I am faced with a quadratic program like so:
$$ \min_x \,\,\, \frac{1}{2}x^TQx+c^Tx \\subject \,to \\x \ge 0 $$
$Q \in \mathbb{S}^{nxn}_{++}\, z,c \in \mathbb{R}^n\, A \in \mathbb{R}^{m x n} \, b \in \mathbb{R}^{m} $
and I want to find the Lagrangian dual for this problem. So, of course I use the Lagrangian function:
$L(x,z)= \frac{1}{2}x^TQx + (c^T - z^T)x$
Now, the textbook that I am using is asking me to show that the Lagrangian dual is $$\max_{y\le c} -\frac{1}{2}y^TQ^{-1}y$$.
I am not understanding how the linear term drops out of the objective function when we go from the Lagrangian function, $L(x,z)$, to the Lagrangian dual problem $-\frac{1}{2}y^TQ^{-1}y$.
Where is the term $c^Tx$ in the Lagrangian dual problem?
The dual function $g$ is defined by $$ g(z) = \inf_x \, L(x,z) = \inf_x \, \frac12 x^T Q x + (c-z)^T x. $$ We can minimize $L(x,z)$ with respect to $x$ by setting the gradient of $L$ with respect to $x$ equal to $0$, which yields \begin{align} & Qx^\star + c - z = 0 \\ \iff & x^\star = Q^{-1}(z - c). \end{align} We find that $$ g(z) = L(x^\star,z) = -\frac12 (z - c)^T Q^{-1}(z - c). $$ Thus, the dual problem is \begin{align} \text{maximize} &\quad -\frac12 (z-c)^T Q^{-1} (z-c) \\ \text{subject to} &\quad z \geq 0. \end{align} Perhaps you got this far on your own. Now just make a change of variables $y = c-z$ to obtain the version of the dual problem given by the textbook.