Theorem: Hermitian matrices have real eigenvalues. Proof: If $Av = \lambda v$ with $v \neq 0$, then $v^*Av = \lambda v^*v$ and also $v^*Av = v^*A^*v = (Av)^*v = \overline{\lambda}v^*v$. So $\lambda = \overline{\lambda}$, meaning $\lambda$ is real.
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.