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Minimal polynomials and invertibility

  1. Mar 1, 2006 #1
    Let [tex]T\in L(V)[/tex]. Let [tex]g(x)\in F[x][/tex] and let [tex]m(x)[/tex] be the minimal polynomial of [tex]T[/tex]. Show that [tex]g(T)[/tex] is invertible [tex]\Leftrightarrow[/tex] [tex]\gcd (m(x),g(x))=1[/tex].

    Backwards is easy. For forwards, suppose I say that [tex]g(T)[/tex] is invertible implies that [tex]g(T)(v)=0 \Rightarrow v=0[/tex] and therefore [tex]g(x)[/tex] prime, therefore it is not divisible, and therefore [tex]\gcd (g,m)=1[/tex]. Is that correct?

    Why is latex not showing up?
    Last edited: Mar 1, 2006
  2. jcsd
  3. Mar 1, 2006 #2
    They had some sort of server disaster yesterday, I take it.

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