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Weighted spectral bounds control the degree without loops

Statement

Let A be a network on a finite nonempty V, n=∣V∣, d0>0, and L=D−A with Dxx=dA(x)=∑yAxy. Put Jxy=1, let I be the identity matrix, and set ΔL=L−d0I+(d0/n)J. Suppose c−d0I⪯ΔL⪯c+d0I, where matrix order is the quadratic-form order on the vertex space. Then the degree without the loop, s(x)=∑y≠xAxy=dA(x)−Axx, satisfies (1+c−)d0−d0/n≤s(x)≤(1+c+)d0−d0/n for every x. The network, vertex-space and matrix-order definitions are Finite symmetric networks and loop removal, Finite vertex Hilbert space in the article convention and Self-adjoint order, norm bounds and diagonal tests.

Facts & Assumptions

[F1]

The network Laplacian has diagonal entry dA(x)−Axx=s(x) (Finite symmetric networks and loop removal).

[F2]

The vertex-space matrix action and inner product use finite coordinate sums (Finite vertex Hilbert space in the article convention).

[F3]

If aI⪯B⪯bI for a Hermitian matrix, then a≤Bxx≤b (Self-adjoint order, norm bounds and diagonal tests).

Proof

Given: The data in the Statement.

Goal: Establish the stated conclusions.

1.1F1F2algebra

The matrix ΔL is real symmetric, hence Hermitian in the vertex-space convention. Its diagonal entry equals s(x)−d0+d0/n, since Ixx=Jxx=1.

2.1F3step 1.1algebra∎

Apply the diagonal order test [F3] to obtain c−d0≤s(x)−d0+d0/n≤c+d0. Adding d0−d0/n proves the pointwise bounds for every vertex. No upper bound on the loop weights was used.

Depends on

Used by · 2

Dependency tree · 2 levels

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