TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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If the characteristic polynomial of an endomorphism splits, some orthonormal basis makes its matrix upper triangular

Statement

Let $V$ be a finite-dimensional real or complex inner product space over $F$ , and let $T : V \to V$ be linear. If the characteristic polynomial of $T$ splits over $F$ , then $V$ has an orthonormal basis in which the matrix of $T$ is upper triangular.

Facts & Assumptions

Given: A finite-dimensional real or complex inner product space $V$ over $F$ and a linear endomorphism $T : V \to V$ whose characteristic polynomial splits over $F$ .

[L1]

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

[L2]

A basis gives an upper-triangular matrix exactly when its successive spans form a complete $T$ -invariant flag (Complete invariant flags are equivalent to upper-triangular matrices).

[L3]

Gram-Schmidt turns a linearly independent list into an orthonormal list with the same successive spans (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans).

Proof

technique · direct

1.1

By [L1], there is a basis $(v_{1}, \dots, v_{n})$ of $V$ in which $[T]$ is upper triangular; equivalently, the flag $0 \subset span (v_{1}) \subset span (v_{1}, v_{2}) \subset \dots \subset span (v_{1}, \dots, v_{n}) = V$ is $T$ -invariant by [L2].

L1L2

2.1

Apply [L3] to $(v_{1}, \dots, v_{n})$ and obtain an orthonormal basis $(e_{1}, \dots, e_{n})$ such that $span (e_{1}, \dots, e_{k}) = span (v_{1}, \dots, v_{k})$ for every $k$ , so the same successive spans are still $T$ -invariant.

L3step 1.1

3.1

Because the orthonormal basis $(e_{1}, \dots, e_{n})$ is adapted to a complete $T$ -invariant flag, [L2] shows that the matrix of $T$ in that basis is upper triangular.

L2step 2.1∎

Depends on

Used by

Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely Theorem

Dependency tree · two levels

14 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

Nicholas Hu, The Schur decomposition (standard reference, not scraped)