An $n\times n$ matrix over $\mathbb F$ is diagonalizable iff it has $n$ linearly independent eigenvectors over $\mathbb F$. Proof: write them as the columns of $P$. Then $AP=PD$ and $P$ is invertible. Conversely, $A=PDP^{-1}$ makes each column of $P$ an eigenvector.
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.