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Transfer two nonbacktracking good events to the required norm bound

Statement

Let G=(V,E) be a finite simple graph on n ordered vertices, with adjacency matrix A. Let J be the n×n all-ones matrix and I the n×n identity matrix. Write d(i) for the degree of vertex i in this order and dmax⁡=max⁡id(i). Let 0<p<1, μ=np>16, and H=(A−p(J−I))/μ. All matrix norms below are Euclidean operator norms on Cn. For a matrix H, its nonbacktracking matrix B is indexed by all ordered vertex pairs and has B(i,j),(k,l)=Hkl1j=k1i≠l. Write ρ(B)=max⁡{∣z∣:z is an eigenvalue of B}. If dmax⁡<8μ and ρ(B)<2, then ∥A−pJ∥<8μ.

Facts & Assumptions

[F1]

The operator norm on bounded linear maps between normed spaces satisfies the triangle inequality and absolute homogeneity (The operator norm is a norm on the space of bounded linear operators). The finite-dimensional operator norm used here is the maximum over the Euclidean unit sphere (The operator norm is zero on the zero domain and otherwise is max_‖ v‖=1‖ Tv‖).

[F2]

For a Hermitian matrix H, set r=max⁡i(∑j∣Hij∣2)1/2 and b=max⁡ij∣Hij∣. If r>0, Theorem 2.2 gives ∥H∥≤rf(ρ(B)/r)+7b, where f(x)=2 for 0≤x≤1 and f(x)=x+x−1 for x≥1.

[F3]

In an ordered finite simple graph, A is symmetric with zero diagonal; row i has d(i) entries equal to one and n−1−d(i) off-diagonal entries equal to zero (The adjacency matrix of a finite simple graph, Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

External input

For a matrix H, its nonbacktracking matrix B is indexed by ordered vertex pairs and has B(i,j),(k,l)=Hkl1j=k1i≠l. Write ρ(B)=max⁡{∣z∣:z is an eigenvalue of B}. For a Hermitian matrix H, set r=max⁡i(∑j∣Hij∣2)1/2 and b=max⁡ij∣Hij∣. If r>0, Theorem 2.2 gives ∥H∥≤rf(ρ(B)/r)+7b, where f(x)=2 for 0≤x≤1 and f(x)=x+x−1 for x≥1.

Proof

Given: The objects and hypotheses in the Statement.

Goal: Prove its conclusions.

1.1F1givenalgebra

Since np>16 and 0<p<1, we have n>0. For any matrix T on Cn, let M be its finite unit-sphere maximum from [F1]. For x≠0, rescaling to x/∥x∥ gives ∥Tx∥≤M∥x∥; the same inequality holds at zero. Thus T is bounded, and its unit-ball supremum equals M, since it is at most M and the unit ball contains the unit sphere. Consequently the triangle inequality and absolute homogeneity in [F1] apply to our finite-dimensional matrix norm. Also ∥Ix∥=1 for every unit vector, so ∥I∥=1.

1.2F2F3givenalgebra

Let r,b be as in [F2] and put ρ=ρ(B). By [F3], H is real symmetric, hence Hermitian, with zero diagonal. Its off-diagonal entries are (1−p)/μ at adjacent vertices and −p/μ otherwise. Hence b≤1/μ and ∑j∣Hij∣2=[d(i)(1−p)2+(n−1−d(i))p2]/μ≤d(i)/μ+p<9, so r<3. For r>0, rf(ρ/r)≤2max⁡{r,ρ}: if ρ≤r it is 2r, and if ρ≥r it is ρ+r2/ρ≤2ρ. Thus [F2] and ρ<2 give ∥H∥<6+7/μ. If r=0, every entry of H vanishes and this bound is immediate.

2.1step 1.1step 1.2algebra∎

Since A−pJ=μH−pI, step 1.1 and step 1.2 give ∥A−pJ∥≤μ∥H∥+p∥I∥<6μ+7+p<6μ+8<8μ, the last comparison using μ>16.

Depends on

Used by · 1

Dependency tree · 2 levels

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External dependencies

Benaych-Georges, Bordenave and Knowles, Spectral radii of sparse random matrices (19 September 2019) · PDF p. 3, Definition 2.1; p. 4, Theorem 2.2

For a Hermitian matrix $H$, set $r=\max_i(\sum_j|H_{ij}|^2)^{1/2}$ and $b=\max_{ij}|H_{ij}|$. If $r>0$, Theorem 2.2 gives $\|H\|\le r f(\rho(B)/r)+7b$, where $f(x)=2$ for $0\le x\le1$ and $f(x)=x+x^{-1}$ for $x\ge1$.

Hypotheses: Finite Hermitian H; the nonbacktracking matrix is defined by the displayed ordered-pair formula. Treat H=0 separately.

Reading limits: Checked the extracted PDF text on pp. 3–4 for Definition 2.1, the matrix norm conventions and Theorem 2.2, including its full deterministic Hermitian hypothesis. The theorem is imported; its proof and the remaining paper were not reconstructed or fully read.

Sources