When Does `min x'Qx + c'x` Have a Finite Minimum?

Citadel GQS Quant Researcher first round: after 30 minutes of resume discussion, a single technical question on the conditions under which an unconstrained quadratic program over `R^n` attains a finite minimum. Requirements Given a real-valued problem: minimize x^T Q x + c^T x x ∈ R^n with real symmetric Q and real c, characterize the conditions on Q and c under which the minimum value is finite. Notes Case 1 — Q positive definite: the problem is strictly convex; minimum exists, is unique, and equals -(1/4) c^T Q^{-1} c attained at x* = -(1/2) Q^{-1} c. Always finite. Case 2 — Q positive semi-definite (PSD but not PD): the problem is convex but not strictly convex. Minimum exists and is finite if and only if c lies in the column space of Q (equivalently, c ⊥ null(Q)). If c has any component in null(Q), walking along that direction drives the objective to -∞. When the condition holds, the minimum is attained on an affine subspace. Case 3 — Q has a negative eigenvalue: the objective is unbounded below. Walking along the corresponding eigenvector direction makes x^T Q x arbitrarily negative, dominating any finite c^T x term. No finite minimum regardless of c. Sanity check: if Q = 0, the objective reduces to c^T x, which is finite (and equal to 0) only if c = 0; otherwise unbounded below by walking along -c. A compact answer: the minimum is finite if and only if Q is PSD and c ∈ range(Q). The interviewer expects both halves of the characterization. Preparation Memorize the closed-form solution and the existence condition for PD vs PSD vs indefinite cases — this is canonical convex-optimization material and appears across quant interviews. Be able to sketch the proof in 90 seconds using the eigen-decomposition Q = V Λ V^T: in the rotated coordinates y = V^T x, the objective decouples into Σ λ_i y_i^2 + d_i y_i where d = V^T c. Each coordinate has a finite minimum iff λ_i > 0, or λ_i = 0 and d_i = 0, or λ_i < 0 (unbounded). State this decomposition fluently. Practice the geometric interpretation: PSD Q makes the objective a paraboloid (possibly degenerate); a negative eigenvalue introduces a saddle-with-downward-direction, and the c term cannot stabilize it. For the follow-up if asked about a constrained version (Ax = b linear constraint), the KKT conditions give a linear system; the constrained problem can have a finite minimum even when Q is not PSD overall, provided Q is PSD on null(A).