ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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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.

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ILU(0) factors of a three-row tridiagonal matrix

Example

Take $A = (\begin{array}{ccc} 4 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 4 \end{array}) .$ In the natural order, fixed-pattern ILU(0) returns $L = (\begin{array}{ccc} 1 & 0 & 0 \\ - 1 / 4 & 1 & 0 \\ 0 & - 4 / 15 & 1 \end{array}), U = (\begin{array}{ccc} 4 & - 1 & 0 \\ 0 & 15 / 4 & - 1 \\ 0 & 0 & 56 / 15 \end{array}),$ and here $L U = A$ exactly.

Facts & Assumptions

Given: The displayed $3 \times 3$ matrix over $Q$ , represented by canonical CSR with strictly increasing column indices in each row, and the fixed natural ordering.

[F1]

Fixed-pattern ILU(0) factorization and application: each lower multiplier is formed by dividing the current work entry by the already computed pivot, and updates are made only at matching positions retained in the input pattern.

[F2]

Compressed sparse row matrices with stored-entry semantics: a canonical CSR record has no repeated column within a row and no stored zero; its row intervals and column indices specify the stored positions, and its value array supplies the corresponding matrix entries, with unstored entries zero.

[F3]

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: matrix multiplication is defined entrywise by the finite sum $(A B)_{i j} = \sum_{k} A_{i k} B_{k j}$ .

[F4]

The rationals form a field: $Q$ with its usual operations is a field.

Verification

1.1givenF2F4algebra

By [F4], the displayed matrix entries are in a field. Its nonzero pattern is the diagonal together with the four adjacent off-diagonal positions. It is canonical, sorted by column in every row, and contains all three diagonal positions, so the algorithm has no input-pattern or missing-diagonal failure.

2.1F1step 1.1algebra

In row 1 the factors are $U_{11} = 4$ and $U_{12} = - 1$ . In row 2 the multiplier is $L_{21} = - 1 / 4$ ; the matching update at column 2 gives $U_{22} = 4 - (- 1 / 4) (- 1) = 15 / 4$ . The row-2 entry at column 3 remains $U_{23} = - 1$ because row 1 has no position at column 3.

3.1F1step 1.1step 2.1algebra

Row 3 has no position at column 1. Its multiplier at column 2 is $L_{32} = (- 1) / (15 / 4) = - 4 / 15$ . The matching update at column 3 gives $U_{33} = 4 - (- 4 / 15) (- 1) = 56 / 15$ . All three pivots are nonzero, so the procedure returns ok with the displayed factors.

4.1F3step 2.1step 3.1algebra∎

By [F3], the product has rows $(4, - 1, 0)$ , $(- 1, 1 / 4 + 15 / 4, - 1)$ , and $(0, - 1, 4 / 15 + 56 / 15)$ , hence $L U = A$ .

Depends on

Used by

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Sources

Yousef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., §10.3.2, printed pp. 307–310 (standard reference, not scraped)
Richard Barrett et al., Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods, 2nd ed., §3.4.2, printed pp. 39–41 (standard reference, not scraped)