I'm currently studying LECTURES ON CHERN-WElL THEORY, in which it defines the trace function on a vector bundle $(M,E)$ by setting $$ \operatorname{tr}(\omega\otimes A) = (\operatorname{tr}A)\omega ,\quad A\in \operatorname{End}(E),\omega \in \mathcal{A}^*(M) $$
but lemma 1.7 said for all $A\in \mathcal{A}^*(M;\operatorname{End}(E))$ and linear connection $\nabla$ on $E$, $$ d\operatorname{tr}A= \operatorname{tr}[\nabla,A] $$
holds.
What confused me is that in the book the lie bracket of $\operatorname{End}(E)$-valued forms is defined while $\nabla$ is NOT a $\operatorname{End}(E)$-valued $1$-form. So, can anyone explain what is this lie bracket $[\nabla,A]$?
I think the auther tried to define a product of two matrixes whose entries are forms and its action on $\Gamma(E)$. In this view the only thing he left is to define the exterior operator on $\mathcal{A}^*(M) \otimes \mathfrak{gl}(n,\mathbb{R}) $ and how it acts on $\Gamma(E)$, since $\nabla=\omega+d$.
If we define this exterior operator(differ from the one on $\mathcal{A}^*(M)$) on $\mathcal{A}^*(M) \otimes \mathfrak{gl}(n,\mathbb{R}) $ by (frame {e_i} of $E$ is already chosen) $$ dA = d(A^i_j)=(dA^i_j),A\in \mathcal{A}^*(M) \otimes \mathfrak{gl}(n,\mathbb{R}) ,\quad de=d(x^ie_i)=(dx^1,\cdots,dx^r)^T,e\in\Gamma(E) $$
then the definition fits all the actions I want on $\operatorname{End}(E)$.