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

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An LDU factorisation isolates the pivot scalars uniquely

Example

The matrix $A = (\begin{array}{cc} 2 & 1 \\ 4 & 5 \end{array})$

has the LDU factorisation

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

and the diagonal pivots $2, 3$ are uniquely determined.

Facts & Assumptions

Given: The displayed matrix $A$ and candidate LDU factors.

[L1]

An LDU factorisation has the displayed unit-lower, diagonal, and unit-upper shape (An LDU factorisation has unit lower-triangular L, diagonal D, and unit upper-triangular U).

[L2]

LDU factorisations with nonzero diagonal pivots are unique (Normalised LU and LDU factorisations with nonzero pivots are unique).

Verification

technique · direct

1.1

Multiplying the right two factors gives $(\begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array}) (\begin{array}{cc} 1 & \frac{1}{2} \\ 0 & 1 \end{array}) = (\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array}),$ and then left multiplication by $(\begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array})$ yields $(\begin{array}{cc} 2 & 1 \\ 4 & 5 \end{array}) = A$ .

givenL1algebra

2.1

The diagonal factor has nonzero entries $2$ and $3$ , so [L2] applies. Any other LDU factorisation of $A$ with nonzero diagonal pivots must therefore have the same diagonal factor and hence the same pivot scalars.

step 1.1L2

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

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Sources

David Bindel, CS 4220: Numerical Analysis, Blocked LU and Cholesky (standard reference, not scraped)