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Answers for “原始矩阵方程 XY+YX=0 究竟在哪里进入模的验证?”

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

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

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

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

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