If symmetric $n \times n $ matrix $S^3=I_n$, does $S=I_n$? Let $S$ be a real symmetric $n \times n$ matrix. Does $S^3=I_n$ imply $S=I_n$?
I started by looking at $S^2=I_n$ and found that this does not imply $S=I_n$, because of the counterexample
\begin{align}\begin{bmatrix} 
-\sin(\alpha) & \cos(\alpha) \\
\cos(\alpha) & \sin(\alpha)
\end{bmatrix} \end{align}
For $S^3$ however, I do not really know how to approach the problem.

By the spectral theorem, $S$ is diagonalisable with real numbers as eigenvalues. From $S^3=I$ every eigenvalue$~\lambda$ of $S$ satisfies $\lambda^3=1$. Conclude.
