Alphabeta Math
LemmaStatement: AI-generatedProof: AI-generated
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Loop coupling preserves the dynamics and centered Laplacian

Statement

Let W be the adjacency matrix of a simple graph on n≥1 vertices and 0≤p≤1. Let B be any diagonal matrix with diagonal in {0,1} and put A=W+B, J=11T, d=np, where 1 is the all-ones vector and I below is the identity matrix. For real phase variables θ=(θ1,…,θn)∈Rn, the equations θ˙i=∑jAijsin⁡(θj−θi) agree pointwise with those for W. More precisely, for any initial data (t0,θ0)∈R×Rn and any interval I containing t0, a function θ:I→Rn is a solution with those initial data for A if and only if it is one for W. Trajectories here mean these solutions on intervals. Writing L(M)=diag⁡(M1)−M, we have L(A)=L(W) and ∥A−pJ∥≤∥W−p(J−I)∥+1. All matrix norms are Euclidean induced operator norms on the finite vertex space.

Facts & Assumptions

Given: The data and hypotheses in the Statement.

Goal: Establish the conclusions in the Statement, including the realization-by-realization Bernoulli specialization.

[F1]

The finite symmetric network convention counts each diagonal entry once in the row sum and defines its Laplacian as the diagonal matrix of row sums minus the adjacency matrix (Finite symmetric networks and loop removal).

[F2]

The finite vertex space has ∥f∥22=∑i∣fi∣2 and a norm satisfying the vector triangle inequality (Finite vertex Hilbert space in the article convention). The induced operator norm of a matrix T is max⁡∥f∥2=1∥Tf∥2 (The operator norm is zero on the zero domain and otherwise is max_‖ v‖=1‖ Tv‖).

[F3]

A first-order system on an open subset of R×Rn and its solutions on intervals with prescribed initial data are understood as in First-order systems, initial value problems, and solutions on intervals. Here the domain is all of R×Rn.

[F4]

Real sine is defined for every real argument by its convergent power series; at zero every term is zero, so sin⁡0=0 (Sine and cosine defined by their real power series).

[F5]

For the probabilistic specialization, let (Ω,w) be a finite probability space and let bi:Ω→{0,1} be mutually independent Bernoulli(p) random variables. Outcomes and probabilities, random variables, Bernoulli laws and mutual independence are defined in Finite probability spaces, outcome weights, events, and event probabilities, Real random variables on finite probability spaces and their finite distributions, Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials and Pairwise and mutual independence of finite-valued random variables, respectively.

Proof

1.1givenF1F4algebra

For any θ∈Rn and each i, the difference between the two vector-field coordinates is Biisin⁡(θi−θi)=Biisin⁡0=0. Also diag⁡(B1)=B, so L(A)=diag⁡(W1)+B−W−B=L(W).

1.2givenF2algebra

For a real diagonal matrix C=diag⁡(c1,…,cn), put m=max⁡i∣ci∣. Every unit vector f satisfies ∥Cf∥22=∑i∣ci∣2∣fi∣2≤m2. A coordinate unit vector at an index attaining m gives equality, so ∥C∥=m. In particular ∥B−pI∥=max⁡i∣Bii−p∣≤1, since Bii∈{0,1} and 0≤p≤1.

2.1F3step 1.1

Both systems have the same domain and the same right-hand side at every (t,θ). For any common interval I and initial data (t0,θ0), the differentiability requirement, initial-value requirement and differential equation for a solution are therefore identical. Their solution sets on that interval coincide; this conclusion does not require a uniqueness theorem.

2.2F2step 1.2algebra

For any two matrices S,T and any unit vector f, the vector triangle inequality gives ∥(S+T)f∥2≤∥Sf∥2+∥Tf∥2≤∥S∥+∥T∥. Taking the maximum over unit vectors proves the operator triangle inequality. Apply it to A−pJ=W−p(J−I)+(B−pI) and use step 1.2 to obtain ∥A−pJ∥≤∥W−p(J−I)∥+1.

3.1F5step 1.1step 2.1step 2.2∎

For every outcome ω∈Ω, the matrix B(ω)=diag⁡(b1(ω),…,bn(ω)) satisfies the deterministic hypotheses. Steps 1.1 and 2.1 therefore show, realization by realization, that adding the independent Bernoulli loops changes neither the Laplacian nor the trajectories with matching initial data on common intervals; step 2.2 also supplies the norm bound for each realization. The deterministic conclusions hold for every binary diagonal choice, regardless of its probability law or dependence. This proves all assertions.

Depends on

Used by · 2

Dependency tree · 2 levels

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