Pipeline-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.

✓ 2 results · all verified · 2 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 2 also cleared it.

Quantum Singular-Value Transformation — Examples

1 · Prerequisites

2 · Summary

The rank-one example verifies the unitary dilation and degree-one QSVT block at every singular value $σ \in [0, 1]$ , including both endpoints. The counterexample takes the zero map and shows that assigning nonzero values to zero-singular-vector pairs makes a cross-space rank-one sum depend on independent kernel-basis choices. The stated odd and even conventions remove that ambiguity.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Parity violation makes singular-value action basis dependent

Statement

A false extension of the odd singular-value formula that assigns a nonzero coefficient to zero-singular-vector pairs depends on arbitrary basis choices. Take $A = 0 : C \to C$ and the scalar function $f (x) = 1$ . Pair the right zero vector $v = 1$ with either left zero vector $u = 1$ or $u = - 1$ . The resulting rank-one sums are respectively $I$ and $- I$ , despite the same zero map $A$ .

Facts & Assumptions

Given: The zero linear map $A : C \to C$ , the scalar value $f (0) = 1$ , and the rank-one map convention of Dirac Notation Conversion.

[F1]

Odd and even singular-value transforms: an odd singular-value transform sums only over positive singular pairs, while an even right transform uses $f (0)$ times the canonical projection onto $\ker A$ .

[F2]

Dirac Notation Conversion: $∣ u ⟩ ⟨ v ∣$ maps $x$ to $⟨ x, v ⟩_{r o o t} u$ .

Proof

1.1givenF2algebra

For $A = 0$ , the entire domain and codomain are zero-singular spaces; both $u = 1$ and $u = - 1$ are unit left-kernel vectors, and $v = 1$ is a unit right-kernel vector. With the first pairing, $∣ 1 ⟩ ⟨ 1 ∣$ sends $x$ to $x$ , so the false zero-pair sum with coefficient $f (0) = 1$ is $I$ .

2.1F1F2step 1.1algebra∎

With the second equally valid left singular basis, $∣ - 1 ⟩ ⟨ 1 ∣$ sends $x$ to $- x$ , so the same false rule gives $- I$ . Since $I \neq - I$ on $C$ , this rule is not basis independent. The valid odd definition avoids the defect because odd polynomials have $f (0) = 0$ and the sum contains only positive singular values; the valid even definition adds $f (0) Π_{\ker A} = I$ , which is canonical and unchanged by the choice of kernel basis.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Rank-one PUE singular-plane calculation

Example

Let $0 \leq σ \leq 1$ , $H = C^{2}$ , $U_{σ} = (\begin{array}{cc} σ & \sqrt{1 - σ^{2}} \\ \sqrt{1 - σ^{2}} & - σ \end{array})$ , and $Π_{R} = Π_{L} = ∣ 0 ⟩ ⟨ 0 ∣$ . The PUE block is the scalar matrix $A = [σ]$ . Use the degree-one phase list $(0, 0)$ , whose scalar polynomial is $P (x) = x$ .

Facts & Assumptions

Given: $σ \in [0, 1]$ , the standard orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩$ of $C^{2}$ , the PUE definition, and the ordered QSVT sequence.

[F1]

PUE singular-space invariant decomposition: for $0 < σ < 1$ , the matrix of the PUE unitary on its singular plane is $W_{R} (σ)$ ; at $σ = 0$ the right kernel maps into the left projector complement, and at $σ = 1$ the singular block is a line.

[F2]

QSVT action on singular subspaces: degree-one QSVT compresses to the odd singular-value transform with coefficient $P (σ)$ , including the zero-kernel case.

Proof

1.1givenF1algebra

Put $s = \sqrt{1 - σ^{2}}$ . Direct multiplication gives $U_{σ}^{*} U_{σ} = U_{σ}^{2} = (\begin{array}{cc} σ^{2} + s^{2} & σ s - σ s \\ σ s - σ s & s^{2} + σ^{2} \end{array}) = I$ , so $U_{σ}$ is unitary. The projection onto $∣ 0 ⟩$ gives $A ∣ 0 ⟩ = Π_{L} U_{σ} ∣ 0 ⟩ = σ ∣ 0 ⟩$ , hence the PUE block is $[σ]$ . Its complementary matrix entry is $s$ , so the ordered basis realizes $W_{R} (σ)$ on the full ambient space: for $0 < σ < 1$ this is the singular plane, at $σ = 0$ it is the zero-kernel/complement block, and at $σ = 1$ it splits into the singular line $C ∣ 0 ⟩$ and residual complement $C ∣ 1 ⟩$ .

2.1F1F2step 1.1algebra∎

The zero phases make both endpoint phase matrices the identity, so the degree-one sequence is $U_{σ}$ itself. Its compressed block is $Π_{L} U_{σ} Π_{R} ∣_{H_{R}} = [σ]$ . The scalar polynomial is the upper-left entry of $W_{R} (x)$ , namely $P (x) = x$ ; therefore for $σ > 0$ the odd transform is $P_{o d d}^{S V} (A) = [P (σ)] = [σ]$ , and for $σ = 0$ the positive singular sum is empty and both blocks are zero. The same formulas give $[1]$ at $σ = 1$ .

Quantum Singular-Value Transformation — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Parity violation makes singular-value action basis dependent

Statement

Facts & Assumptions

Proof

Rank-one PUE singular-plane calculation

Example

Facts & Assumptions

Proof

Sources