1
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ $T$ 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
  • 1
    $\begingroup$ $a$ is a root of a polynomial iff $(x-a)$ divides that polynomial. $\endgroup$ – Prahlad Vaidyanathan Oct 23 '13 at 16:12
4
$\begingroup$

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$~\mu_T$.

$\endgroup$
0
$\begingroup$

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 $m(0)=0$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.