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A finite reversible transition operator is self-adjoint on weighted L2

Statement

For a finite reversible chain $(X, P, π)$ , equip $C^{X}$ with $⟨ f, g ⟩_{π} : = \sum_{x} π_{x} f_{x} \overline{g_{x}}$ . Then $P$ is self-adjoint for this inner product, and the mean-zero subspace

$H_{0} : = {f : \sum_{x} π_{x} f_{x} = 0} = {f : ⟨ f, 1 ⟩_{π} = 0}$

is invariant under $P$ . The restriction $P ∣_{H_{0}}$ is therefore a well-defined endomorphism and is self-adjoint for the restricted inner product.

Facts & Assumptions

Given: A finite set $X$ , a row-stochastic $P$ , and a strictly positive stationary probability vector $π$ satisfying detailed balance $π_{x} P_{x y} = π_{y} P_{y x}$ .

[F1]

Finite reversible Markov chain with positive stationary law: $π_{x} > 0$ , $π P = π$ , and $π_{x} P_{x y} = π_{y} P_{y x}$ for every $x, y$ .

[F2]

Real and complex inner product spaces, with the inner product linear in the first argument: the inner product is linear in the first argument, conjugate-linear in the second, conjugate-symmetric, and positive definite.

[F3]

Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space: an endomorphism is self-adjoint when it equals its adjoint.

Proof

1.1F1F2algebra

Positivity of every $π_{x}$ makes $⟨ f, f ⟩_{π} = \sum_{x} π_{x} ∣ f_{x} ∣^{2}$ nonnegative and zero only when $f = 0$ ; the finite weighted sum is sesquilinear and conjugate-symmetric, so it is an inner product.

2.1F1F3step 1.1algebra

For arbitrary $f, g$ , expand and use detailed balance: $⟨ P f, g ⟩_{π} = \sum_{x, y} π_{x} P_{x y} f_{y} \overline{g_{x}} = \sum_{x, y} π_{y} P_{y x} f_{y} \overline{g_{x}} = \sum_{y} π_{y} f_{y} \overline{(P g)_{y}} = ⟨ f, P g ⟩_{π}$ . Thus $P$ is self-adjoint.

3.1F1step 2.1algebra∎

Stationarity gives $\sum_{x} π_{x} (P f)_{x} = \sum_{y} (\sum_{x} π_{x} P_{x y}) f_{y} = \sum_{y} π_{y} f_{y}$ . Hence a mean-zero $f$ has mean-zero image, so $H_{0}$ is invariant; restricting the identity in step 2.1 to $H_{0}$ proves the restriction is self-adjoint.

Depends on

Used by

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms, Chapter 18 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, Chapter 8 (standard reference, not scraped)