TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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For endomorphisms $S$ and $T$ of one finite-dimensional vector space, $\det (S T) = \det (S) \det (T)$

Statement

If $S, T : V \to V$ are linear operators on one finite-dimensional vector space over a field, then

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

Facts & Assumptions

Given: $S, T, V$ as in the statement.

[L1]

On a positive-dimensional space, an operator scales every alternating top-degree form by its determinant, and that scalar is unique (On a positive-dimensional space, $\det (T)$ is the unique scalar by which $T$ scales every alternating top-degree form).

[L2]

The determinant is basis independent (The determinant of a linear operator is independent of the chosen ordered basis).

[F1]

In dimension zero, the operator determinant is defined to be $1$ (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space).

Proof

technique · direct

1.1

If $\dim V = 0$ , all three determinants in the formula are $1$ .

1.2

Suppose $\dim V = n \geq 1$ , and let $ω$ be any alternating $n$ -linear form. Applying [L1] to $S$ and then to $T$ gives $ω (S T v_{0}, \dots, S T v_{n - 1}) = \det (S) \det (T) ω (v_{0}, \dots, v_{n - 1}) .$

2.1

The uniqueness clause of [L1], applied to $S \circ T$ , identifies the scaling scalar in step 1.2 as $\det (S \circ T)$ . Together with step 1.1, this proves the formula in every finite dimension.

step 1.1step 1.2L1L2∎

Depends on

Used by

Determinant is a group homomorphism GL(V)→ F^×, and det(T⁻¹)=det(T)⁻¹ Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed. (standard reference, not scraped)

For endomorphisms SS and TT of one finite-dimensional vector space, det⁡(ST)=det⁡(S)det⁡(T)det(ST)=det(S)det(T)