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Math Help - relatively prime

  1. #1
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    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.
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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 (m,f) was not one, this would say that m\mid f which would imply that 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 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.
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