= Solution
In coordinates, expanding $d\alpha$ and the <Lie bracket of vector fields> shows
$$
d\alpha(X,Y)=X(\alpha(Y))-Y(\alpha(X))-\alpha([X,Y]).
$$
The terms differentiating components of $X,Y$ cancel against the bracket, leaving $(\partial_i\alpha_j-\partial_j\alpha_i)X^iY^j$. This is the <exterior derivative of a one-form evaluated on vector fields>.
A <connection on a vector bundle> is a linear map $\nabla:\Gamma(E)\to\Omega^1(E)$ satisfying $\nabla(fs)=df\otimes s+f\nabla s$; evaluation on $X$ defines $\nabla_Xs$. Its <curvature form of a connection> is $d_A^2$, locally $F_A=dA+A\wedge A$. Direct expansion, using the displayed exterior-derivative identity, gives
$$
\boxed{F_A(X,Y)s=\nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s.}
$$
The derivative terms on a scalar multiplying $s$ cancel, and the expression is also linear over smooth functions in $X,Y$. Thus it is an alternating tensor with values in $\operatorname{End}(E)$, namely an element of $\Omega^2(\operatorname{End}(E))$.
The <dual connection> and <tensor product connection> induce the <endomorphism bundle connection> on $E^*\otimes E$:
$$
(\widetilde\nabla_X\phi)(s)=\nabla_X(\phi s)-\phi(\nabla_Xs).
$$
Expanding twice cancels the cross terms and gives
$$
([\widetilde\nabla_X,\widetilde\nabla_Y]\phi)(s)
=[\nabla_X,\nabla_Y](\phi s)-\phi([\nabla_X,\nabla_Y]s).
$$
Here $\phi,s$ are local smooth sections; derivatives are not defined for isolated fiber elements without extensions. Subtract the corresponding $\widetilde\nabla_{[X,Y]}$ identity to get the <curvature of an endomorphism bundle connection>:
$$
\boxed{F_{\operatorname{End}(A)}(X,Y)\phi=[F_A(X,Y),\phi].}
$$
This vanishes for all $\phi$ exactly when each $F_A(X,Y)$ is central in the full matrix algebra, hence scalar. For rank $r>0$, $\omega=r^{-1}\operatorname{tr}F_A$ is a smooth two-form and
$$
\boxed{F_{\operatorname{End}(A)}=0\quad\Longleftrightarrow\quad F_A=\omega\,\operatorname{id}_E.}
$$
Conversely scalar curvature commutes with every endomorphism. This is the <scalar-curvature criterion for a flat endomorphism connection>; it permits a nonflat connection on $E$ itself.
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