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.

Innovation, gain, and posterior errors follow from joint covariance perturbations

Statement

Let $S = S^{T} \in R^{p \times p}$ satisfy $x^{T} S x \geq μ ∥ x ∥_{2}^{2}$ for all $x \in R^{p}$ , with $μ > 0$ . Let $B \in R^{n \times p}$ , $D = D^{T} \in R^{n \times n}$ , and let symmetric $E_{S}$ and matrices $E_{B}, E_{D}$ satisfy $∥ E_{S} ∥_{F} \leq e_{S} < μ, ∥ E_{B} ∥_{F} \leq e_{B}, ∥ E_{D} ∥_{F} \leq e_{D} .$ Put $\hat{S} = S + E_{S}$ , $\hat{B} = B + E_{B}$ , $\hat{D} = D + E_{D}$ , $Φ = D - B S^{- 1} B^{T}$ , and $\hat{Φ} = \hat{D} - \hat{B} {\hat{S}}^{- 1} {\hat{B}}^{T}$ . If $b \geq ∥ B ∥_{2}$ , define $β = (μ - e_{S})^{- 1}$ . Then $\hat{S} ⪰ (μ - e_{S}) I ≻ 0$ and $∥ {\hat{S}}^{- 1} ∥_{2} \leq β, ∥ {\hat{S}}^{- 1} - S^{- 1} ∥_{2} \leq μ^{- 1} β e_{S},$ $∥ \hat{B} {\hat{S}}^{- 1} - B S^{- 1} ∥_{2} \leq e_{B} β + b μ^{- 1} β e_{S},$ and $∥ \hat{Φ} - Φ ∥_{2} \leq e_{D} + 2 b e_{B} β + b^{2} μ^{- 1} β e_{S} + e_{B}^{2} β .$

Facts & Assumptions

Given: The symmetric matrices, the lower bound $μ$ , the perturbation bounds, and the finite-dimensional real operator and Frobenius norms.

[F1]

The operator norm is the maximum Euclidean stretch of a unit vector. The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||

[F2]

Compatible operator norms are submultiplicative. The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2

[F3]

The operator norm is the largest singular value and is at most the Frobenius norm. Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison

[F4]

For real vectors, Cauchy–Schwarz follows by expanding $0 \leq ∥ x - t y ∥_{2}^{2}$ at $t = (x^{T} y) / ∥ y ∥_{2}^{2}$ when $y \neq 0$ ; the $y = 0$ case is immediate. Expanding $∥ x + y ∥_{2}^{2}$ then gives the Euclidean triangle inequality. By [F1], evaluating each unit vector gives the triangle inequality for the operator norm.

Proof

Given: The hypotheses in the Statement.

1.1F1F3F4givenalgebra

For every $x$ , [F3] and [F4] give $∣ x^{T} E_{S} x ∣ \leq ∥ x ∥_{2} ∥ E_{S} x ∥_{2} \leq ∥ E_{S} ∥_{2} ∥ x ∥_{2}^{2} \leq e_{S} ∥ x ∥_{2}^{2}$ , hence $x^{T} \hat{S} x = x^{T} S x + x^{T} E_{S} x \geq (μ - e_{S}) ∥ x ∥_{2}^{2} .$ Thus $\hat{S}$ is positive definite. Also $∥ S^{- 1} ∥_{2} \leq μ^{- 1}$ and $∥ {\hat{S}}^{- 1} ∥_{2} \leq (μ - e_{S})^{- 1} = β$ : if $y = S x$ (or $y = \hat{S} x$ ), then [F4] gives $∥ x ∥_{2} ∥ y ∥_{2} \geq x^{T} y$ and the corresponding quadratic lower bound gives the stated inverse estimate.

2.1F2step 1.1algebra

The resolvent identity is ${\hat{S}}^{- 1} - S^{- 1} = - S^{- 1} E_{S} {\hat{S}}^{- 1} .$ Using [F2] and step 1.1 gives $∥ {\hat{S}}^{- 1} - S^{- 1} ∥_{2} \leq μ^{- 1} e_{S} β$ .

3.1F2F3step 1.1step 2.1algebra

Splitting the gain difference yields $\hat{B} {\hat{S}}^{- 1} - B S^{- 1} = E_{B} {\hat{S}}^{- 1} + B ({\hat{S}}^{- 1} - S^{- 1}) .$ The operator norm is bounded by the Frobenius norm for $E_{B}$ , so [F2] and steps 1.1–1.2 imply the displayed gain bound.

4.1F2F3F4step 1.1step 2.1algebra∎

Expanding the Schur difference gives $\hat{Φ} - Φ = E_{D} - E_{B} {\hat{S}}^{- 1} B^{T} - B {\hat{S}}^{- 1} E_{B}^{T} - B ({\hat{S}}^{- 1} - S^{- 1}) B^{T} - E_{B} {\hat{S}}^{- 1} E_{B}^{T} .$ Apply the operator-norm triangle inequality from [F4], [F2], the Frobenius-to-operator bound [F3], and the inverse estimates of steps 1.1–1.2. The five terms are bounded by $e_{D}$ , $e_{B} β b$ , $b β e_{B}$ , $b^{2} μ^{- 1} β e_{S}$ , and $e_{B}^{2} β$ , respectively, which proves the stated posterior bound.

Depends on

Used by

The square-root filter has an explicit conditioned finite-horizon error bound Theorem

Dependency tree · two levels

10 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

Simo Särkkä, Bayesian Filtering and Smoothing (standard reference, not scraped)
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)