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 Axμx2

Statement

Let F{R,C}, let AMn(F), let μF, and let xFn satisfy x2=1. Put r:=Axμx. Then

η(A,μ,x)=r2.

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

Facts & Assumptions

Given: A unit vector xFn, a scalar μF, a matrix AMn(F), and the residual r=Axμx, where F{R,C}.

[L1]

The residual and the normwise backward error are defined by r=Axμx and η(A,μ,x)=inf{E2:(A+E)x=μx} (The residual r=Axμx and the normwise backward error of an approximate eigenpair).

[L2]

The induced operator norm satisfies Ey2E2y2 (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

[L3]

Cauchy--Schwarz gives x,yx2y2 (Cauchy–Schwarz: u,vuv, with equality exactly for linearly dependent vectors).

Proof

technique · direct
1.1

If (A+E)x=μx, then Ex=(Axμx)=r. Because x2=1, [L2] gives E2Ex2=r2. So every admissible perturbation has norm at least r2.

L1L2algebra
1.2

Define E:=rx. Then Ex=r(xx)=r, so (A+E)x=μx. For any unit vector y, [L3] gives Ey2=r2xyr2, with equality at y=x. Hence E2=r2.

L1L3constructalgebra
2.1

Step 1.1 gives the lower bound and step 1.2 attains it, so η(A,μ,x)=r2.

step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

9 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