prove that if T is invertible transformation there is polynomial $p$ such that $T^{-1} = p(T) $

Question

I know how to prove this using Hamilton.C but something doesn't make sense to me.

if I assume that there is such polynomial p(x), so p(T)T = I .

then looking at these polynomials I get: p(x)x = 1 so deg p(x) + deg (x) = deg (1)

and degree of polynomial is >= 0 , but deg (1) = 0 so how is it possible? what is not right with what I'm doing? thanks

Answer 1

$p(T)T=I$ does not imply $p(x)x=1$.

The evaluation map $E: k[x] \to L(V)$ given by $E(p)=p(T)$ is not injective. There are non-zero polynomials that are sent to zero; $p(T)T-I$ is one of them. The characteristic polynomial is another (that's the Cayley–Hamilton theorem).

That phenomenon is common. Consider $\omega=(-1+\sqrt3)/2$. Then $\omega^2+\omega+1=0$ and so $\omega(-\omega-1)=1$ and $\omega^{-1}=-\omega-1$, but $x(-x-1)=1$ does not hold as a polynomial equality.