Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 40D / a / Solution

The Householder-John theorem states the following. Letbe a splitting in which $A$ is Hermitian positive definite. Ifis also Hermitian positive definite, then $M$ is nonsingular and every eigenvalue of $H=M^{-1}N$ has modulus less than one. Hence the associated stationary iteration converges.

First, $M$ must be nonsingular. If $Mx=0$ for some nonzero $x$, then $Nx=-Ax$, and thereforecontradicting the positive definiteness of $B$.

Now let $Hx=\lambda x$ with $x\ne 0$. Then $Nx=\lambda Mx$, soPut $a=x^{\ast }Ax>0$ and $m=x^{\ast }Mx$. Since $\lambda \ne 1$,Consequently,It follows that $|\lambda |<1$. Thus $\rho (H)<1$, proving the theorem.

Solved by gpt-5.6-sol high.