ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The real quarter-turn is normal and appears as a single 2x2 block in the real normal classification

Example

The quarter-turn

$A = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$

is normal on $R^{2}$ , has complex eigenvalues $\pm i$ , and in the standard orthonormal basis already appears as the $2 \times 2$ block from the real normal classification.

Facts & Assumptions

Given: The quarter-turn matrix $A$ acting on $R^{2}$ with the standard inner product.

[L1]

In an orthonormal basis, normality is equivalent to commuting with the transpose (In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose).

[L2]

Real normal operators are orthogonally block-diagonalisable with $1 \times 1$ and $2 \times 2$ rotation-scaling blocks (A real normal endomorphism is orthogonally block-diagonalisable with 1x1 real blocks and 2x2 rotation-scaling blocks).

Verification

technique · direct

1.1

Direct multiplication gives $A^{T} A = A A^{T} = I_{2}$ , so $A$ is normal by [L1]. Its characteristic polynomial is $x^{2} + 1$ , so over $C$ its eigenvalues are $i$ and $- i$ .

L1algebra

2.1

The matrix $A$ itself has the form $(\begin{array}{cc} a & b \\ - b & a \end{array})$ with $a = 0$ and $b = - 1 \neq 0$ , so it is exactly one of the $2 \times 2$ blocks allowed by [L2].

L2step 1.1∎

Depends on

Used by

FALSE: Every normal operator is diagonalisable over its base field False statement
FALSE: If <Tv,v> is nonnegative for every v, then T is automatically self-adjoint False statement

Dependency tree · two levels

11 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

Nicholas Hu, The Schur decomposition (standard reference, not scraped)