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*}