What is the minimal polynomial of $T^{-1}$? > Suppose $T$ is an invertible linear operator on a finite dimensional vector space with minimal polynomial $p_T=x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$. Is it true that the minimal polynomial of $T^{-1}$ is given by $p_{T^{-1}}=x^m+\frac{1}{a_0}(a_{1}x^{m-1}+...+a_{m-1}x+1)$? I think the answer is yes, but I could not find an online source to corroborate this (although I can find references for the the characteristic polynomial of $T^{-1}$). I also posted this as a question with my work a while back,but didn't really get the answer I was looking for: Minimal Polynomial of Inverse

Lemma: Given a non-constant polynomial $$ p(x) = x^m+a_{m-1}x^{m-1}+\ldots+a_0,\qquad a_0\
eq 0$$ then $\xi\
eq 0$ is a root of $p(x)$ iff $\frac{1}{\xi}$ is a root of the reciprocal polynomial $$ q(x) = a_0 x^m + a_1 x^{m-1} +\ldots + 1. $$

Proof: $p(\xi)=0$ implies $\frac{p(\xi)}{\xi^m} = q\left(\frac{1}{\xi}\right)=0$ and $\xi\to\frac{1}{\xi}$ is an involution.

If $T$ is an invertible linear operator and $p,q$ are the minimal polynomials of $T,T^{-1}$ is is not difficult to show that $\text{deg } p = \text{deg } q$ since both $\text{deg } p \leq \text{deg } q$ and $\text{deg } p \geq \text{deg } q$ have to hold. Then the claim follows from the above Lemma.