TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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A simple eigenvalue and a gauge-fixed right eigenvector admit local $C^{1}$ branches in the underlying real matrix space

Statement

Let $A_{0}$ be a square matrix with simple eigenvalue $λ_{0}$ , and choose compatible eigenvectors $x_{0}, y_{0}$ normalized by $y_{0}^{*} x_{0} = 1$ . Then, in a neighborhood of $A_{0}$ inside the underlying real matrix space, there exist unique $C^{1}$ maps $A \mapsto λ (A)$ and $A \mapsto x (A)$ such that

$A x (A) = λ (A) x (A), y_{0}^{*} x (A) = 1,$

with $λ (A_{0}) = λ_{0}$ and $x (A_{0}) = x_{0}$ .

Facts & Assumptions

Given: A base matrix $A_{0}$ , a simple eigenvalue $λ_{0}$ , and normalized compatible eigenvectors $x_{0}, y_{0}$ .

[L1]

For a simple eigenvalue, one may normalize compatible left and right eigenvectors by $y_{0}^{*} x_{0} = 1$ (For a simple eigenvalue, left and right eigenvectors pair nontrivially and may be normalized by $y^{*} x = 1$ ).

[L2]

The parametrized implicit-function theorem gives a unique local $C^{1}$ solution once the derivative in the solved-for variables is invertible (The parametrized implicit function theorem with $C^{k}$ regularity).

Proof

technique · direct

1.1

Consider the real map $F (A, λ, x) = ((A - λ I) x, y_{0}^{*} x - 1)$ . Its derivative in $(λ, x)$ at $(A_{0}, λ_{0}, x_{0})$ is $(μ, h) \mapsto ((A_{0} - λ_{0} I) h - μ x_{0}, y_{0}^{*} h)$ . If this derivative vanishes, then left-multiplying the first component by $y_{0}^{*}$ gives $- μ y_{0}^{*} x_{0} = 0$ , hence $μ = 0$ by [L1]. Then $(A_{0} - λ_{0} I) h = 0$ and $y_{0}^{*} h = 0$ , so $h$ is a multiple of $x_{0}$ whose pairing with $y_{0}$ is zero; therefore $h = 0$ . Thus the derivative is injective. Because domain and codomain have the same real dimension, it is invertible.

L1givenalgebra

2.1

The hypotheses of [L2] now apply to $F$ at $(A_{0}, λ_{0}, x_{0})$ . Therefore there are neighborhoods and unique $C^{1}$ maps $A \mapsto λ (A)$ and $A \mapsto x (A)$ solving $F (A, λ (A), x (A)) = 0$ . Those equations are exactly $A x (A) = λ (A) x (A)$ and $y_{0}^{*} x (A) = 1$ , with the required base values.

L2step 1.1∎

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Sources

Alan Edelman and Steven G. Johnson, Matrix Calculus for Machine Learning and Beyond (standard reference, not scraped)
David Bindel, CS 6210: Matrix Computations - Perturbation theory (standard reference, not scraped)

A simple eigenvalue and a gauge-fixed right eigenvector admit local C1C1 branches in the underlying real matrix space