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Spectral bound for a symmetric lazy walk killed on a marked set

Statement

Let X be a finite set of size n≥2, and let P=(I+Q)/2 where Q is a symmetric row-stochastic matrix. Let u=n−1/21X. Assume the second-largest eigenvalue of P is at most 1−δ, where 0<δ≤1. For M⊆X, put ε=∣M∣/n, U=X∖M, and let ΠU be the coordinate orthogonal projection defined by ΠUex=ex for x∈U and ΠUex=0 for x∉U. Let PU=ΠUPΠU restricted to vectors supported on U. Then P has spectrum in [0,1] and

∥PU∥≤1−δε.

The empty compression has norm 0 by convention.

Facts & Assumptions

Given: The finite symmetric lazy transition matrix P=(I+Q)/2, its uniform unit vector u, the marked set M, and the second-eigenvalue bound.

[F1]

Finite row-stochastic kernel: Qxy≥0 and ∑yQxy=1 for every row.

[F2]

Marked sets, stationary mass, and walk-search tasks: in the uniform case ε=∣M∣/n.

[F5]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: on a nonzero domain, ∥T∥=max⁡∥z∥=1∥Tz∥; the zero-domain norm is 0.

[F6]

Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors: ∣⟨v,w⟩∣≤∥v∥∥w∥ in a real or complex inner-product space.

Proof

1.1F1F3F4F6algebra

For any vector z∈CX, weighted Cauchy–Schwarz, obtained by applying [F6] to (Qxy)y and (Qxyzy)y, gives ∣(Qz)x∣2≤∑yQxy∣zy∣2. Summing over x and using symmetry, so that every column sum is also 1, yields ∥Qz∥2≤∥z∥2. Since Q is self-adjoint by its real symmetric matrix and [F4], [F3] gives a real orthonormal eigenbasis; the contraction inequality forces every eigenvalue of Q into [−1,1]. Thus P=(I+Q)/2 is self-adjoint with spectrum in [0,1]. Its row sums are 1, so Pu=u.

1.2F2F6givenalgebra

If U=∅, then PU=0 and the conclusion follows from [F5]. Otherwise choose a unit vector z supported on U and write z=αu+v with v⊥u. [F6] applied to ⟨z,u⟩=⟨z,ΠUu⟩ gives ∣α∣2≤∥ΠUu∥2=∣U∣/n=1−ε, so ∥v∥2=1−∣α∣2.

2.1F3givenstep 1.2algebra

The orthogonal complement u⊥ is invariant under P by self-adjointness and Pu=u. Its eigenvalues are at most 1−δ by the stated second-eigenvalue hypothesis. Therefore ⟨z,Pz⟩=∣α∣2+⟨v,Pv⟩≤∣α∣2+(1−δ)∥v∥2≤1−δε.

3.1F3F4F5step 1.1step 2.1∎

The compression PU is positive semidefinite because ⟨z,PUz⟩=⟨z,Pz⟩≥0 for every vector supported on U, by the nonnegative spectrum in step 1.1. It is self-adjoint, so its operator norm is its largest eigenvalue; equivalently it is the maximum of its unit-vector Rayleigh quotients. Step 2.1 bounds each such quotient by 1−δε, proving the result.

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