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For $\omega =1$, the Jacobi method iteration matrix $T=I-D^{-1}A$ has zero diagonal, nonnegative axial entries $1/5$, and diagonal-neighbour entries $1/20$. Its row sums are at most one; they are strictly below one at grid points next to the Dirichlet boundary. Its graph is connected by the positive axial entries.

To prove the strict spectral radius bound, let $Tz=\lambda z$ with $z\ne 0$ and choose a component of maximum magnitude $M>0$. Thenso $|\lambda |\le 1$. If equality held, every positive-weight neighbour of a maximal component would also have modulus $M$, and that row would have sum one. Propagating along axial edges reaches a boundary-adjacent row, whose sum is strictly less than one, a contradiction. This also covers a one-point interior grid, for which $T=0$. HenceThe standard convergence theorem for a stationary linear iteration says it converges from every initial vector exactly when its iteration matrix has spectral radius below one. Since $A$ is invertible by (i), the error satisfies $e^{(r)}=T^r{e}^{(0)}\to 0$ andWeak row diagonal dominance alone would not prove convergence; the connection to a strictly deficient boundary row supplies the required strictness.

Order the interior grid values into a vector. The coefficient matrix has diagonal $4+c{h}^2$ and off-diagonal entries $-1$ at horizontal and vertical neighbours, with boundary values already zero. Every row is strictly diagonally dominant because $c>0$. The standard convergence theorem for strict diagonal dominance gives convergence of both the Jacobi method and the Gauss-Seidel method. The matrix is also symmetric positive definite.

For $D=(4+c{h}^2)I$, the weighted Jacobi method is $u^{(r+1)}=u^{(r)}+\omega D^{-1}(b-A{u}^{(r)})$, with iteration matrix $H_{\omega }=I-\omega D^{-1}A$. The sine-product grid eigenvectors giveThese weighted Jacobi eigenvalues for a square Dirichlet grid determine convergence. All eigenvalues of $A$ are positive. Thus for fixed $h$ convergence requires $0<\omega <2(4+c{h}^2)/a_{\max }$, where $a_{\max }=8{\cos }^2(\pi h/2)+c{h}^2$. As $h\downarrow 0$, this upper bound tends to $1$ from above. Any fixed $\omega >1$ therefore fails on sufficiently fine grids; $\omega \le 0$ also fails. Consequently $0<\omega \le 1$ is necessary for all mesh sizes, and it is sufficient, since $a_{\max }<2(4+c{h}^2)$ for every finite grid.

Choose a matrix splitting $A=M-N$ with $M$ invertible. Then $Ax=b$ is equivalent toand the associated stationary iterative method for a linear system isFor the Jacobi method, write $A=D+L+U$ with $D$ diagonal and $L,U$ strictly lower and upper triangular, and takeThusThe iteration converges for every initial vector exactly when the spectral radius of its iteration matrix satisfies

Let $D=\mathrm{diag}(A)$. The Jacobi method isso $H=I-D^{-1}A$ and $v=D^{-1}b$. If $A$ is strictly diagonally dominant, every Gershgorin disc of $H$ is centered at zero and has radiusEvery eigenvalue therefore has modulus below one, so $\rho (H)<1$ and part (i) proves convergence.

Writewhere $D=\mathrm{diag}(a_1,\dots ,a_n)$ and $L,U$ are the strict lower and upper triangular parts. The Jacobi method iswith iteration matrix

Here $A$ is symmetric, so $U=L^T$. Regard the iteration as the splittingLetBecause $A$ is tridiagonal, changing alternating signs reverses the sign of every off-diagonal entry while preserving every diagonal entry. ThereforeSince $S^T=S=S^{-1}$ and $A$ is positive definite, $SAS$ is positive definite. The Householder-John theorem now givesThe Jacobi iterates therefore converge to the unique solution $A^{-1}b$ for every starting vector, which is the jacobi convergence for a symmetric positive-definite tridiagonal matrix.

