Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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For a unit vector x, the smallest perturbation making (μ,x) an exact eigenpair has spectral norm ∥Ax−μx∥2

Statement

Let F∈{R,C}, let A∈Mn(F), let μ∈F, and let x∈Fn satisfy ∥x∥2=1. Put r:=Ax−μx. Then

η(A,μ,x)=∥r∥2.

In particular, the rank-one perturbation E:=−rx∗ attains the infimum.

Facts & Assumptions

Given: A unit vector x∈Fn, a scalar μ∈F, a matrix A∈Mn(F), and the residual r=Ax−μx, where F∈{R,C}.

[L1]

The residual and the normwise backward error are defined by r=Ax−μx and η(A,μ,x)=inf⁡{∥E∥2:(A+E)x=μx} (The residual r=Ax−μx and the normwise backward error of an approximate eigenpair).

[L2]

The induced operator norm satisfies ∥Ey∥2≤∥E∥2∥y∥2 (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

[L3]

Cauchy--Schwarz gives ∣⟨x,y⟩∣≤∥x∥2∥y∥2 (Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors).

Proof

technique · direct
1.1L1L2algebra

If (A+E)x=μx, then Ex=−(Ax−μx)=−r. Because ∥x∥2=1, [L2] gives ∥E∥2≥∥Ex∥2=∥r∥2. So every admissible perturbation has norm at least ∥r∥2.

1.2L1L3constructalgebra

Define E:=−rx∗. Then Ex=−r(x∗x)=−r, so (A+E)x=μx. For any unit vector y, [L3] gives ∥Ey∥2=∥r∥2 ∣x∗y∣≤∥r∥2, with equality at y=x. Hence ∥E∥2=∥r∥2.

2.1step 1.1step 1.2∎

Step 1.1 gives the lower bound and step 1.2 attains it, so η(A,μ,x)=∥r∥2.

Depends on

Used by

Dependency tree · two levels

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Sources