Finite-budget Jacobi stationary iteration

Input $n\ge 1$, $A\in M_n(\mathbb{C})$, $b,x^{(0)}\in {\mathbb{C}}^n$, and an iteration budget $m\in {\mathbb{N}}_0$. Check dimensions; on a mismatch return invalid-input. Form $D,L,U$ by Diagonal, lower, and upper parts in the stationary-iteration convention. If some diagonal entry of $A$ is zero, return pivot-failure(i) for the least such index $i$.

Otherwise set $M=D$ and $N=L+U$, so $A=M-N$. For each $k=0,\dots ,m-1$, compute the diagonal solve

$D{x}^{(k+1)}=(L+U)x^{(k)}+b,$

equivalently, simultaneously for every $i$,

$x_i^{(k+1)}=\frac{b_i-{\sum }_{j\mathrm{\ne }i}{a}_{ij}{x}_j^{(k)}}{a_{ii}}\mathrm{.}$

Return completed with $x^{(m)}$. The solve is the unique diagonal instance of Forward and backward substitution for triangular linear systems when the pivot check succeeds; Forward and backward substitution are correct, unique, and quadratic in scalar operations gives its well-definedness. A zero budget returns $x^{(0)}$. The procedure reports no convergence or accuracy certificate.