= Solution
Apply the <matrix norm induced by an energy norm> and set $T=A^{1/2}C(\omega)A^{-1/2}=I-A^{1/2}B^{-1}A^{1/2}$. Multiplication gives
$$
T^TT=I-A^{1/2}\bigl[B^{-1}+B^{-T}-B^{-T}AB^{-1}\bigr]A^{1/2}.
$$
The bracket factors as $B^{-T}(B+B^T-A)B^{-1}$. Because $B=D/\omega+L$ and $A=D+L+L^T$,
$$
B+B^T-A=\left(\frac2\omega-1\right)D.
$$
Thus $T^TT=M(\omega)$ for the stated <matrix>. In particular $M(\omega)$ is a <positive semidefinite matrix>. The Euclidean <operator norm> squared is the largest <eigenvalue> of $T^TT$, and all those <eigenvalues> are nonnegative. Consequently
$$
\boxed{\|C(\omega)\|_A=\|T\|_2=\rho(M(\omega))^{1/2}.}
$$
This identity is the algebraic step in the <energy-norm proof of convergence of SOR>; <spectral radius> equals the largest <eigenvalue> here because $M(\omega)=T^TT$.
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