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

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For a Hermitian simple eigenvalue, one may take $y = x$ and the first-order formulas simplify accordingly

Statement

If $A (t)$ is a differentiable Hermitian matrix path, and $λ (t), x (t)$ is a differentiable simple eigenpair branch with $∥ x (t) ∥_{2} = 1$ , then one may choose the phase locally so that $x (t)^{*} x^{'} (t) = 0$ . With that choice,

$λ^{'} (t) = x (t)^{*} A^{'} (t) x (t), x^{'} (0) = - S A^{'} (0) x (0) .$

Facts & Assumptions

Given: A differentiable Hermitian matrix path $A (t)$ and a differentiable simple unit eigenvector branch $x (t)$ .

[L1]

For a simple eigenpair, the eigenvalue derivative is $y^{*} A^{'} x$ , and in a fixed gauge the eigenvector derivative is $- S A^{'} x$ (Along a differentiable matrix path, a simple eigenvalue satisfies $λ^{'} = y^{*} A^{'} x$ under the normalization $y^{*} x = 1$ , In a fixed gauge, the derivative of a simple right eigenvector is obtained by applying the reduced resolvent to the perturbation).

Proof

technique · direct

1.1

If $A (t) = A (t)^{*}$ and $A (t) x (t) = λ (t) x (t)$ , then taking adjoints shows $x (t)^{*} A (t) = λ (t) x (t)^{*}$ . Thus the same unit eigenvector can serve as both left and right eigenvector. Multiplying $x (t)$ by a unit complex phase if necessary imposes the gauge $x (t)^{*} x^{'} (t) = 0$ .

givenalgebra

2.1

Substitute $y = x$ into the formulas summarized in [L1]. This gives $λ^{'} (t) = x (t)^{*} A^{'} (t) x (t)$ and, in the chosen gauge, $x^{'} (0) = - S A^{'} (0) x (0)$ .

L1step 1.1∎

Depends on

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Sources

Alan Edelman and Steven G. Johnson, Matrix Calculus for Machine Learning and Beyond (standard reference, not scraped)

For a Hermitian simple eigenvalue, one may take y=xy=x and the first-order formulas simplify accordingly