My question is same as this MSE post but I want to use direct properties of Lie derivative and trace to prove (I know another proof using this fact that pullback map commutes with contraction)
$$\operatorname{tr}(\mathscr{L}_X S) = \mathscr{L}_X \operatorname{tr}S.$$
To do this I have calculated the following:
$$\operatorname{tr}(\mathscr{L}_X S)=g\Big((\mathscr{L}_X S)e_i,e_i\Big)=g\Big(\mathscr{L}_X Se_i,e_i\Big)-g\Big(S\mathscr{L}_X e_i,e_i\Big)=\mathscr{L}_Xg\Big( Se_i,e_i\Big)-(\mathscr{L}_Xg)( Se_i,e_i)-g(Se_i,\mathscr{L}_Xe_i)-g\Big(S\mathscr{L}_X e_i,e_i\Big)=\mathscr{L}_X \operatorname{tr}S-(\mathscr{L}_Xg)( Se_i,e_i)-g(Se_i,\mathscr{L}_Xe_i)-g\Big(S\mathscr{L}_X e_i,e_i\Big) $$
But I couldn't cancel out the extra terms. Any tiny hint (suffices)?
$\newcommand\tr{\operatorname{tr}}$By the linearity of trace, it suffices to prove the identity for $S = \theta\otimes v$, where $\theta$ is a $1$-form and $v$ is a vector field. First, observe that \begin{align*} \mathcal{L}_X(\theta\otimes v) &= (\mathcal{L}_X\theta)\otimes v + \theta\otimes (\mathcal{L}_Xv)\\ \tr (\theta\otimes v) &= \langle\theta, v\rangle. \end{align*} Therefore, \begin{align*} \tr\mathcal{L}_X(\theta\otimes v) &= \tr((\mathcal{L}_X\theta)\otimes v + \theta\otimes (\mathcal{L}_Xv)\\ &=\tr((\mathcal{L}_X\theta)\otimes v) + \tr(\theta\otimes (\mathcal{L}_Xv))\\ &= \langle \mathcal{L}_X\theta, v\rangle + \langle \theta, \mathcal{L}_Xv\rangle\\ &= X\langle \theta,v\rangle\\ &= \mathcal{L}_X(\tr \theta\otimes v) \end{align*}