 Solution

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.