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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 has Jordan form

Statement

Every endomorphism of a finite-dimensional vector space over an algebraically closed field has a Jordan canonical form over that field, including the endomorphism of the zero space.

Facts & Assumptions

Given: A finite-dimensional vector space over an algebraically closed field F and an endomorphism T.

[L1]

Every nonconstant polynomial over an algebraically closed field has a root in that field (An algebraically closed field: every nonconstant polynomial has a root in the field).

[L2]

If a is a root of p∈F[x], then p=(x−a)q for some q∈F[x] (Factor theorem over a commutative ring).

[L3]

An endomorphism has Jordan form over F exactly when its characteristic polynomial splits over F (Jordan form over the base field exists exactly when the characteristic polynomial splits).

Proof

technique · induction on $\deg\chi_T$
1.1

If deg⁡χT=0, then χT=1 and [L3] gives the empty Jordan form.

baseL3
1.2

The induction is on the degree of an arbitrary nonzero polynomial over F, not only of a characteristic polynomial, since the factor produced below need not itself be one. If the degree is positive, [L1] supplies a root a∈F and [L2] writes χT=(x−a)q with deg⁡q smaller; the induction hypothesis applies to q in that strengthened form and factors it into linear factors, so χT splits.

L1L2ih
2.1

Fact [L3] now gives a Jordan canonical form for T, completing the induction.

step 1.1step 1.2L3discharge-induction∎

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