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Bauer-Fike bounds eigenvalue motion by the eigenvector condition number

Statement

Let $n \geq 1$ , let $A = V D V^{- 1} \in C^{n \times n}$ be diagonalizable, where $D = diag (λ_{1}, \dots, λ_{n})$ , and let $E \in C^{n \times n}$ . Every $μ \in σ (A + E)$ satisfies

$dist (μ, σ (A)) \leq κ_{2} (V) ∥ E ∥_{2}, κ_{2} (V) = ∥ V ∥_{2} ∥ V^{- 1} ∥_{2} .$

No estimate independent of $V$ is asserted for a nonnormal diagonalization.

Facts & Assumptions

Given: A diagonalization $A = V D V^{- 1}$ , a perturbation $E$ , and an eigenvalue $μ$ of $A + E$ .

[F1]

Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism: an eigenvalue has a nonzero eigenvector.

[F2]

The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2: the operator norm is submultiplicative.

[F3]

A finite matrix within unit distance of the identity is invertible: if $∥ X ∥_{2} < 1$ , then $I + X$ is invertible.

Proof

1.1givenF1algebra

Similarity by the invertible $V$ preserves singularity, so $σ (A) = {λ_{1}, \dots, λ_{n}}$ . If $μ \in σ (A)$ , the distance is zero and the claim holds. Otherwise let $d = \min_{j} ∣ λ_{j} - μ ∣ > 0$ and suppose for contradiction that $d > κ_{2} (V) ∥ E ∥_{2}$ . Since $D - μ I$ is diagonal, $∥ (D - μ I)^{- 1} ∥_{2} = 1 / d$ .

2.1givenF2F3step 1.1

By [F2], $X = (D - μ I)^{- 1} V^{- 1} E V$ satisfies $∥ X ∥_{2} \leq d^{- 1} ∥ V^{- 1} ∥_{2} ∥ E ∥_{2} ∥ V ∥_{2} < 1$ . The Neumann lemma [F3] makes $I + X$ invertible.

3.1step 2.1F1algebra∎

Similarity gives $V^{- 1} (A + E - μ I) V = (D - μ I) + V^{- 1} E V = (D - μ I) (I + X)$ , which is invertible by step 2.1. This contradicts $μ \in σ (A + E)$ , so the supposition was false and $d \leq κ_{2} (V) ∥ E ∥_{2}$ .

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Sources

Youcef Saad, Numerical Methods for Large Eigenvalue Problems, 2nd ed., §3.2.1 (standard reference, not scraped)
James W. Demmel, Applied Numerical Linear Algebra, Theorem 4.5 (standard reference, not scraped)