PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Complete invariant flags are equivalent to upper-triangular matrices

Statement

Let $B = (v_{1}, \dots, v_{n})$ be an ordered basis of $V$ , and put $V_{j} = span (v_{1}, \dots, v_{j})$ , with $V_{0} = 0$ . Then $[T]_{B}^{B}$ is upper triangular if and only if every $V_{j}$ is $T$ -invariant. Equivalently, upper-triangular bases are exactly the bases adapted to complete invariant flags $0 = V_{0} \subset V_{1} \subset \dots \subset V_{n} = V, \dim V_{j} = j .$

Facts & Assumptions

Given: An endomorphism $T : V \to V$ and an ordered basis $B = (v_{1}, \dots, v_{n})$ .

[L1]

Triangularisability means that the matrix of $T$ in some ordered basis is upper triangular (Triangularisable endomorphisms and simultaneous triangularisability).

[L2]

A subspace $W$ is $T$ -invariant when $T (W) \subseteq W$ (Invariant subspaces, restrictions, and induced quotient operators).

[L3]

The $j$ -th matrix column is the coordinate column of $T (v_{j})$ in the ordered basis (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

Proof

technique · direct

1.1L1L2L3

If the matrix is upper triangular, its $j$ -th column has no nonzero entry below row $j$ , so $T (v_{j}) \in V_{j}$ ; linearity then gives $T (V_{j}) \subseteq V_{j}$ for every $j$ .

2.1step 1.1L1L2L3∎

Conversely, if every $V_{j}$ is invariant, then $T (v_{j}) \in V_{j}$ , so the $j$ -th matrix column has zero entries below row $j$ and the matrix is upper triangular; the statements include the empty flag for $V = 0$ and the flag $0 \subset V$ in dimension one.

Depends on

Dependency tree · 2 levels

Sources

S. Axler, Linear Algebra Done Right, 4th ed., Result 5.39