Proposition. Eliminating equations and identifying kernels [fdip-001W]
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  • October 23, 2026

Let \(R\) be a commutative unital ring and choose \(M=X\oplus Y\) with components \(Q(x,y)=(Ax+By,Ex+Cy)\). Assume \(C:Y\to Y\) is an isomorphism and set \(S=A-BC^{-1}E\). For every \(u\in X,v\in Y\), \[Q(x,y)=(u,v)\] is equivalent to the two equations \[y=C^{-1}(v-Ex),\qquad Sx=u-BC^{-1}v.\] No invertibility of \(S\) is required. In particular, the maps \[\begin {aligned} J_S:\ker S&\longrightarrow \ker Q,& x&\longmapsto (x,-C^{-1}Ex),\\ \pi _X:\ker Q&\longrightarrow \ker S,& (x,y)&\longmapsto x \end {aligned}\] are inverse \(R\)-linear maps. Here each kernel is the submodule of elements sent to zero.

Proof. 

  • October 23, 2026

The second component of \(Q(x,y)=(u,v)\) is \(Ex+Cy=v\); since \(C\) is invertible, it is equivalent to \(y=C^{-1}(v-Ex)\). Substitution in \(Ax+By=u\) gives \[Ax+BC^{-1}(v-Ex)=Sx+BC^{-1}v=u.\] Reversing these steps proves the full equivalence.

For any \(x\in X\), direct substitution gives \[Q(x,-C^{-1}Ex)=(Sx,0).\] Thus \(J_S\) sends \(\ker S\) into \(\ker Q\). If \(Q(x,y)=0\), the proved equivalence gives \(Sx=0\) and \(y=-C^{-1}Ex\). Therefore \(\pi _XJ_Sx=x\) and \(J_S\pi _X(x,y)=(x,y)\) on the stated kernels. Both maps are restrictions of linear maps, so this is a module isomorphism, not merely a bijection of sets.

[boyd2004convex, Appendix C.4.1, pp. 672-673] gives real linear-system elimination. The direct argument above extends it to modules. After exchanging the source variables to \((y,x)\) and the right side to \((v,u)\), its first-pivot equations become the displayed formulas. The kernel correspondence does not assume that the whole system is invertible.

The pivot hypothesis cannot be deduced from invertibility of \(Q\). Over a nonzero ring take \(X=Y=R\) and \(Q(x,y)=(y,x)\). Then \(Q^2=I\), but its component \(C=0:R\to R\) is not an isomorphism. These formulas cannot invert that prescribed pivot.

Zero summands cause no exception. If \(Y=0\), its unique endomorphism is its identity and is invertible; then \(S=A\) and elimination changes nothing. If \(X=0\), \(S\) is the unique invertible endomorphism of zero and \(Q=C\). Over the zero ring every unital module is zero, because \(x=1x=0x=0\); all statements remain valid.