Minimal polynomials and invertibility

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The discussion centers on the relationship between the invertibility of the operator \( g(T) \) and the minimal polynomial \( m(x) \) of a linear operator \( T \in L(V) \). It is established that \( g(T) \) is invertible if and only if the greatest common divisor \( \gcd(m(x), g(x)) = 1 \). The reasoning provided confirms that if \( g(T) \) is invertible, then \( g(x) \) must be prime and not divisible by \( m(x) \), leading to the conclusion that \( \gcd(g, m) = 1 \). The discussion also briefly mentions a technical issue with LaTeX rendering.

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Treadstone 71
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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?
 
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Treadstone 71 said:
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?

They had some sort of server disaster yesterday, I take it.

-Dan
 

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