Understanding Hyperbolic Householder Transformations

Question

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 $\textbf{z}\,J\,\textbf{z}^T>0$.

Answer 1

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 $\textbf{z}\Theta = [1\;\;\;0]$.