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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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Every symmetric bilinear form on a finite-dimensional space over a field of characteristic not 2 has an orthogonal basis

Statement

Let V be finite-dimensional over a field of characteristic not 2. Every symmetric bilinear form B on V admits a basis whose distinct vectors are pairwise orthogonal for B.

Facts & Assumptions

Given: A finite-dimensional F-vector space V, char⁡F≠2, and a symmetric bilinear form B.

[L1]

In characteristic not 2, a symmetric bilinear form is recovered from qB(v)=B(v,v) by B(u,v)=12bqB(u,v) (If char⁡F≠2, quadratic forms and symmetric bilinear forms correspond by q(v)=B(v,v) and B(u,v)=12bq(u,v)).

[L2]

A subspace of a finite-dimensional space is finite-dimensional, and an independent subset extends without Choice to a basis (If dim⁡FV=n and U is a linear subspace of V, then U is finite-dimensional, dim⁡FU≤n, and dim⁡FU=n if and only if U=V).

[L3]

Proof

technique · induction on $n=\dim V$
1.1

If n=0, the empty basis is orthogonal. If B=0, any basis is orthogonal.

baseL3given
1.2

Assume n>0, B≠0, and the theorem below dimension n. By [L1], choose v∈V with B(v,v)≠0. Put v⊥={w:B(w,v)=0}.

ihL1choose
2.1

Every w∈V has the decomposition w=av+z, where a=B(w,v)B(v,v)−1 and z=w−av∈v⊥. If av∈v⊥, then aB(v,v)=0, so a=0. Hence V=Fv⊕v⊥.

step 1.2algebra
3.1

Since v∉v⊥, this subspace is proper and [L2] gives dim⁡v⊥<n. The restricted form is symmetric, so the induction hypothesis gives it an orthogonal basis; adjoining v gives an orthogonal basis of V.

step 1.2step 2.1ihL2L3
4.1

The base cases and induction step prove the theorem, including degenerate forms and the zero space.

step 1.1step 3.1discharge-induction∎

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