A linear operator $T:V\to V$ is nilpotent if $T^k=0$ for some positive integer $k$. The least such $k$ is its nilpotency index (for nonzero spaces). Example: $T(x,y)=(y,0)$ satisfies $T^2=0$ but $T\ne0$. Its only eigenvalue is zero.
Like how this is organised?
Build your own knowledge tree on knowen.md: notes, decisions and tasks in one connected markdown graph, shared with your team and your AI agents.