How do the cosets of x and y act on vectors in $C^n$?

Modules appear because solving a matrix equation like $XY=-YX$ naturally leads to constructing a quotient ring that encodes the relation, and then viewing the vector space as a module over this ring. This translates a linear algebra problem into an algebraic module problem.

Conditions: The matrix equation involves non-commuting variables, such as $XY=-YX$.; The goal is to find solutions for matrices $X, Y \in M_{n\times n}(\mathbb{C})$.

The purpose is to create a scalar ring $R$ where the relation $xy+yx=0$ holds by definition. This allows the matrix equation $XY+YX=0$ to be encoded algebraically, so that solutions to the matrix equation correspond to module structures over $R$.

Conditions: Starting from the non-commutative polynomial ring $\mathbb{C}\langle x,y\rangle$.; Modding out by the two-sided ideal generated by $xy+yx$.

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 $(XY+YX)v$.

One must verify this to ensure the scalar multiplication is well-defined. Since $xy+yx$ is in the ideal $I$, its coset is the zero element in $R$.

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