Exercise 9.B.1

Proof. Choose an orthonormal basis of ${ℝ}^3$ that puts the matrix of $S$ in the form given by 9.36. Since $M(S)$ is a $3$-by-$3$ matrix, one of the diagonal blocks is a $1$-by-$1$ matrix containing $1$ or $-1$. Hence $\mathit{Sx}=x$ or $\mathit{Sx}=-x$ for some nonzero vector $x$ in the chosen basis. Applying $S$ again in the two cases gives $S^{2}x=x$, as desired.

Geometrically speaking, an isometry on ${ℝ}^3$ is a rotation about an axis, perhaps with a reflexion through a plane orthogonal to the axis. Hence an isometry on ${ℝ}^3$ either sends the vectors in the axis to themselves or to their reflexion. If it’s the first case, we already have what we wanted to prove, if it’s the second, applying the isometry again sends the reflexions of the vectors back to the vectors themselves. □