Let $(M,g)$ be a $(m+1)$-dimensional Riemannian manifold with Levi-Civita connection $\nabla$. 


The *Ricci curvature* can be viewed as a differential operator $\text{Rc}:\Gamma(S^2\_+M)\rightarrow\Gamma(S^2M)$ from the space of metrics on $M$ to the space of symmetric tensors on $M$ given by $g\mapsto\text{Rc}(g)$. 


The *formal adjoint* of the divergence is the map ${\delta}\_g^\*:\Gamma(TM)\rightarrow\Gamma(S^2M)$ given locally by
\begin{align\*}
({\delta}\_g^\*X)\_{ij}&=\frac{1}{2}\big[g(\nabla\_iX,\partial\_j)+g(\partial\_i,\nabla\_jX)\big]\\
&=\frac{1}{2}(L\_Xg)\_{ij}
\end{align\*}
where we identify vector fields and covector fields via the musical isomorphisms. The *Lie derivative* with respect to a vector field $X\in\Gamma(TM)$ of a symmetric tensor is the map $L\_X:\Gamma(S^2M)\rightarrow\Gamma(S^2M)$ given locally by
\begin{align\*}
(L\_Xr)\_{ij}&=\nabla\_X(r\_{ij})-r\big([X,\partial\_i],\partial\_j\big)-r\big(\partial\_i,[X,\partial\_j]\big)\\
&=(\nabla\_Xr)\_{ij}+(\nabla\_iX)^kr\_{kj}+(\nabla\_jX)^kr\_{ki}
\end{align\*}
where the *Lie bracket* $[X,Y]=\nabla\_XY-\nabla\_YX$. Now set
\begin{align\*}
r(g)=\text{Rc}\_g+mg\in\Gamma(S^2M).
\end{align\*}
We get a map $r:\Gamma(S^2\_+M)\rightarrow\Gamma(S^2M)$ given by $g\mapsto r(g)=\text{Rc}(g)+mg$, sometimes called the *Einstein operator* since if $r(g)=0$ then $\text{Rc}(g)=-mg$


**Question**: Does the linearisation $2Dr\_g({\delta}^\*\_gX)$ of $r$ at $g$ in the direction of $2{\delta}^\*\_gX$ equal
\begin{align\*}
2Dr\_g({\delta}^\*\_gX)=Dr\_g(L\_Xg)=L\_Xr(g)?
\end{align\*}



Linearisation of Einstein operator
Let $\Phi\_s$ denote the one parameter family of diffeomorphisms generated by $X$ and let $g(s) = \Phi\_s^\* g$. Note that $r(\Phi\_s^\* g) = \Phi\_s^\* r(g)$ by definition, so you have that the right hand side of your equation is exactly 
$$ L\_X r(g) = \lim\_{s\to 0} \frac{\Phi\_s^\* r(g) - r(g)}{s} .$$


On the other hand, the left hand side of your expression is 
$$ \lim\_{s \to 0} \frac{r(g(s)) - r(g)}{s} = D r\_g (\dot{g}) $$
by definition of "linearisation". And we have that $\dot{g} = \frac{d}{ds} g(s) | \_{s = 0} = L\_X g$ by construction. So yes, the two sides are equal. 


