CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A zero stationary weight makes the full-state similarity undefined

Statement refuted

For every finite chain and every stationary probability vector $π$ , including one with zero coordinates, the full-state matrix $diag (\sqrt{π}) P diag (1 / \sqrt{π})$ is defined and equals the discriminant.

Facts & Assumptions

Given: A finite row-stochastic kernel $P$ and a stationary probability vector $π$ , as defined by Finite row-stochastic kernel and Invariant and stationary distribution for a Markov kernel.

[F1]

Finite row-stochastic kernel: each row of $P$ has nonnegative entries summing to one.

[F2]

Invariant and stationary distribution for a Markov kernel: on a finite state space, stationarity is $π (y) = \sum_{x} π (x) P_{x y}$ .

[F3]

Szegedy row-preparation map and discriminant: the discriminant is the endomorphism $D : = T^{*} S T$ .

Counterexample

Take $X = {0, 1}$ ,

$P = (\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array}), π = (1, 0) .$

1.1F1F2givenalgebra

Each row of $P$ is a probability vector, and $π P = (1, 0) = π$ , so $π$ is stationary.

1.2givenalgebra

The second coordinate has $π_{1} = 0$ , so the diagonal matrix $diag (1 / \sqrt{π_{0}}, 1 / \sqrt{π_{1}})$ is undefined over $R$ ; hence the proposed full-state similarity matrix is not defined.

2.1F3givenstep 1.1step 1.2∎

The proposed assertion identifies the full-state expression with the discriminant $D$ of [F3] and requires the displayed matrix to be defined for every stationary probability vector. Step 1.2 shows it is undefined for this valid stationary vector, so the universal assertion is false. A restriction to the positive support is a different statement.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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)