TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A commuting split family is simultaneously triangularisable

Statement

Let $A$ be a family of pairwise commuting endomorphisms of a finite-dimensional $F$ -vector space $V$ . If $χ_{S}$ splits over $F$ for every $S \in A$ , then $A$ is simultaneously triangularisable. The family may be empty.

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ and a pairwise commuting family $A \subseteq {End}_{F} (V)$ such that every $χ_{S}$ splits over $F$ .

[L1]

An endomorphism whose characteristic polynomial splits is triangularisable ( $T$ is triangularisable iff its minimal polynomial splits iff its characteristic polynomial splits).

[L2]

If $W$ is invariant under $S$ , then $χ_{S} = χ_{S ∣_{W}} χ_{\overset{ˉ}{S}}$ (For invariant $W$ , $χ_{T} = χ_{T ∣_{W}} χ_{\overset{ˉ}{T}}$ ).

[L3]

A basis is upper triangular for an operator exactly when its initial spans form an invariant flag (Complete invariant flags are equivalent to upper-triangular matrices).

[L4]

Invariance makes every quotient operator well defined and linear (Invariance makes the induced quotient operator well defined and linear, with $π T = \overset{ˉ}{T} π$ ).

[L5]

A quotient basis lifts after a basis of the subspace to an adapted basis of the whole space (A quotient basis lifts to a basis adapted to $W$ ).

Proof

technique · induction

1.1base

If $\dim V = 0$ , the empty basis simultaneously triangularises every family.

1.2L1L2chooseih

Assume $\dim V > 0$ and the theorem in smaller dimensions; if $A$ is empty or all its members are scalar, any basis works, while otherwise choose a nonscalar $A \in A$ , use [L1] to obtain a nonzero proper eigenspace $E$ , observe that every $S \in A$ preserves $E$ because $S A = A S$ , and use [L2] plus induction on $E$ to obtain a common eigenvector $v \in E$ .

2.1step 1.2L2L4ih

Put $W = F v$ ; it is invariant under every $S \in A$ , the induced quotient operators commute by direct evaluation on cosets, and [L2] shows each quotient characteristic polynomial splits, so induction gives a common triangular basis of $V / W$ .

3.1step 1.2step 2.1L3L5discharge-induction∎

Lift that quotient basis after $v$ by [L5]; its initial spans are invariant for every $S$ by the quotient construction, so [L3] makes every representing matrix upper triangular in the same basis, and this also covers the empty and all-scalar branches.

Depends on

Used by

FALSE: Every endomorphism has a commuting diagonal-plus-nilpotent decomposition over its base field False statement

Dependency tree · two levels

16 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

S. Axler, Linear Algebra Done Right, 4th ed., Section 5C (standard reference, not scraped)
K. Hoffman and R. Kunze, Linear Algebra, 2nd ed., Section 6.4 (standard reference, not scraped)