D-005 — Linear algebra (§3) + chain assembly (§6); task 0.3 fulfilled

Date: 2026-08-17 | Session: 001 | Tasks: 0.2d + 0.3 | Status: complete Source: fulltext pp. 5–6 (§3), p. 11 (§6). Re-derived independently.

Lemma 3.1 (Inertia under pull-back). [VERIFIED]

Q₀ Hermitian on ℂᵐ, A : ℂᵈ → ℂᵐ linear ⟹ n₊(AQ₀A) ≤ n₊(Q₀). Proof re-derived: if AQ₀A is positive definite on U ⊂ ℂᵈ, then A|_U is injective (Ax = 0, x ∈ U ⟹ xAQ₀Ax = 0 ⟹ x = 0) and Q₀ is positive definite on A(U); hence dim U = dim A(U) ≤ n₊(Q₀). Taking U maximal: n₊(A*Q₀A) ≤ n₊(Q₀). ✔ (This is Sylvester's law of inertia for congruence; A need not be invertible — injectivity on U suffices.)

Lemma 3.2 (Rank–trace inequality) — TASK 0.3 INDEPENDENT REPRODUCTION

Statement. P, Q Hermitian d×d, P ⪰ 0, rank P ≤ r, n₊(Q) ≤ b ⟹ r ≥ 2 tr P + 4 tr Q − 4b − ‖P+Q‖²_HS. (3.1)

My reproduction. Jordan decomposition Q = Q⁺ − Q⁻, Q± ⪰ 0, Q⁺Q⁻ = 0, rank Q⁺ = n₊(Q) ≤ b. Expand: ‖P+Q‖²_HS = ‖P‖² + ‖Q⁺‖² + ‖Q⁻‖² + 2tr(PQ⁺) − 2tr(PQ⁻) − 2tr(Q⁺Q⁻). tr(PQ⁺) ≥ 0 (trace of product of two PSD matrices: tr(PQ⁺) = tr(P^{1/2}Q⁺P^{1/2}) ≥ 0 ✔); tr(Q⁺Q⁻) = 0 ✔. Drop the ≥0 term: ‖P+Q‖² ≥ ‖P‖² − 2tr(PQ⁻) + ‖Q⁻‖² + ‖Q⁺‖². () Von Neumann trace inequality: tr(PQ⁻) ≤ Σᵢ pᵢnᵢ (p₁≥⋯≥p_d ≥0 eigenvalues of P, n₁≥⋯≥n_d≥0 of Q⁻). ⟹ first three terms of () satisfy ‖P‖² − 2tr(PQ⁻) + ‖Q⁻‖² ≥ Σᵢ(pᵢ − nᵢ)². () Split at r (pᵢ = 0 for i > r since rank P ≤ r): Σᵢ(pᵢ−nᵢ)² ≥ Σ_{i≤r}(pᵢ−nᵢ)² + Σ_{i>r} nᵢ². Scalar inequality x² ≥ 2x − 1 at x = pᵢ−nᵢ (i≤r): Σ_{i≤r}(pᵢ−nᵢ)² ≥ 2Σ_{i≤r}(pᵢ−nᵢ) − r = 2tr P − r − 2Σ_{i≤r}nᵢ ≥ 2tr P − r − 4Σ_{i≤r}nᵢ (nᵢ ≥ 0). And Σ_{i>r}nᵢ² ≥ 0 ≥ −4Σ_{i>r}nᵢ. Since tr Q⁻ = Σ_{all i} nᵢ: () ≥ 2tr P − r − 4tr Q⁻. () For Q⁺: positive eigenvalues q₁..q_{b′}, b′ ≤ b; scalar (x−2)² ≥ 0, i.e. x² ≥ 4x − 4: ‖Q⁺‖² = Σ qⱼ² ≥ Σ(4qⱼ − 4) = 4tr Q⁺ − 4b′ ≥ 4tr Q⁺ − 4b. () Insert (),(*) into (): ‖P+Q‖² ≥ 2tr P − r − 4tr Q⁻ + 4tr Q⁺ − 4b = 2tr P − r + 4tr Q − 4b (tr Q = tr Q⁺ − tr Q⁻). Rearranged: r ≥ 2tr P + 4tr Q − 4b − ‖P+Q‖². ∎ [VERIFIED — every inequality step independently justified]

