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?
Best Answer
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)