Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-30
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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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An invertible matrix can fail unpivoted LU at the first pivot

Statement refuted

Every invertible square matrix has an unpivoted unit-lower LU factorisation.

Facts & Assumptions

Given: The matrix A=(0110).

[L1]

A square matrix has an unpivoted unit-lower LU factorisation with nonzero pivots exactly when every leading principal minor is nonzero (A square matrix has an unpivoted unit-lower LU factorisation exactly when all leading principal minors are nonzero).

Counterexample

technique · direct
1.1givenalgebra

The determinant of A is −1, so A is invertible. Its first leading principal minor is the 1×1 determinant Δ1=0.

2.1step 1.1L1

By [L1], the vanishing of Δ1 forbids an unpivoted unit-lower LU factorisation with nonzero pivots. Therefore invertibility alone does not suffice.

3.1step 2.1∎

Step 2.1 refutes the statement.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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