Equality case (checked): P = Π₁, Q = 2Π₂, orthogonal projections Π₁ ⊥ Π₂, ranks r, b: tr P = r, tr Q = 2b, ‖P+Q‖² = tr Π₁ + 4tr Π₂ = r + 4b. RHS = 2r + 8b − 4b − (r + 4b) = r ✔ equality. Tightness map (where the proof can lose): (i) dropping 2tr(PQ⁺) ≥ 0 — lossy unless P ⊥ Q⁺; (ii) von Neumann — lossy unless P, Q⁻ simultaneously diagonalizable with aligned order; (iii) x² ≥ 2x−1 — tight only at pᵢ−nᵢ = 1; (iv) x² ≥ 4x−4 — tight only at qⱼ = 2. ⟹ The certificate is tight exactly for eigenvalue pattern {1}_P, {2}_Q: this is why simple zeros (charge 1) and off-line pairs (charge... matching v-norms ≈ 1, pairs at signature (1,1) scaled by 2) saturate it. Any improvement must exploit a regime where these four inequalities are NOT simultaneously tight — key input for task 1.1.

Remark 3.3 (variational view, checked): minimize U ↦ ‖P+UQU‖²_HS over the (compact) unitary group; first-order condition tr([P,UQU]H) = 0 ∀ Hermitian H ⟹ [P, UQU*] = 0 ⟹ simultaneous diagonalization ⟹ (3.1) reduces to scalar (x−1)² ≥ 0, x² ≥ 0, (x−2)² ≥ 0 in the three sign cases. Setting Q = 0 and optimizing the scalar parameter recovers the Cauchy–Schwarz rank bound rank P ≥ (tr P)²/‖P‖²_HS. ✔ [Note: Lemma 3.2 with Q≠0 is NOT implied by C–S; the 4tr Q − 4b term is the genuinely new content — it charges off-line mass at rate 4 instead of losing it entirely.]

Genealogy (checked against §1.2): (3.1) is the matrix form of m² ≥ 2m − 1 (per-zero: multiplicity m costs m² in ‖G̃‖², earns 2m−1... i.e. m² − (2m−1) = (m−1)² ≥ 0); with simple zeros on the rank side and multiple ones at flat charge 4 it recovers m² ≥ 3m − 2 (m² − (3m−2) = (m−1)(m−2) ≥ 0 ✔). The flat charge 4 for Q-eigenvalues is what treats off-line pairs and multiple zeros uniformly.

Lemma 3.4 (Weyl). [VERIFIED]

A, E Hermitian, ‖E‖ ≤ θ ⟹ n₊(A) ≥ #{i : λᵢ(A+E) > θ}. Courant–Fischer: λᵢ(A) ≥ λᵢ(A+E) − ‖E‖ > θ − θ = 0 for each counted i. ✔ (Role: lets inertia statements survive the o(1) tail Ẽ; insurance for the chain, used implicitly when passing from G̃+Ẽ to G̃.)

Chain assembly (§6) — full verification

Setup: P₁ := (aL²)⁻¹ Σ_{ρ∈on, m_ρ=1} v_ρv_ρᵀ (simple on-line), Q′ := G̃ − P₁.

