Let $A$ be a matrix with real eigenvalues such that $A^2$ is similiar to A. What are the possible eigenvalues of $A$?
I know that similiar matrix have the same eigenvalues, and also, eigenvalue square is the eigenvalue of $A^2$
So i would have: $\lambda$v = $\lambda^2$v = $A^2$v = $A$v
I don't know what to do after, i'm done???
Similar matrices do have the same eigenvalues. If $u$ is an eigenvector, then $$A^2u = \lambda^2 u$$ There is a matrix $P$ such that $PA^2P^{-1}=A$. Then $$AP(u)=\lambda^2P(u)$$ so $\lambda^2$ is an eigenvalue of $A$ with eigenvector $P(u)$. We can iterate this: $\lambda^4$, $\lambda^8$, etc. are all eigenvalues of $A$. Since there are only finitely many, this must be a cycle. But this is only possible when the eigenvalues have absolute value $0$ or $1$, so they must be $0$, $1$, or $-1$.