Let $A \in M_3(\mathbb{R})$ be a unitary matrix. Show that $A^2$ has $1$ as an eigenvalue.
My try: Since A is unitary that means that its column vectors are an orthogonal base for $\mathbb{R}$, so A is similar to $I_3$. That means that $A^2$ is also similar to $I_3$. Since similar matrices have the same characteristic polynomial and $1$ is an eigenvalue for $I_3$, $1$ is an eigenvalue for $A^2$.
Is the proof okay?
Eigenvalues of orthogonal matrices have unit modulus and come in complex conjugate pairs. If the dimension is odd, it is not possible to have them all in pairs so at least one of them must be real. The real eigenvalue must be either +1 or -1. When raised to the second power, either one of those will become +1.