Proposition: Exact affine cancellation criterion

Equation (eq:repair) has a solution if and only if $[D]=0$ in ${\mathcal{V}}_{\mathrm{def}}/\mathrm{im}B$. If $c_{\ast }$ is one solution, all solutions are $c_{\ast }+\ker B$.

For a linear probe map $T:{\mathcal{V}}_{\mathrm{def}}\to {\mathbb{C}}^m$, the tested equation

does not generally imply the full equation. It does imply the full equation if $T$ is injective on $\mathrm{span}\{D\}+\mathrm{im}B$.

Proof. The repair equation is equivalent to $-D\in \mathrm{im}B$. Since $\mathrm{im}B$ is a vector space, this is equivalent to $D\in \mathrm{im}B$, proving the first assertion. Subtracting two solutions proves the affine description.

If the tested equation holds, $D+Bc\in \ker T$. Injectivity on the displayed subspace forces $D+Bc=0$. Without injectivity, take ${\mathcal{V}}_{\mathrm{def}}={\mathbb{C}}^2$, $B=0$, $D=(0,1)$, and $T(x,y)=x$. The tested equation holds but the full one fails.