Characteristic Polynomial not needing an Ordered Basis of Eigenvectors

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Linear Algebra 4$_{th}$ ed. defines the characteristic polynomial for a linear operator as follows:

Let T be a linear operator on a n-dimensional vector space V with order basis, $\beta$. We define the characteristic polynomial, $f(t)$ of T to the characteristic polynomial of $A=[T]_\beta$. That is, $f(t)=|A-tI_n|$

Does $\beta$ need to be a basis composed of eigenvectors? If not, why not? When you need a regular ordered basis of V vs. an ordered basis of eigenvectors.

Thank you!