特徵值的連續性

$\mathbf{f}(\mathbf{x})=\begin{bmatrix} f_1(\mathbf{x})\\ \vdots\\ f_m(\mathbf{x}) \end{bmatrix}$

$n\ge 1$，考慮

$p(t)=a_nt^n+a_{n-1}t^{n-1}+\cdots+a_1t+a_0$

$q(t)=b_nt^n+b_{n-1}t^{n-1}+\cdots+b_1t+b_0$

$\displaystyle \min_{\sigma}\max_{i=1,\ldots,n}\vert\lambda_i-\mu_{\sigma(i)}\vert<\epsilon$

$S^{-1}AS=\Lambda=\begin{bmatrix} \lambda_1&~&~\\ ~&\ddots&~\\ ~&~&\lambda_n \end{bmatrix}$

$\displaystyle \min_{\lambda_i}\vert\mu-\lambda_i\vert\le\kappa(S)\Vert E\Vert$

$\displaystyle \Vert S\Vert=\max_{\mathbf{x}\neq\mathbf{0}}\frac{\Vert S\mathbf{x}\Vert}{\Vert\mathbf{x}\Vert}$

$(\mu I-A)^{-1}(\mu I-B)= (\mu I-A)^{-1}(\mu I-A-E)=I-(\mu I-A)^{-1}E$

$1\le\Vert(\mu I-A)^{-1}E\Vert=\Vert S(\mu I-\Lambda)^{-1}S^{-1}E\Vert\le\Vert S\Vert\cdot\Vert (\mu I-\Lambda)^{-1}\Vert\cdot\Vert S^{-1}\Vert\cdot\Vert E\Vert$

$\displaystyle 1\le\kappa(S)\Vert E\Vert\max_{i}\vert\mu-\lambda_i\vert^{-1}=\kappa(S)\Vert E\Vert\left(\min_{i}\vert\mu-\lambda_i\vert\right)^{-1}$

[1] Roger A. Horn and Charles R. Johnson, Matrix Analysis, Cambridge University Press, 1985.