ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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.

Two commuting non-scalar matrices simultaneously diagonalised in one explicit basis

Example

The matrices

$A = (\begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array}), B = (\begin{array}{cc} 3 & 1 \\ 0 & 4 \end{array})$

commute and are simultaneously diagonalised by the ordered basis $((1, 0), (1, 1))$ .

Facts & Assumptions

Given: The displayed matrices and $P = (\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array})$ .

[L1]

Pairwise commuting diagonalisable endomorphisms have a common eigenbasis (A family of diagonalisable endomorphisms of a finite-dimensional space is simultaneously diagonalisable if and only if its members commute pairwise).

Verification

technique · direct

1.1algebra

With $P^{- 1} = (\begin{array}{cc} 1 & - 1 \\ 0 & 1 \end{array})$ , direct multiplication gives $A = P diag (1, 2) P^{- 1}$ and $B = P diag (3, 4) P^{- 1}$ .

2.1step 1.1L1algebra∎

Thus the columns of $P$ , namely $(1, 0)$ and $(1, 1)$ , are a common eigenbasis. The two diagonal matrices commute, so their conjugates commute as well, agreeing with [L1].

Depends on

A family of diagonalisable endomorphisms of a finite-dimensional space is simultaneously diagonalisable if and only if its members commute pairwise

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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.