LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Two commuting complex 2x2 matrices are simultaneously triangularizable

Statement

Let $A, B \in M_{2} (C)$ commute. There is a one-dimensional subspace of $C^{2}$ invariant under both $A$ and $B$ . Equivalently, in a basis beginning with a nonzero vector in that subspace, both matrices are upper triangular.

Statement references

Bénard, Florens, and Rodau, §2.2

Tillmann, §§4.2–4.4

Facts & Assumptions

Given: Two commuting complex $2 \times 2$ matrices $A, B$ .

[F1]

Every complex $2 \times 2$ matrix has an eigenvalue and a nonzero eigenspace because its characteristic polynomial has a complex root.

Proof

1.1F1cases

If $A$ is scalar, choose an eigenvector of $B$ using [F1]; its span is preserved by $A$ and $B$ .

1.2F1givenalgebra

If $A$ is not scalar, choose an eigenvalue $λ$ of $A$ . Its eigenspace $E_{λ}$ is nonzero and has dimension one, since a two-dimensional eigenspace would force $A = λ I$ . For $v \in E_{λ}$ , commutation gives $A (B v) = B (A v) = λ B v$ , so $B v \in E_{λ}$ ; hence this line is invariant under both matrices.

2.1step 1.1step 1.2algebra∎

A basis beginning with a nonzero vector in the common invariant line makes both matrices upper triangular.

Used by

SL2 character variety and peripheral eigenvalues Definition
Figure-eight A-polynomial calculation Example
Rationality of the adjoint-twisted slope Lemma

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Léo Bénard, Vincent Florens, and Adrien Rodau, A slope invariant and the A-polynomial of knots, §2.2 (standard reference, not scraped)
Stephan Tillmann, Varieties associated to 3-manifolds, §§4.2–4.4 (standard reference, not scraped)