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.

A factor perturbation gives an explicit Gram-matrix error

Statement

Let $G, E \in R^{m \times n}$ and set $\hat{G} = G + E$ . Then $∥ \hat{G} {\hat{G}}^{T} - G G^{T} ∥_{F} \leq 2 ∥ G ∥_{F} ∥ E ∥_{F} + ∥ E ∥_{F}^{2} .$ The estimate is valid for all ranks, including $G = 0$ and rank-deficient factors.

Facts & Assumptions

Given: Real matrices $G, E$ of the same finite dimensions.

[F1]

The Frobenius norm is the Euclidean norm of the matrix entries, so it is invariant under transpose. The Frobenius norm $∥ A ∥_{F} = (\sum_{i, j} ∣ a_{i j} ∣^{2})^{1 / 2}$ on real or complex matrices

[F2]

The operator norm is the maximum stretch of a unit vector, and the Frobenius norm is the square root of the sum of squared singular values; therefore $∥ A ∥_{2} \leq ∥ A ∥_{F}$ for every real matrix $A$ . The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison

Proof

Given: The hypotheses in the Statement.

1.1givenalgebra

Expanding $\hat{G} {\hat{G}}^{T} = (G + E) (G + E)^{T}$ gives $\hat{G} {\hat{G}}^{T} - G G^{T} = G E^{T} + E G^{T} + E E^{T} .$

2.1F1F2step 1.1algebra∎

For any compatible matrices $A, B$ , applying the definition of the operator norm to each column gives $∥ A B ∥_{F}^{2} = \sum_{j} ∥ A B e_{j} ∥_{2}^{2} \leq ∥ A ∥_{2}^{2} \sum_{j} ∥ B e_{j} ∥_{2}^{2} = ∥ A ∥_{2}^{2} ∥ B ∥_{F}^{2} \leq ∥ A ∥_{F}^{2} ∥ B ∥_{F}^{2} .$ Thus $∥ G E^{T} ∥_{F}, ∥ E G^{T} ∥_{F} \leq ∥ G ∥_{F} ∥ E ∥_{F}$ and $∥ E E^{T} ∥_{F} \leq ∥ E ∥_{F}^{2}$ . The triangle inequality applied to step 1.1 proves the stated estimate.

Depends on

Used by

Dependency tree · two levels

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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)