Let $T \in GL(V) \subset End(V)$, where $V$ is finitely dimensional vector space over field. Suppose the minimal polynomial $\mathcal{m}_T$ and it's characteristic polynomial $\chi_T$ coincide. What can we say about $\mathcal{m}_{T^{-1}}$?
The only thing I can come up with is that $V \cong \mathbb{K}[x]/( \mathcal{m}_T)$ as a $\mathbb{K}[x]$ -module so it has a basis in which the matrix is not that bad. But I really don't want to find the inverse matrix and hope there is a more clever way to do it.
Assume without loss of generality that $\ m_T\ $ and $\ m_{T^{-1}}\ $ are both monic, $\ \deg m_T=n\ $, $\ \deg m_{T^{-1}}=r\ $, and $$ m_T(x)= \sum_{k=0}^n a_k x^k\ . $$ Then multiplying the equation $$ T^n+\sum_{k=0}^{n-1}a_k T^k=0 $$ by $\ T^{-n}\ $ gives $$ a_n+\sum_{k=0}^{n-1} a_k \left(T^{-1}\right)^{n-k}=0\ , $$ so $\ m_{T^{-1}}(x)\ $ must divide $\ x^nm_T\left(\frac{1}{x}\right)\ $. Conversely $\ m_T(x)\ $ must also divide $\ x^rm_{T^{-1}} \left(\frac{1}{x}\right)\ $. It follows that $\ r=n\ $ , and $\ m_{T^{-1}}(x)=$$a_0^{-1} x^n m_T\left(\frac{1}{x}\right)\ $.