Exercise 1.10.1

Answers

$S - λM = [\begin{matrix} 5 - λ & 4 \\ 4 & 5 - 4 λ \end{matrix}]$ . Solving $\det (S - λM) = 0$ , we find the eigenvalues: $λ_{1} = \frac{25 + \sqrt{481}}{8}, λ_{2} = \frac{25 - \sqrt{481}}{8}$ .
$H = M^{- \frac{1}{2}} S M^{- \frac{1}{2}} = [\begin{matrix} 5 & 2 \\ 2 & \frac{5}{4} \end{matrix}]$ , Solving for eigenvalues we have the same eigenvalues as above: $λ_{1} = \frac{25 + \sqrt{481}}{8}, λ_{2} = \frac{25 - \sqrt{481}}{8}$ .
Solve for $(S - λM) x = 0$ , we find that $x_{1} = [\begin{matrix} \frac{15 + \sqrt{481}}{8} \\ 1 \end{matrix}]$ and $x_{2} = [\begin{matrix} \frac{15 - \sqrt{481}}{8} \\ 1 \end{matrix}]$ . It’s clear that $x^{T} x \neq 0$ . But $x_{1}^{T} M x_{2} = [\begin{matrix} \frac{15 + \sqrt{481}}{8} & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & 4 \end{matrix}] [\begin{matrix} \frac{15 - \sqrt{481}}{8} \\ 1 \end{matrix}] = 0$
Solve for $(H - λI) y = 0$ , we find that $y_{1} = [\begin{matrix} \frac{15 + \sqrt{481}}{16} \\ 1 \end{matrix}]$ and $y_{2} = [\begin{matrix} \frac{15 - \sqrt{481}}{16} \\ 1 \end{matrix}]$ . So $y_{1}^{T} y_{2} = 0$ .

niuers

2020-03-20 00:00