LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-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.

Invertibility and a norm bound for the pseudo-arclength bordered matrix

Statement

Let $n \geq 1$ , let $A : R^{n + 1} \to R^{n}$ be linear of rank $n$ , and let $t \in \ker A$ have $∥ t ∥_{2} = 1$ . Set $E = t^{⊥}$ and let $Q = (A ∣_{E})^{- 1} : R^{n} \to E$ . For a unit vector $τ \in R^{n + 1}$ , form the square bordered matrix $B = [\begin{array}{c} A \\ τ^{T} \end{array}] .$ Then $B$ is invertible exactly when $τ^{T} t \neq 0$ . If $0 < c \leq 1$ , $∣ τ^{T} t ∣ \geq c$ , and $K_{A} \geq 0$ satisfies $∥ Q u ∥_{2} \leq K_{A} ∥ u ∥_{2}$ for all $u \in R^{n}$ , then $∥ B^{- 1} ∥_{o p} \leq \sqrt{K_{A}^{2} (1 + c^{- 1})^{2} + c^{- 2}} .$

Here $t$ is the current unit tangent and $τ$ is the normal used in the corrector hyperplane. The lower bound $c$ is the explicit transversality margin; as it tends to zero this estimate need not remain bounded.

Facts & Assumptions

Given: The dimensions, rank, unit kernel vector $t$ , orthogonal complement $E = t^{⊥}$ , unit vector $τ$ , and bound $∣ τ^{T} t ∣ \geq c > 0$ in the Statement.

[F1]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a linear map from a finite-dimensional domain, the domain dimension is the sum of the kernel and image dimensions.

[F2]

In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ : for a subspace $W$ of a finite-dimensional inner-product space, $\dim W + \dim W^{⊥} = \dim V$ .

[F3]

If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ : a subspace of a finite-dimensional space has dimension at most the ambient dimension, and equal dimension forces equality. The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial: a linear map is injective exactly when its kernel is zero. The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ : the coordinate space $R^{n}$ has dimension $n$ .

[F4]

Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ : every Euclidean linear map has a finite operator-norm bound.

[F5]

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

[F6]

Rank and nullity of a linear map with finite-dimensional domain: for a linear map with finite-dimensional domain, rank is the dimension of its image.

Proof

Given: The hypotheses in the Statement.

1.1F1F3F6given

By [F1], [F3], and [F6], $\dim \ker A = (n + 1) - n = 1$ . Since $t$ is a unit vector in the kernel, $\ker A = span {t}$ .

2.1F2F3step 1.1

Apply [F2] to $W = \ker A$ in $R^{n + 1}$ to obtain $\dim E = (n + 1) - 1 = n$ . The codomain $R^{n}$ also has dimension $n$ by [F3].

3.1F1F3step 2.1givenalgebra

If $x = α t$ also lies in $E = t^{⊥}$ , then $0 = t^{T} x = α ∥ t ∥_{2}^{2} = α$ , so $E \cap \ker A = {0}$ . Thus $A ∣_{E}$ is injective by [F3]. Its domain and codomain both have dimension $n$ , so [F1] and [F3] make it surjective. Thus $Q$ is well-defined.

4.1givenstep 1.1step 3.1construct

For $(u, s) \in R^{n} \times R$ , write a prospective solution as $v = e + α t$ with $e \in E$ . The first row equation $A v = u$ forces $e = Q u$ , and the last row equation forces $α = \frac{s - τ^{T} Q u}{τ^{T} t} .$ If $τ^{T} t \neq 0$ , these formulas give one solution for every $(u, s)$ , so $B$ is invertible. If $τ^{T} t = 0$ , then $B t = 0$ with $t \neq 0$ , so $B$ is singular.

5.1F4F5step 4.1algebra

Under $∣ τ^{T} t ∣ \geq c$ , [F5] gives $∣ α ∣ \leq c^{- 1} (∣ s ∣ + K_{A} ∥ u ∥_{2}) .$ Consequently $∥ v ∥_{2} \leq K_{A} (1 + c^{- 1}) ∥ u ∥_{2} + c^{- 1} ∣ s ∣ .$ Applying [F5] to the two scalar coefficients of $\sqrt{∥ u ∥_{2}^{2} + ∣ s ∣^{2}}$ gives the stated operator-norm bound for $B^{- 1}$ .

6.1step 4.1step 5.1∎

Step 4.1 proves the exact invertibility criterion, and step 5.1 proves the quantitative inverse bound under the stated angular margin.

Depends on

Used by

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Sources

Eugene L. Allgower and Kurt Georg, Numerical Continuation Methods: An Introduction, Springer Series in Computational Mathematics 13 (1990) (standard reference, not scraped)
K. A. Cliffe, A. Spence, and S. J. Tavener, The numerical analysis of bifurcation problems with application to fluid mechanics, Acta Numerica 9 (2000), 39–131 (standard reference, not scraped)