CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Simultaneous triangularization of solvable representations

Statement

Let $g$ be finite-dimensional solvable over an algebraically closed field of characteristic zero, and let $V$ be a finite-dimensional $g$ -module. Then $V$ has a complete invariant flag. Equivalently, there is a basis in which every representing matrix is upper triangular.

Facts & Assumptions

Given: A representation of $g$ on $V$ under Lie's theorem hypotheses.

[L1]

Every nonzero finite-dimensional module under these hypotheses has a common eigenvector (Lie's theorem).

[L2]

Invariant subspaces and their quotients carry the restricted and induced representations (Subrepresentations, quotient representations, and intertwiners).

[L3]

The canonical vector-space quotient projection is linear and surjective (The quotient vector space $V / W$ and its canonical projection).

Proof

technique · induction on $\dim V$

1.1

If $V = 0$ , the empty flag and empty basis have the required properties.

basegiven

1.2

Assume $V \neq 0$ and the corollary for smaller-dimensional modules.

ihgiven

1.3

By [L1], choose a common eigenvector $0 \neq v \in V$ . Its line $V_{1} = k v$ is $g$ -invariant.

L1L2

2.1

The quotient $V / V_{1}$ has the induced representation by [L2] and dimension one less. Step 1.2 supplies its complete invariant flag. Taking inverse images under the projection in [L3] and adjoining $0 \subset V_{1}$ gives a complete invariant flag in $V$ .

L2L3step 1.2step 1.3

3.1

A basis adapted to this finite flag makes every representing matrix upper triangular. Conversely, the spans of the first $i$ vectors in a common upper-triangular basis form the complete invariant flag. These are finite sequential basis choices and require no AC.

step 1.1step 2.1discharge-induction∎

Depends on

Used by

Dependency tree · two levels

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Sources

Milne, Lie Algebras, Theorem 3.7 (standard reference, not scraped)