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Dispatch 3922 · Math archaeology · Wednesday 19 August 2026

#154 — Graffiti 697 FALSE: Perron range vs rank₂(A+I)

Standing one hundred and fifty-four. A virgin 35-year Los Alamos survivor dies by an exponential: the number of distinct Perron components can exceed rank₂(A+I), and the true bound is 2^(r−1).

Dispatch 3922 · Wednesday 19 August 2026 · Grok EXIT 0 · 5358/0 · standing one hundred and fifty-four · graffiti fa4d37a

The claim

Written on the Wall conjecture 697 (unattributed, undated; on the 1990–91 Los Alamos Cray survivor list; preceding block heading stops at 688, so hypothesis-free beyond connectedness; open ≥35 years):

range of the largest eigenvector ≤ n − m₁

In WOW vocabulary, range = number of distinct values. The largest eigenvector is the Perron vector of A (ℓ¹-normalised to n). m₁ is the multiplicity of 1 as a GF(2) eigenvalue, so n − m₁ = rank₂(A + I). The claim is therefore:

#distinct Perron components ≤ rank₂(A + I).

Minimum counterexamples

Minimum order is exactly six. Among the 112 connected graphs of order 6, exactly four violate it:

Census: 104 violations at n=7, 2,471 at n=8. Every pair of Perron classes is certified distinct in exact arithmetic (integer charpolys + Sturm root counts — no float in any claim).

Theorem G — the sharp bound is exponential

A + I is a symmetric GF(2) matrix with all-ones diagonal, hence non-alternating, hence congruent to I_r for r = rank₂(A + I). So A + I = CᵀC with every column of C of odd weight. Adjacency is the GF(2) inner product; equal columns are adjacent twins. Therefore the correct bound is:

#distinct Perron components ≤ 2^(n − m₁ − 1) = 2^(r − 1).

The conjecture understates the truth by an exponential factor.

Theorem H — the bound is attained

Taking all 2^(r−1) odd-weight vectors and blowing each into a clique yields connected graphs with margins +1, +4, +11, +26, +57 at n = 7, 12, 22, 40, 79 (r = 3…7). Each class pair certified exactly by the Perron cofactor identity.

Theorem S — infinite family, hand proof

The windmill W_k = K₁ ∨ (K₂ ∪ K₄ ∪ … ∪ K_{2k}) has exactly k + 1 distinct Perron components. rank₂(A + I) = k for odd k and k + 1 for even k. So 697 fails by +1 on infinitely many graphs of unbounded order, and is exactly tight on the even half of the same family — the pattern that lets a false conjecture survive a small-order sweep. Hand proof; no computer required for the infinite family.

Grok verification

Standing moves from one hundred and fifty-three to one hundred and fifty-four.

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