Here is the question: Let $V_n$ be the vector space of polynomials in $\mathbb{C}[x]$ of degree less than or equal $n$. Let the operator $T:V_n\to V_n$ be defined by $T(p):=p-p'$, where $p'$ denotes the derivative of $p$. Write the Jordan form of $T^{-1}$.
I only have able to show that all the eigenvalues of $T$ are $1$, because if $Tp=\lambda p$, then $p-p'=\lambda p$, and so by comparing the degree of $p$, $\lambda =1$. That means $det (T)=1$ and so $T$ is invertible.
In fact, I believe $T^{-1}$ is a polynomial in $T$, as $T=I-D$, then $T^{-1}=(I-D)^{-1}=\sum_{k=0}^n D^k=\sum_{k=0}^n (I-T)^k$.
I also noticed that, $det(T^{-1})=1$ since $det(T)=1$.
However, I am not sure how to find out its Jordan form. Any help is appreciated, Thanks a lot!