
This algebra (and other cyclic division algebras) have found their way to multiantenna radio communications. The discoverers call it the Golden code. IIRC this code is in the hyperWLAN standard, but I won't trust my memory 100% here.
Another division algebra that reasonably often shows up in lattice constructions (see Conway & Sloane) is the ring of Icosians. It is really a subalgebra of the usual Hamiltonian quaternions, where in place of $\mathbf{R}$ we restrict the coordinates to the field $\mathbf{Q}(\sqrt5)$. The resulting algebra has its non-trivial Hasse invariants only at the infinite places, so its discriminant is small, and hence it is destined to yield dense lattices and such.
A: Anything that involves general homothecies (scalar multiples of the identity viewed as linear operators), or multi-linearity (in particular determinants) will not go through over division rings (or maybe only in some some much weakened form; for instance homothecies by central scalars can still be considered). Anything involving eigenvalues and eigenspaces falls into this category (an eigenspace is a subspace where a linear operator acts as a homothecy). As an example of a theorem that won't be available, the Cayley-Hamilton theorem (indeed the characteristic polynomial of a finite-dimensional linear operator over a division ring is not even defined).
A: By Schur's lemma, whenever $M$ is an irreducible module over some ring $R$, then $\operatorname{End}(M)$ is a division ring.
