Write
$$
\Theta^\mu{}_\nu
=\frac12R^\mu{}_{\nu\rho\sigma}f^\rho\wedge f^\sigma.
$$
For a <diagonal curvature operator>, $\Theta^{\mu\nu}$ contains only $f^\mu\wedge f^\nu$. Consequently $R^\mu{}_{\nu\rho\sigma}$ can be nonzero only when the plane indexed by $\rho,\sigma$ is the same as the plane indexed by $\mu,\nu$. In the contraction
$$
R_{\nu\sigma}=R^\mu{}_{\nu\mu\sigma},
$$
this condition cannot hold for $\nu\ne\sigma$. Therefore
$$
\boxed{R_{\nu\sigma}=0\quad\text{for }\nu\ne\sigma},
$$
which is the statement that a <diagonal curvature operator implies diagonal Ricci tensor>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 4 / c / Solution

The curvature operator is diagonal in an orthonormal coframe $\{e^a\}$ when every <curvature 2-form> satisfies $\mathcal R^{ab}=K_{ab}e^a\wedge e^b$. Its matrix on the corresponding basis of <2-forms> is then diagonal.

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 Incoming links: Diagonal curvature operator