relatively prime

• Dec 11th 2012, 06:26 PM
relatively prime
Let $\displaystyle \sigma$ be a linear transform on a vector field $\displaystyle V$. Suppose $\displaystyle m(x)$ is the minimal polynomial of $\displaystyle \sigma$, $\displaystyle f(x)$ is a polynomial. Show that if $\displaystyle f(\sigma)$ is invertible, then $\displaystyle m(x),f(x)$ is relatively prime. That is, $\displaystyle (m(x),f(x))=1.$
• Dec 11th 2012, 06:37 PM
Drexel28
Re: relatively prime
You know that the minimal polynomial is irreducible. So, if $\displaystyle (m,f)$ was not one, this would say that $\displaystyle m\mid f$ which would imply that $\displaystyle f(\sigma)=0$, yeah?
• Dec 11th 2012, 08:51 PM
My mistake, I was being silly. The key is that the eigenvalues of $\displaystyle f(\sigma)$ are $\displaystyle \{f(\lambda)\}$ where $\displaystyle \{\lambda\}$ are the eigenvalues of $\displaystyle \sigma$. Now, note that since $\displaystyle f(\sigma)$ is invertible it has no zero eigenvalues, and so $\displaystyle f(\lambda)\ne 0$ for all $\displaystyle \lambda$--I think you can conclude from that.