# relatively prime

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