Theorem: Finite-bank separation with coverage

Choose $a_1,\dots ,a_k\in \mathfrak{A}$ such that their evaluated vectors span $T(\mathfrak{A})$, and set

There is a weight vector $w\in {\mathbb{R}}^m$ with

if and only if $s\notin \mathrm{col}(M)$. Every such $w$ gives a test

annihilating the entire ambiguity $\mathfrak{A}$.

If the evaluated coverage $\mathrm{col}(M)=T(\mathfrak{A})$ has not been proved, the same matrix calculation certifies blindness only to the listed generators.

Proof. The orthogonal complement of $\mathrm{col}(M)$ is $\ker M^{\mathsf{T}}$. A vector in this complement has nonzero pairing with $s$ exactly when $s$ is not in $\mathrm{col}(M)$. For any $a\in \mathfrak{A}$, coverage gives $T(a)\in \mathrm{col}(M)$, so

Without coverage, this reasoning applies only to the known span.