Recently, I am reading a paper about eigenvalue problems. Consider the following problem, which occurs at the first page of the paper.
\begin{align} \text{minimize}\quad &x^TAx \\ \text{subject to}\quad & x^Tx = 1 \\ & c^Tx = 0 \tag{$\bigstar$} \end{align} where $A$ is a real symmetric matrix and in addition, $c^Tc = 1$. Its corresponding Lagrangian can be written as $$ L(x, \lambda, \mu) = x^TAx - \lambda\cdot(x^Tx - 1) + 2\mu \cdot c^Tx $$ Letting its first derivative equal to $0$, we have $$ Ax - \lambda\cdot x + \mu \cdot c = 0 $$ Multiplying the LHS of the equation by $c^T$ and using the fact $c^Tc = 1$, I obtain $$ \mu = -c^TAx + \color{red}{\lambda\cdot c^Tx} \tag{$\spadesuit$} $$ But as shown in the second page of the paper, the paper obtains $$ \mu = -c^TAx \tag{$\clubsuit$} $$ If I understand correctly, it uses the constraint $c^Tx = 0$ ($\bigstar$). Why can the constraint be used here when minimizing the Lagrangian? Should not Lagrangian functions be minimized over the domain of $x$ rather than the feasible region to obtain the dual?
The $x$ in the Lagrangian can be any value. However the "subject to" constraints tell us that the domain over which you are to minimize $L$ are only those that satisfy $c^Tx = 0$ and $x^Tx = 1$, i.e. all $x$ in the intersection of the unit $n$-sphere and the plane $c^Tx = 0$. After you have derived the Lagrangian equations
$$\frac{\partial L}{\partial \mu} = 2c^Tx = 0,~~~\frac{\partial L}{\partial \lambda} = 1 - x^Tx = 0,~~~\frac{\partial L}{\partial x} = 2(Ax - \lambda x + \mu c) = 0$$
you are now free to use $c^Tx = 0$ (the first equation) to simplify the other equations as you are looking for a solution that satisfy all of these equations at the same time. The only thing you cannot do is to apply them in the Lagrangian itself before deriving the Lagrangian equations.