Consider $f(x) = \frac{1}{2}x^{T}Qx - c^{T}x$.
Under what conditions on $Q$ does $f$ have a stationary point, but no local maxima or minima?
I need help refining my thoughts here. I don't think I am quite on the right path.
Note that a stationary point exists when $Qx = c^{T}$ and $x = Q^{-1}c^{T}$ is a stationary point when $Q$ is invertible.
Suppose that $Q$ is not invertible. Then det $Q = 0$ and so $Q$ cannot be positive definite. Thus, if there is a stationary point, it will not be a local maxima or minima.
May someone share their thoughts? I feel that I am missing something.
The gradient of $f$ is $Qx-c$, written as a column ($x$ and $c$ are column vectors). So the necessary and sufficient condition for the existence of a stationary point is $c\in \operatorname{ran}Q$. This is always fulfilled if $Q$ is invertible, but it may be fulfilled also for some $(Q,c)$ pairs with noninvertible $Q$.
The Hessian of $f$ is $Q$. Since the Hessian is constant (all higher order derivatives are zero), the second derivative test tells the whole story: the critical point is a local minimum if $Q$ is positive semidefinite, a local maximum if $Q$ is negative semidefinite, and a saddle point if $Q$ is indefinite.
To summarize, the conditions are: $Q$ is indefinite and its range contains $c$.