 Solution

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.