= Solution
Write the <product state> as $|\Psi\rangle=|a\rangle\otimes|b\rangle$. The two <partial inner products> are
$$
|\psi'_j\rangle=(\langle\psi_j|a\rangle)|b\rangle=c_j|b\rangle,\qquad j=1,2.
$$
Both <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 $\lambda_1|\psi'_1\rangle=\lambda_2|\psi'_2\rangle$, and the multipliers are not both zero. If both $c_j$ vanish, choose $(\lambda_1,\lambda_2)=(1,0)$.
Thus \b[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>.
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