CCFU Proof 14 — Null Vector: Q(φ,1) = 0

2605165677067

CCFU Proof 14 — Null Vector: Q(φ,1) = 0

05/16/2026

Given.
Let φ = (1+√5)/2, so φ² = φ + 1.

C₂ characteristic polynomial: λ² − λ − 1 = 0
Associated quadratic form (eigenvector convention):
    Q(x,y) = x² − xy − y²

Companion matrix:
    A₂ = [[1, 1],
          [1, 0]]

Eigenvector verification.
v = (φ, 1) is an eigenvector of A₂ for eigenvalue φ:
    A₂(φ,1) = (φ+1, φ) = (φ², φ) = φ(φ,1).  ∎

Convention note.
In Proof 9, Q is written as Q(x,y) = y² − xy − x²
(state-pair convention). Here Q(x,y) = x² − xy − y²
(eigenvector convention). These are coordinate-swapped
versions of the same characteristic quadratic structure.

Claim.
    Q(φ, 1) = 0

Proof.
    Q(φ, 1) = φ² − φ·1 − 1²
            = φ² − φ − 1
            = 0          [since φ² − φ − 1 = 0]  ∎

Note.
The eigenvector (φ,1) of A₂ is a null vector of the quadratic
form Q associated with the characteristic polynomial of C₂.
This is Q-null, not Minkowski-null. The structural correspondence
— that φ is null in its own quadratic form — is an observation,
not a physical claim.