An $n$-dimensional Riemannian manifold $(M,g)$ is an Einstein soliton if there exist a vector field $\zeta$ and a real constant $\lambda$ such that $\DeclareMathOperator{\Ric}{Ric} \DeclareMathOperator{\Div}{div}$ $$ \Ric+\frac{1}{2}L_\zeta g=\lambda g +\frac{r}{2}g, $$ where $L_{\zeta}$ stands for the Lie derivative operator in the direction of $\zeta$, $\Ric$ is the Ricci curvature and $r$ is the scalar curvature of $g$.
Denote by $\eta=i_\zeta g$ the dual $1$-form of $\zeta$ and define the $(1, 1)$-tensor field $F$ by $$ g(FX,Y):=\frac{1}{2}d\eta(X,Y), $$ for any $X,Y$ smooth vector fields of $M$.
Then how we obtain following two identities from above equations: $$ \nabla\zeta=F-Q +\left(\lambda +\frac{r}{2}\right)I, $$ where $\nabla$ is the Levi-Civita connection of $g$, $Q$ is the Ricci operator defined by $g(QX, Y) := \Ric(X, Y)$, for $X, Y \in \mathfrak{X}(M)$ and $I$ is the identity endomorphism on the set of vector fields on $M$.
And the second one is \begin{align} \Div(\zeta)&=\frac{2n\lambda+(n-2)r}{2} \\ \Div(F\zeta)&=-\|F\|-\sum_{i=1}^ng((\nabla_{E_i}F)E_i,\zeta) \end{align}