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For a three-by-three matrix with diagonal entries one and every off-diagonal entry $\alpha$, the Jacobi method iteration matrix has eigenvalue $-2\alpha$ on the all-ones vector and eigenvalue $\alpha$ with multiplicity two on its orthogonal complement. Its spectral radius is $2|\alpha |$, so Jacobi iteration converges exactly when $|\alpha |<1/2$.

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.