ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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.
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A complex symmetric matrix can be nonzero, square to zero, and fail to be normal

Example

The complex matrix

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

is symmetric, nonzero, nilpotent of index $2$ , and not normal.

Facts & Assumptions

Given: The complex matrix $A$ above.

[L1]

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

Verification

technique · direct

1.1

The matrix is symmetric because $A^{T} = A$ , and direct multiplication gives $A^{2} = 0$ while $A \neq 0$ , so $A$ is a nonzero nilpotent matrix of index $2$ .

algebra

2.1

Its conjugate transpose is $A^{*} = (\begin{array}{cc} 1 & - i \\ - i & - 1 \end{array})$ , and direct multiplication gives $A^{*} A = (\begin{array}{cc} 2 & 2 i \\ - 2 i & 2 \end{array})$ but $A A^{*} = (\begin{array}{cc} 2 & - 2 i \\ 2 i & 2 \end{array})$ . Hence $A^{*} A \neq A A^{*}$ , so $A$ is not normal by [L1].

L1algebra∎

Depends on

In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose

Used by

FALSE: Every complex symmetric matrix is unitarily diagonalisable False statement

Dependency tree · two levels

4 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)