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A simple eigenpair has a regularity-matched gauge-fixed parameter branch

Statement

Let k≥1 and let A(λ)∈Cd×d be a Ck matrix path. Let ζ0 be a simple eigenvalue of A(0), and choose compatible left and right eigenvectors p0,q0 with p0∗q0=1. Then there are unique local Ck functions ζ(λ),q(λ) such that A(λ)q(λ)=ζ(λ)q(λ),p0∗q(λ)=1,ζ(0)=ζ0,q(0)=q0. There is also a unique local Ck left eigenvector p(λ) with A(λ)∗p(λ)=ζ(λ)‾p(λ),p(λ)∗q(λ)=1,p(0)=p0. Moreover, ζ′(0)=p0∗A′(0)q0.

Facts & Assumptions

Given: A, ζ0, p0, q0 and the hypotheses in the Statement.

[F1]

Compatible left and right eigenvectors for a simple eigenvalue: A(0)q0=ζ0q0 and p0∗A(0)=ζ0p0∗.

[F2]

For a simple eigenvalue, left and right eigenvectors pair nontrivially and may be normalized by y∗x=1: compatible left and right eigenvectors for a simple eigenvalue can be normalized by p0∗q0=1.

[F3]

The parametrized implicit function theorem with Ck regularity: a Ck equation with an invertible derivative in the solved variables has a unique local Ck solution branch for every k≥1.

[F4]

A simple eigenvalue and a gauge-fixed right eigenvector admit local C1 branches in the underlying real matrix space: the simple eigenvalue and the same gauge-fixed eigenvector have unique local C1 branches.

[F5]

Proof

Given: The data and hypotheses in the Statement.

1.1F1F2givenalgebra

Regard complex vector spaces as real spaces of twice the dimension and define Φ(λ,ζ,q):=((A(λ)−ζI)q, p0∗q−1). This is Ck by the matrix path hypothesis. Its derivative in (ζ,q) at (0,ζ0,q0) sends (a,v) to ((A(0)−ζ0I)v−aq0, p0∗v). If this derivative is zero, left multiplication of the first component by p0∗ gives −a=0 by [F1] and [F2]; then v is in the eigenspace spanned by q0, and p0∗v=0 forces v=0. Thus the derivative is injective and, its finite-dimensional domain and codomain having equal real dimension, invertible.

2.1F3F4step 1.1

Apply [F3] to Φ=0 in the solved variables (ζ,q). It gives unique Ck branches satisfying the eigenvalue equation, gauge, and base values. By uniqueness they agree with the C1 branches in [F4].

2.2F1F2F3step 1.1algebra

Apply the same implicit-function construction as step 1.1 to A(λ)∗ at eigenvalue ζˉ0, with gauge q0∗p(λ)=1. The same injectivity computation, with q0,p0 interchanged and conjugated, gives a unique Ck left eigenvector branch p~(λ); its eigenvalue is ζ(λ)‾ by uniqueness of the simple local branch.

3.1F1F2step 2.1step 2.2givenalgebra

Set c(λ)=p~(λ)∗q(λ). It is Ck and c(0)=1, so it stays nonzero after shrinking the interval. Define p(λ)=p~(λ)/c(λ)‾; then p(λ)∗q(λ)=c(λ)/c(λ)=1 and its eigenvector equation is unchanged. If p^ is another such local branch, rescale it by d(λ)=q0∗p^(λ), which is nonzero near zero and equals 1 at zero. Then p^/d satisfies the fixed gauge q0∗(p^/d)=1 and solves the same adjoint eigenpair equation, so IFT uniqueness from step 2.2 gives p^/d=p~. The normalization p^∗q=1 then forces d=1/c‾, hence p^=p.

4.1F1F2F5step 2.1algebra∎

Differentiate A(λ)q(λ)=ζ(λ)q(λ) at zero and left-multiply by p0∗. The terms containing q′(0) cancel by [F1], and p0∗q0=1 by [F2], leaving ζ′(0)=p0∗A′(0)q0, also the interface stated in [F5].

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