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Suppose ${A_i}\in {\mathbb{C}^{n \times n}},(i = 0,1,2....m)$ and ${\rm{P(}}x {\rm{) = }}{{\rm{A}}_m}{x ^m} + .....{A_1}x + {A_0}$ is a matrix polynomial.

Is this true that the roots of $det(P(x))=0$ are Continuous?

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  • can you clarify the question? $det(P(x))$ is just some polynomial of degree $\leq mn^2$, so as a function in $x$ it is continuous. What do you mean that its roots are continuous? – hunter Oct 03 '15 at 19:26
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1 Answers1

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$\det P(x)$ is a polynomial in $x$. Now, it is known that the roots of a polynomial vary continuously with the polynomial's coefficients (e.g. see here). Hence the roots of the polynomial equation $\det P(x)=0$ vary continuously with the entries of $A_0$ up to $A_m$.

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