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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 $(y+I)v=Yv$.

The matrix equation enters at the final step of the verification computation, where the expression $(XY+YX)v$ is evaluated. Because $X$ and $Y$ are assumed to solve $XY+YX=0$, this expression equals the zero vector, confirming that the ideal generator acts trivially.

Conditions: $X,Y \in M_{n\times n}(\mathbb{C})$.; $XY+YX=0$.; The verification expands $(xy+yx+I)v$ into $(XY+YX)v$.

The matrix equation enters at the final step of the verification computation, where the expression $(XY+YX)v$ is evaluated. Because $X$ and $Y$ are assumed to solve $XY+YX=0$, this expression equals the zero vector, confirming that the ideal generator acts trivially.

Conditions: $X,Y \in M_{n\times n}(\mathbb{C})$.; $XY+YX=0$.; The verification expands $(xy+yx+I)v$ into $(XY+YX)v$.