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Diskwise Neumann certificates bound a complex matrix resolvent

Statement

Let n≥1, let Γ be a nonempty finite union of closed line segments in C, let A⊆Cn×n be nonempty, and let A0∈Cn×n. Each segment of Γ is assigned to one of finitely many disks Dj={z:∣z−zj∣≤rj} whose radius is rj≥0 and whose disk contains both endpoints of that segment. For each disk there are Rj∈Cn×n and nonnegative bounds ej,δA,sj satisfying ej≥∥I−Rj(zjI−A0)∥∞,C,δA≥sup⁡A∈A∥A−A0∥∞,C,sj≥∥Rj∥∞,C. If qj=ej+sj(rj+δA)<1 for each j, then every z∈Γ and A∈A make zI−A invertible, and ∥(zI−A)−1∥∞,C≤max⁡jsj1−qj. The theorem bounds the resolvent on the supplied contour cover; it does not count enclosed eigenvalues or certify a nonnormal eigenvalue disk from a residual.

Facts & Assumptions

Given: Γ,A,A0,Dj,Rj,ej,δA,sj and qj as in the statement.

[F1]

A defect of complex infinity norm at most q<1 has inverse norm at most (1−q)−1 (A complex infinity-norm defect bounds the inverse norm).

[F2]

The complex infinity matrix norm is compatible with products (The complex infinity matrix norm is compatible and submultiplicative).

[F3]

A product of square matrices is nonsingular only if each factor is nonsingular, by rank-nullity (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T).

[F4]

The complex infinity matrix norm is the maximum absolute row sum, so ∥I∥∞,C=1 (The complex induced infinity matrix norm).

Proof

1.1givenalgebra

Each segment of Γ has both endpoints in its assigned disk. A disk is convex, so the segment between its endpoints lies in the disk; the finite assignments therefore cover all of Γ.

1.2givenF1F2F4algebra

Fix a disk, z∈Dj, and A∈A. The exact identity I−Rj(zI−A)=I−Rj(zjI−A0)−Rj((z−zj)I−(A−A0)) and [F2] give ∥I−Rj(zI−A)∥∞,C≤ej+sj(∣z−zj∣+∥A−A0∥∞,C)≤qj<1. [F1] applies to M=Rj(zI−A)=I−E with E=I−M, giving ∥M−1∥∞,C≤(1−qj)−1. Here [F4] gives ∥(z−zj)I∥∞,C=∣z−zj∣, and the row-sum norm's triangle inequality and compatibility [F2, F4] justify the defect estimate.

2.1givenF1F2F3F4step 1.1step 1.2algebra∎

By [F4], ∥I∥∞,C=1. Since M=Rj(zI−A) is nonsingular, both square factors are nonsingular by [F3]. Then (zI−A)−1=M−1Rj, so [F2] gives ∥(zI−A)−1∥∞,C≤sj/(1−qj). The disk cover from step 1.1 yields the maximum bound on all of Γ, uniformly in A.

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