Understanding Hyperbolic Householder Transformations Note 1: This post is a continuation of a previous post on Householder transformations. I'm using this post to document my understanding. Please provide your valuable comments. 
Note 2: All vectors in the following discussion are row vectors contrary to the popular convention. This choice is due to other factors based on my reading of this book.
Let $\textbf{z}$ denote an arbitrary row vector in $n$-dimensional Euclidean space. Suppose that this vector corresponds to a point on a hyperbola. We wish to transform $\textbf{z}$ to the form $\alpha\textbf{e}_0$ such that the point $\alpha\textbf{e}_0$ is also on the same hyperbola as $\textbf{z}$. Let the Hyperbolic Householder matrix, $\Theta$ be the desired transformation matrix which is $J$-orthogonal and involutary. In other words, 
$\begin{eqnarray}
\Theta\,J\,\Theta^T = \Theta^T\,J\,\Theta = J\;\;\;\;\textrm{where, }\;J=I_p\oplus -I_q,\;\;\;p\geq 1,\;\;\;q\geq 1.
\end{eqnarray}$
Here, $\oplus$ is the direct sum operator. Observe that, $JJ^T=\textbf{I}$. Hence, the goal is to find a $\Theta$ such that 
$\begin{eqnarray}
\textbf{z}\,\Theta=\alpha\textbf{e}_0.
\end{eqnarray}$
How would you go about finding $\Theta$? 
Aside: This transformation from $\textbf{z}$ to $\alpha\textbf{e}_0$ is not always possible. Based on the previous equation, we notice that
$\begin{eqnarray}
\textbf{z}\,\Theta\,J\,\Theta^T\textbf{z}^T=\alpha\textbf{e}_0\,J\,\textbf{e}_0^T\alpha=\alpha^2\;\;\;\textit{i.e.},\;\;\;\textbf{z}\,J\,\textbf{z}^T=\alpha^2.
\end{eqnarray}$
Since, $\alpha^2$ is positive, we require that $\textbf{z}\,J\,\textbf{z}^T>0$. If this condition is not met, then we should look for a mapping from $\textbf{z}$ to $\alpha\textbf{e}_{n-1}$ such that $\textbf{z}\,J\,\textbf{z}^T=-\alpha^2$. To keep things simple, let's assume that $\textbf{z}\,J\,\textbf{z}^T>0$.  
 A: The row vectors, $\textbf{z}$ and $\alpha\textbf{e}_0$ are depicted in Figure 1 below. Suppose now that the difference $\textbf{z}-\sqrt{\textbf{z}\,J\,\textbf{z}^T}\textbf{e}_0$ is $\textbf{g}$, i.e., $\textbf{g}:=\textbf{z}-\sqrt{\textbf{z}\,J\,\textbf{z}^T}\textbf{e}_0$. This vector is shown in yellow in Figure 1.

Notice that $\textbf{z}$ and $\sqrt{\textbf{z}\,J\,\textbf{z}^T}\textbf{e}_0$ are not going to be equal in length unless $\textbf{z}$ happens to lie in the same direction as $\textbf{e}_0$, which is the trivial case. Following the insight gained based on the procedure for deriving the Householder transformation matrix, let us form the vector, $(\textbf{z}-0.5\textbf{g})$. This is depicted in Figure 2 below. 

The vector $(\textbf{z}-0.5\textbf{g})$ is not orthogonal to $\textbf{g}$ as in the derivation of the Householder transformation matrix. However, let's rotate $(\textbf{z}-0.5\textbf{g})$ w.r.t $J$ and see what happens. The result is shown in Figure 3 below. 

The vectors $(\textbf{z}-0.5\textbf{g})$ and $(\textbf{z}-0.5\textbf{g})\,J$ are equal in length and are points on a circle. To see this, notice that 
$\begin{eqnarray}
||\textbf{z}-0.5\textbf{g}||^2 = (\textbf{z}-0.5\textbf{g})JJ^T(\textbf{z}-0.5\textbf{g})^T = ||(\textbf{z}-0.5\textbf{g})J||^2.
\end{eqnarray}$
Consequently, the vectors $(\textbf{z}-0.5\textbf{g})$ and $(\textbf{z}-0.5\textbf{g})\,J$ are two equal sides of an isosceles triangle with $\beta\textbf{e}_0$ being the altitude from origin to the third side of this triangle, for some $\beta$. In fact, the point $\sqrt{\textbf{z}\,J\,\textbf{z}^T}\textbf{e}_0$ appears to be the orthocenter of this triangle (I think that this point needs a stronger argument). 
Utilizing the fact that an altitude in a triangle is perpendicular to the side opposite to the vertex from which it is drawn, we see that the altitude from the point $(\textbf{z}-0.5\textbf{g})$, which is in the same direction as $\textbf{g}$ is perpendicular to $(\textbf{z}-0.5\textbf{g})\,J$. See Figure 4 below. 

