How can we show that trace and Lie derivative of a $(1,1)$-tensor commute? That is, I want to prove that $$ \operatorname{tr}(L_X S) = L_X \operatorname{tr}S, $$ where $S$ is a $(1,1)$-tensor. This is Exercise 2.5.10 of Riemannian Geometry (Third Edition), by Peter Petersen.
My idea is to consider the definition of a trace of a linear operator $L : V \to V$ (where $V$ is a vector space) given by $$ \operatorname{tr}L = \sum_{i=1}^n g(L(e_i),e_i), $$ where the $e_i$'s form an orthonormal basis.
Also, Exercise 2.5.9 (the exercise preceding this one) asks to demonstrate that $\operatorname{tr} (\nabla_X S) = \nabla_X \operatorname{tr}S$, and perhaps this may be used in the proof of the current problem?
I also wanted to type change the $(1,1)$-tensor $S$ into a $(0,2)$-tensor $\hat S$ via the fomula $\hat S(X,Y) = g(S(X),Y)$, so that I can use Proposition 2.1.2 of Petersen:
If $X$ is a vector field and $T$ a $(0,k)$-tensor on a manifold $M$, then $$ (L_X T)(Y_1,\ldots,Y_k)=D_X(T(Y_1,\ldots,Y_k))-\sum_{i=1}^k T(Y_1,\ldots,L_X Y_i,\ldots,Y_k). $$
Can I request any thoughts on this matter?
Hint We can write the Lie derivative of a tensor $S$ of type $(1, 1)$ in terms of any torsion-free connection (the Levi-Civita connection $\nabla$ of the metric certainly works) as $$\mathcal L_X S = \nabla_X S - (\nabla X) \circ S + S \circ \nabla X .$$
(Note, by the way, that the trace identity $\operatorname{tr} \mathcal L_X S = \mathcal L_X \operatorname{tr} S$ only involves the Lie derivative and trace, and so in particular it is independent of the metric. Correspondingly, we can write the trace of an endomorphism $L : V \to V$ in terms of any basis $(e_a)$ as $\sum_{a = 1}^n \epsilon^a(L(e_a))$, where $(\epsilon^a)$ is the basis of $V^*$ dual to $(e_a)$.)