Definitions:
${A_j},{\Delta _j} \in {\mathbb{C}^{n \times n}},(j = 0,1,2....m)$
${\rm{P(}}\lambda {\rm{) = }}{{\rm{A}}_m}{\lambda ^m} + .....{A_1}\lambda + {A_0}$ is a matrix polynomial, and $\lambda $ is a complex variable.$\lambda$ is eigenvalue of $P(\lambda )$ if $\det P(\lambda ) = 0$
${P_\Delta }(\lambda ) = ({A_m} + {\Delta _m}){\lambda ^m} + ....... + ({A_1} + {\Delta _1}){\lambda ^1} + ({A_0} + {\Delta _0})$
$\left\| {{\Delta _j}} \right\| \le \varepsilon ,(j = 0,1,2....m)$ for any subordinate matrix norm.
- ${w_j} \ge 0,(j = 0,1....,m)$
$A = \left\{ {{P_\Delta }(\lambda ):\left\| {{\Delta _j}} \right\| \le \varepsilon {w_j}} \right\}$
${\xi _\varepsilon } = \min \left\{ {\left| {\det ({A_m} + {\Delta _m})} \right|:\left\| {{\Delta _m}} \right\| \le \varepsilon {w_m}} \right\}$
Suppose that $P(\lambda )$ has $(nm)$ eigenvalue. we know $A$ is compact set.
Why does there is ${M_\varepsilon } > 0$ such that for any ${{P_\Delta }(\lambda ) \in A}$ and for any $\lambda \in \mathbb{C}$ with $\left| \lambda \right| > {M_\varepsilon }$ then we have ${\left| {\det {P_\Delta }(\lambda ) - \det ({A_m} + {\Delta _m}){\lambda ^{nm}}} \right| < {\xi _\varepsilon }\left| {{\lambda ^{nm}}} \right|}$?
