TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13

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On a positive-dimensional space, $\det (T)$ is the unique scalar by which $T$ scales every alternating top-degree form

Statement

Let $V$ be $n$ -dimensional over a field $F$ , with $n \geq 1$ , and let $T : V \to V$ be linear. For every alternating $n$ -linear form $ω : V^{n} \to F$ ,

$ω (T v_{0}, \dots, T v_{n - 1}) = \det (T) ω (v_{0}, \dots, v_{n - 1}) .$

Moreover, $\det (T)$ is the unique scalar having this property for every such $ω$ and every $n$ -tuple.

Facts & Assumptions

Given: $V, F, n, T$ and $ω$ as in the statement.

[L1]

For any ordered basis $B$ , every alternating $n$ -linear form satisfies $η (v_{0}, \dots, v_{n - 1}) = η (B) \det M_{B} (v_{0}, \dots, v_{n - 1})$ (An alternating top-degree form is determined by its value on one ordered basis).

[F1]

In an ordered basis $B$ , $\det (T) = \det ([T]_{B})$ (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).

[L2]

The matrix determinant is alternating, multilinear in the columns, and $\det (I_{n}) = 1$ (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring).

Proof

technique · direct

1.1

Fix an ordered basis $B$ . Define $ω_{T} (v_{0}, \dots, v_{n - 1}) : = ω (T v_{0}, \dots, T v_{n - 1})$ ; this is an alternating $n$ -linear form.

given

1.2

Define $δ_{B} (v_{0}, \dots, v_{n - 1}) : = \det M_{B} (v_{0}, \dots, v_{n - 1})$ . It is alternating and $n$ -linear, and $δ_{B} (B) = 1$ .

2.1

Applying [L1] to $ω_{T}$ and then to $ω$ gives $ω_{T} (v_{0}, \dots, v_{n - 1}) = ω_{T} (B) \det M_{B} (v_{0}, \dots, v_{n - 1})$ and $ω_{T} (B) = ω (B) \det ([T]_{B})$ .

step 1.1L1

3.1

Combining the two formulas in step 2.1 and using [F1] and [L1] once more gives $ω_{T} = \det (T) ω$ .

step 2.1F1L1algebra

4.1

If a scalar $c$ has the stated scaling property, evaluate it for $δ_{B}$ at $B$ . The left side is $\det ([T]_{B})$ , so $c = \det ([T]_{B}) = \det (T)$ .

step 1.2F1∎

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Sources

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

On a positive-dimensional space, det⁡(T)det(T) is the unique scalar by which TT scales every alternating top-degree form