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

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The normwise condition number of a simple eigenvalue is $∥ x ∥_{2} ∥ y ∥_{2} / ∣ y^{*} x ∣$

Statement

Let $λ$ be a simple eigenvalue of $A$ , and let $x, y$ be compatible nonzero right and left eigenvectors. Then

$κ (λ; A) = \frac{∥ x ∥_{2} ∥ y ∥_{2}}{∣ y^{*} x ∣} .$

In particular, for the normalization $y^{*} x = 1$ one has $κ (λ; A) = ∥ x ∥_{2} ∥ y ∥_{2}$ .

Facts & Assumptions

Given: A simple eigenvalue $λ$ of $A$ and compatible nonzero vectors $x, y$ .

[F1]

The condition number is $κ (λ; A) = \sup_{∥ H ∥_{2} = 1} ∣ D λ (A) [H] ∣$ (The normwise condition number of a simple eigenvalue).

[L1]

For any differentiable branch normalized by $y^{*} x = 1$ , $D λ (A) [H] = y^{*} H x$ (Along a differentiable matrix path, a simple eigenvalue satisfies $λ^{'} = y^{*} A^{'} x$ under the normalization $y^{*} x = 1$ ).

Proof

technique · direct

1.1F1L1algebra

Normalize first so that $y^{*} x = 1$ . Then [L1] gives $∣ D λ (A) [H] ∣ = ∣ y^{*} H x ∣ \leq ∥ y ∥_{2} ∥ H ∥_{2} ∥ x ∥_{2}$ by Cauchy-Schwarz. Taking the supremum over $∥ H ∥_{2} = 1$ in [F1] yields $κ (λ; A) \leq ∥ x ∥_{2} ∥ y ∥_{2}$ .

2.1F1L1step 1.1constructalgebra∎

Let $u = y / ∥ y ∥_{2}$ and $v = x / ∥ x ∥_{2}$ , and define $H = u v^{*}$ . Then $∥ H ∥_{2} = 1$ and $y^{*} H x = ∥ y ∥_{2} ∥ x ∥_{2}$ . Hence [F1] and [L1] give the reverse inequality, so $κ (λ; A) = ∥ x ∥_{2} ∥ y ∥_{2}$ under the normalization $y^{*} x = 1$ . Undoing the normalization inserts the factor $∣ y^{*} x ∣^{- 1}$ and gives the general formula.

Depends on

Used by

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Dependency tree · two levels

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Sources

David Bindel, CS 6210: Matrix Computations - Perturbation theory (standard reference, not scraped)
G. W. Stewart and Ji-guang Sun, Matrix Perturbation Theory (standard reference, not scraped)

The normwise condition number of a simple eigenvalue is ∥x∥2∥y∥2/∣y∗x∣∥x∥2​∥y∥2​/∣y∗x∣