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

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Engel's theorem for strictly upper triangular matrices

Example

Let $n_{n} (k)$ act on $k^{n}$ by the standard action of strictly upper triangular matrices. Every element acts nilpotently, $e_{1}$ is a common zero vector, and the standard flag realizes the conclusion of Engel triangularization.

Facts & Assumptions

Given: A field $k$ , an integer $n \geq 1$ , the standard basis $e_{1}, \dots, e_{n}$ of $V = k^{n}$ , and the inclusion action of $n_{n} (k) \subseteq g l (V)$ .

[L1]

Engel triangularization produces a flag $0 = V_{0} \subset V_{1} \subset \dots \subset V_{n} = V$ with $g V_{i} \subseteq V_{i - 1}$ for a nil representation (Engel triangularization theorem).

[L2]

For $n \geq 2$ , $n_{n} (k)$ is the strictly upper triangular matrix Lie algebra and is nilpotent of class $n - 1$ (Strictly upper triangular matrices form a nilpotent Lie algebra).

Verification

technique · direct

1.1

Set $V_{i} = span (e_{1}, \dots, e_{i})$ for $0 \leq i \leq n$ . If $A$ is strictly upper triangular, its $j$ th column has nonzero entries only in rows smaller than $j$ , so $A e_{j} \in V_{j - 1}$ . Therefore $A (V_{i}) \subseteq V_{i - 1}$ for every $i$ . Iterating gives $A^{n} (V) = A^{n} (V_{n}) \subseteq V_{0} = 0$ , so every $A \in n_{n} (k)$ is a nilpotent operator.

givenalgebra

2.1

Taking $i = 1$ in step 1.1 gives $A e_{1} = 0$ for every $A \in n_{n} (k)$ , so $e_{1} \neq 0$ is a common zero vector. The full chain $0 = V_{0} \subset V_{1} \subset \dots \subset V_{n} = V$ has $\dim V_{i} = i$ and the containment required by [L1], so this explicit standard flag realizes Engel triangularization.

L1step 1.1

3.1

If $n = 1$ , then $n_{1} (k) = 0$ , the sole operator is zero, and the flag $0 \subset k e_{1}$ satisfies the same calculation. For $n \geq 2$ , [L2] additionally identifies the acting Lie algebra as nilpotent of class $n - 1$ , although elementwise nilpotence of this particular representation was proved directly in step 1.1. All basis and flag data are explicit and finite, so no choice principle is used.

L2step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Milne, Lie Algebras, Engel's theorem and the flag algebra n(F) (standard reference, not scraped)