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The Householder-John theorem states: if $A=M-N$ is Hermitian positive definite and $M^{\ast }+N$ is also Hermitian positive definite, then $M$ is invertible and

Now let $A=D+L+L^T$ be real symmetric positive definite and tridiagonal. For the Jacobi splitting, $M=D$ and $N=-(L+L^T)$. Positive definiteness of $A$ implies every diagonal entry is positive, so $D$ is positive definite. Introduce the alternating-sign diagonal matrixTridiagonality givesSince $S$ is orthogonal, $SAS$ is positive definite whenever $A$ is. Both hypotheses of the Householder-John theorem hold, and hence the Jacobi iteration matrix satisfies $\rho (M^{-1}N)<1$. Therefore the Jacobi convergence for a symmetric positive-definite tridiagonal matrix follows.

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.