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**HallsofIvy** A slightly different proof, without using the minimal polynomial (directly):

If $\displaystyle \lambda$ is an eigenvalue of R then there exist non-zero v such that $\displaystyle Rv= \lambda v$. Apply R to both sides of that equation- $\displaystyle R^2v= R(\lambda v)= \lambda Rv= \lambda(\lambda v)= \lambda^2 v$. But it is given that $\displaystyle R^2= I$ so we are saying that $\displaystyle v= \lambda^2 v$ so that $\displaystyle (1- \lambda^2)v= 0$. Since $\displaystyle v\neq 0$, we must have $\displaystyle 1- \lambda^2= 0$ and, from that, $\displaystyle \lambda= 1$ or $\displaystyle \lambda= -1$.