TITLE: Concrete cases where $YX=qXY$
QUESTION [7 upvotes]: I was reading Kassel on the quantum plane and he defines an $R$-point on this plane as a pair of $X$, and $Y$ elements of the non commutative algebra $R$ such that $$YX=qXY,$$
with $q$ invertible.
Can anyone give me a concrete example of such algebra $R$? 
Is there a matrix algebra that could fit this example?
Thank you in advance
Edit. I found that if we take R as the Heisenberg Algebra then
$$X=\left(\begin{array}{ccc}
0 & a & b\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}\right),\,Y=\left(\begin{array}{ccc}
0 & a & c\\
0 & 0 & 1/q\\
0 & 0 & 0
\end{array}\right),$$ is an $R$-point on the quantum plane. If you have any other concrete example, please write :)

REPLY [3 votes]: A "concrete" algebra where this happens is the "algebra of functions of the quantum plane": let $V$ be the space of functions $\mathbb N_0\to \mathbb C$ and define two elements of $\operatorname{End}_\mathbb C(V)$ as follows: $x$ is the endomorphism that shifts a function, so that $(xf)(n) = f(n+1)$, and $y$ is such that $(yf)(n) = q^n f(n)$. Then $yx=qxy$, and the subalgebra generated by $x,y$ is indeed isomorphic to $\mathbb C\langle x,y\mid yx-qxy\rangle$. If you know some Spanish, check exercise 4.5 here.