# Show the existence of a polynomial $f$ such that $T^{-1} = f(T)$ for an invertible linear operator $T.$

Came across this question and have had difficulty approaching it so any help is greatly appreciated.

Let $T$ be an invertible linear operator on a finite dimensional vector space $V$ over a field $F$. Prove that there exists a polynomial $f$ over $F$ such that $T^{-1} = f(T)$.

-
The characteristic polynomial of $T$ has a non-zero constant coefficient. Use that to express $T^{-1}$ as a linear combination of powers of $T.$ Addendum: $A^{-1} = f(A)$ is actually used in some algorithms to find the inverse (or solution of a system) when $A$ is sparse (hence matrix-vector product is relatively-cheap operation). –  user2468 Aug 17 '12 at 1:02

Let $p(x)$ be the characteristic polynomial of $T$. Since $T$ is invertible 0 is not an eigenvector so $p(x)$ has non-zero constant term. Write $p(x) = x^n + a_{n-1} x^{n-1} + \ldots + a_0$. But any linear transformation satisfies its characteristic polynomial, i.e. $p(T) = 0$. Thus $$T^n + a_{n-1} T^{n-1} + \cdots + a_1 T + a_0 = 0$$ which gives $$T(\frac{-1}{a_0}(T^{n-1} + a_{n-1}T^{n-2} + \cdots + a_1)) = Id$$ so $T^{-1} = \frac{-1}{a_0}(T^{n-1} + a_{n-1}T^{n-2} + \cdots + a_1)$.