Quasi-nilpotent operator with finite-dimensional kernel

Question

Can we find an operator $T$ on a Banach space with the following properties?

a) $\sigma_p(T)=\sigma(T)=\{0\}$.

b) $\ker T$ is of finite dimension.

c) $TX$ is a closed and finite-codimensional subspace of $X$?

Answer 1

Let $X = \mathbb{C}^2$. The matrix $$ T = \left( \begin{array}{cc} 0 \ & 1 \\ 0 \ & 0 \end{array} \right) $$ has double eigenvalue $0$, and in finite dimensions the point spectrum and spectrum are identical, so we have $\sigma_p(T)=\sigma(T)=\{0\}$. It is easy to see that the kernel of $T$ is the linear span of the first basis vector. In particular it is finite dimensional. Consequently, $TX$ has co-dimension one in $X$, and is trivially closed.