Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck pass
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.

Strict row diagonal dominance guarantees Jacobi and Gauss–Seidel convergence

Statement

Let n≥1 and A=(aij)∈Mn(R) satisfy strict row diagonal dominance:

∣aii∣>∑j≠i∣aij∣(i=1,…,n).

Then the Jacobi and fixed-order Gauss–Seidel iterations defined by Finite-budget Jacobi stationary iteration and Finite-budget Gauss–Seidel stationary iteration converge in exact arithmetic to the unique solution of Ax=b for every b,x(0)∈Rn.

Facts & Assumptions

Given: n≥1, a real matrix A=(aij) satisfying the displayed strict row diagonal-dominance inequalities, any b,x(0)∈Rn, and the fixed order 1,…,n.

[A1]

For each row i, ∣aii∣>∑j≠i∣aij∣.

[L1]

The diagonal/lower/upper convention has A=D−L−U, with L strictly lower and U strictly upper (Diagonal, lower, and upper parts in the stationary-iteration convention).

[F1]

Jacobi uses the stationary split MJ=D, NJ=L+U and solves Dx(k+1)=(L+U)x(k)+b (Finite-budget Jacobi stationary iteration).

[F2]

Fixed-order Gauss–Seidel uses MG=D−L, NG=U and solves (D−L)x(k+1)=Ux(k)+b (Finite-budget Gauss–Seidel stationary iteration).

[L2]

If a finite-dimensional triangular system has nonzero diagonal, forward or backward substitution gives its unique solution (Forward and backward substitution are correct, unique, and quadratic in scalar operations).

[L3]

For a stationary split A=M−N with invertible M, the iteration matrix is G=M−1N and the update is x(k+1)=Gx(k)+M−1b (Stationary iteration from a matrix splitting A=M−N).

[L4]

Such an iteration converges to the unique solution for every right-hand side and every start if and only if ρ(G)<1 (A stationary splitting converges for every start if and only if its iteration matrix has spectral radius below 1).

Proof

1.1A1F1F2L2algebra

By [A1], every aii≠0. Thus D and the lower-triangular matrix D−L have nonzero diagonal, so the solves in [F1] and [F2] are uniquely defined by [L2].

2.1F1F2L3step 1.1algebra

By [F1]–[F2] and [L3], the iteration matrices are GJ=D−1(L+U) and GG=(D−L)−1U.

3.1A1L1step 2.1algebra

For each row i, set qi=∑j≠i∣aij∣/∣aii∣ and qJ=max⁡iqi<1. Since Lij=−aij below the diagonal and Uij=−aij above it, ∑j∣(GJ)ij∣=qi. Hence for any eigenpair GJz=λz with z≠0, choose i with ∣zi∣=max⁡j∣zj∣; the row equation gives ∣λ∣∣zi∣≤qi∣zi∣≤qJ∣zi∣, so ∣λ∣<1 and ρ(GJ)<1.

3.2A1L1step 2.1algebra

Let GGz=λz with z≠0, and choose i with ∣zi∣=max⁡j∣zj∣=:R. The equation Uz=λ(D−L)z in row i is ∑j>iUijzj=λ(aiizi−∑j<iLijzj), so ∣λ∣ ∣aiizi−∑j<iLijzj∣≤∑j>i∣Uij∣∣zj∣. By the reverse triangle inequality and [A1], the left factor is at least ∣λ∣(∣aii∣−∑j<i∣aij∣)R, while the right side is at most ∑j>i∣aij∣R. Therefore ∣λ∣≤max⁡i∑j>i∣aij∣∣aii∣−∑j<i∣aij∣<1; the denominators are positive and each strict inequality follows from [A1]. Thus ρ(GG)<1.

4.1L4step 3.1step 3.2∎

The bounds in steps 3.1 and 3.2 and [L4] imply that both methods converge to the unique solution of Ax=b for every b and every initial vector.

Depends on

Used by

Dependency tree · two levels

13 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