A derivation on tensors is a map $T \mapsto D T$ that preserves the type of the tensor $T$; is linear; commutes with contractions; and satisfies the product rule $$ D\left(T_1 \otimes T_2\right)=\left(D T_1\right) \otimes T_2+T_1 \otimes D T_2 $$
Now for $v_1,\cdots,v_s\in V$, $\phi^1,\cdots,\phi_t\in V^*$ and $L\in End(V)$, we define \begin{aligned} &L\left(v_1 \otimes \cdots \otimes v_s \otimes \phi_1 \otimes \cdots \otimes \phi_t\right)\\ &=L\left(v_1\right) \otimes \cdots \otimes v_s \otimes \phi_1 \otimes \cdots \otimes \phi_t\\ &+\cdots\\ &+v_1 \otimes \cdots \otimes L\left(v_s\right) \otimes \phi_1 \otimes \cdots \otimes \phi_t\\ &-v_1 \otimes \cdots \otimes v_s \otimes\left(\phi_1 \circ L\right) \otimes \cdots \otimes \phi_t\\ &-\cdots\\ &-v_1 \otimes \cdots \otimes v_s \otimes \phi_1 \otimes \cdots \otimes\left(\phi_t \circ L\right) . \end{aligned}
And it's easy to see that any $(1,1)$-tensor $L$ defines a derivation on tensors.
In Riemannian Geometry of Peter Petersen, he gives a proposition as follows:
If $L\in \mathcal{so}(V)$, i.e., $L$ is skew-adjoint, then $L$ commutes with type change of temsors.
I'm trying to show this conclusion through coordinates for $(0,1)-tensors$ as follows:
Suppose $T=a_{i}x^i$, $L=b_{i}^{j}x_j\otimes x^i$, where $b_i^j=-b_j^i$. We denote the type change operator by $C$, then \begin{aligned} C(L(T))&=C(-a_ib_k^ix^k)=-a_ib_m^ig^{mk}x_k.\\ L(C(T))&=L(a_ig^{ik}x_k)=a_ig^{ik}b_k^jx_j. \end{aligned} I think I can't show they are the same. Can anyone explain it and give a proof for general tensors?