Diagonal curvature operator implies diagonal Ricci tensor

If ${\mathcal{R}}^{ab}$ is proportional to $e^a\wedge e^b$ for every pair, then a curvature component can be nonzero only when its two index pairs describe the same coordinate two-plane. The contraction $R_{bd}=R^a{}_{bad}$ therefore vanishes for $b\ne d$, so the Ricci tensor is diagonal in the same frame.