ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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.

Lagrange polynomials give the three eigenspace projections of a diagonalisable endomorphism

Example

Let $T$ be diagonalisable with distinct eigenvalues $λ_{0}, λ_{1}, λ_{2}$ . Put

$e_{i} (x) = \prod_{j \neq i} \frac{x - λ_{j}}{λ_{i} - λ_{j}} .$

Then $E_{i} = e_{i} (T)$ is projection onto $E_{λ_{i}} (T)$ along the other eigenspaces; the $E_{i}$ are pairwise orthogonal idempotents and $E_{0} + E_{1} + E_{2} = I$ .

Facts & Assumptions

Given: A diagonalisable $T$ with the three displayed distinct eigenvalues.

[L1]

Distinct eigenspaces give a direct sum of the whole space (An endomorphism is diagonalisable exactly when $V = ⨁_{i < r} E_{λ_{i}} (T)$ for some finite list of distinct scalars $λ_{i}$ ).

[L2]

Primary projections are polynomial expressions in $T$ obtained from the relevant congruences (Each projection in the primary decomposition is a polynomial in the endomorphism).

Verification

technique · direct

1.1

The denominators are nonzero, and substitution gives $e_{i} (λ_{j}) = 1$ for $i = j$ and $0$ otherwise. Hence $e_{i} (T)$ is identity on $E_{λ_{i}} (T)$ and zero on the other eigenspaces.

L1L2algebra

2.1

The action on every summand in [L1] now gives $E_{i}^{2} = E_{i}$ , $E_{i} E_{j} = 0$ for $i \neq j$ , and $\sum_{i} E_{i} = I$ .

step 1.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 43 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Keith Conrad, The Minimal Polynomial and Some Applications, Remark 4.12 (standard reference, not scraped)