= Solution
The <Householder-John theorem> states the following. Let
$$
A=M-N
$$
be a splitting in which $A$ is Hermitian positive definite. If
$$
B=M^*+N
$$
is 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 therefore
$$
x^*Bx=x^*(M^*+N)x=-x^*Ax<0,
$$
contradicting the positive definiteness of $B$.
Now let $Hx=\lambda x$ with $x\ne0$. Then $Nx=\lambda Mx$, so
$$
Ax=(1-\lambda)Mx.
$$
Put $a=x^*Ax>0$ and $m=x^*Mx$. Since $\lambda\ne1$,
$$
m=\frac{a}{1-\lambda}.
$$
Consequently,
$$
0<x^*Bx
=\overline m+\lambda m
=a\left(\frac1{1-\overline\lambda}
+\frac{\lambda}{1-\lambda}\right)
=a\frac{1-|\lambda|^2}{|1-\lambda|^2}.
$$
It follows that $|\lambda|<1$. Thus $\rho(H)<1$, proving the theorem.
Back to article page
