Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 39A / a / Solution

The Householder-John theorem says that, for a splitting $A=M-N$ with $A$ Hermitian positive definite, positivity of $H=M^{\ast }+N=M^{\ast }+M-A$ implies $M$ is invertible and $\rho (M^{-1}N)<1$. Invertibility follows since $Mx=0$ for $x\ne 0$ would give $x^{\ast }Hx=-x^{\ast }Ax<0$.

For clarity, the spectral-radius conclusion follows from a useful energy identity. Put $T=M^{-1}N=I-M^{-1}A$. Direct multiplication givesIf $Tv=\lambda v$, thenThe eigenvalue one is impossible because $A$ is invertible. Both quadratic forms are positive, so $|\lambda |<1$. The splitting iteration $M{x}^{(k+1)}=N{x}^{(k)}+b$ propagates its error by $T$; spectral radius less than one therefore gives convergence from every initial vector.