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Strict row diagonal dominance guarantees Jacobi and Gauss–Seidel convergence
Statement
Let and satisfy strict row diagonal dominance:
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 for every .
Facts & Assumptions
Given: , a real matrix satisfying the displayed strict row diagonal-dominance inequalities, any , and the fixed order .
For each row , .
The diagonal/lower/upper convention has , with strictly lower and strictly upper (Diagonal, lower, and upper parts in the stationary-iteration convention).
Jacobi uses the stationary split , and solves (Finite-budget Jacobi stationary iteration).
Fixed-order Gauss–Seidel uses , and solves (Finite-budget Gauss–Seidel stationary iteration).
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).
For a stationary split with invertible , the iteration matrix is and the update is (Stationary iteration from a matrix splitting ).
Such an iteration converges to the unique solution for every right-hand side and every start if and only if (A stationary splitting converges for every start if and only if its iteration matrix has spectral radius below ).
Proof
By [A1], every . Thus and the lower-triangular matrix have nonzero diagonal, so the solves in [F1] and [F2] are uniquely defined by [L2].
By [F1]–[F2] and [L3], the iteration matrices are and .
For each row , set and . Since below the diagonal and above it, . Hence for any eigenpair with , choose with ; the row equation gives , so and .
Let with , and choose with . The equation in row is , so . By the reverse triangle inequality and [A1], the left factor is at least , while the right side is at most . Therefore ; the denominators are positive and each strict inequality follows from [A1]. Thus .
The bounds in steps 3.1 and 3.2 and [L4] imply that both methods converge to the unique solution of for every and every initial vector.
Depends on
- Diagonal, lower, and upper parts in the stationary-iteration convention
- Finite-budget Jacobi stationary iteration
- Finite-budget Gauss–Seidel stationary iteration
- Stationary iteration from a matrix splitting $A=M-N$
- A stationary splitting converges for every start if and only if its iteration matrix has spectral radius below $1$
- Forward and backward substitution are correct, unique, and quadratic in scalar operations
Used by
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Sources
- Yousef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., §4.2.3 (standard reference, not scraped)
- David Bindel, Matrix Computations (CS 6210), Cornell University, Fall 2012, Lecture 31 (standard reference, not scraped)
- Rutgers University MATH 574, Lecture Notes, §3 Matrix iterative methods (standard reference, not scraped)