rankOneEnd_goldenOperator_sq
plain-language theorem explainer
A nondegenerate rank-one endomorphism, after normalization by the scalar μ = ℓ(v), induces a golden operator G satisfying G² = G + I. Anyone citing the algebraic projector-to-golden step in reciprocal cost geometry would use this. The proof is a one-line application of the general normalized-projector golden equation to the rank-one square identity.
Claim. Let $\ell : V \to \mathbb{R}$ be linear and $v \in V$ with $\ell(v) \neq 0$. Write $A$ for the rank-one endomorphism $x \mapsto \ell(x)\, v$, and let $P = \ell(v)^{-1} A$ be its normalization. Then the golden operator $G = \varphi P + (1-\varphi)(I-P)$ satisfies $G^{2} = G + I$.
background
This module is the algebraic core of the paper route "Golden and Metallic Structures on Hessian Manifolds": once a projector $P$ satisfies $P^{2}=P$, the almost-product $F=2P-I$ obeys $F^{2}=I$, and the golden operator $G=\varphi P+(1-\varphi)(I-P)$ obeys $G^{2}=G+I$. No full Hessian manifold is formalized here; only endomorphisms of a real module.
The rank-one endomorphism $A:x\mapsto\ell(x),v$ is the algebraic shape of the rank-one Hessian projector used on that route. Normalizing by the scalar $\mu=\ell(v)$ produces $P=\mu^{-1}A$. Upstream, any operator with $A^{2}=\mu A$ and $\mu\neq 0$ already yields a golden operator after that normalization. The missing input is exactly the square law for rank-one maps, which supplies $A^{2}=\mu A$ with $\mu=\ell(v)$.
proof idea
One-line term proof. Apply the general lemma that normalizing any endomorphism with $A^{2}=\mu A$ and $\mu\neq 0$ produces a golden operator satisfying $G^{2}=G+I$, feeding it the rank-one square identity $A^{2}=(\ell v),A$ together with the nondegeneracy hypothesis $\ell v\neq 0$. No further algebraic expansion is needed.
why it matters
This is the rank-one specialization of the projector-to-golden step that the forcing stack needs. Downstream it is packaged into the projector-golden certificate (any cost-induced normalized projector carries the golden-operator equation) and into the multi-coordinate Phase 4 result that the golden operator of the $n$-dimensional recognition-cost Hessian satisfies $G^{2}=G+I$ (paper §4, eq. (6.7)).
In the Recognition Science chain this is the algebraic face of T6: $\varphi$ is forced as the self-similar fixed point once a genuine projector appears. The module status is zero sorry and zero new axiom, so the certificate is fully discharged on the algebraic side.
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