Therefore, 
$\begin{eqnarray}
&\phantom{\Rightarrow}&(\textbf{z}-0.5\textbf{g})J\textbf{g}^T=0 \\
&\Rightarrow& \textbf{z}J\textbf{g}^T-0.5\textbf{g}J\textbf{g}^T=0 \\
&\Rightarrow& 0.5 =(\textbf{g}J\textbf{g}^T)^{-1}\textbf{z}J\textbf{g}^T.
\end{eqnarray}$
From Figure 1, we observe that 
$\begin{eqnarray}
\sqrt{\textbf{z}\,J\,\textbf{z}^T}\textbf{e}_0 &=& \textbf{z}-\textbf{g}\\
&=& \textbf{z}-2\times 0.5\textbf{g}\\
&=& \textbf{z}-2(\textbf{g}J\textbf{g}^T)^{-1}\textbf{z}J\textbf{g}^T\textbf{g} \;\;\;\;\;\textrm{    (where, we replaced 0.5 with } (\textbf{g}J\textbf{g}^T)^{-1}\textbf{z}J\textbf{g}^T \textrm{    from before)}\\
&=& \textbf{z}\;(\,\textbf{I}-2(\textbf{g}J\textbf{g}^T)^{-1}J\textbf{g}^T\textbf{g}).
\end{eqnarray}$
From the last equation, we can read off $\Theta=(\,\textbf{I}-2(\textbf{g}J\textbf{g}^T)^{-1}J\textbf{g}^T\textbf{g})$ as the required $J$-orthogonal, involutary matrix that transforms $\textbf{z}$ to the form $\sqrt{\textbf{z}\,J\,\textbf{z}^T}\textbf{e}_0$.
Numerical Example: Let $\textbf{z}=[\sqrt{2}\;\;\;1]$ be a point on the hyperbola, $z_1^2-z_2^2=1$. It is easy to verify that 
$J = \begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}.$
The goal is to find a Hyperbolic Householder matrix that transforms $\textbf{z}$ to the point $[1\;\;\;0]$. For this example, $\sqrt{\textbf{z}\,J\,\textbf{z}^T}=1$. 
Now, 
$\begin{eqnarray}
\textbf{g}=\textbf{z}-\sqrt{\textbf{z}\,J\,\textbf{z}^T}\textbf{e}_0 = [\sqrt{2}-1\;\;\;1].
\end{eqnarray}$
Let's take a detour. We can verify for this example that, 
$\begin{eqnarray}
\textbf{z}-0.5\textbf{g}=[\sqrt{2}\;\;\;1]-0.5 [\sqrt{2}-1\;\;\;1]=[0.5\sqrt{2}+0.5\;\;\;0.5]
\end{eqnarray}$
and as a consequence, 
$\begin{eqnarray}
(\textbf{z}-0.5\textbf{g})J=[0.5\sqrt{2}+0.5\;\;\;0.5]\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix} = [0.5\sqrt{2}+0.5\;\;\;-0.5]
\end{eqnarray}$. 
Finally,
$\begin{eqnarray}
(\textbf{z}-0.5\textbf{g})J\textbf{g}^T=[0.5\sqrt{2}+0.5\;\;\;-0.5]\begin{bmatrix}
\sqrt{2}-1 \\
1
\end{bmatrix} = 0.5((\sqrt{2})^2-1)-0.5 = 0
\end{eqnarray}$. 
The vectors in this example can be visualized as shown in Figure 5 below. 

Back to our original goal of finding the Hyperbolic Householder transformation matrix. By substituting $\textbf{g} = [\sqrt{2}-1\;\;\;1]$ in $\Theta=(\,\textbf{I}-2(\textbf{g}J\textbf{g}^T)^{-1}J\textbf{g}^T\textbf{g})$, we see that 
$\Theta = \begin{bmatrix}
\sqrt{2}& 1 \\
-1 & -\sqrt{2}
\end{bmatrix}.$
We can verify that $\textbf{z}\Theta = [1\;\;\;0]$. 