With zero boundary values eliminated, the diagonal of $L_h$ is $-4I/h^2$. The Jacobi method is thereforewhich averages the four neighboring components. The sine addition identity shows that the supplied product vectors are eigenvectors, withThe discrete sine vectors form a basis, so these are all eigenvalues. Their largest modulus isattained at the lowest and highest sine frequencies. Thus the asymptotic error reduction factor per iteration is $\cos (\pi h)$. If the convergence rate is expressed logarithmically, $R=-\log \rho (J)\sim {\pi }^2{h}^2/2$ as $h\to 0$. A fixed error reduction therefore takes $O(h^{-2})$ Jacobi sweeps. For the special grid $m=2$, $J=0$ and one sweep solves its single interior equation.

Update the unknowns in row order, immediately using newly computed components. The component equations areThus $G=\left(\begin{array}{ccc}0& 1& -1\\ 0& -1/2& 0\\ 0& -3/4& 1/2\end{array}\right)$ and $c=(-1,3/2,1/4{)}^T$. Using only old values in all three row equations would instead describe the Jacobi method, not the Gauss-Seidel method.

On a rectangular grid with zero Dirichlet boundary conditions, the nine-point Poisson equation matrix has diagonal $10/3$, axial off-diagonal entries $-2/3$, and diagonal-neighbor entries $-1/6$. Its quadratic form is the sum of squared differences over internal links, weighted by $2/3$ or $1/6$, plus the weighted squares of values linked to the fixed zero boundary. The connected axial grid forces a zero quadratic form to have all entries zero. This proves positive-definite matrix in every ordering. The Jacobi method converges because its nonnegative iteration matrix has row sums at most one, with a strict deficit at boundary-adjacent vertices; a hypothetical unit-modulus eigenvector would propagate its maximum modulus to such a deficit row, which is impossible.

The periodic second-difference circulant matrix has constant null space, so solvability requires $\sum b_j=0$. Its Jacobi method has eigenvalues $\cos (2\pi j/n)$: for odd $n$ it converges to some solution, while for even $n$ an alternating mode oscillates unless absent from the error. Lexicographic Gauss-Seidel iteration converges to some solution for every compatible right-hand side and every initial vector. For $K=-A=D+L+U$, a coordinate sweep decreases the energy $e^{T}Ke/2$ by the sum of squared coordinate changes. This excludes all nonconstant unit-modulus eigenmodes. The constant unit eigenvalue is simple, since the invariant left vector $w=(0,1,\dots ,1,2{)}^T$ has $w^{T}1=n\ne 0$. Neither method is a strict contraction to a unique solution without fixing the constant nullspace.

For $A={\mu }^{-1}I+U-U^T$, with $U$ the four-dimensional unit superdiagonal, the Jacobi method has eigenvalues $2i\mu \cos (j\pi /5)$, $j=1,\dots ,4$. The Gauss-Seidel method has two zero eigenvalues and $-{\mu }^2(3\pm \sqrt{5})/2$. Therefore both methods converge for every starting vector exactly when $0<|\mu |<(\sqrt{5}-1)/2$. This equality of convergence ranges follows from their spectra, not from positive-definiteness of the nonsymmetric coefficient matrix.

With relaxation parameter $\omega$, weighted Jacobi updates byIts iteration matrix is $H_{\omega }=I-\omega D^{-1}A$.

Equivalently, the weighted Jacobi method takes a weighted average of the old vector and its full Jacobi method update: $x^{(k+1)}=(1-\omega )x^{(k)}+\omega x_{\mathrm{Jacobi}}^{(k+1)}$. If $Bv=\lambda v$, its iteration matrix multiplies that eigenvector by $1-\omega +\omega \lambda$. For the diagonal-4 Jacobi iteration for a three-variable tridiagonal system, the factors are $1-\omega$ and $1-\omega \pm \omega \sqrt{2}/4$; a Poisson equation smoothing parameter must not be inferred from this different matrix.