LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-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.

A small shifted Jacobian has an explicit inverse bound

Statement

Let $d \geq 1$ , $B \in R^{d \times d}$ , $h > 0$ , $γ > 0$ , and $0 \leq q < 1$ satisfy $∥ γ h B ∥_{2} \leq q$ , where $∥ \cdot ∥_{2}$ is the matrix norm induced by the Euclidean vector $2$ -norm, as in The matrix norm induced by a published vector p-norm. Then $A = I_{d} - γ h B$ is invertible and $∥ A^{- 1} ∥_{2} \leq \frac{1}{1 - q} .$ In particular, if $∥ γ h B ∥_{2} \leq \frac{1}{2}$ , then $∥ A^{- 1} ∥_{2} \leq 2$ .

Facts & Assumptions

Given: The finite-dimensional real vector space $R^{d}$ , its Euclidean norm, and the stated $B, h, γ, q$ .

[F1]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: the Euclidean norm obeys the triangle inequality and is homogeneous.

[F2]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a linear map on a finite-dimensional vector space, dimension equals nullity plus rank.

[F3]

Invertible Euclidean linear maps: a square Euclidean linear map is invertible when it has a two-sided linear inverse.

[F4]

Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1: $∥ B x ∥_{2} \leq ∥ B ∥_{2} ∥ x ∥_{2}$ and $∥ B C ∥_{2} \leq ∥ B ∥_{2} ∥ C ∥_{2}$ .

Proof

Given: The data in the Statement.

1.1givenF1F4algebra

For every $x \in R^{d}$ , [F4] gives $∥ B x ∥_{2} \leq ∥ B ∥_{2} ∥ x ∥_{2}$ , and the triangle inequality in [F1] gives $∥ A x ∥_{2} = ∥ x - γ h B x ∥_{2} \geq ∥ x ∥_{2} - γ h ∥ B x ∥_{2} \geq (1 - q) ∥ x ∥_{2}$ . Since $1 - q > 0$ , $A x = 0$ forces $x = 0$ , so $A$ is injective.

2.1step 1.1F2F3algebra

The domain and codomain of $A$ both have dimension $d$ . By [F2] and step 1.1, $rank A = d$ , so its image is all of $R^{d}$ . Thus $A$ is bijective and has the inverse of [F3].

3.1step 2.1step 1.1algebra∎

For any $z \in R^{d}$ , let $x = A^{- 1} z$ . Step 1.1 yields $(1 - q) ∥ A^{- 1} z ∥_{2} \leq ∥ z ∥_{2}$ ; taking the supremum over $∥ z ∥_{2} = 1$ proves the inverse bound. Substitution of $q = 1 / 2$ gives the final claim.

Depends on

Used by

Dependency tree · two levels

35 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

WIAS Berlin, Numerical Methods for Stiff Ordinary Differential Equations, §2.4 Rosenbrock Methods (standard reference, not scraped)
Daniel Kressner, Advanced Numerical Analysis, §3.5 Rosenbrock methods (standard reference, not scraped)