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

1.1givenalgebra

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 $x (t)^{*} x^{'} (t) = 0$ .

2.1L1step 1.1∎

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

Depends on

Used by

Dependency tree · two levels

7 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

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