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A B-dual residual bounds distance to the spectrum of a Hermitian-definite pencil

Statement

Let $n \geq 1$ , let $F \in {R, C}$ , and let $A, B \in F^{n \times n}$ be Hermitian with $B$ positive definite. For $x \in F^{n} ∖ {0}$ and $μ \in R$ , put $r = A x - μ B x$ . Then $dist (μ, σ (A, B)) \leq \frac{∥ r ∥_{B^{- 1}}}{∥ x ∥_{B}},$ where $σ (A, B)$ is the finite generalized spectrum and $dist (μ, S) : = \min_{λ \in S} ∣ μ - λ ∣$ . Equivalently, the interval with centre $μ$ and radius $∥ r ∥_{B^{- 1}} / ∥ x ∥_{B}$ contains at least one generalized eigenvalue.

Facts & Assumptions

Given: A Hermitian-definite pencil $(A, B)$ , a nonzero vector $x$ , and a real scalar $μ$ .

[F1]

The B-inner product, B-norm, and dual residual norm: for $B = L L^{*}$ , $∥ x ∥_{B} = ∥ L^{*} x ∥_{2}$ and $∥ r ∥_{B^{- 1}} = ∥ L^{- 1} r ∥_{2}$ .

[F2]

A Hermitian-definite pencil has a real spectrum and a B-orthonormal eigenbasis: $C = L^{- 1} A L^{- *}$ has an orthonormal eigenbasis and its eigenvalues are the generalized eigenvalues of $(A, B)$ .

[F3]

Regular matrix pencils and homogeneous generalized eigenvalues: $σ (A, B)$ is the set of finite ratios $α / β$ from homogeneous roots with $β \neq 0$ .

Proof

1.1givenF1algebra

Let $y = L^{*} x$ and $C = L^{- 1} A L^{- *}$ . Since $L^{- 1} B x = L^{*} x = y$ , the transformed residual is $L^{- 1} r = L^{- 1} (A x - μ B x) = (C - μ I) y .$ By [F1], $∥ y ∥_{2} = ∥ x ∥_{B}$ and $∥ L^{- 1} r ∥_{2} = ∥ r ∥_{B^{- 1}}$ .

2.1F2F3step 1.1algebra∎

Let $u_{1}, \dots, u_{n}$ be the orthonormal eigenbasis of $C$ from [F2], with real eigenvalues $λ_{1}, \dots, λ_{n}$ . By [F3], these are exactly the finite generalized eigenvalues in $σ (A, B)$ . Write $y = \sum_{i} c_{i} u_{i}$ . Then $∥ (C - μ I) y ∥_{2}^{2} = \sum_{i} ∣ c_{i} ∣^{2} ∣ λ_{i} - μ ∣^{2} \geq dist (μ, {λ_{i}})^{2} \sum_{i} ∣ c_{i} ∣^{2} = dist (μ, σ (A, B))^{2} ∥ y ∥_{2}^{2} .$ Taking square roots and applying step 1.1 proves the bound.

Depends on

Used by

A mass-stiffness pencil has real B-orthogonal modes and a residual enclosure Example

Dependency tree · two levels

13 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

Peter Arbenz, Lecture Notes on Solving Large Scale Eigenvalue Problems, §11.4 (standard reference, not scraped)
LAPACK Users' Guide, Generalized Symmetric Definite Eigenproblems (standard reference, not scraped)