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LemmaStatement: AI-generatedProof: AI-generated
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A complete weighted network distinguishes a loop cap from spectral control

Statement

On V={1,2} take A=2J, d0=4, α=1/1000, c−=−1/100 and c+=1/100, where J is the 2×2 all-ones matrix. Degrees and Laplacians have the network convention Finite symmetric networks and loop removal, and matrix norms and order have the vertex-space convention Finite vertex Hilbert space in the article convention. Both ∥A−(d0/2)J∥≤αd0 and c−d0I⪯L−d0I+(d0/2)J⪯c+d0I hold, and c−>−1, 0<α<1/5, c−<0<c+. Yet dmax⁡=4>(1+c+)d0+1−d0/2=76/25. The loop-dependent bound of Loop-dependent weighted degree envelope remains valid. This example concerns this degree bound only; it makes no assertion that a synchronization theorem fails.

Facts & Assumptions

[F1]

Degrees are row sums and the network Laplacian is D−A (Finite symmetric networks and loop removal); finite matrix action and norms use Finite vertex Hilbert space in the article convention.

[F2]

Common positive rescaling preserves the normalized centered-matrix inequalities (Covariance under a common positive rescaling of weights).

[F3]

The upper degree envelope with arbitrary loops has ℓmax⁡ in place of a fixed unit cap (Loop-dependent weighted degree envelope).

Proof

Given: The data in the Statement.

Goal: Establish the stated conclusions.

1.1F1F2algebra

Begin with A=J on two vertices and d0=2. Both row degrees are 2, so L=2I−J and both centered matrices are zero. Apply [F2] with λ=2: the asserted A=2J, d0=4 have degrees 4, L=4I−2J, and again both residuals are zero. A zero norm satisfies the positive adjacency bound, and −(1/25)I⪯0⪯(1/25)I follows by testing quadratic forms. All strict sign inequalities in the Statement follow immediately from the chosen rational numbers.

2.1F1F3step 1.1algebra∎

The proposed unit-cap expression is (101/100)4+1−2=76/25<4. In contrast, here ℓmax⁡=2, so [F3] gives dmax⁡≤(101/100)4−2+2=101/25, which is true. Thus the failure comes exactly from substituting a unit loop cap for an arbitrary nonnegative loop, despite exact centered spectral residuals.

Depends on

Used by · 1

Dependency tree · 2 levels

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