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Answers for “Check OK 计算证明了什么?”

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Meet the concept

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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 RR. It shows that the generator of the ideal, xy+yxxy+yx, acts as zero on every vector because the matrices XX and YY satisfy XY+YX=0XY+YX=0.

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

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Know when to use it

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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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Understand why

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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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Find a method

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The coset of xx acts as left multiplication by the matrix XX, and the coset of yy acts as left multiplication by the matrix YY. This defines the scalar multiplication for the RR-module structure on Cn\mathbb{C}^n.

Conditions: X,Y∈Mn×n(C)X, Y \in M_{n\times n}(\mathbb{C}) satisfy XY+YX=0XY+YX=0.; R=C⟨x,y⟩/⟨xy+yx⟩R = \mathbb{C}\langle x,y\rangle / \langle xy+yx\rangle.; v∈Cnv \in \mathbb{C}^n.

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