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

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If $k > \dim V$ , then $Λ^{k} V = 0$

Statement

If $V$ is finite-dimensional with $\dim V = n$ and $k > n$ , then $Λ^{k} V = 0$ .

Facts & Assumptions

Given: A finite-dimensional vector space $V$ with $\dim V = n$ and a degree $k > n$ .

[L1]

The equality $\dim V = n$ means that $V$ has a basis with $n$ elements (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

[L2]

In a space with a spanning set of $n$ elements, every linearly independent subset has at most $n$ elements (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ ).

[L3]

In a finite-dimensional vector space, a decomposable wedge is nonzero exactly when its vectors are linearly independent (In a finite-dimensional vector space, a decomposable wedge is nonzero exactly when its vectors are linearly independent).

[L4]

The exterior power $Λ^{k} V$ is the quotient of $V^{\otimes k}$ generated by decomposable wedges $v_{1} \land \dots \land v_{k}$ (The $k$ th exterior power as the tensor-power quotient by repeated-vector relations).

Proof

technique · direct

1.1

If some $k$ -tuple $(v_{1}, \dots, v_{k})$ in $V$ were linearly independent, then its image set would be a linearly independent subset of $V$ with $k$ elements. That contradicts [L1] and [L2] because $k > n$ . So every $k$ -tuple in $V$ is linearly dependent.

L1L2algebra

2.1

By [L3], step 1.1 forces every decomposable wedge $v_{1} \land \dots \land v_{k}$ to vanish.

L3step 1.1

3.1

By [L4], every element of $Λ^{k} V$ is a linear combination of decomposable wedges, so step 2.1 forces every element of $Λ^{k} V$ to be zero.

L4step 2.1

4.1

Hence $Λ^{k} V = 0$ whenever $k > n$ .

step 3.1∎

Depends on

Used by

Bases and dimensions of exterior powers of ℝ², ℝ³, and ℝ⁴ Example

Dependency tree · two levels

23 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 k>dim⁡Vk>dimV, then ΛkV=0ΛkV=0