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 $V$?
 A: It means that for $V$ and $W$ $n$-dimensional vector spaces over a field $F$, and $A$ and $B$ $n$ by $n$ matrices with entries in the field $F$ there are ordered bases $\alpha = (\alpha_i)_{i \in \{1,\dots,n\}}$ and $\alpha' = (\alpha'_i)_{i \in \{1,\dots,n\}}$ of $V$ and and $\beta = (\beta_i)_{i \in \{1,\dots,n\}}$ and $\beta' = (\beta'_i)_{i \in \{1,\dots,n\}}$ of $W$, and a linear map $T : V \to W$ such that $A$ is the matrix $T_{\alpha}^{\beta}$ and $B$ is the matrix $T_{\alpha'}^{\beta'}$ where for ordered bases $q$ of $V$ and $r$ of $W$ for vectors $x \in V$ and $y \in W$ $x_q$ denotes the $q$-coordinates of $x$ and $y_r$ denotes the $r$-coordinates of $y$, and $T_{q}^{r}$ the matrix such that for all $x \in V$ $T_q^r x_q = (T(x))_r$. So $A$ and $B$ represent the same linear map, and so $A$ represents $B$.
To be more specific, let $P$ be a matrix such that $B = P^{-1} A P$, and $T$ a linear map and $\alpha$ and $\beta$ ordered bases so that $A = T_{\alpha}^{\beta}$. Then $B = T_{\alpha P}^{\beta P}$ where $\alpha P$ is the ordered basis with $i$th vector $\sum_j p_{j,i} \alpha_j$, $\beta P$ is the ordered basis with $i$th vector $\sum_j p_{j,i} \beta_j$, and $p_{i,j}$ is the $i,j$ entry of matrix $P$.
A: 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.
