Problem: Show that the definition of the characteristic polynomial of a liner operator on a finite-dimensional vector space V is independent of the choice of basis for V .
Let T be a liner operator on a finite-dimensional vector space V and let α and β be two ordered bases for V . Then there exists an invertible matrix Q such that $[T]_\alpha=Q^{-1}[T]_\beta Q$
Question: why does it claim that "Then there exists an invertible matrix Q such that $[T]_\alpha=Q^{-1}[T]_\beta Q$"?
where $[T]_α$ and $[T]_β$ are matrix representations of T with respect to the ordered bases α and β, respectively. Since $[T]_α$ and $[T]_β$ have the same characteristic polynomial. Hence the definition of the characteristic polynomial of a liner operator on a finite-dimensional vector space V is independent of the choice of basis for V.
Best Answer
I would start by noticing that $[\alpha]_{\mathcal{B}} = P[\alpha]_{\mathcal{B}'}$, where $P$ is the change of basis matrix from $\mathcal{B}$ to $\mathcal{B}'$. We do also have that $[T\alpha]_{\mathcal{B}} = [T]_{\mathcal{B}}[\alpha]_{\mathcal{B}}$, by definition. Besides that, the following relation holds $[T\alpha]_{\mathcal{B}} = P[T\alpha]_{\mathcal{B}'}$
Combining these relations, one has \begin{align*} [T\alpha]_{\mathcal{B}} = [T]_{\mathcal{B}}[\alpha]_{\mathcal{B}} = [T]_{\mathcal{B}}P[\alpha]_{\mathcal{B}'} = P[T\alpha]_{\mathcal{B}'} \Longrightarrow [T\alpha]_{\mathcal{B}'} = P^{-1}[T]_{\mathcal{B}}P[\alpha]_{\mathcal{B}'} \Longrightarrow [T]_{\mathcal{B}'} = P^{-1}[T]_{\mathcal{B}}P \end{align*} as desired.
Based on such results, we conclude the characteristic polynomial of a linear operator independs of the basis we choose to write its matrix representation. Indeed, one has \begin{align*} \det([T]_{\mathcal{B'}} - \lambda I) = \det(P^{-1}[T]_{\mathcal{B}}P - \lambda P^{-1}IP) = \det(P^{-1}([T]_{\mathcal{B}} - \lambda I)P) = \det([T]_{\mathcal{B}} - \lambda I) \end{align*} once we have that $\det(X^{-1}) = [\det(X)]^{-1}$. Hopefully it helps.