False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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FALSE: If <Tv,v> is nonnegative for every v, then T is automatically self-adjoint

Statement

If $⟨ T v, v ⟩$ is nonnegative for every vector $v$ , then $T$ is automatically self-adjoint.

Facts & Assumptions

Given: The quarter-turn matrix $A = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ on $R^{2}$ .

[L1]

The quarter-turn is not self-adjoint, because it is normal of the $2 \times 2$ rotation type rather than diagonal over $R$ (The real quarter-turn is normal and appears as a single 2x2 block in the real normal classification).

Refutation

technique · direct

1.1

For every $v = (x, y)^{T}$ , one has $A v = (- y, x)^{T}$ and therefore $⟨ A v, v ⟩ = - x y + x y = 0 \geq 0$ . Thus the hypothesis of the claim holds for $A$ .

algebra

2.1

By [L1], the same operator $A$ is not self-adjoint. Hence the displayed inequality alone does not force self-adjointness, and the claim is false.

L1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)