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

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For a simple eigenvalue, left and right eigenvectors pair nontrivially and may be normalized by $y^{*} x = 1$

Statement

Let $λ$ be a simple eigenvalue of $A$ , and let $x, y \neq 0$ be compatible right and left eigenvectors. Then $y^{*} x \neq 0$ . Consequently, after rescaling either vector, one may impose the normalization

$y^{*} x = 1.$

Facts & Assumptions

Given: A simple eigenvalue $λ$ of $A$ and compatible nonzero vectors $x, y$ with $A x = λ x$ and $y^{*} A = λ y^{*}$ .

[F1]

Compatible left and right eigenvectors for a simple eigenvalue satisfy the displayed equations above (Compatible left and right eigenvectors for a simple eigenvalue).

Proof

technique · direct

1.1F1assume-contraalgebra

Assume for contradiction that $y^{*} x = 0$ . Then $x \in \ker y^{*}$ . Also $y^{*} (A - λ I) = 0$ , so $range (A - λ I) \subseteq \ker y^{*}$ . Because $λ$ is simple, $rank (A - λ I) = n - 1$ and $\dim \ker y^{*} = n - 1$ , hence $range (A - λ I) = \ker y^{*}$ . Therefore $x = (A - λ I) z$ for some $z$ .

2.1contradiction: simplicitydischarge-contradictionstep 1.1algebra∎

Step 1.1 gives $(A - λ I)^{2} z = 0$ while $(A - λ I) z = x \neq 0$ , so $z$ starts a Jordan chain of length $2$ for $λ$ . That contradicts the simplicity of $λ$ . Hence $y^{*} x \neq 0$ . Scaling $x$ by $(y^{*} x)^{- 1}$ yields the normalization $y^{*} x = 1$ .

Depends on

Compatible left and right eigenvectors for a simple eigenvalue

Used by

Dependency tree · two levels

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Sources

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

For a simple eigenvalue, left and right eigenvectors pair nontrivially and may be normalized by y∗x=1y∗x=1