CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29

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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.
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Determinant multiplicativity follows from the top exterior power

Statement

For endomorphisms $S, T$ of an $n$ -dimensional vector space $V$ with $n \geq 1$ ,

$\det (S \circ T) = \det (S) \det (T) .$

Facts & Assumptions

Given: Endomorphisms $S, T$ of an $n$ -dimensional vector space $V$ , $n \geq 1$ .

[L1]

The top exterior power acts by the determinant: $Λ^{n} T = \det (T) \cdot id$ (On $Λ^{n} V$ , the induced map $Λ^{n} T$ is multiplication by $\det T$ ).

[L2]

Exterior powers preserve composition: $Λ^{n} (S \circ T) = Λ^{n} S \circ Λ^{n} T$ (Exterior powers are functorial).

Proof

technique · direct

1.1L2

By [L2], $Λ^{n} (S \circ T) = Λ^{n} S \circ Λ^{n} T$ .

2.1L1step 1.1

Substituting the scalar actions of [L1] gives $\det (S \circ T) \cdot id = \det (S) \det (T) \cdot id$ on $Λ^{n} V$ .

3.1step 2.1∎

Since $Λ^{n} V$ is one-dimensional and nonzero, comparing the two scalars gives $\det (S \circ T) = \det (S) \det (T)$ .

Depends on

Used by

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

Keith Conrad, Exterior Powers (standard reference, not scraped)