If $V^{\mu}$ is a killing vector, then is $∇_μ V^{\mu} = 0$?

Question

Working with the Levi-Civita Connection and a symmetric metric I want to show that if $V^{\mu}$ is a killing vector, then $∇_μ V^{\mu} = 0$.

I am not sure how to show this fact, but I believe it to be true. I proceeded as follows: $$∇_μ V^{\mu} = g^{\lambda\mu}∇_μ V_{\lambda} = -g^{\lambda\mu}∇_{\lambda} V_{\mu}$$ Then I got stuck. It it not obvious to me how to proceed.

Answer 1

Hint: The defining property of the Killing vector is $$\nabla_iV_j+\nabla_jV_i=0$$ A contraction by $g^{ij}$ will lead to your result as $2\nabla_iV^i=0$.