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The Check OK computation proves that the proposed scalar multiplication is well-defined and does not contradict the defining relation of the quotient ring $R$. It shows that the generator of the ideal, $xy+yx$, acts as zero on every vector because the matrices $X$ and $Y$ satisfy $XY+YX=0$.

Conditions: $R = \mathbb{C}\langle x,y\rangle / \langle xy+yx\rangle$.; $X,Y$ solve $XY+YX=0$.; The module action is defined by $(x+I)v=Xv$ and $(y+I)v=Yv$.

The Check OK computation proves that the proposed scalar multiplication is well-defined and does not contradict the defining relation of the quotient ring $R$. It shows that the generator of the ideal, $xy+yx$, acts as zero on every vector because the matrices $X$ and $Y$ satisfy $XY+YX=0$.

Conditions: $R = \mathbb{C}\langle x,y\rangle / \langle xy+yx\rangle$.; $X,Y$ solve $XY+YX=0$.; The module action is defined by $(x+I)v=Xv$ and $(y+I)v=Yv$.