There exists a polynomial $p \in \mathbb{C}[z]$ such that $T^{-1} = p(T).$ [duplicate]

Let $V$ be a vector space over $\mathbb{C}$, and let $T$ be an invertible linear operator on $V$ . Prove that there exists a polynomial $p \in \mathbb{C}[z]$ such that $T^{-1} = p(T).$

In the finite-dimensional case, we can prove by using the Cayley-Hamilton theorem with the help of characteristic polynomial. But as we don't have any such thing as characteristic polynomial in infinite-dimensional case, we can not use that idea. How to prove this in infinite-dimensional case ? Thanks in advance.

Solution 1:

This is not true in general. If $$\tag1T^{-1}=p(T),$$ then $q(T)=0$, where $q(t)=tp(t)-1$. In particular, any eigenvalue of $T$ has to be a root of $q$. It follows that any operator with infinitely many eigenvalues cannot satisfy $(1)$.

For any easy example of an operator with infinitely many eigenvalues, let $V=\mathbb C[x]$, the space of polynomials, and $T$ the linear operator induced by $$ Tx^n=nx^n. $$ Then, by definition, every positive integer is an eigevalue for $T$, and so $T$ cannot satisfy $(1)$ for any $p$.