Every operator on a nonzero finite-dimensional complex vector space has an eigenvalue, because its positive-degree characteristic polynomial has a complex root. More generally this holds over an algebraically closed field. It fails over $\mathbb R$, as shown by a $90^\circ$ rotation of $\mathbb R^2$.
Parents
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.

