Extremize $\| \mathrm A \mathrm x \|_2$ subject to $\| \mathrm B \mathrm x \|_2 = 1$ Problem:

Suppose $A, B \in \mathbb{R}^{n \times n}$.  Formulate a condition for vectors $\vec{x} \in \mathbb{R}^n$
  to be critical points of $\|A\vec{x}\|_2$ subject to $\|B\vec{x}\|_2 = 1$.  Also, give an alternative expression for the value of $\|A\vec{x}\|_2$ at these critical points, in terms of a Lagrange multiplier for this optimization problem.

Thoughts:
Set up Lagrange:
$$\Lambda(\vec{x},\lambda) = \|A\vec{x}\|_2 - \lambda(\|B\vec{x}\|_2 -1)$$
Take the partial of $x_i$
$$\frac{\delta\Lambda}{\delta x_i} \Lambda= \frac{\sum_jA_{ji}x_i}{\|A\vec{x}\|_2} - \lambda\frac{\sum_jB_{ji}x_i}{\|B\vec{x}\|_2}$$
Assuming I did the differentiation correctly, which I'm not sure about.
Also:
$$\frac{\delta\Lambda}{\delta \lambda} \Lambda \implies \|B\vec{x}\|_2 = 1$$
But I'm not sure what to do at this point.
 A: We have
$$\begin{array}{ll} \text{extremize} & \| \mathrm A \mathrm x \|_2\\ \text{subject to} & \| \mathrm B \mathrm x \|_2 = 1\end{array}$$
Let $\mathrm y := \mathrm B \mathrm x$. Assuming that $\mathrm B$ is invertible, then we have an optimization problem in $\mathrm y \in \mathbb R^n$
$$\begin{array}{ll} \text{extremize} & \| \mathrm A \mathrm B^{-1} \mathrm y \|_2\\ \text{subject to} & \| \mathrm y \|_2 = 1\end{array}$$
The maximum is
$$\| \mathrm A \mathrm B^{-1} \mathrm y \|_2 \leq \| \mathrm A \mathrm B^{-1} \|_2 \underbrace{\| \mathrm y \|_2}_{=1} = \| \mathrm A \mathrm B^{-1} \|_2 = \sqrt{\lambda_{\max} (\mathrm B^{-T} \mathrm A^T \mathrm A \mathrm B^{-1})}$$
The maximizer, $\mathrm y_{\max}$, is found at the intersection of the eigenspace associated with the maximum eigenvalue of $\mathrm B^{-T} \mathrm A^T \mathrm A \mathrm B^{-1}$ with the unit Euclidean sphere. Note that $\mathrm x_{\max} = \mathrm B^{-1} \mathrm y_{\max}$. 
The minimum and the minimizer are now easy to find.
A: Squaring the $2$-norms, we obtain the quadratically constrained quadratic program (QCQP)
$$\begin{array}{ll} \text{extremize} & \| \mathrm A \mathrm x \|_2^2\\ \text{subject to} & \| \mathrm B \mathrm x \|_2^2 = 1\end{array}$$
We define the Lagrangian as follows
$$\mathcal L (\mathrm x, \lambda) := \| \mathrm A \mathrm x \|_2^2 - \lambda ( \| \mathrm B \mathrm x \|_2^2 - 1 ) = \mathrm x^T \mathrm A^T \mathrm A \mathrm x - \lambda ( \mathrm x^T \mathrm B^T \mathrm B \mathrm x - 1)$$
Taking the partial derivatives and finding where they vanish, we obtain
$$(\mathrm A^T \mathrm A - \lambda \, \mathrm B^T \mathrm B) \, \mathrm x = \mathrm 0_n \qquad \qquad \qquad \mathrm x^T \mathrm B^T \mathrm B \mathrm x = 1$$
Note that the former is a homogeneous linear system. As we are interested in non-zero solutions,
$$\det (\mathrm A^T \mathrm A - \lambda \, \mathrm B^T \mathrm B) = 0$$
Hence, we find the spectrum of the linear matrix pencil $\mathrm A^T \mathrm A - \lambda \, \mathrm B^T \mathrm B$, which gives us at most $n$ critical points on the ellipsoid $\mathrm x^T \mathrm B^T \mathrm B \mathrm x = 1$.
