Linear transformation is invertible if and only if $x$ does not divide its minimal polynomial Let $T : V \to V$ be a linear transformation of a finite dimensional vector space over a field $\mathbb{F}$ to itself. Prove that $T$ is invertible if and only if $x$ does not divide the minimal polynomial $m(x)$.
I don't really understand how to do this at all, so could someone show me a proof I can try to understand! 
 A: If $T$ satisfies any polynomial equation with non-zero (and therefore invertible) constant term, say $a_dT^d+\cdots+a_1T+a_0I=0$ with $a_0\neq 0$, you can rewrite this as
$$
  I=-\frac1{a_0}(a_dT^d+\cdots+a_1T)=-\frac1{a_0}(a_dT^{d-1}+\cdots+a_1T^0)T,
$$
which implies that $T$ is invertible; in particular this is the case if the minimal polynomial has a nonzero constant term. This part does not use the minimality of the minimum polynomial, but the other part does.
Suppose the minimal polynomial$~\mu_T$ has zero constant term (or equivalently that it is divisible by $X$), say
$$
  \mu_T=a_dX^d+\cdots+a_1X
$$
so that
$$
  0=a_dT^d+\cdots+a_1T=(a_dT^{d-1}+\cdots+a_1T^0)T.
$$
Now the expression in parentheses cannot be $0$ since it is a polynomial of degree$~d-1<\deg\mu_T$ in$~T$, but then $T$ cannot be invertible, for in that case the right hand side could not become$~0$.
One proves similarly that more generally $T-\lambda\operatorname{id}$ for some scalar$~\lambda$ is non-invertible (in other words that $\lambda$ is an eigenvlaue of$~T$) if and only if $X-\lambda$ divides$~\mu_T$.
A: Resolution reached in comments: 


*

*An operator in  a finite-dimensional space is invertible iff $0$ is not its eigenvalue.

*Eigenvalues are precisely the zeros of the minimal polynomial.

*A polynomial $m$ is divisible by $x$ iff $m(0)=0$. 

