Similar matrices have same engenvalues $\implies$ we can define characteristic polynomial for any basis 
This is from Linear Algebra - Hoffman and Kunze. 
If $B$ is a matrix and $A$ is it's similar, what it means to say that $A$ represents $B$ in some ordered basis for $V$?

Given a linear operator $T $ on a finite dimensional vector space, we can find its matrix representation with respect to a chosen basis. But there is an infinite number of choices for a basis, so an infinite number of 'A's. To find the eigenvalyes, which of these do we choose? The good news is that it does not matter, the result (eigenvalues) will be the same as the above lemma shows. Hence the result in your question title follows.
