TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Along a differentiable matrix path, a simple eigenvalue satisfies $λ^{'} = y^{} A^{'} x$ under the normalization $y^{} x = 1$

Statement

Let $A (t)$ be differentiable, and let $λ (t)$ be a simple eigenvalue with differentiable compatible eigenvectors $x (t), y (t)$ normalized by $y (t)^{*} x (t) = 1$ . Then

$λ^{'} (t) = y (t)^{*} A^{'} (t) x (t) .$

Facts & Assumptions

Given: A differentiable matrix path $A (t)$ , a differentiable simple eigenpair branch $λ (t), x (t), y (t)$ , and the normalization $y (t)^{*} x (t) = 1$ .

[L1]

Simple eigenpairs admit local differentiable branches after gauge fixing (A simple eigenvalue and a gauge-fixed right eigenvector admit local $C^{1}$ branches in the underlying real matrix space).

Proof

technique · direct

1.1

Differentiate the eigenvalue equation $A (t) x (t) = λ (t) x (t)$ : $A^{'} (t) x (t) + A (t) x^{'} (t) = λ^{'} (t) x (t) + λ (t) x^{'} (t) .$ Left-multiply by $y (t)^{*}$ . Since $y (t)^{*} A (t) = λ (t) y (t)^{*}$ , the terms with $x^{'} (t)$ cancel.

givenL1algebra

2.1

Step 1.1 leaves $y (t)^{*} A^{'} (t) x (t) = λ^{'} (t) y (t)^{*} x (t)$ . The normalization $y (t)^{*} x (t) = 1$ therefore gives $λ^{'} (t) = y (t)^{*} A^{'} (t) x (t)$ .

step 1.1algebra∎

Depends on

Used by

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Sources

David Bindel, CS 6210: Matrix Computations - Perturbation theory (standard reference, not scraped)

Along a differentiable matrix path, a simple eigenvalue satisfies λ′=y∗A′xλ′=y∗A′x under the normalization y∗x=1y∗x=1