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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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The dual family of a finite basis is a basis of the dual space, with the same dimension

Statement

If B=(b1,,bn) is a basis of a finite-dimensional F-vector space V, then its dual family B=(b1,,bn) is a basis of V. Consequently dimV=dimV=n.

Facts & Assumptions

Given: A finite basis B=(b1,,bn) of V and its dual family.

[L3]

The dimension of a finite-dimensional space is the cardinality of any finite basis (Finite-dimensional vector space, and its dimension dimFV; infinite-dimensional means having no finite basis).

Proof

technique · direct
1.1

If i=1ncibi=0, evaluation at bj and [L1] give cj=0 for every j; hence the dual family is linearly independent.

L1algebra
1.2

For fV, set g=i=1nf(bi)bi. If v=jajbj as in [L2], then g(v)=jajf(bj)=f(v), so f=g and the dual family spans V.

L1L2algebra
2.1

Steps 1.1 and 1.2 make B a basis, and [L3] gives dimV=n=dimV. For n=0, both sums are empty and the unique functional on the zero space is zero, so the same proof applies.

step 1.1step 1.2L3

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 69 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources