 Solution

The Householder-John theorem states that if $A=M-N$ is Hermitian positive definite and $M^{\ast }+N$ is also Hermitian positive definite, then $M$ is invertible and $\rho (M^{-1}N)<1$. The associated stationary iteration $x^{(k+1)}=M^{-1}N{x}^{(k)}+M^{-1}b$ therefore converges from every starting vector. For designing a method, choose a splitting with easily solved $M$ and verify the positive-definiteness condition on $M^{\ast }+N$; it is a sufficient certificate and does not require computing every iteration-matrix eigenvalue.

For completeness, put $C=M^{\ast }+N=M^{\ast }+M-A$. If $Mv=0$ for $v\ne 0$, then $v^{\ast }Cv=-v^{\ast }Av<0$, a contradiction. If $M^{-1}Nv=zv$, then $Av=(1-z)Mv$. The case $z=1$ contradicts invertibility of $A$. Taking scalar products and using Hermitian symmetry giveshence $|z|<1$. The usual spectral-radius convergence criterion now proves the stated conclusion.

For the given scheme take the matrix splittingSince $A$ is real symmetric, $U=L^T$, and since it is positive definite, every diagonal entry in $D$ is positive. Thusis positive definite exactly for $0<\omega <2$. The theorem applies, and the iteration matrix has spectral radius below one. The unique solution $x_{\ast }=A^{-1}b$ is its fixed point, so $x^{(k)}-x_{\ast }=(M^{-1}N{)}^k(x^{(0)}-x_{\ast })\to 0$. ThereforeThe scheme is successive over-relaxation; at $\omega =1$ it becomes the Gauss-Seidel method.