1. ## Minimal polynomial problem

Prove that if T is invertible and the degree of $minP_{T} = m$
then
$minP_{T^{-1}}(x) = cx^{m}minP_{T}(x^{-1})$

where $c=minP_{T}(0)^{-1}$

2. Regardless of getting the actual proof of this problem, I am simply just having trouble seeing why this would be true in the first place.

3. Originally Posted by linalguser3
Prove that if T is invertible and the degree of $minP_{T} = m$
then
$minP_{T^{-1}}(x) = cx^{m}minP_{T}(x^{-1})$

where $c=minP_{T}(0)^{-1}$
Fact: suppose $p(x)=x^m + c_{m-1}x^{m-1} + \cdots + c_1x + c_0$ is irreducible over some field. then $q(x)=c_0^{-1}x^m p(x^{-1})=x^m + c_1c_0^{-1}x^{m-1} + \cdots + c_{m-1}c_0^{-1}x + c_0^{-1}$ is irreducible too.

Proof: suppose $q(x)=(x^r + a_{r-1}x^{r-1} + \cdots + a_0)(x^s + b_{s-1}x^{s-1} + \cdots + b_0).$ then $c_0^{-1}x^{-m}p(x)=q(x^{-1})=(x^{-r} + a_{r-1}x^{-r+1}+ \cdots + a_0)(x^{-s} + b_{s-1}x^{-s+1} + \cdots + b_0).$ therefore:

$p(x)=(x^r + a_1a_0^{-1}x^{r-1} + \cdots + a_0^{-1})(x^s + b_1b_0^{-1}x^{s-1} + \cdots + b_0^{-1}),$ which is impossible because $p(x)$ is irreducible.

using what i just proved, you should be able to solve your problem quite easily!