这是一份ed.ac爱丁堡格大学MATH10082作业代写的成功案

$$
\eta_{2} \equiv\left||M|^{-1 / 2} F|M|^{-1 / 2} \mid\right|_{2}<1
$$
then
$$
|\sin \Theta|_{F} \leqslant|Z| \frac{\eta_{\mathrm{F}}}{\sqrt{1-\eta_{2}}} / \min {\hat{\lambda} \in A, \hat{i} \in i} \frac{|\lambda-\hat{\lambda}|}{\sqrt{\hat{\lambda} \hat{\lambda}}} \text {, } $$ where $\eta{\mathrm{F}} \equiv\left||M|^{-1 / 2} F|M|^{-1 / 2} \mid\right|_{\mathrm{F}}$, and $Z$ is $J$-unitary.$$
|\sin \Theta|_{F}=|S|_{F} \leqslant|W|_{F} / \min _{\lambda \in A, \hat{i} \in i} \frac{|\hat{\lambda}-\hat{\lambda}|}{\sqrt{|\lambda \hat{\lambda}|}} .
$$

MATH10082 COURSE NOTES ：

To derive the absolute bound, multiply $r=(A-\hat{\lambda} I) \hat{x}$ by $Y^{}$ and use $Y^{} A=Y^{} A$, $$ Y^{} \hat{x}=(A-\hat{\lambda} I)^{-1} Y^{} r . $$ With $P$ being the orthogonal projector onto $\mathscr{S}^{\perp}=\operatorname{range}(Y)$ one gets $$ P \hat{x}=\left(Y^{\dagger}\right)^{} Y^{} \hat{x}=\left(Y^{\dagger}\right)^{}(\Lambda-\hat{\lambda} I)^{-1} Y^{*} r .
$$
From $|\hat{x}|_{2}=1$ follows
$$
|\sin \Theta|_{2}=|P \hat{x}|_{2} \leqslant \kappa(Y)\left|(\Lambda-\hat{\lambda} I)^{-1}\right|_{2}|r|_{2} .
$$
To derive the relative bound, we will use the absolute bound. Multiply $\left(D_{1} A D_{2}\right) \hat{x}=\hat{\hat{x}}$ by $D_{1}^{-1}$ and set $z \equiv D_{2} \hat{x} /\left|D_{2} \hat{x}\right|$,
$$
A z=\hat{\lambda} D_{1}^{-1} D_{2}^{-1} z .
$$
The residual for $\hat{\lambda}$ and $z$ is
$$
f \equiv A z-\hat{\lambda} z=\hat{\lambda}\left(D_{1}^{-1} D_{2}^{-1}-I\right) z=\hat{\lambda}\left(D_{1}^{-1}-D_{2}\right) \hat{x} /\left|D_{2} \hat{x}\right|_{2} .
$$
Hence
$$
|f|_{2} \leqslant|\hat{\lambda}| \alpha_{2}, \quad|f|_{2} \leqslant|\hat{\lambda}| \alpha_{1} /\left|D_{2} \hat{x}\right|_{2} .
$$
The idea is to first apply the absolute bound to the residual $f$ and then make an adjustment from $z$ to $\hat{x}$. Since $f$ contains $\hat{\lambda}$ as a factor we will end up with a relative bound.