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

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On $Λ^{n} V$ , the induced map $Λ^{n} T$ is multiplication by $\det T$

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional vector space of dimension $n \geq 1$ . Then $Λ^{n} V$ is one-dimensional with basis $e_{1} \land \dots \land e_{n}$ for any ordered basis $(e_{1}, \dots, e_{n})$ , and

$Λ^{n} T = \det (T) \cdot {id}_{Λ^{n} V} .$

For $n = 0$ , $Λ^{0} T = {id}_{F} = \det (T) \cdot {id}_{F}$ by convention.

Facts & Assumptions

Given: An endomorphism $T$ of an $n$ -dimensional vector space $V$ , $n \geq 1$ , and an ordered basis $B = (e_{1}, \dots, e_{n})$ .

[L1]

The matrix of $Λ^{n} T$ in the top wedge basis has the single entry $\det A_{[n], [n]} = \det A$ (In basis-wedge coordinates, the matrix of $Λ^{k} T$ is the signed matrix of $k$ -minors).

[L2]

The determinant of an operator is $\det (T) = \det ([T]_{B})$ (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).

Proof

technique · direct

1.1

By [L1] with $k = n$ , the wedge basis of $Λ^{n} V$ is the single vector $e_{1} \land \dots \land e_{n}$ , so $Λ^{n} V$ is one-dimensional and the matrix of $Λ^{n} T$ is the $1 \times 1$ matrix with entry $\det A$ .

1.2

By [L2], that entry is $\det (T)$ .

2.1

Hence $Λ^{n} T (e_{1} \land \dots \land e_{n}) = \det (T) (e_{1} \land \dots \land e_{n})$ , and one-dimensionality makes $Λ^{n} T = \det (T) \cdot id$ .

step 1.1step 1.2

3.1

For every decomposable top wedge $v_{1} \land \dots \land v_{n}$ , one has $v_{1} \land \dots \land v_{n} = c (e_{1} \land \dots \land e_{n})$ for a unique scalar $c$ by step 1.1, so step 2.1 gives

$Λ^{n} T (v_{1} \land \dots \land v_{n}) = c Λ^{n} T (e_{1} \land \dots \land e_{n}) = \det (T) (v_{1} \land \dots \land v_{n}) .$

[step 1.1, step 2.1, algebra]

4.1

Steps 2.1 and 3.1 establish the claimed scalar action in positive dimension, and the conventions cover $n = 0$ .

step 2.1step 3.1∎

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)

On ΛnVΛnV, the induced map ΛnTΛnT is multiplication by det⁡TdetT