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Jacobi iteration on a strictly row-diagonally-dominant system

Example

Take

$A = (\begin{array}{cc} 5 & 1 \\ 1 & 4 \end{array}), b = (\binom{2}{3}), x^{(0)} = (\binom{0}{0}) .$

Both rows are strictly diagonally dominant. The Jacobi iteration matrix is $G_{J} = (\begin{array}{cc} 0 & - 1 / 5 \\ - 1 / 4 & 0 \end{array})$ , with eigenvalues $\pm 1 / \sqrt{20}$ , and the exact solution is $x_{*} = (5 / 19, 13 / 19)^{T}$ . The strict-dominance theorem therefore guarantees convergence from every start to this solution.

Facts & Assumptions

Given: The displayed $A$ , $b$ , and initial vector.

[F1]

Jacobi uses $D = diag (A)$ and the simultaneous update $D x^{(k + 1)} = (L + U) x^{(k)} + b$ (Finite-budget Jacobi stationary iteration).

[L1]

Strict row diagonal dominance guarantees Jacobi convergence to the unique solution for every start (Strict row diagonal dominance guarantees Jacobi and Gauss–Seidel convergence).

Verification

1.1givenalgebra

The row inequalities are $5 > 1$ and $4 > 1$ . Direct multiplication gives $A (5 / 19, 13 / 19)^{T} = (2, 3)^{T}$ , so the displayed vector is the exact solution.

2.1F1L1step 1.1algebra∎

By [F1], $G_{J} = - D^{- 1} (A - D) = (\begin{array}{cc} 0 & - 1 / 5 \\ - 1 / 4 & 0 \end{array})$ , whose characteristic polynomial is $λ^{2} - 1 / 20$ . Thus $ρ (G_{J}) = 1 / \sqrt{20} < 1$ ; [L1] gives convergence from every initial vector to the verified solution.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Yousef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., §§4.1–4.2.3 (standard reference, not scraped)