TheoremStatement: AI-adaptedProof: 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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Jordan form over the base field exists exactly when the characteristic polynomial splits

Statement

An endomorphism of a finite-dimensional $F$ -vector space has a Jordan canonical form over $F$ if and only if its characteristic polynomial splits into linear factors over $F$ . This includes the zero space, where the characteristic polynomial is $1$ and the Jordan form is empty.

Facts & Assumptions

Given: An endomorphism $T$ of a finite-dimensional $F$ -vector space $V$ .

[L1]

If $χ_{T}$ splits, $V$ is the direct sum of invariant generalised eigenspaces $G_{λ}$ , and $(T - λ I) ∣_{G_{λ}}$ is nilpotent (Split characteristic polynomials decompose into generalised eigenspaces of the algebraic multiplicities).

[L2]

Every finite-dimensional nilpotent endomorphism has a basis of Jordan strings (Every finite-dimensional nilpotent endomorphism has a basis of Jordan strings).

[L3]

The characteristic polynomial of a block triangular matrix is the product of the characteristic polynomials of its diagonal blocks, including empty blocks (The characteristic polynomial of a block upper- or lower-triangular matrix is the product of the characteristic polynomials of its diagonal blocks).

Proof

technique · direct

1.1L1L2construct

Suppose $χ_{T}$ splits. For every $G_{λ}$ from [L1], apply [L2] to $N_{λ} = (T - λ I) ∣_{G_{λ}}$ ; its nilpotent strings are Jordan strings for $T$ at $λ$ , and concatenating their bases across the direct sum gives a Jordan basis of $V$ .

1.2L3algebra

Conversely, if $T$ has a Jordan basis, its matrix is block diagonal with blocks $J_{m} (λ)$ for $λ \in F$ ; [L3] gives $χ_{T} = \prod (x - λ)^{m}$ , which splits over $F$ .

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove both directions; the empty direct sum and empty product prove the zero-space case.

Jordan form over the base field exists exactly when the characteristic polynomial splits

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources