CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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 $F_{2}$ , let $B$ on $F_{2}^{2}$ have matrix $A = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ .

[L1]

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

[L2]

In characteristic $2$ , alternating forms are symmetric (Alternating forms are skew-symmetric; the converse holds when $char F \neq 2$ , while in characteristic $2$ alternating forms are symmetric).

[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 $v = (x, y)$ , $B (v, v) = x y + y x = 2 x y = 0$ , so $B$ is alternating and hence symmetric by [L1] and [L2]. Its matrix has determinant $- 1 = 1$ in $F_{2}$ , 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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Sources

K. Conrad, Bilinear Forms, §§4–5 (standard reference, not scraped)

In characteristic 22, a symmetric bilinear form need not have an orthogonal basis