proof of roots of characteristic polynomial are eigenvalues How do I prove the 2 directions of this statement?


A root $r$ of characteristic polynomial provides equation $det(M - r Id)=0$ so $M-r Id$ hasn't a full rank and there exist a nontrivial vector $\tilde{v}$ with $0 =(M-rId)\tilde{v}= M\tilde{v}-r\tilde{v}$ so $M\tilde{v}= r\tilde{v}$ and $r$ is by definition an eigenvalue. Walking backwards alongside my arguments provides the proof in other direction.
