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.
Best Answer
I try to formulate the proof of Wedderburn's theorem.
Lemma1: Suppose A is a simple F-algebra, $\mathcal{I},\mathcal{J}$ are two minimal left ideal of A, then there exists $\alpha \in A$, s.t $\mathcal{I}= \mathcal{J}\alpha$, thus $\mathcal{I}\cong \mathcal{J}$ holds.
Proof: Noted that $\mathcal{I}A, \mathcal{J}A$ are two-sided ideal of A. Since A has no proper two-sided ideal, if $\mathcal{I},\mathcal{J}$ are not trivial, then $\mathcal{I}A = \mathcal{J}A =A$, so we can deduce that $\mathcal{IJ}A = \mathcal{I}A=A$ and $\mathcal{IJ}\neq 0$. Let $\alpha \in J$, then $\mathcal{I}\alpha \subset \mathcal{J}$. Since $\mathcal{J}$ is a minimal left ideal and $ \mathcal{I}\alpha\neq {0}$, then $\mathcal{I}\alpha = \mathcal{J}$, then $\mathcal{I}\cong \mathcal{J}$.
Lemma2: Suppose A is F-algebra, then $End_A(A) \cong A^{op}$.
Proof: By defining a right multiplication action.
Lemma3:(suggested by @rschwieb) Suppose $M_{i},N_{i},\: i = 1,2,...,n$ are left R-module, then $$ Hom_R( \oplus_{i=1}^n M_i, N )\cong \oplus_{i=1}^n Hom_R(M_i, N) $$ $$ Hom_R(M, \oplus_{i=1}^n N_i )\cong \oplus_{i=1}^n Hom_R(M, N_i) $$
Proof: By induction. When n=2, it is obvious.
Back to the proof of Wedderburn's theorem: Since A is a simple algebra, we can write $A = \oplus_{i=1}^n A_i$, where $A_i$ are the simple A-module. In other words, $A_i$ are the minimal left ideals of A. By lemma 1, the minimal left ideals of simple algebra A are isomorphic to each other. Thus, $A = \oplus_{i=1}^n A_i \cong \oplus_{i=1}^n A_1$. By lemma 2, we have $A^{op}\cong End_A(A)=End_A(\oplus_{i=1}^n A_1)$, where $End_A(A_1)$ is well-defined since $A_1$ is the left ideal of A. With the help of lemma 3, $$ End_A(\oplus_{i=1}^n A_1) = Hom_A(\oplus_{i=1}^n A_1, \oplus_{i=1}^n A_1)\cong \oplus_{i=1}^n \oplus_{i=1}^n Hom_A(A_1,A_1) $$ $$= \oplus_{i=1}^n \oplus_{i=1}^n End_A(A_1) $$ By installing every item into the entry of a matrix, we have $$ \oplus_{i=1}^n \oplus_{i=1}^n End_A(A_1)\cong M_n( End_A(A_1)) $$ By Schur's lemma, since A1 is a simple A-module, $End_A(A_1)$ is a division ring, which is denoted by D. Namely, we have $A^{op} \cong M_n(D)$. Then, $A \cong M_n(D)^{op}\cong M_n(D) $, as we desired, where n is the number of minimal left ideals of A and $D= End_A(A_1)$.