TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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$T$ is triangularisable iff its minimal polynomial splits iff its characteristic polynomial splits

Statement

Let $T$ be an endomorphism of a finite-dimensional vector space $V$ over $F$ . The following are equivalent:

$T$ is triangularisable over $F$ ;
the minimal polynomial $μ_{T}$ is a product of linear factors in $F [x]$ ;
the characteristic polynomial $χ_{T}$ is a product of linear factors in $F [x]$ .

For $V = 0$ , both polynomials are $1$ , the empty product, and the empty basis triangularises $T$ .

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ and an endomorphism $T : V \to V$ .

[L1]

An ordered basis gives an upper-triangular matrix exactly when its initial spans form a complete invariant flag (Complete invariant flags are equivalent to upper-triangular matrices).

[L2]

For a $T$ -invariant subspace $W$ , $χ_{T} = χ_{T ∣_{W}} χ_{\overset{ˉ}{T}}$ (For invariant $W$ , $χ_{T} = χ_{T ∣_{W}} χ_{\overset{ˉ}{T}}$ ).

[L3]

A monic irreducible polynomial divides $μ_{T}$ if and only if it divides $χ_{T}$ (The minimal and characteristic polynomials have exactly the same monic irreducible factors).

[L4]

The eigenvalues of $T$ over $F$ are exactly the roots in $F$ of $χ_{T}$ (For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$ ).

[L5]

A basis of a subspace followed by representatives of a quotient basis is a basis of the whole space (A quotient basis lifts to a basis adapted to $W$ ).

Proof

technique · induction

1.1L1L3

If $T$ is triangularisable, then $x I - [T]$ is upper triangular and its determinant is the product of its diagonal entries $x - a_{j j}$ , so $χ_{T}$ splits; by [L3], $χ_{T}$ splits exactly when $μ_{T}$ splits.

1.2base

If $\dim V = 0$ , the empty basis and the polynomial $1$ give all three conditions.

1.3L4chooseih

Assume $\dim V > 0$ , that $χ_{T}$ splits, and that the reverse implication holds in smaller dimensions; choose a root $λ \in F$ of $χ_{T}$ , then [L4] supplies a nonzero eigenvector $v$ , and $W = F v$ is a one-dimensional invariant subspace.

2.1step 1.3L1L2L5ih

By [L2], $χ_{T} = (x - λ) χ_{\overset{ˉ}{T}}$ , so $χ_{\overset{ˉ}{T}}$ splits; the induction hypothesis triangularises $\overset{ˉ}{T}$ on $V / W$ , and [L5] lifts its triangular basis after $v$ to a basis whose initial spans are $T$ -invariant, so [L1] triangularises $T$ .

3.1step 1.1step 1.2step 2.1discharge-induction∎

Step 1.1 gives $1 \Rightarrow 3 \Leftrightarrow 2$ , while steps 1.2-2.1 give $3 \Rightarrow 1$ in every finite dimension, completing all three equivalences.

Depends on

Used by

Dependency tree · two levels

23 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., Result 5.44 (standard reference, not scraped)
K. Hoffman and R. Kunze, Linear Algebra, 2nd ed., Theorem 5 in Section 6.4 (standard reference, not scraped)

TT is triangularisable iff its minimal polynomial splits iff its characteristic polynomial splits