Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated
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Perturbing a shifted Jacobian perturbs its inverse explicitly

Statement

Let d≥1, h,γ>0, and J,J~∈Rd×d. Set A=Id−γhJ and A~=Id−γhJ~. If both matrices are invertible, then A~−1−A−1=γh A~−1(J~−J)A−1. Consequently, if ∥A−1∥2≤K and ∥A~−1∥2≤K~, then ∥A~−1−A−1∥2≤γhKK~∥J~−J∥2. Here the induced matrix norm is the one defined in A small shifted Jacobian has an explicit inverse bound.

Facts & Assumptions

Given: The stated matrices and inverse bounds.

[F1]

Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1: the induced Euclidean matrix norm is submultiplicative: ∥AB∥2≤∥A∥2∥B∥2.

[F2]

A small shifted Jacobian has an explicit inverse bound: if ∥γhB∥2≤q<1, then I−γhB is invertible and its inverse norm is at most (1−q)−1.

Proof

Given: The data in the Statement.

1.1givenalgebra

Multiplying out the right-hand side gives A~−1(A−A~)A−1=A~−1AA−1−A~−1A~A−1=A~−1−A−1. Also A−A~=γh(J~−J), proving the identity.

2.1step 1.1F1algebra

Apply the induced-norm inequality in [F1] twice to the identity in step 1.1 and then use the two inverse bounds. This gives the claimed estimate.

3.1F2step 2.1algebra∎

If ∥γhJ∥2≤q<1 and ∥γhJ~∥2≤q~<1, [F2] supplies K=(1−q)−1 and K~=(1−q~)−1, so the estimate is explicit under these sufficient smallness conditions.

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