CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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If $\dim V = n$ , then $\dim Λ^{k} V = (\binom{n}{k})$

Statement

If $V$ is finite-dimensional over $F$ with $\dim V = n$ , then for every $0 \leq k \leq n$ ,

$\dim Λ^{k} V = (\binom{n}{k}) .$

Facts & Assumptions

Given: A finite-dimensional vector space $V$ with $\dim V = n$ .

[L1]

The increasing-index wedges of an ordered basis form a basis of $Λ^{k} V$ , indexed by the $k$ -element subsets (Increasing-index wedges of a basis form a basis of $Λ^{k} V$ ).

[L2]

The dimension of a finite-dimensional vector space is the size of any of its bases (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

[L3]

The binomial coefficient is $(\binom{n}{k}) = ∣ [n]^{k} ∣$ , the number of $k$ -element subsets of an $n$ -element set (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

Proof

technique · direct

1.1

Since $\dim V = n$ , there is an ordered basis $(e_{1}, \dots, e_{n})$ of $V$ by [L2].

2.1

By [L1], the set of wedges $e_{I}$ over the $k$ -element subsets $I$ of ${1, \dots, n}$ is a basis of $Λ^{k} V$ .

L1step 1.1

3.1

By [L3] there are exactly $(\binom{n}{k})$ such subsets, and by [L2] the dimension equals that count.

L2L3step 2.1∎

Depends on

Used by

The oriented unit volume form Definition
Bases and dimensions of exterior powers of ℝ², ℝ³, and ℝ⁴ Example

Dependency tree · two levels

28 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)

If dim⁡V=ndimV=n, then dim⁡ΛkV=(nk)dimΛkV=(kn​)