Transform a sphere to an ellipse using a matrix In Lay's Linear Algebra and Its Applications textbook, he has this example:

I'd like to know how to use this matrix to transform the equation of the sphere to the equation of the ellipse. Any suggestions?
Update: Just to make my question more clear, I can do the problem in the example, but that is not my question posted here. I'd like to know how to start with the equation $x^2+y^2+z^2=1$ and use the matrix $A$ to discover the equation of the ellipse.
 A: If you are familiar with Singular Value Decomposition already, here is how you can approach this problem. If not, you may want to skip to the solution and re-visit once you've done SVD.
$$ A = \begin{bmatrix} 4 && 11 && 14 \\ 8 &&7 && -2\end{bmatrix}$$
can be decomposed as $$A = U\Sigma V^*$$ where 
$$ \Sigma = \begin{bmatrix} 6\sqrt10 && 0 && 0 \\0&& 3\sqrt10&& 0 \end{bmatrix}$$
$$ V^* = \frac{1}{3}\begin{bmatrix} 1 && -2 && -2 \\ 2 && -1 && 2 \\2 &&2 &&-1 \end{bmatrix}$$
($U$ is a bit harder to calculate)
Here is the geometrical interpretation: $V^*$ is a co-ordinate transformation matrix (rotation and reflections), while $\Sigma$ is a scaling matrix. This gives us a clue to work in an alternate co-ordinate system that appears in in $V^*$.
A further clue is that the singular values of this matrix are $6\sqrt10$ and $3\sqrt10$. These are the amounts of 'scalings' our unit vectors are going to receive.
Solution. We choose to work in the co-ordinate system given by the orthonormal unit vectors: 
$$ e_1 = (\frac{1}{3}, \frac{2}{3}, \frac{2}{3}) $$
$$ e_2 = (\frac{-2}{3}, \frac{-1}{3}, \frac{2}{3}) $$
$$ e_3 = (\frac{-2}{3}, \frac{2}{3}, \frac{-1}{3}) $$
We will continue to represent these vectors and the matrix $A$ in terms of the original standard basis $ (1,0,0), (0,1,0), (0,0,1)$.
We get
$$ Ae_1 = (18,6) $$
$$ Ae_2 = (3,-9) $$
$$ Ae_3 = (0,0) $$
Indeed these are the points shown in your figure.
A general point given by the vector $ xe_1+ye_2+ze_3$ is mapped to $$ (18x+3y, 6x-9y)$$
Now coming to the unit sphere, pick any circle on the sphere parallel to the $e_1 - e_2$ plane. Parameterize it as $(r\cos \theta, r\sin\theta, z)$. We have:
$$ (r\cos \theta, r\sin\theta, z) \mapsto (18r\cos \theta+3r\sin\theta, 6r\cos\theta-9r\sin\theta) $$
$$ = (6\sqrt10 r\cos(\alpha-\theta),3\sqrt10 r\sin(\alpha-\theta))$$
(after a bit of rearrangement).
This is the equation of the ellipse. $r=1$ gives the outermost ellipse. Thus the unit sphere maps to an elliptical disk. For example, the point $(0,0,1)$ maps to $(0,0)$.
A: Observe that $\|Ax\|^2=(Ax)^T(Ax)=x^T(A^TA)x$. Now $A^TA$ is a simmetric, positive definite three-by-three matrix. Therefore it is diagonalizable. Let $D$ be its diagonal form. In that case the maximum of $\|Dx'\|$ for $\|x'\|=1$ correspond to the maximal diagonal element of $D$. This is quite easy to show. But the maximal diagonal element of $D$ is the maximal eigenvalue of $A^TA$. So the answer to your question is the maximal eigenvalue of $A^TA$. 
