Alphabeta Math
AlgorithmAlgorithm: Literature-sourcedProof: Not applicablePipeline-generatedjudge pass (gpt-6-luna)
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.

Core-SVD recompression with rank and error tolerances

Definition

Input: A field F∈{R,C}, positive integers m,n, a finite factor list Us∈Fm×rs and Vs∈Fn×rs representing M=∑sUsVs∗, a rank cap 0≤k≤min⁡(m,n), and a Frobenius tolerance δ≥0. The conjugate transpose M∗ and Frobenius norm are those of Finite tensor arrays and the Frobenius inner product.

Procedure: Concatenate the factors as U=[U1 ⋯ Up] and V=[V1 ⋯ Vp], so M=UV∗. Take compact singular value decompositions U=QURU and V=QVRV, where the columns of QU,QV are orthonormal and span the respective column spaces. Form the core C=RURV∗ and compute an SVD C=PΣQ∗. Let its positive singular values be s1≥⋯≥st>0. Find the smallest q∈{0,…,min⁡(k,t)} for which eq:=(∑j>qsj2)1/2≤δ. If no such q exists, return tolerance-not-met. Otherwise return success with factors U^=QUP[:,1:q]Σq and V^=QVQ[:,1:q], where Σq=diag⁡(s1,…,sq), and the exact discarded-tail certificate eq; when q=0, use empty m×0 and n×0 factors.

The compact decompositions retain only positive singular directions. If the factor list is empty or its represented matrix is zero, then t=0 and the procedure returns the rank-zero representation because δ≥0. Any SVD choices in repeated singular subspaces are permitted. No result is returned when the requested cap cannot certify the tolerance.

Depends on

Used by

Dependency tree · two levels

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Sources