Showing that the Lie bracket of two Killing fields on a Riemannian manifold is again a Killing field using the Killing equation I've read that the Lie bracket of two Killing fields $X$ and $Y$ on a Riemannian  manifold $M$ are again a Killing field so I thought it might be a good exercise to try to prove this using the Killing equation. 
We want to show that $$\langle \nabla_{Z}[X,Y],W \rangle =-\langle \nabla_{W}[X,Y],Z \rangle$$ for any two (smooth) vector fields $Z$ and $W$ on $M$. Here is my attempt:
$\langle \nabla_{Z}[X,Y],W \rangle= \langle \nabla_{Z}(\nabla_{X}Y-\nabla_YX),W\rangle$ (using symmetry)
$=\langle \nabla_{Z} \nabla_X Y,W \rangle -\langle \nabla_{Z} \nabla_Y X,W \rangle$ (using linearity of the connection)
$=Z\langle \nabla_XY,W \rangle-\langle \nabla_XY,\nabla_ZW\rangle-Z\langle \nabla_YX,W \rangle +\langle \nabla_YX,\nabla_ZW \rangle$ (compatibility of the metric)
At this point I thought about using the fact that $X$ and $Y$ satisfy the Killing equation and applying it to the terms being acted on by $Z$ but this doesn't appear to work. Maybe I'm going down the wrong road here but does anyone have a proof of this using the Killing equation?
 A: How does $\nabla[X,Y]$ relate to $\nabla X$ and $\nabla Y$?  We have to show that $\nabla[X,Y]$ is skew-adjoint if both $\nabla X$ and $\nabla Y$ are. So we will relate $\nabla[X,Y]$ with $[\nabla X,\nabla Y]$, which we know will also be skew-adjoint. It follows from Killing's equation the Kostant formula $\nabla^2X(Z,W) + R(X,Z)W = 0$ for $X$ Killing. So, let's compute $$\begin{align} [\nabla X, \nabla Y](Z) &= \nabla X(\nabla_ZY) - \nabla Y(\nabla_ZX) \\ &= \nabla_{\nabla_ZY}X - \nabla_{\nabla_ZX}Y \\ &=\nabla_Z\nabla _YX  - \nabla^2X(Z,Y) - \nabla_Z\nabla_XY + \nabla^2 Y(Z,X) \\ &= \nabla_Z\nabla_YX + R(X,Z)Y - \nabla_Z\nabla_XY - R(Y,Z)X \\ &= \nabla_Z[Y,X] - R(Z,X)Y - R(Y,Z)X \\ &= -(\nabla[X,Y])(Z) + R(X,Y)Z.\end{align}$$Hence $$\nabla[X,Y] = R(X,Y) - [\nabla X,\nabla Y]$$is the difference of two skew-adjoint operators, hence skew-adjoint as well. We conclude that $[X,Y]$ is Killing.
A: I haven't tried to prove it like you but I suspect you will have to use Bianchi equation at some point.
The easiest way is to use the characterisation based on the Lie derivative.
A vector field $X$ is a Killing vector field if and only if
$\mathcal{L}_X g=0,$
where $g$ is the metric of your Riemannian manifold.
So for $X,Y$ Killing vector fields, you have to check that
$$\mathcal{L}_{[X,Y]} g=0.$$
You will have to use that $\mathcal{L}_{X} g=\mathcal{L}_{Y} g=0$ and the Jacobi identity.
