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Hermitian eigenvalues move by at most the spectral norm perturbation

Statement

Let $A, E \in C^{n \times n}$ be Hermitian, and order their real eigenvalues decreasingly as $λ_{1} (A) \geq \dots \geq λ_{n} (A)$ , $λ_{1} (E) \geq \dots \geq λ_{n} (E)$ , and likewise for $A + E$ . Then for every $1 \leq k \leq n$ ,

∣ λ_{k} (A + E) - λ_{k} (A) ∣ \leq ∥ E ∥_{2} .

Facts & Assumptions

Given: Hermitian $A, E$ and the ordered eigenvalues in the statement.

[F1]

Realification duplicates each Hermitian eigenvalue and preserves operator 2-norms (Realification preserves Hermitian spectra, norms, and congruence).

[F2]

Weyl's two-sided eigenvalue inequalities hold for real self-adjoint sums (Weyl inequalities bound the eigenvalues of a sum of self-adjoint endomorphisms).

[F3]

The operator norm is induced by the Euclidean vector norm (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

Proof

1.1givenF1F2

Realify $A$ and $E$ . Their eigenvalues in decreasing order repeat each complex eigenvalue twice, so the $2 k$ -th eigenvalue of $R (A)$ is $λ_{k} (A)$ and similarly for $E$ and $A + E$ . By [F2] with indices $i = 2 k, j = 1$ , the upper Weyl inequality gives $λ_{2 k} (R (A + E)) \leq λ_{2 k} (R (A)) + λ_{1} (R (E))$ . With $i = 2 k, j = 2 n$ , the lower inequality gives $λ_{2 k} (R (A + E)) \geq λ_{2 k} (R (A)) + λ_{2 n} (R (E))$ .

2.1givenF1F3step 1.1algebra∎

For a unit eigenvector $u$ of the Hermitian $E$ , $∣ λ ∣ = ∥ E u ∥_{2} \leq ∥ E ∥_{2}$ by [F3]. Hence $- ∥ E ∥_{2} \leq λ_{2 n} (R (E))$ and $λ_{1} (R (E)) \leq ∥ E ∥_{2}$ by [F1]. Step 1.1 places $λ_{k} (A + E) - λ_{k} (A)$ between these two bounds, proving the result.

Depends on

Used by

Near-orthonormal residual data enclose all ordered Hermitian eigenvalues Theorem

Dependency tree · two levels

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Sources

Nicholas J. Higham, Numerical Linear Algebra and Matrix Analysis, MIMS EPrint 2015.103, §§5.2 and 5.4 (standard reference, not scraped)