Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 40C / c / Solution

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.