CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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A nondegenerate indefinite symmetric form can have $W \cap W^{⊥} \neq 0$

Statement refuted

Every nondegenerate symmetric bilinear form $B$ on a finite-dimensional real vector space satisfies $V = W \oplus W^{⊥_{B}}$ for every subspace $W$ .

Facts & Assumptions

Given: On $R^{2}$ , the symmetric form $B (x, y) = x_{0} y_{0} - x_{1} y_{1}$ , the vector $v = (1, 1)$ , and $W = R v$ .

[L1]

A bilinear form is nondegenerate when its radical is zero; for a matrix form this is equivalent to its representing matrix having full rank (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[L2]

The orthogonal-decomposition theorem requires a positive-definite inner product and then gives $V = W \oplus W^{⊥}$ (For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ ).

Counterexample

technique · counterexample

1.1

The matrix of $B$ is $diag (1, - 1)$ , whose determinant is $- 1$ , so [L1] makes $B$ nondegenerate. It is indefinite because $B ((1, 0), (1, 0)) = 1$ and $B ((0, 1), (0, 1)) = - 1$ .

L1algebra

1.2

Yet $B (v, v) = 1 - 1 = 0$ . Therefore $B (v, a v) = 0$ for every $a v \in W$ , so the nonzero vector $v$ lies in both $W$ and $W^{⊥_{B}}$ . Thus their sum is not direct.

algebra

2.1

This does not contradict [L2]: positivity, not merely nondegeneracy, is the hypothesis that forces the orthogonal direct sum.

step 1.1step 1.2L2∎

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A nondegenerate indefinite symmetric form can have W∩W⊥≠0W∩W⊥=0