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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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Diagonal criterion: if vi,wi0 and vi,wj=0 for ij, then v1,,vm are linearly independent

Statement

Let F be a field, let V be an F-vector space, and let ,:V×VF be a bilinear form. Suppose vectors v1,,vm,w1,,wmV satisfy

vi,wi0andvi,wj=0  for ij.

Then v1,,vm are linearly independent.

Facts & Assumptions

Given: a field F, a vector space V over F, a bilinear form , on V, and vectors v1,,vm,w1,,wmV satisfying the displayed hypotheses.

[F1]

A bilinear form is linear in each variable separately (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms).

[F2]

A field has no zero divisors (Field).

Proof

technique · direct
1.1

Suppose i=1mcivi=0. Pairing with wj and using linearity in the first variable from [F1] gives i=1mcivi,wj=0.

F1assume-contra
2.1

Every term with ij vanishes by hypothesis, so this reduces to cjvj,wj=0. Since vj,wj0 and a field has no zero divisors by [F2], we get cj=0.

F2step 1.1
3.1

The index j was arbitrary, so every coefficient is 0. Hence v1,,vm are linearly independent.

step 2.1discharge-contradiction

Remarks

  • The argument uses only bilinearity and the diagonal pattern. No positivity and no nondegeneracy is involved, which is why the criterion works over F2 as well as over R.

Depends on

Used by

Dependency tree · two levels

17 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