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Thread: relatively prime

  1. #1
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    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, $\displaystyle (m(x),f(x))=1.$
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    MHF Contributor Drexel28's Avatar
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    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?
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    Re: relatively prime

    Why minimal polynomial is irreducible? It of course could be reducible? Do you see the Jordan canonical form of a matrix...
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    MHF Contributor Drexel28's Avatar
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    Re: relatively prime

    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.
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