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Local leading-eigenvector stability under a spectral gap

Statement

Let $p \geq 2$ , let $A \in R^{p \times p}$ be symmetric, and order its eigenvalues as $λ_{1} (A) \geq \dots \geq λ_{p} (A)$ . Suppose its leading eigenvalue is simple, with gap $g : = λ_{1} (A) - λ_{2} (A) > 0,$ and let $v$ be a unit leading eigenvector. Let $E \in R^{p \times p}$ be symmetric, put $ϵ : = ∥ E ∥_{o p}$ , and assume $0 \leq ϵ < g / 2$ . If $\tilde{v}$ is any unit leading eigenvector of $A + E$ , then the leading eigenvalue of $A + E$ is simple and $∣ λ_{1} (A + E) - λ_{1} (A) ∣ \leq ϵ, \sin ∠ (v, \tilde{v}) : = \sqrt{1 - (v^{T} \tilde{v})^{2}} \leq \frac{ϵ}{g - ϵ} .$ The bound concerns the line spanned by the eigenvector, so it is unchanged by either sign choice. It is a local eigengap bound; it does not assert a dimension-free Davis--Kahan theorem or a finite-sample inference procedure.

Facts & Assumptions

Given: A real symmetric $p \times p$ matrix $A$ , $p \geq 2$ , a simple leading eigenvalue with gap $g > 0$ , a unit leading eigenvector $v$ , a real symmetric perturbation $E$ with operator norm $ϵ < g / 2$ , and a unit leading eigenvector $\tilde{v}$ of $A + E$ .

[F1]

Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis: a real symmetric matrix has an orthonormal eigenbasis with real eigenvalues.

[F2]

Courant-Fischer min-max principle for self-adjoint endomorphisms on finite-dimensional real inner product spaces: for a symmetric operator, the ordered eigenvalues are characterized by the minimum-over-subspaces and maximum-over-subspaces Rayleigh-quotient formulas.

[F3]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: $∥ E ∥_{o p} = \max_{∥ x ∥_{2} = 1} ∥ E x ∥_{2}$ .

[F4]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: the Euclidean norm is induced by $x^{T} y$ , and $∣ x^{T} y ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ .

Proof

technique · direct

1.1F2F3F4givenalgebra

For every unit vector $x$ , [F3] and [F4] give $∣ x^{T} E x ∣ \leq ∥ x ∥_{2} ∥ E x ∥_{2} \leq ϵ$ . The top-eigenvalue Rayleigh characterization in [F2] therefore gives $λ_{1} (A + E) \leq λ_{1} (A) + ϵ$ , while evaluating the perturbed Rayleigh quotient at $v$ gives $λ_{1} (A + E) \geq v^{T} (A + E) v \geq λ_{1} (A) - ϵ$ . This proves the eigenvalue bound.

2.1F1F2F3step 1.1algebra

The minimum--maximum formula in [F2], with the $(p - 1)$ -dimensional subspace $v^{⊥}$ , gives $λ_{2} (A + E) \leq \max_{∥ x ∥_{2} = 1, x ⊥ v} x^{T} (A + E) x \leq λ_{2} (A) + ϵ$ , since every such $x$ has $x^{T} A x \leq λ_{2} (A)$ . Step 1.1 gives $λ_{1} (A + E) \geq λ_{1} (A) - ϵ$ , hence $λ_{1} (A + E) - λ_{2} (A + E) \geq g - 2 ϵ > 0$ ; the perturbed leading eigenvalue is simple. Also $λ_{1} (A + E) - λ_{2} (A) \geq g - ϵ > 0$ .

3.1F1F3step 2.1algebra∎

Let $P = I - v v^{T}$ , the orthogonal projection onto $v^{⊥}$ , and write ${\tilde{λ}}_{1} = λ_{1} (A + E)$ . Projecting $(A + E) \tilde{v} = {\tilde{λ}}_{1} \tilde{v}$ onto $v^{⊥}$ gives $({\tilde{λ}}_{1} I - A ∣_{v^{⊥}}) P \tilde{v} = P E \tilde{v}$ . In an orthonormal eigenbasis of $A ∣_{v^{⊥}}$ , every eigenvalue of ${\tilde{λ}}_{1} I - A ∣_{v^{⊥}}$ is at least ${\tilde{λ}}_{1} - λ_{2} (A) \geq g - ϵ$ by step 2.1. Therefore $(g - ϵ) ∥ P \tilde{v} ∥_{2} \leq ∥ P E \tilde{v} ∥_{2} \leq ϵ$ . Indeed, $z = P z + (I - P) z$ is an orthogonal decomposition, so $∥ P z ∥_{2} \leq ∥ z ∥_{2}$ ; then [F3] and $∥ \tilde{v} ∥_{2} = 1$ give the final inequality. Since $∥ P \tilde{v} ∥_{2}^{2} = 1 - (v^{T} \tilde{v})^{2}$ , division by $g - ϵ > 0$ proves the angle bound.

Depends on

Used by

A simple population eigengap gives root-n sample PCA eigenspace consistency Corollary
A vanishing eigengap allows a fixed eigenvector rotation Counterexample

Dependency tree · two levels

31 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

Online Bayesian Data Analysis, Lecture Notes 8: Learning Signal Representations (11 April 2016) (standard reference, not scraped)
Soumendu Sundar Mukherjee, Multivariate Analysis, Lecture 8: Principal Component Analysis (standard reference, not scraped)