Generating $\hat{u}$ and $\tilde{u}$ from $\hat{s}$ and $\tilde{s}$

Having generated $\tilde{s}$ and $\hat{s}$ in the ways that have been described, the algorithm sets $s$ to a linear combination of these vectors, but the choice is not restricted to $\pm \hat{s}$ or $\pm \tilde{s}$ as suggested in the introduction of this chapter(unless $\tilde{s}$ and $\hat{s}$ are nearly or exactly parallel). Instead, the vectors $\hat{u}$ and $\tilde{u}$ of unit length are found in the span of $\tilde{s}$ and $\hat{s}$ that satisfy the condition $\hat{u}^T \tilde{u} = 0$ and $\hat{u}^T H \tilde{u} = 0$. The final $s$ will be a combination of $\hat{u}$ and $\tilde{u}$.
If we set:

$$\begin{cases}G = \tilde{s} & \\ V = \hat{s} & \end{cases}$$

We have

$$\begin{cases}\tilde{u}= \cos(\theta) G+ \sin(\theta) V \\ \... ...es}\quad \quad \text{ we have directly }\hat{u}^T \tilde{u}= 0$$ (5.8)

We will now find $\theta$ such that $\hat{u}^T H \tilde{u} = 0$:

$$\hat{u}^T H \tilde{u} = 0$$

$$\Leftrightarrow (-\sin(\theta) G+ \cos(\theta) V)^T H (\cos(\theta) G+ \sin(\theta) V) =0$$

$$\Leftrightarrow (\cos^2(\theta)-\sin^2(\theta)) V^T H G + (G^T H G - V^T H V) \sin(\theta) \cos(\theta) =0$$

Using

$$\begin{cases}\sin(2 \theta) = 2\sin(\theta) \cos(\theta) \\ \co... ...in^2(\theta) \\ tg (\theta) = \frac{\sin(\theta)}{\cos(\theta)} \end{cases}$$

We obtain

$$(V^T H G) \cos(2 \theta) + \frac{G^T H G- V^T H V}{2} \sin(2 \theta)=0$$

$$\Leftrightarrow \theta = \frac{1}{2} arctg (\frac{2 V^T H G}{V^T H V - G^T H G})$$ (5.9)

Using the value of $\theta$ from Equation 5.10 in Equation 5.9 give the required $\hat{u}$ and $\tilde{u}$.