This is a simple question but I do not understand how to do it and I'm not sure if this is the right place to ask. Consider the optimization problem
\begin{align}
\text{minimize } &\quad x^T P x \\
\text{subject to} & \quad Ax-b \leq 0.
\end{align}
Why does the dual problem contain $P^{-1}$ ?
The dual function is $g(\lambda) = -\frac14 \lambda^T A P^{-1} A^T \lambda – b^T \lambda$.
Here's my work: The Lagrangian is
$$ L(x,\lambda) = x^TPx + \lambda^T(Ax -b)).
$$
The dual function is
$$
g(\lambda) = \inf_x L(x,\lambda) .
$$
Best Answer
I'll assume that $P$ is symmetric positive definite. The Lagrangian is
$$ L(x, \lambda) = x^T P x + \lambda^T (Ax - b). $$
We can minimize $L$ with respect to $x$ by setting the gradient of $L$ with respect to $x$ equal to $0$: $$ 2Px + A^T \lambda = 0 \implies x = -\frac12 P^{-1} A^T \lambda. $$
Plugging in for $x$, we find that the dual function is
\begin{align} g(\lambda) &= (-\frac12 P^{-1} A^T \lambda)^T P (-\frac12 P^{-1} A^T \lambda) + \lambda^T A (-\frac12 P^{-1} A^T \lambda) - \lambda^T b \\ &= \frac14 \lambda^T A P^{-1} P P^{-1} A^T \lambda - \frac12 \lambda^T A P^{-1} A^T \lambda - \lambda^T b \\ &= -\frac14 \lambda^T A P^{-1} A^T \lambda - \lambda^T b. \end{align}