CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Every finite-dimensional endomorphism over an algebraically closed field is triangularisable

Statement

Let $F$ be algebraically closed. Every endomorphism of a finite-dimensional $F$ -vector space is triangularisable over $F$ .

Facts & Assumptions

Given: An algebraically closed field $F$ , a finite-dimensional $F$ -vector space $V$ , and an endomorphism $T : V \to V$ .

[L1]

An endomorphism is triangularisable exactly when its characteristic polynomial splits into linear factors over its base field ( $T$ is triangularisable iff its minimal polynomial splits iff its characteristic polynomial splits).

[L2]

In an algebraically closed field, every nonconstant polynomial has a root in the field (An algebraically closed field: every nonconstant polynomial has a root in the field).

[L3]

If $f (a) = 0$ over a commutative ring, then $x - a$ divides $f$ (Factor theorem over a commutative ring).

Proof

technique · induction

1.1

A monic polynomial of degree $0$ is $1$ , hence is the empty product of linear factors.

base

1.2

Let $f \in F [x]$ be monic of positive degree and assume every monic polynomial of smaller degree splits; [L2] gives a root $a$ , and [L3] writes $f = (x - a) q$ with $q$ monic of degree one less, so the induction hypothesis makes $q$ , and therefore $f$ , split.

L2L3ih

2.1

Apply steps 1.1-1.2 to the monic polynomial $χ_{T}$ ; it splits over $F$ , so [L1] triangularises $T$ , with $V = 0$ covered by the degree-zero base case.

step 1.1step 1.2L1discharge-induction∎

Depends on

Used by

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Sources

S. Axler, Linear Algebra Done Right, 4th ed., Result 5.47 (standard reference, not scraped)