Choosing value of ω for SOR I am learning about successive overrelaxation, and I'm wondering if there is an intuitive reason as to why ω must be between 0 and 2. I know that the method will not converge is ω is not on this interval, but I'm wondering if anyone can give an explanation of why this makes sense.
 A: I shall try to give you an intuitive idea of why $\omega \in (0,2)$ is essential.
There are many different ways of stating the SOR iteration, but for the purpose of answering your question I will use the following form
\begin{equation}
x^{(k+1)} = (1 - \omega) x^{(k)} + \omega D^{-1} \left[ b  - Lx^{(k+1)} - Ux^{(k)} \right],
\end{equation}
which is based on the splitting of $A$ as
\begin{equation}
A = D + L + U,
\end{equation}
where $D$ is diagonal, $L$ is strictly lower triangular and $U$ is strictly upper triangular. 
Now consider the extreme situation where we are trying to solve the scalar equation
\begin{equation}
a x = 0, \quad a \not = 0.
\end{equation} 
Then $L$ and $U$ are both zero and the iteration collapses to
 \begin{equation}
x^{(k+1)} = (1 - \omega) x^{(k)}
\end{equation}
from which we deduce that
\begin{equation}
x^{(k)} = (1- \omega)^{k-1} x^{(1)}.
\end{equation}
Now, if $\omega \not \in (0,2)$ and $x^{(1)} \not = 0$, then $x^{(k+1)}$ will grow exponentially and it is only if $\omega \in (0,2)$ that we have convergence to zero. 
In summary, even in the case of $n=1$ we cannot hope for convergence unless $\omega \in (0,2)$, and, as a general principle, increasing the dimension of a problem tends to provided numerical algorithms with more, rather than fewer, opportunities to misbehave.
You will have no trouble extending these above idea to the case where $A = D$ is a diagonal matrix and the right hand side $b$ has components which are zeros.
I hope this helps.
