= Solution
The <Householder-John theorem> says that, for a splitting $A=M-N$ with $A$ Hermitian positive definite, positivity of $H=M^*+N=M^*+M-A$ implies $M$ is invertible and $\rho(M^{-1}N)<1$. Invertibility follows since $Mx=0$ for $x\ne0$ would give $x^*Hx=-x^*Ax<0$.
For clarity, the spectral-radius conclusion follows from a useful energy identity. Put $T=M^{-1}N=I-M^{-1}A$. Direct multiplication gives
$$
A-T^*AT=(I-T)^*H(I-T).
$$
If $Tv=\lambda v$, then
$$
(1-|\lambda|^2)v^*Av=|1-\lambda|^2v^*Hv.
$$
The <eigenvalue> one is impossible because $A$ is invertible. Both quadratic forms are positive, so $|\lambda|<1$. The splitting iteration $Mx^{(k+1)}=Nx^{(k)}+b$ propagates its error by $T$; <spectral radius> less than one therefore gives \b[convergence from every initial vector].
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