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For second-qubit inputs $|0⟩$ and $|1⟩$, c gives control expectation values $\mathrm{Re}{U}_{00}$ and $\mathrm{Re}{U}_{11}$. Preparing the maximally mixed state $I/2$ is equivalent to choosing these inputs independently with equal probability and discarding the choice. Linearity of the density operator evolution and quantum measurement probabilities therefore givesHenceThe factor two is essential: the Hadamard test directly estimates the normalized trace. More generally, a target density operator $\rho$ gives $⟨Z⟩=\mathrm{Re}\mathrm{Tr}(\rho U)$. This argument never requires knowledge of the eigenvectors of $U$.

Yes. Replace the target qubit by a register of $k$ qubits and implement the controlled unitary gateThe algebra in c uses only normalization and unitarity, so for an arbitrary pure register quantum state $|\varphi ⟩$ it still gives $⟨Z⟩=\mathrm{Re}⟨\varphi |U|\varphi ⟩$. For a mixed register quantum state $\rho$, linearity gives $\mathrm{Re}\mathrm{Tr}(\rho U)$. With $\rho =I/2^k$,Only the control qubit needs to be pure. A uniformly randomized computational basis preparation supplies the target maximally mixed state. This is the principle of trace estimation by a maximally mixed Hadamard test. The extension is algebraic; efficiently implementing controlled-$U$ still requires a known suitable gate construction. Moreover, estimating the unnormalized trace to fixed additive precision requires increasingly precise estimation of the normalized signal as $k$ grows. There is no automatic efficient reconstruction of all entries of $U$.