P₁ ⪰ 0, rank P₁ ≤ s₁, tr P₁ ≤ s₁ (Lemma 2.1, m_ρ=1: each ‖v_ρ‖² ≤ aL²). ✔
Q′ = pull-back of ⊕_{on, m≥2} m_ρ|x|² ⊕ ⊕_pairs [[0,m_ρ],[m_ρ,0]] — the pair block [[0,m],[m,0]] has eigenvalues ±m ✔ (one positive per pair); on-line multiple zeros contribute one positive direction each ⟹ n₊(Q′) ≤ s₂ + p (Lemma 3.1). ✔
Apply Lemma 3.2 with r = s₁, b = s₂ + p, tr Q′ = tr G̃ − tr P₁: rank P₁ ≥ 2tr P₁ + 4(tr G̃ − tr P₁) − 4(s₂+p) − ‖G̃‖² = 4tr G̃ − 2tr P₁ − 4(s₂+p) − ‖G̃‖².
Key inequality: tr P₁ + 2(s₂+p) ≤ s₁ + 2s₂ + 2p ≤ N(I′) (tr P₁ ≤ s₁; m_ρ ≥ 2 per multiple on-line point; m ≥ 1 per off-line zero, 2 zeros per pair). ✔ ⟹ −2tr P₁ − 4(s₂+p) ≥ −2N(I′).
⟹ rank P₁ ≥ 4tr G̃ − 2N(I′) − ‖G̃‖²_HS = 4N(I′) − 2N(I′) − R(ψ)N + o(N) = (2 − R(ψ) − o(1))N (Prop 4.2 + Thm 5.7 + N(I′) = N(I) + O(√T L)). ✔
Rank side: N^s_0(I) = s₁ − O(√T L) ≥ rank P₁ − O(√T L) (Cor 4.5). ✔ ⟹ (1.2): N^s_0 ≥ (2 − R(ψ) − o(1))N. ψ₀: 2 − 4/3 = 2/3 ∎ (i). (ii): from the display before the key inequality: 3tr P₁... rearranged form: 3s₁ + 4(s₂+p) ≥ 4tr G̃ − ‖G̃‖² (using tr P₁ ≤ s₁ to go from −2tr P₁ to ≥ −2s₁: check — rank P₁ ≥ 4tr G̃ − 2tr P₁ − 4(s₂+p) − ‖G̃‖² and rank P₁ ≤ s₁, tr P₁ ≤ s₁ ⟹ s₁ ≥ 4tr G̃ − 2s₁ − 4(s₂+p) − ‖G̃‖² ⟹ 3s₁ + 4(s₂+p) ≥ 4tr G̃ − ‖G̃‖² ✔); subtract s₁ + 2s₂ + 2p ≤ N(I′): 2(s₁+s₂+p) ≥ (3 − R(ψ) − o(1))N; N_d(I′) ≥ s₁+s₂+2p ≥ s₁+s₂+p ⟹ N_d ≥ ½(3 − R(ψ) − o(1))N. ψ₀: ½(3−4/3) = 5/6 ∎ (ii). [VERIFIED] Cumulative (0,T): dyadic summation, rate O(log log T/log T) (Remark 6.1). ✔ Theorem B: χ-twisted inputs, no poles for χ≠χ₀, chain unchanged. ✔

TASK 0.3 SIGN-OFF

Lemma 3.2 reproduced independently with all four loss-points mapped and the equality case confirmed. The inequality is elementary (von Neumann + two scalar inequalities) but the STRUCTURE — charging Q at rate 4 with cap 4b — is the paper's load-bearing idea.

OBSERVATIONS FOR THE ATTACK

The four tightness conditions are the blueprint for improvement. Beating 2/3 requires either (i) data that makes tr(PQ⁺) > 0 usable (correlation between simple-zero vectors and off-line/multiple mass), (ii) eigenvalue distributions of P, Q⁻ that break von Neumann alignment, or (iii)/(iv) non-saturating eigenvalues — i.e. moments beyond the second. All three need NEW prime-side input (support > 1 or higher moments) — consistent with §7.2, now understood mechanistically.
The chain uses rank P₁ ≤ s₁ AND tr P₁ ≤ s₁ (both saturated by simple zeros). Multiple on-line zeros waste (m_ρ−1) trace AND occupy Q′-budget at rate... a double zero on the line costs 2 in N but returns 0 rank: the worst case for the certificate. Bounding multiple-zero density by OTHER means (e.g. derivatives of ξ, Levinson-style) would tighten tr P₁ + 2(s₂+p) ≤ N — a cross-track hook (Track A ↔ classical mollifier methods).