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For a qubit control and finite-dimensional target with unitary operator V, the controlled unitary gate $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$ is entangling if and only if V is not scalar. If $V=e^{i\theta }I$, the controlled operation is a phase gate on the control tensored with the target identity and preserves every product state.

Otherwise the spectral theorem supplies two orthonormal eigenvectors with distinct eigenvalues ${\lambda }_0,{\lambda }_1$. Set $|b⟩=(|v_0⟩+|v_1⟩)/\sqrt{2}$ and use product input $|+⟩|b⟩$. The two conditional target vectors $|b⟩$ and $V|b⟩$ are linearly independent because their eigenbasis coefficients would otherwise require ${\lambda }_0={\lambda }_1$. The conditional-vector product-state criterion shows that their coherent controlled superposition is entangled. Thus even a target gate arbitrarily close to scalar can be entangling, although the resulting entanglement need not be large.

By the spectral theorem, the nonscalar unitary V has orthonormal eigenvectors $|v_0⟩,|v_1⟩$ with distinct eigenvalues ${\lambda }_0\ne {\lambda }_1$. Choose $|b⟩=(|v_0⟩+|v_1⟩)/\sqrt{2}$. Then $V|b⟩=({\lambda }_0|v_0⟩+{\lambda }_1|v_1⟩)/\sqrt{2}$ is not proportional to $|b⟩$.

On the product input $|+⟩\otimes |b⟩$, the controlled unitary gate producesIts two conditional target vectors are linearly independent, so the conditional-vector product-state criterion proves that the output is entangled. Equivalently, in the control computational basis and target eigenbasis its coefficient matrix is $\frac{1}{2}\left(\begin{array}{cc}1& 1\\ {\lambda }_0& {\lambda }_1\end{array}\right)$, whose determinant is $({\lambda }_1-{\lambda }_0)/4\ne 0$. Thus U is an entangling gate. The excluded scalar case would only apply a phase gate to the control and could not entangle the factors; this is the controlled-unitary entangling criterion.

Write the product state as $|\Psi ⟩=|a⟩\otimes |b⟩$. The two partial inner products areBoth vectors therefore lie in the same one-dimensional vector subspace, possibly with one or both equal to zero. If $(c_1,c_2)\ne (0,0)$, choose ${\lambda }_1=c_2$ and ${\lambda }_2=c_1$; then ${\lambda }_1|{\psi }_1^{\prime }⟩={\lambda }_2|{\psi }_2^{\prime }⟩$, and the multipliers are not both zero. If both $c_j$ vanish, choose $({\lambda }_1,{\lambda }_2)=(1,0)$.

Thus the conditional vectors are linearly dependent, with a nontrivial pair of multipliers in every case. The printed existential condition does not explicitly exclude $(0,0)$, which would make it vacuous for arbitrary vectors; the argument establishes the stronger intended statement. These unnormalized vectors need not themselves be valid unit state vectors, and a zero vector corresponds to a measurement outcome of probability zero. This is the forward direction of the conditional-vector product-state criterion.