determinant is one of the eigenvalue of a 2x2 matrix, geometrical implication? I got this special 2x2 matrix in the middle of deriving some other stuff, looks it's a bit unique,
$$
\begin{bmatrix}
1 -x^2/(1+z) & xy/(1+z) \\
xy/(1+z) & 1 -y^2/(1+z) \\
\end{bmatrix}
$$
where, 
$$
x^2+y^2+z^2=1,   
-1<x,y,z<1
$$
It can be shown eigenvalues are constant 1 and z, respectively, while determinant is z, is there any special geometrical meaning of this matrix? 
 A: Since $(x,y,z)$ is a point on a sphere, let switch to spherical coordinates.
$(x,y,z):\begin{cases}
x=\cos(\theta)\sin(\phi)\\
y=\sin(\theta)\sin(\phi)\\
z=\cos(\phi)\end{cases}$
$1+z=1+\cos(\phi)=2\cos^2(\frac {\phi}2)$
Thus the matrix becomes 
$A=\begin{bmatrix} 
1-\dfrac{x^2}{1+z} & \dfrac{xy}{1+z}\\
\dfrac{xy}{1+z} & 1-\dfrac{x^2}{1+z}\\
\end{bmatrix}=\begin{bmatrix}
1-2\cos(\theta)^2\sin(\frac {\phi}2)^2 & 2\cos(\theta)\sin(\theta)\sin(\frac {\phi}2)^2\\
2\cos(\theta)\sin(\theta)\sin(\frac {\phi}2)^2 & 1-2\sin(\theta)^2\sin(\frac {\phi}2)^2\\
\end{bmatrix}$
Let's now consider the rotation matrix $R(\theta)$ and the dilatation matrix $D(z)$
$R(\theta)=\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}\qquad D(z)=\begin{bmatrix} z & 0 \\ 0 & 1\end{bmatrix}$
A bit of calculation and trigonometric formulas are involved but you can verify that 
$$\begin{align}A&=R^{-1}DR\\\\A&=R(-\theta)D(z)R(\theta)\end{align}$$
Although we have broken $A$ into its core elements, I still have hard time figuring out the general meaning of such a geometric transformation in $2$D-space.
