Exercise 6.11.6

If $U$ and $T$ are two rotations, we have $\det <!--\; FUNCTION\; APPLICATION\; -->(\mathit{UT})=\det <!--\; FUNCTION\; APPLICATION\; -->(U)\det <!--\; FUNCTION\; APPLICATION\; -->(T)=1$. Hence by Theorem 6.47, $\mathit{UT}$ contains no reflection. If $V$ could be decomposed by three one-dimensional subspaces, all of them are identities, thus $\mathit{UT}$ is an identity mapping. Otherwise, $V$ must be decomposed into one one-dimensional and one two-dimensional subspace. Thus $\mathit{UT}$ is a rotation on the two-dimensional subspace and is an identity on the one-dimensional space. Hence, $\mathit{UT}$ must be a rotation.