Is $v$ necessarily an eigenvector of the linear map $\varphi\circ\varphi$?
Solution 1:
You almost have it. $\varphi^2(v)=\varphi(\varphi(v))=\varphi(\lambda v)= \lambda \varphi( v)=\lambda \lambda v=\lambda^2 v$. The eigenvalue is $\lambda^2$
You almost have it. $\varphi^2(v)=\varphi(\varphi(v))=\varphi(\lambda v)= \lambda \varphi( v)=\lambda \lambda v=\lambda^2 v$. The eigenvalue is $\lambda^2$