Alphabeta Math
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.

✓ 3 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. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Hierarchical low rank matrices and error propagation — Examples

1 · Prerequisites

2 · Summary

The examples compute a rank-one separated Cauchy block, show why an inadmissible identity block is kept dense when a subunit spectral error is required, and exhibit two accepted local low-rank update truncations whose accumulated error destroys positive definiteness.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

A nearfield identity block requires full rank for subunit spectral error

Example

Let τ={1,2} have coordinates 0 and 1, choose leaf threshold b=2, and use the block τ×τ. Its distance from itself is zero, so it is inadmissible and terminates as a dense near block. For I2, no rank-one approximation has operator-norm error below one.

Facts & Assumptions

Given: The cluster τ, its two coordinates, the threshold b=2, and the identity matrix I2.

[F1]

Finite cluster trees and geometric admissibility: a pair is admissible only at positive distance, and the block recursion stops when both clusters are leaves.

[F2]

H-matrix block representation: near block leaves are stored as full dense matrices.

[F3]

The best rank-at-most-k approximation in operator norm is the rank-k truncation of a singular value decomposition: every rank-at-most-one matrix has operator-norm error at least the second singular value of the identity matrix.

[F4]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: the operator norm is the maximum vector stretch over unit vectors.

Verification

1.1givenF1F2algebra

Since dist⁡(τ,τ)=0, (τ,τ) is not admissible. The cluster has size b and is a leaf, so its block pair is terminal and near by [F1]; [F2] therefore stores this block densely.

2.1givenF3F4algebra∎

The singular values of I2 are 1,1. Applying [F3] with rank cap one shows ∥I2−B∥2≥1 for every rank-at-most-one B, where [F4] identifies this norm as the maximum unit-vector stretch. Thus replacing this dense near block by a rank-at-most-one matrix cannot meet any spectral error tolerance strictly below one.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Two accepted local truncations can destroy positive definiteness after accumulation

Example

Let e2=(0,1)T, M=diag⁡(1,−1/50), and D1=D2=(3/50)e2e2∗. The exact sum A=M+D1+D2=diag⁡(1,1/10) is positive definite. If each rank-one update is separately recompressed to rank zero with tolerance δ=3/50 before assembly, both local truncations are accepted and the assembled approximation is M, which is indefinite. Each local error 3/50 is below the smaller diagonal coefficient 1/10 of A. The example concerns truncation of low-rank updates before their sum is assembled; it does not claim that every SVD truncation of a completed positive-definite H-matrix destroys positivity.

Facts & Assumptions

Given: The matrices M,D1,D2, tolerance δ=3/50, and rank cap k=0 for the two update compressions.

[F1]

Core-SVD recompression with rank and error tolerances: the procedure accepts a rank-zero output when the discarded singular-value tail is at most δ.

[F2]

Core-SVD recompression is best at the declared Frobenius tolerance: the returned rank-zero approximation has Frobenius error exactly its discarded singular-value tail.

[F3]

Finite tensor arrays and the Frobenius inner product: the Frobenius norm is computed from the squared moduli of all matrix entries.

Verification

1.1givenalgebra

The exact sum is A=diag⁡(1,−1/50+3/50+3/50)=diag⁡(1,1/10). For every nonzero z=(z1,z2)T, z∗Az=∣z1∣2+(1/10)∣z2∣2>0, so A is positive definite.

1.2givenF1F2F3algebra

Each Di has rank one and its only positive singular value is 3/50, so its rank-zero discarded tail and Frobenius error are both 3/50=δ by [F3]. Thus [F1] accepts each update as zero, and [F2] gives its exact local error.

2.1givenstep 1.1step 1.2algebra∎

After both updates are replaced by zero, the sum is M. Its quadratic form at e2 is e2∗Me2=−1/50<0, so it is not positive definite. The accumulated perturbation is D1+D2=(3/25)e2e2∗, whose action in this direction exceeds either local error; hence the two separate local tolerances do not preserve definiteness of the assembled sum.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A rank-one approximation of a separated Cauchy block

Example

Take x1=0,x2=1 in [0,1] and y1=2,y2=3 in [2,3]. For Kij=1/(yj−xi) and rank order r=1, the first geometric term is K~ij=1/yj. It gives a rank-one approximation with an exactly computable nonzero residual.

Facts & Assumptions

Given: The four points and the order r=1.

[F1]

A separated Cauchy block has a geometric rank bound: for a=1,b=2,c=1, the rank-one approximation has entries 1/yj and each residual entry has absolute value at most 1/2.

[F2]

Finite tensor arrays and the Frobenius inner product: the Frobenius norm is the square root of the sum of squared entry moduli.

Verification

1.1givenF1algebra

The exact block is K=(1/21/311/2). The rank-one approximation is K~=(1/21/31/21/3)=(1,1)T(1/2,1/3), so its rank is one.

2.1F1F2step 1.1algebra∎

The residual is K−K~=(001/21/6), hence ∥K−K~∥F=1/4+1/36=10/6≤1=2⋅2 (1/2)/(2−1) by [F2]. This attains the stated finite block bound at the entry (2,1).

Sources