LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge 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.

QR prediction preserves the canonical covariance factor

Statement

Let $P = L L^{T}$ with $L \in R^{d \times r}$ and let $Q = L_{Q} L_{Q}^{T}$ with $L_{Q} \in R^{n \times s}$ , where zero columns and rank deficiency are allowed. For any real $F \in R^{n \times d}$ , form $G = [F L L_{Q}]$ and append zero columns if needed so that the transpose has at least $n$ rows. In exact arithmetic, if $G^{T} = U R$ is a reduced QR factorization with $U^{T} U = I$ , then $L^{-} : = R^{T}$ satisfies

L^{-} (L^{-})^{T} = F P F^{T} + Q .

Facts & Assumptions

Given: The factor inputs, matrix $F$ , zero-padding convention, and an exact reduced QR factorization of the transposed padded array.

[F1]

For a matrix with at least as many rows as columns, successive Householder transformations produce a reduced QR factorization whose orthogonal factor has orthonormal columns. Successive Householder or Givens transformations produce full and reduced QR factorisations with the standard dense operation counts

Proof

Given: The hypotheses in the Statement.

1.1givenalgebra

The padded zero columns do not change the Gram matrix, so $G G^{T} = [F L L_{Q}] [F L L_{Q}]^{T} = F L L^{T} F^{T} + L_{Q} L_{Q}^{T} = F P F^{T} + Q .$ The cross terms are absent because the two factors occupy disjoint column blocks.

2.1F1step 1.1algebra∎

By [F1], $G^{T} = U R$ and $U^{T} U = I$ . Thus $L^{-} (L^{-})^{T} = R^{T} R = R^{T} U^{T} U R = G G^{T} = F P F^{T} + Q .$ No inverse or full-rank assumption was used, so the identity includes zero and rank-deficient factors.

Depends on

Successive Householder or Givens transformations produce full and reduced QR factorisations with the standard dense operation counts

Used by

Dependency tree · two levels

4 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

Kevin S. Tracy, A Square-Root Kalman Filter Using Only QR Decompositions (standard reference, not scraped)
M. A. Salzmann, Some Aspects of Kalman Filtering, Technical Report 140 (standard reference, not scraped)
Simo Särkkä, Bayesian Filtering and Smoothing (standard reference, not scraped)