The following question is question no. 5.27 on p. 282 of Boyd & Vendenberghe's Convex Optimization (Cambridge University Press, 7th Printing, 2009):
Consider the equality contrained least-squres problem $$ \begin{align*} \text{minimize}\quad&\|Ax-b\|_2^2\\ \text{subject to}\quad& Gx=h \end{align*} $$ where $A \in \mathbb{R}^{m\times n}$ with $\text{rank}(A) = n$, and $G \in \mathbb{R}^{p\times n}$ with $\text{rank} G = p$.
Give the KKT conditions, and derive expressions for the primal solution $x^*$ and the dual solution $\nu^*$.
The following is the beginning of a solution to this problem that I found on the Internet:
The Lagrangian is $$ \begin{align*} L(x,\nu) &= \|Ax-b\|_2^2 + \nu^T(Gx-h)\\ &= x^TA^TAx + (G^T\nu - 2A^Tb)^Tx - \nu^Th, \end{align*} $$ with minimizer $x = -(1/2)(A^TA)^{-1}(G^T\nu-2A^Tb)$. The dual function is $$ g(\nu) = -(1/4)(G^T\nu-2A^Tb)^T(A^TA)^{-1}(G^T\nu-2A^Tb)-\nu^Th $$
I don't understand why $x$ is $L$'s minimizer. If $x$ were the global minimizer of $L$ then, yes, I would understand why it must be the expression above. However, when we calculate the dual function $g$, we are supposed to minimize only over the feasible set of the primal problem (see excerpt below from p. 216 of the textbook), i.e. in this case over the set $\mathcal{D} = \{x :\ Gx=h\}$.
