We prove Broyden's theorem by induction on the dimension. We assume $| \alpha | < 1$. Using the induction hypothesis for $Q _ { + }$ (resp. for $Q _ { - }$), we denote by $x _ { + } > 0$ (resp. $x _ { - } > 0$) a vector of $\mathbb { R } ^ { n - 1 }$ and $S _ { + }$ (resp. $S _ { - }$) the sign diagonal matrix, such that $Q _ { + } x _ { + } = S _ { + } x _ { + }$, resp. $Q _ { - } x _ { - } = S _ { - } x _ { - }$. We set
\begin{itemize}
  \item $\eta _ { + } = - \frac { \left( x _ { + } \mid q \right) } { \alpha + 1 } , \quad \eta _ { - } = - \frac { \left( x _ { - } \mid q \right) } { \alpha - 1 }$
  \item $z _ { + } = \binom { x _ { + } } { \eta _ { + } } , \quad z _ { - } = \binom { x _ { - } } { \eta _ { - } }$
  \item $S ^ { + } = \left( \begin{array} { c c } S _ { + } & 0 \\ 0 & + 1 \end{array} \right) , \quad S ^ { - } = \left( \begin{array} { c c } S _ { - } & 0 \\ 0 & - 1 \end{array} \right)$
\end{itemize}
Show using question 1(a) that in the case where $S _ { + } \neq S _ { - }$ then one of the pairs $(z _ { + } , S ^ { + })$ or $(z _ { - } , S ^ { - })$ satisfies Broyden's theorem.