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

Requirements

Consider the real-valued optimization problem:

 minimize x^T Q x + c^T x
     x ∈ R^n

where Q is real and symmetric and c is real. Describe the conditions on Q and c that ensure the objective has a finite minimum value.

Notes

• Case 1 — If Q is positive definite, the objective is strictly convex, so it has one unique minimizer with value -(1/4) c^T Q^{-1} c, attained at x* = -(1/2) Q^{-1} c. The value is finite for every c.
• Case 2 — If Q is positive semi-definite but not positive definite, the objective is convex without being strictly convex. A finite minimum exists exactly when c belongs to the column space of Q (equivalently, c ⊥ null(Q)). Any component of c in null(Q) permits movement in that direction that sends the objective to $-\infty$. When the condition is satisfied, the minimum is attained across an affine subspace.
• Case 3 — If Q possesses a negative eigenvalue, the objective is unbounded below. Moving along the associated eigenvector can make x^T Q x arbitrarily negative, overwhelming the finite linear term c^T x. Thus no finite minimum exists for any c.
• Sanity check: with Q = 0, the objective becomes c^T x, which is finite (and equals 0) exactly when c = 0; otherwise, moving in the -c direction makes it unbounded below.
• In compact form, the minimum is finite if and only if Q is PSD and c ∈ range(Q). Both parts of this characterization are expected.

Preparation

• Know the closed-form minimizer and the existence criterion for the PD, PSD, and indefinite cases — these are standard convex-optimization facts that recur in quantitative interviews.
• Be prepared to outline the proof within 90 seconds using the eigendecomposition $Q = V \Lambda V^T$: after rotating to y = V^T x, the objective separates as $\sum \lambda_i y_i^2 + d_i y_i$, with d = V^T c. Each coordinate has a finite minimum when $\lambda_i > 0$, when λ_i = 0 and d_i = 0, or when $\lambda_i < 0$ (unbounded). Explain this decomposition smoothly.
• Rehearse the geometric view: PSD Q produces a paraboloid that may be degenerate, whereas a negative eigenvalue creates a saddle with a downward direction that the c term cannot prevent.
• If asked about the constrained variant with a linear constraint (Ax = b), mention that the KKT conditions produce a linear system; the constrained problem may still have a finite minimum even if Q is not globally PSD, as long as Q is PSD on null(A).