= Solution
Let another <local frame> be $e'=eA$ with $A:U\to GL_r(\mathbb R)$ smooth. Since the coefficient columns satisfy $u=Au'$, the Leibniz rule gives the <change of frame of a vector-bundle connection>:
$$
\theta'=A^{-1}\theta A+A^{-1}dA.
$$
The <curvature form of a connection> is linear over <smooth functions> in its section argument, so its <matrix> transforms homogeneously:
$$
\Theta'=A^{-1}\Theta A.
$$
Consequently
$$
\operatorname{tr}\Theta'=\operatorname{tr}(A^{-1}\Theta A)=\operatorname{tr}\Theta.
$$
The entries of $A$ are functions, of degree zero, so ordinary cyclic invariance of <trace> applies without a graded sign. The local forms therefore glue to a globally defined two-form $\alpha=\operatorname{tr}\Theta$, the <trace of vector-bundle curvature>.
Apply the <Bianchi identity> proved in part (b):
$$
d\alpha=\operatorname{tr}(\Theta\wedge\theta)-\operatorname{tr}(\theta\wedge\Theta).
$$
For the first term,
$$
\operatorname{tr}(\Theta\wedge\theta)=\sum_{i,j}\Theta_{ij}\wedge\theta_{ji}
=\sum_{i,j}\theta_{ji}\wedge\Theta_{ij}
=\sum_{i,j}\theta_{ij}\wedge\Theta_{ji}
=\operatorname{tr}(\theta\wedge\Theta).
$$
The middle interchange has sign $(-1)^{2\cdot1}=1$, and the next equality simply renames $i,j$. Thus
$$
\boxed{\alpha=\operatorname{tr}\Theta\text{ is frame-independent and }d\alpha=0.}
$$
One can also check closedness locally: the off-diagonal terms in $\operatorname{tr}(\theta\wedge\theta)=\sum_{i,j}\theta_{ij}\wedge\theta_{ji}$ cancel in pairs, while the diagonal terms vanish. Hence $\alpha=d(\operatorname{tr}\theta)$ in each frame, and $d^2=0$ gives the same conclusion.
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