= Solution
The <matrix splitting> produces $Mx^{(k+1)}=Nx^{(k)}+b$. If the iterates converge to $z$, passing to the limit yields $Mz=Nz+b$, hence $Az=b$. Thus any finite limit solves the system; when $A$ is nonsingular it is the unique solution.
The <iteration matrix> is $G=M^{-1}N=I-M^{-1}A$. For $M$ symmetric positive definite, its positive-definite square root gives
$$
\boxed{M^{1/2}GM^{-1/2}=I-M^{-1/2}AM^{-1/2},}
$$
which is symmetric when $A$ is symmetric. Let $\theta_i$ be the <eigenvalues> of $\widetilde A=M^{-1/2}AM^{-1/2}$. In its orthonormal eigenbasis the transformed error is multiplied by $1-\theta_i$ at each step. Therefore convergence to the unique solution for every starting vector is equivalent to
$$
\boxed{|1-\theta_i|<1\text{ for all }i\quad\Longleftrightarrow\quad0<\theta_i<2\text{ for all }i.}
$$
Sufficiency follows because each scalar mode tends to zero. Necessity follows by starting with a nonzero error in any mode whose multiplier has modulus at least one. Equivalently the condition says $\boxed{A\text{ and }2M-A\text{ are positive definite}}$, by congruence of their quadratic forms.
Under the usual nonsingularity assumption, a symmetric $A$ that is not positive definite has a negative quadratic direction. Its congruent $\widetilde A$ then has a negative <eigenvalue>, making an error multiplier $1-\theta_i>1$. A starting error in that mode diverges, proving the final claim in that setting.
If singular matrices are permitted, the last claim needs qualification. A zero <eigenvalue> gives a persistent error mode, but the iterates may still converge to another solution of a compatible singular system. For example $A=0$, $M=I$, $b=0$ makes every iterate constant and already a solution, despite $A$ not being positive definite. Thus literal nonconvergence cannot be deduced from a zero eigenvalue alone.
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