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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, $q\ne 1$, so scalar commutativity gives $(1-q)XY=0$. Hence the scalar points are the union of the two coordinate axes:

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 $XY=q^{-1}YX$,Therefore $Y{H}_{\lambda }(X)\subseteq H_{q^{-1}\lambda }(X)$, interpreting a missing eigenvalue as the zero space. In particular the linear subspaces $V_X={⨁}_{\lambda \ne 0}{H}_{\lambda }(X)$ and $H_0(X)$ are invariant under both matrices. If $\lambda \ne 0$, the numbers $\lambda ,q^{-1}\lambda ,\dots ,q^{-n}\lambda$ are distinct. A nonzero vector $Y^{n}v$ with $v\in H_{\lambda }(X)$ would force all $n+1$ intervening generalized eigenspaces to be nonzero, impossible in dimension $n$. Thus $Y^n{V}_X=0$: $Y$ 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: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:Here $V_X,V_Y$ mean sums of generalized eigenspaces, not merely ordinary eigenspaces; that distinction is necessary for non-diagonalizable points.