= Solution
Take an affinely parametrized <null geodesic congruence>, $\nabla_kk=0$. Choose the auxiliary <null vector> and a transverse orthonormal screen basis to undergo <parallel transport> along each ray. For the unprojected tensor $C_{ab}=\nabla_bk_a$, the product rule and the <Ricci identity> yield
$$
\begin{aligned}
k^c\nabla_cC_{ab}
&=\nabla_b(k^c\nabla_ck_a)-(\nabla_bk^c)(\nabla_ck_a)
+k^c[\nabla_c,\nabla_b]k_a\\
&=-(\nabla_bk^c)(\nabla_ck_a)-R_{acbd}k^ck^d.
\end{aligned}
$$
Here the <Riemann curvature tensor> convention is the one in Question 1; its last-index antisymmetry converts the covector commutator to the displayed curvature term. Because $C_{ab}k^a=C_{ab}k^b=0$, inserting the <screen-space projector> in the product term adds only terms killed by these contractions. Parallel propagation makes differentiation commute with the screen projection. Therefore, in screen indices $A,B$,
$$
\frac{dB_{AB}}{d\lambda}=-B_{AC}B^C{}_B-R_{AcBd}k^ck^d.
$$
This is the optical matrix evolution equation. Its trace has curvature contraction $q^{ab}R_{acbd}k^ck^d=R_{cd}k^ck^d$: the additional terms in $q^{ab}-g^{ab}$ vanish by curvature antisymmetry. Write $B=(\theta/2)q+\sigma+\omega$ using the <null expansion>, <null shear>, and <null twist>. Symmetric--antisymmetric cross terms have zero trace, $\sigma$ is trace-free, and
$$
\operatorname{tr}(B^2)=\frac12\theta^2+\sigma_{ab}\sigma^{ab}-\omega_{ab}\omega^{ab}.
$$
The minus sign in the last trace comes from $\omega_{ab}=-\omega_{ba}$ and the positive screen metric. Taking the trace therefore derives the <Null Raychaudhuri equation>:
$$
\boxed{\frac{d\theta}{d\lambda}=-\frac12\theta^2-\sigma_{ab}\sigma^{ab}+\omega_{ab}\omega^{ab}-R_{ab}k^ak^b.}
$$
For a nonaffine tangent obeying $\nabla_kk=\kappa k$, the same product-rule derivation adds $+\kappa\theta$ on the right; the projected derivative of $\kappa$ multiplies $k_a$ and vanishes. In $D$ dimensions the quadratic <null expansion> term is $-\theta^2/(D-2)$.
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