Lemma. Every spoke order has an invertible detector image [cn-0034]

Status:

Statement. For every half-order matrix [cn-0011] \(S_\prec \), the image \(\Phi (S_\prec )\) under the fixed rank detector [cn-0032] is invertible.

Proof. The natural order. Put \(H(p,q)=1_{q<p}\). Then \(S=\tfrac 12\mathbf 1\mathbf 1^T-H-\tfrac 12 I_n\). Since additive and diagonal terms are killed [cn-0033], \(\Phi (S)=-\Phi (H)\). Subtract \(1\) from every row of \(H\) with \(p>s+1\), obtaining \(H'\). This row operation changes \(H\) by an additive matrix, so \(\Phi (H')=\Phi (H)\).

Off the diagonal, \(H'\) is supported exactly on the two triangles [cn-0028]: its value is \(+1\) on the lower triangle and \(-1\) on the upper triangle. Let \(A_Q\) have columns \(h_{pq}\) for those \(s^2\) positions. Then

\[\Phi (S)=-A_Q D_Q A_Q^T,\qquad D_Q=\operatorname {diag}\bigl (H'(p,q)(x_p-x_q)w_pw_q\bigr )_{(p,q)\in Q_s}.\]

Two-triangle interpolation [cn-0029] makes \(A_Q\) invertible; all diagonal entries of \(D_Q\) are nonzero. Their product is invertible, regardless of its signature.

An arbitrary order. List its indices as \(\pi (1)\prec \cdots \prec \pi (n)\) and apply the argument to the distinct node tuple \(x_{\pi (1)},\ldots ,x_{\pi (n)}\). This only reindexes the sum defining \(\Phi (S_\prec )\): the node polynomial \(F\) and the original linear map are unchanged. The interpolation theorem required distinct nodes, not increasing numerical order. Thus one fixed detector works at every vertex.

Proof dependencies
Half-order matrices [cn-0011]; Two-triangle pattern [cn-0028]; Interpolation basis [cn-0029]; Detector [cn-0032]; Annihilation [cn-0033]
Provenance
OpenAI, 23 September 2026 [cn-0042], Lemma 4.2(iii), equation (4.6); reconstructed as a reading account on 9 October 2026. Lean source: HalfOrderInvertibility.