= Solution
For $0<\omega<2$, the coefficient $\gamma=2/\omega-1$ is positive. Moreover
$$
G=A^{1/2}B^{-T}DB^{-1}A^{1/2}
$$
is a <positive-definite matrix>: for a nonzero <vector> $z$, put $y=B^{-1}A^{1/2}z\ne0$; then $z^TGz=y^TDy>0$. Hence every <eigenvalue> of $M=I-\gamma G$ is strictly below one. Part (d) also gives $M=T^TT$, so these <eigenvalues> are nonnegative. Therefore
$$
\boxed{\|C(\omega)\|_A=\sqrt{\rho(M(\omega))}<1,\qquad0<\omega<2.}
$$
The <energy-norm proof of convergence of SOR> thus makes the iteration a strict <contraction mapping> in this norm. Its <forward error> satisfies $e^{(j+1)}=C(\omega)e^{(j)}$, so it converges geometrically to zero.
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