# characteristic polynomial independent of the choice of basis proof question

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.

• The matrix $Q$ is just the change-of-basis matrix. If $\alpha$ is the standard basis, then the columns of $Q$ are just the basis vectors of $\beta$.
– Nick
Commented Feb 25, 2020 at 3:28

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.