Kernel of nilpotent operators on infinite dimensional vector spaces

Question

Suppose $T$ is a nilpotent linear transformations on an infinite dimensional vector space. By nilpotent I mean that $T^k=0$ for some $k$. Does it follow that the kernel of $T$ must be infinite dimensional?

Answer 1

Yes. Observe that for each $k$ there is a short exact sequence $$0\to \ker(T^k)\to \ker(T^{k+1})\stackrel{T^k}\to T^k(\ker(T^{k+1}))\to 0$$ and that $T^k(\ker(T^{k+1}))\subseteq \ker(T)$. So $\dim \ker(T^{k+1})\leq \dim \ker(T^k)+\dim \ker(T)$, and by induction on $k$ we see $\dim \ker(T^k)\leq k\dim\ker(T)$ for all $k$. In particular, if $\ker(T)$ is finite dimensional, then $\ker(T^k)$ is finite-dimensional for all $k$. But $T^k=0$ for some $k$, so this impossible.