# Minimal polynomials and invertibility

1. Mar 1, 2006

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

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

Why is latex not showing up?

Last edited: Mar 1, 2006
2. Mar 1, 2006

### topsquark

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

-Dan