Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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In characteristic 2, a symmetric bilinear form need not have an orthogonal basis

Statement refuted

Refuted claim: Every symmetric bilinear form over every field has an orthogonal basis.

Facts & Assumptions

Given: Over F2, let B on F22 have matrix A=(0110).

[L1]

The ring F2 is a field of characteristic 2 (For every prime p, the two operations on Z/p make it a field).

[L3]

A finite-dimensional bilinear form is nondegenerate exactly when its representing matrix is invertible (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

Counterexample

technique · contradiction from the displayed form
1.1

For v=(x,y), B(v,v)=xy+yx=2xy=0, so B is alternating and hence symmetric by [L1] and [L2]. Its matrix has determinant 1=1 in F2, so it is nondegenerate by [L3].

L1L2L3algebra
2.1

If (u,v) were an orthogonal basis, alternation would give B(u,u)=B(v,v)=0 and orthogonality would give B(u,v)=B(v,u)=0. The matrix in that basis would be zero, contradicting nondegeneracy.

step 1.1L3algebra
3.1

Thus this symmetric form has no orthogonal basis, and the characteristic-not-2 hypothesis in the diagonalization theorem is essential.

step 2.1

Depends on

Used by

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Dependency tree · next 3 levels

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Sources