ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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.

A full PLU factorisation can be computed explicitly by hand

Example

For $A = (\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{array}),$

partial pivoting gives

$P = (\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}), L = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{1}{2} & \frac{1}{2} & 1 \end{array}), U = (\begin{array}{ccc} 2 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}),$

and these satisfy $P A = L U$ .

Facts & Assumptions

Given: The displayed matrix $A$ and the candidate factors $P, L, U$ .

[L1]

Every square matrix admits a PLU factorisation, and partial pivoting records the row swaps in a permutation matrix (Every real or complex square matrix admits a PLU factorisation, and the first failed pivot marks the singular boundary, Permutation matrices, partial pivoting, and the pivot-growth factor).

Verification

technique · direct

1.1

The largest entry in modulus in the first column is $2$ in row $2$ , so the first pivot swap sends that row to the top. The first elimination multipliers are $0$ for the new second row and $1 / 2$ for the new third row, producing the intermediate matrix $(\begin{array}{ccc} 2 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & \frac{1}{2} & \frac{3}{2} \end{array}) .$ The second pivot is already the entry $1$ in row $2$ , so no further swap is needed, and eliminating the $(3, 2)$ entry with multiplier $1 / 2$ gives the displayed $U$ . Recording the two nonzero multipliers in the permuted row order gives the displayed $L$ .

givenL1algebra

2.1

Direct multiplication gives $L U = (\begin{array}{ccc} 2 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{array}) = P A .$ Hence the displayed matrices are a correct PLU factorisation.

step 1.1algebra

3.1

Steps 1.1-2.1 verify the example.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Section 2.6 (standard reference, not scraped)