Prove that $\det(I_n-TAT^{-1})=\det(I_n-A)$

Question

Prove that $\det(I_n-TAT^{-1})=\det(I_n-A)$ for any matrix $T\in\mathbb R_{n\times n}$ with $\det(T)\ne0$.

Answer 1

Let $T,A \in \operatorname{GL}_n$.

We know that $\operatorname{det}(AB)=\operatorname{det}(A)\operatorname{det}(B)$. Therefore:

$\operatorname{det}(I_n - TAT^{-1}) = \operatorname{det}(TT^{-1} - TAT^{-1}) = \operatorname{det}(T)\operatorname{det}(I_n - A)\operatorname{det}(T^ {-1}) \\ = \operatorname{det}(TT^{-1})\operatorname{det}(I_n - A)$

which delivers the desired result.