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 $V$?

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 $P$.
