For all $\omega \neq 0$, $\rho(L_{\omega}) \ge |1 - \omega|$, where $L_{\omega}$ is the SOR matrix Let $A = (a_{ij}) \in M_n(\Bbb C)$ be invertible, such that $a_{ii} \neq 0$ for all $i$. Split $A$ into $D - E - F$, where $D$ is the diagonal of $A$, $E$ is the strict lower triangular part of $-A$ and $F$ is the strict upper triangular part of $-A$. 
Let $\omega \in \Bbb C \setminus \{0\}$, and define:
$$L_{\omega} = \left(\frac1{\omega} D - E\right)^{-1} \left(\frac{1-\omega}{\omega} D + F\right)$$

It is required to prove that $\rho(L_{\omega}) \ge |1-\omega|$

Where $\rho(M)$ denotes the spectral radius of a matrix $M$.
Until now, I have no good ideas. I tried to prove the constraint on each eigenvalue by considering the characteristic polynomial,
$$P_{L_{\omega}} (x) = \det\left( xI - \left(\frac1{\omega} D - E\right)^{-1} \left(\frac{1-\omega}{\omega} D + F\right) \right) \\ = \ldots \\ = \frac{\omega^n}{\prod a_{ii}} \det \left(\frac{x - 1 + \omega}{\omega} D - Ex - F\right) $$
This seems to lead nowhere.
I tried to prove it for the case $n=2$ to have a better understanding of the situation, however I couldn't. Just in case it helps, I obtained the following as the characteristic polynomial for the case $n=2$.
$$x^2 - \frac{\omega^2}{a_{11}a_{22}}a_{12}a_{21}x + (1-\omega)^2$$
Finally, I have no relevant theorems in mind; all the theorems regarding the spectral radius, which are known to me, are just possible upper bounds.
Source of the claim (page $27$).
 A: For  $\omega \in \Bbb C \setminus \{0\}$ you have :
\begin{gather*}\det\left(L_{\omega}\right ) & =&  \det\left(\left(\frac1{\omega} D - E\right)^{-1} \left(\frac{1-\omega}{\omega} D + F\right) \right) \\ & = &\frac{\det\left(\frac{1-\omega}{\omega} D + F\right)}{\det \left(\frac1{\omega} D - E\right) }
\end{gather*}
But your decomposition is such that E and F are strictly triangular so we end up with :
\begin{gather*}
\det \left(\frac1{\omega} D - E\right) = \frac{1}{\omega^{n}}\prod a_{ii}\\
\det \left(\frac{1-\omega}{\omega} D + F\right) = \frac{\left(1 - \omega \right )^{n}}{\omega^{n}}\prod a_{ii}\\
\end{gather*}
That gives us : $\det\left(L_{\omega}\right ) = \left(1 - \omega \right )^{n}$
But you also know by definition of the spectral radius that $\rho(L_{\omega}) = \displaystyle \max_{i} |\lambda_i|$.
To the power of $n$ it gives :
\begin{gather*}
\rho(L_{\omega})^{n} &\ge& \prod_{i} |\lambda_i|\\
 &\ge& |\det\left(L_{\omega}\right )|\\
 &\ge& |1 - \omega|^{n}
\end{gather*}
And from that follows $\rho(L_{\omega}) \ge |1 - \omega|$.
