= Solution
An $R$-point means a unital <algebra homomorphism over a field> $\alpha:k_q[x,y]\to R$, as in <points of a noncommutative algebra>. It is specified by $X=\alpha(x)$ and $Y=\alpha(y)$ satisfying $YX=qXY$. For $R=\mathbb C$ and $q$ not a <root of unity>, $q\ne1$, so scalar commutativity gives $(1-q)XY=0$. Hence the \b[scalar points are the union of the two coordinate axes]:
$$
\boxed{\{(z,0):z\in\mathbb C\}\cup\{(0,w):w\in\mathbb C\}.}
$$
For matrix-valued <points of a noncommutative algebra>, put $H=\mathbb C^n$ and let $H_\lambda(X)=\ker(X-\lambda I)^n$ be a <generalized eigenspace>. From $XY=q^{-1}YX$,
$$
(X-q^{-1}\lambda I)^nY=q^{-n}Y(X-\lambda I)^n.
$$
Therefore $YH_\lambda(X)\subseteq H_{q^{-1}\lambda}(X)$, interpreting a missing <eigenvalue> as the zero space. In particular the <linear subspaces> $V_X=\bigoplus_{\lambda\ne0}H_\lambda(X)$ and $H_0(X)$ are invariant under both matrices. If $\lambda\ne0$, the numbers $\lambda,q^{-1}\lambda,\ldots,q^{-n}\lambda$ are distinct. A nonzero vector $Y^nv$ with $v\in H_\lambda(X)$ would force all $n+1$ intervening <generalized eigenspaces> to be nonzero, impossible in dimension $n$. Thus $Y^nV_X=0$: \b[$Y$ is nilpotent on $V_X$], although $X$ is invertible there.
Apply the same argument with $X,Y$ exchanged, using $XY=q^{-1}YX$. The sum $V_Y$ of the nonzero $Y$ <generalized eigenspaces> is invariant under both matrices, and $X$ is nilpotent on it. Consequently $V_Y\subseteq H_0(X)$. Conversely, since $Y$ is nilpotent on $V_X$, its nonzero <generalized eigenspaces> all lie in $H_0(X)$. Decompose the restriction of $Y$ to that invariant space into its nonzero and zero <generalized eigenspaces>:
$$
H_0(X)=V_Y\oplus U,\qquad U=H_0(X)\cap H_0(Y).
$$
The zero <generalized eigenspace> of $Y$ is $X$-invariant by the scaling identity, so $U$ is invariant under both matrices. On $U$, both matrices have only zero <eigenvalues>, hence are <nilpotent matrices>. This proves the <spectral decomposition of a quantum-plane point>:
$$
\boxed{\mathbb C^n=V_X\oplus V_Y\oplus U.}
$$
Here $V_X,V_Y$ mean sums of <generalized eigenspaces>, not merely ordinary <eigenspaces>; that distinction is necessary for non-diagonalizable points.
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