It follows from these equations that $$ \rho_{n}=\frac{\operatorname{det} F(w)}{d_{2}} \text { and } \mu_{n}=\frac{\operatorname{det} F(w)}{d_{1}} . $$ Hence $\rho_{n}=\rho_{n-1}\left(1-\varphi_{n} \psi_{n}\right)$ and $\mu_{n}=\mu_{n-1}\left(1-\varphi_{n} \psi_{n}\right)$. If we define (this is the same $w_{0}$ as in formula (3.4.15)) $$ w_{0}=-\sum_{j=2}^{n-1} f_{n j} \nu_{j}=\overline{-\sum_{j=2}^{n-1} f_{1 j} \mu_{j}} $$ then $\delta_{n}=w-w_{0}, \Delta_{n}=\overline{w-w_{0}}$ and $$ \left|w-u_{0}\right|^{2}=\varphi_{n} \psi_{n} \rho_{n-1} \mu_{n-1} . $$
$$ T=\left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right) $$ we take it as the representation of a transformation with respect to the standard basis $T=\operatorname{Rep} \varepsilon_{2}, \mathcal{\varepsilon}{2}(t)$ and we look for a basis $B=\left\langle\vec{\beta}{1}, \vec{\beta}{2}\right\rangle$ such that $$ \operatorname{Rep}{B, B}(t)=\left(\begin{array}{cc} \lambda_{1} & 0 \ 0 & \lambda_{2} \end{array}\right) $$ that is, such that $t\left(\vec{\beta}{1}\right)=\lambda{1} \vec{\beta}{1}$ and $t\left(\vec{\beta}{2}\right)=\lambda_{2} \vec{\beta}{2}$. $$ \left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right) \vec{\beta}{1}=\lambda_{1} \cdot \vec{\beta}{1} \quad\left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right) \vec{\beta}{2}=\lambda_{2} \cdot \vec{\beta}_{2} $$
证明 .
We are looking for scalars $x$ such that this equation $$ \left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right)\left(\begin{array}{l} b_{1} \ b_{2} \end{array}\right)=x \cdot\left(\begin{array}{l} b_{1} \ b_{2} \end{array}\right) $$ has solutions $b_{1}$ and $b_{2}$, which are not both zero. Rewrite that as a linear system. $$ (3-x) \cdot b_{1}+\begin{array}{r} 2 \cdot b_{2}=0 \ (1-x) \cdot b_{2}=0 \end{array} $$
