 Solution

The Killing equation gives ${\nabla }_{\mu }{K}^{\mu }=0$ and ${\nabla }_{[\mu }{K}_{\lambda ]}={\nabla }_{\mu }{K}_{\lambda }$. Commuting derivatives once and contracting the Ricci identity givesTherefore, using the contracted Bianchi identity and symmetry of $R_{\mu \beta }$,Since ${\nabla }_{[\mu }{K}_{\lambda ]}$ is antisymmetric, swapping the two contracted index names shows that the left side equals its derivative-antisymmetrized form:Now set $A_{\mu \lambda }={\nabla }_{[\mu }{K}_{\lambda ]}$. The right side is the contracted commutator $[{\nabla }^{\mu },{\nabla }^{\lambda }]A_{\mu \lambda }$. By the curvature commutator on a covariant tensor, it is a sum of contractions of the symmetric Ricci tensor with the antisymmetric tensor $A$, and hence vanishes. Thusso the Killing vector preserves scalar curvature.