CCFU Proof 11 — Lorentz Decomposition A₄ = B·K

2605165673496

CCFU Proof 11 — Lorentz Decomposition A₄ = B·K

05/16/2026

Given.
spec(A₄) = {φ, 1/φ, +1, −1}
A₄ᵀG₄A₄ = G₄  [from Proof 4]
sig(G₄) = (3,1)  [from Proof 4]
[Dependency: Proof 4]
Note: an explicit companion-matrix realization of A₄ is given in Theory #15c / CCFU Paper I, but this proof uses only the spectral data and invariant form established in Proof 4.

A₄ and all derived matrices share the same eigenbasis,
so all spectral arguments are exact.

Define.
B is the spectral square root of A₄² in the shared eigenbasis,
with spec(B) = {φ, 1/φ, 1, 1}. Set K = B⁻¹A₄.

Claim.
(a) B is a pure boost: BᵀG₄B = G₄, spec(B) = {φ, 1/φ, 1, 1}.
(b) K is an involution: K² = I.
(c) KᵀG₄K = G₄.
(d) A₄ = B · K.

Proof of (a).
spec(A₄²) = {φ², 1/φ², 1, 1}. All eigenvalues positive,
so the spectral square root exists. spec(B) = {φ, 1/φ, 1, 1}.
Since all eigenvalue pairs are reciprocal (φ·(1/φ) = 1, 1·1 = 1),
B preserves G₄: BᵀG₄B = G₄.  ∎

Proof of (b).
In the shared eigenbasis:
    spec(K) = spec(B⁻¹A₄)
            = {φ/φ, (1/φ)/(1/φ), 1/1, (−1)/1}
            = {1, 1, 1, −1}
Therefore spec(K²) = {1, 1, 1, 1}.
Since K² is diagonalizable with all eigenvalues 1: K² = I.  ∎

Proof of (c).
KᵀG₄K = (B⁻¹A₄)ᵀG₄(B⁻¹A₄)
       = A₄ᵀ(B⁻¹)ᵀG₄B⁻¹A₄
       = A₄ᵀG₄A₄       [since BᵀG₄B = G₄ ⟹ (B⁻¹)ᵀG₄B⁻¹ = G₄]
       = G₄  ∎

Proof of (d).
B · K = B · B⁻¹ · A₄ = A₄  ∎