Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Universal property of the finite-dimensional exterior power

Statement

Let A:Vk→W be an alternating k-linear map into a real vector space W. Then there is a unique linear map

A~:⋀kV→W

such that

A~(v1∧⋯∧vk)=A(v1,…,vk)

for all v1,…,vk∈V.

Facts & Assumptions

Given: An alternating k-linear map A:Vk→W.

[F1]

The kth exterior power is the dual space ⋀kV=Alt⁡k(V)∗, and v1∧⋯∧vk is the evaluation functional ω↦ω(v1,…,vk) (The finite-dimensional exterior power of vectors).

[L1]

The wedges of a basis form a basis of Alt⁡k(V), so the corresponding decomposable k-vectors span ⋀kV by duality (Wedge monomials in a dual basis form a basis).

Proof

technique · direct
1.1L1givenchooseconstruct

Choose a basis e1,…,en of V. Define A~ on the spanning set of decomposable wedges by A~(ei1∧⋯∧eik):=A(ei1,…,eik) for i1<⋯<ik, and extend linearly. This is possible because [L1] gives a basis indexed by those increasing tuples.

2.1F1step 1.1algebra

For arbitrary v1,…,vk, expand each vj in the chosen basis. Multilinearity of A and of the wedge, together with alternation on both sides, reduce both expressions to the same signed sum over increasing k-tuples. Hence A~(v1∧⋯∧vk)=A(v1,…,vk).

3.1L1step 2.1

If L:⋀kV→W is another linear map with the same property, then L and A~ agree on every decomposable basis wedge, hence on all of ⋀kV by linearity and [L1]. So A~ is unique.

4.1step 1.1step 2.1step 3.1∎

Therefore every alternating k-linear map factors uniquely through ⋀kV.

Depends on

Used by

Dependency tree · two levels

6 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