# Math Help - relatively prime

1. ## 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, $(m(x),f(x))=1.$

2. ## 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?

3. ## Re: relatively prime

Why minimal polynomial is irreducible? It of course could be reducible? Do you see the Jordan canonical form of a matrix...

4. ## Re: relatively prime

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.