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Suppose

  • ${A_j},\,{\Delta _j} \in {\mathbb C^{n \times n}},\quad\big(\,j = 0,\,1,\,2,\,\ldots,\,m\,\big)$
  • ${P_\Delta }\left(\lambda\right) = \left({A_m} + {\Delta _m}\right){\lambda ^m} + \, \cdots \, + \left({A_1} + {\Delta _1}\right){\lambda ^1} + \left({A_0} + {\Delta _0}\right)$ is a matrix polynomial, and $\lambda $ is a complex variable.

Can we say that

$$\lim_{\lambda \to \infty } \dfrac{{{{\left\lvert\, \lambda\, \right\rvert}^{ mn}}}}{{\left\lvert\, {\det \big({P_\Delta }\left(\lambda\right)\big) - \det \big({A_m} + {\Delta _m}\big){\lambda ^{mn}}} \,\right\rvert}} = \infty $$

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1 Answer 1

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Yes, because $$ \frac{\det(P_\Delta(\lambda))-\det(A_m+\Delta_n)\lambda^{mn}}{\lambda^{mn}} = \det(Q_\Delta(\lambda))-\det(A_m+\Delta_m), \tag{$*$} $$ where $$ Q_\Delta(\lambda) = \frac{P_\Delta(\lambda)}{\lambda^m} = (A_m+\Delta_m) + \sum_{j=0}^{m-1} (A_j+\Delta_j)\lambda^{j-m}; $$ since $\lim_{\lambda\to\infty} Q_\Delta(\lambda) = A_m+\Delta_m$ and $\det$ is continuous, the expression ($*$) tends to $0$.

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