optimality of quadratic programming problems Suppose we have a general quadratic programming problem:
\begin{align}
\min_{x}\,\,&c^Tx+\frac{1}{2}x^TQx,\\
\mbox{s.t.}\,\,& Ax=b,\\
&x\geq0,
\end{align}
where $Q$ is positive semi-definite, $A$ has full column rank.
Suppose the problem is feasible. 
Is it possible to prove that: if the problem is lower bounded, then there must be a feasible solution $x^{\ast}$ which optimizes the problem?
 A: Let us consider the problem without the non-negativity constraint:
\begin{align}
\min_{x}\,\,&f(x):=c^Tx+\frac{1}{2}x^TQx,\\
\mbox{s.t.}\,\,& Ax=b
\end{align}
where $Q$ is positive semi-definite, $A$ has full column rank.
Any solution to the linear problem $Ax=b$ can be written as $x=x_p+Zy$, where $x_p$ is a particular solution to the problem (which we know exists since the QP is feasible by assumption), $Z$ is a base of $Ker(A)$ and $y$ a vector. Substituting $x$ into the objective function, we can transform the original constrained QP into an unconstrained QP in $y$. The equivalent QP thus reads:
\begin{align}
\min_{y}\,\,&f(x_p)+\frac{1}{2}y^TZ^TQZy+c^TZy+x_p^TQZy
\end{align}
The first order necessary condition for this problem is then:
\begin{equation}
(Z^TQ^TZ)y=-Z^T(Qx_p+c)
\end{equation}
At this point, there are two possibilities:


*

*Either $(Z^TQ^TZ)$ is positive definite and all is good, the above system has a unique solution $y^*$ given by:
\begin{equation}
y^*=-(Z^TQ^TZ)^{-1}Z^T(Qx_p+c)
\end{equation}
from which we retrieve a unique optimal $x^*$ as:
\begin{equation}
x^*=x_p+Zy^*
\end{equation}

*Or $(Z^TQ^TZ)$ is positive semi-definite and we must be more careful. Remember from linear algebra that a linear system $Px=q$ has solutions if and only if $q$ is orthogonal to $Ker(P^T)$. In our case, this implies that the system 
\begin{equation}
(Z^TQ^TZ)y=-Z^T(Qx_p+c)
\end{equation}
has solutions (and thus so does the QP) if and only if $Z^T(Qx_p+c)$ is orthogonal to $Ker(Z^TQ^TZ)$. Suppose this is not the case, then there exists a vector $y$ such that:
\begin{align}
y^T Z^T(Qx_p+c)&\neq 0\\
(Z^TQ^TZ)y&=0
\end{align}
which also holds for any vector $w=\alpha y$, where $\alpha\in\mathbb{R}$ ($\alpha\neq 0$).
Plugging this $w$ into the objective of the unconstrained QP yields an objective function of the form:
\begin{align}
f(x_p)+\frac{\alpha^2}{2}y^TZ^TQZy+\alpha c^TZy+\alpha x_p^TQZy&=f(x_p)+\alpha y^TZ^T(Qx_p+c)
\end{align}
and since $y^TZ^T(Qx_p+c)\neq 0$, we could let the value of the objective function fall arbitrarily low by letting $\alpha\to\infty$ or $\alpha\to-\infty$ (depending on the sign of $y^TZ^T(Qx_p+c)$ ), violating the boundedness assumption.
It follows that the system 
\begin{equation}
(Z^TQ^TZ)y=-Z^T(Qx_p+c)
\end{equation}
has solutions and so does the QP, with an optimal objective value of $f(x_p)$, where $x_p$ is any solution to $Ax=b$.


Going back to the initial problem with the non-negativity constraint, we see that the previous reasoning does not readily apply because by letting $\alpha\to\infty$ or $\alpha\to-\infty$, the resulting vector $x=x_p+\alpha Zy$ might violate the non-negativity constraint. There is thus a little more work to be done, but I believe this is the right path...
A: The problem is in my opinion the lag of positivity in Q.
Usually You would start by stating that if the problem is bounded from below we can find a minimizing sequence $(x_{n})_{n\in N}$, such that
$$\forall n: x_{n}\hbox{ is feasible, that is: } x_{n}\geq0\wedge Ax_{n}=b \ \forall n\in N$$
Now the question of will the $x_{n}$ converge, has to be answered:
Assuming we have this minimizing sequence we know that there is $C>0$ such that
$$
c^{T}x_{n}+\tfrac{1}{2}x_{n}^{T}Qx_{n}\leq C\ \forall n\in N.
$$
If we could now argue that $x_{n}$ is uniformly bounded (for instance if $Q$ would be positiv definite, we had the postulated property, also if the kernel of $Q$ would be compensated by the equality constraint), we could use Bolzano Weierstrass, to conclude that there is a subsequnce converging to $x_{\infty}$). Then showing that the limit will also be feasible You are almost done. Nevertheless this is not possible here...but maybe it is helpful?
