Characteristic Polynomial Independent From the Choice of Basis I want to prove that definition of characteristic polynomial of a linear operator on a finite-dimensional vector space $V$ is independent of the choice of basis for $V$.
Proof>>
Choose two different basis for V, $\beta, \beta '$.
Let $Q = [I_v]_{\beta'}^\beta$. Then, $[T]_{\beta'} = Q^{-1
}[T]_\beta Q  $ ,which means that $[T]_\beta$ and $[T]_{\beta'}$ are similar. 
Since similar matrices are having same characteristic polynomial, 
$det([T]_\beta-\lambda I$) = $det([T]_{\beta '}-\lambda I$).
How's this proof? 
 A: If $A$ is the matrix of linear transformation with respect to some basis, and if $B$ is the matrix of the same linear transformation with respect to some other basis then $B=P^{-1}AP$ for some invertible matrix $P$,
Now the characteristic polynomial with respect to B is $det(P^{-1}AP-\lambda I)$,
now using formulas of determinants prove that this equals $det(A-\lambda I)$ for all values of $\lambda$, hence your question is solved. 
Last part one may use that $det(XY)=det(YX)$
$det(A-\lambda I)$=$det(I(A-\lambda I))$=$det(PP^{-1}(A-\lambda I))$
=$det(P(A-\lambda I)P^{-1})$=$det(P^{-1}AP-\lambda I)$
(Proved)
A: I think it might be more intuitive this way:
Let $A \colon V \rightarrow V $ be a linear operator. Let $e = \{e_1,\ldots , e_n\}$ and $f = \{f_1,\ldots , f_n\}$ be bases for $V$. Let $[A]_e$ and $[A]_{f}$ be the matrices of the linear operator $A$ in the given respective bases. Then there exists an invertible matrix $C$ such that $[A]_f = C^{-1} [A]_e C$. 
Let $P_A (\lambda)$ represent the characteristic polynomial. Then 
$$P_{[A]_f }(\lambda) = \textrm{det } \left( {[A]_f \space - \lambda I}\right) = \textrm{det } \left( C^{-1} [A]_e C \space - \lambda I\right) = \textrm{det }\left( C^{-1} [A]_e C \space - \lambda C^{-1}C\right)$$
Now if we factor out $C$ then factor out $C^{-1}$ we get:
$$\textrm{det } \left( C^{-1} ([A]_e \space - \lambda I) C\right) = \textrm{det } C^{-1} \cdot P_{[A]_e }(\lambda) \cdot \textrm{det} \space C =  \textrm{det } C^{-1} \cdot \textrm{det} \space C \cdot P_{[A]_e }(\lambda) \\= \textrm{det } C^{-1}C\cdot \chi_{[A]_e }(\lambda) =  P_{[A]_e }(\lambda) $$
In other words, we have shown that $$P_{[A]_f }(\lambda) =  P_{[A]_e }(\lambda) $$
