TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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

Dependency tree · two levels

8 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

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)