A linear operator is diagonalizable over $\mathbb F$ if it admits an ordered basis of eigenvectors over $\mathbb F$. Its matrix in that basis is diagonal. This is equivalent to its matrix in any basis being similar over $\mathbb F$ to a diagonal matrix. The choice of field matters.
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.