Concrete cases where $YX=qXY$ 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 :)
 A: The simplest thing to do is just to define and algebra with an element like that:
$R=F\langle X, Y\rangle/(YX-qXY)$ where, say, $q\in F$.
In $M_2(\mathbb R)$, setting $X=Y=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ gives you an example of $XY=qYX$ for any $q\in\mathbb R$ at all.
A: This is a little generalization of the one you found. This pair works.
$$X=\left(\begin{array}{ccc}
1 & a & b\\
0 & q & q(a d + c - c q)\\
0 & 0 & q q
\end{array}\right),\,Y=\left(\begin{array}{ccc}
0 & 1 & c\\
0 & 0 & d\\
0 & 0 & 0
\end{array}\right).
$$
A: 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.
