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!

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    $\begingroup$ $T$ is invertible iff $0$ is not an eigen-value. What do eigen-values have to do with roots of the minimal polynomial? $\endgroup$ – Prahlad Vaidyanathan Oct 23 '13 at 15:46
  • $\begingroup$ All eigenvalues of T are roots of the minimal polynomial I think? Do you have a proof of this? $\endgroup$ – user101293 Oct 23 '13 at 15:57
  • $\begingroup$ Okay got my proof of eigenvalues being roots of the minimal polynomial but how does it relate to x not dividing the minimal polynomial $\endgroup$ – user101293 Oct 23 '13 at 16:08
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    $\begingroup$ $a$ is a root of a polynomial iff $(x-a)$ divides that polynomial. $\endgroup$ – Prahlad Vaidyanathan Oct 23 '13 at 16:12

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 with 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$~\mu_T$.


Resolution reached in comments:

  1. An operator in a finite-dimensional space is invertible iff $0$ is not its eigenvalue.
  2. Eigenvalues are precisely the zeros of the minimal polynomial.
  3. A polynomial $m$ is divisible by $x$ iff $m(0)=0$.

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