TITLE: Minimal Polynomial of Inverse
QUESTION [7 upvotes]: Suppose $T$ has minimal polynomial $x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ and $T$ is invertible (hence $a_0\not=0$). Is it true that the minimal polynomial of $T^{-1}$ is $\frac{1}{a_0}(1+a_{m-1}x+...+a_1x^{m-1})+x^m$?

My thought was that since $T^m+a_{m-1}T^{m-1}+...+a_1T+a_0I=0$, we have
$I+a_{m-1}T^{-1}+...+a_1T^{m-1}+a_0T^{-m}=0\implies\frac{1}{a_0}(I+a_{m-1}T^{-1}+...+a_1T^{m-1})+T^{-m}=0$.
So the minimal polynomial of $T^{-1}$ divides $\frac{1}{a_0}(1+a_{m-1}x+...+a_1x^{m-1})+x^m$ and so (degree of minimal polynomial of $T^{-1}$)$\leq m=$(degree of minimal polynomial of $T$). On the other hand, if $x^i+b_{i-1}x^{i-1}+...+b_1x+b_0$ is the minimal polynomial of $T^{-1}$ (so $b_0\not=0$), then $0=T^{-i}+b_{i-1}T^{-i+1}+...+b_1T^{-1}+b_0I\implies 0=I+b_{i-1}T+...+b_1T^{i-1}+b_0T^i$ so the minimal polynomial of $T$ divides $b_0x^i+b_1x^{i-1}+...+b_{i-1}x+1$ and so $m=$(degree of minimal polynomial of $T$)$\leq$(degree of minimal polynomial of $T^{-1}$
Hence, degree of minimal polynomial of $T^{-1}$ is $m$ and $\frac{1}{a_0}(1+a_{m-1}x+...+a_1x^{m-1})+x^m$ is a monic polynomial of degree $m$ annihilating $T^{-1}$, so it is the minimal polynomial of $T^{-1}$.
Does this look correct? Thank you

REPLY [5 votes]: Yes, it is correct.
Minor correction: $$I+a_{m-1}T^{-1}+ \ldots +a_1T^{\color{red}{-(}m-1\color{red}{)}}+a_0T^{-m}=0$$
To make the proof slightly shorter.
Note that in the first half of the proof, you have proven that if $S$ is invertible ,then the degree of minimal polynomial of $S^{-1}$ is less than or equal to the degree of minimal polynomial of $S$.
Replacing $S=T$, we have
The degree of minimal polynomial of $T^{-1}$ is less than equal to the degree of minimal polynomial of $T$.
Replacing $S$ with $T^{-1}$ and recognizing that $(T^{-1})^{-1}=T$, we can conclude directly that 
The degree of minimal polynomial of $T$ is less than equal to the degree of minimal polynomial of $T^{-1}$.
Hence their minimal polynomials share the same degree.