Exercise 6.11.17

I think here we won’t say identity is a rotation. Otherwise, the identity mapping could be decomposed into $n$ identity mapping. Also, we need some facts. That is, if $W_i$ is a subspace with dimension one in the decomposition, then $T_{W_i}$ could not be a rotation since $T_{W_i}(x)$ should be either $x$ or $-x$. Hence every ration has the dimension of its subspace two.

1.

By the Corollary after Theorem 6.47 we know that there is at most one reflection in the decomposition. To decompose a space with dimension $n$ by rotations, there could be only $\frac{1}{2}(n-1)$ rotations.

2.

Similarly, there is at most one reflection. If there’s no reflection, then there’re at most $\frac{1}{2}n$ rotations. If there’s one reflection, there at most

$$\lfloor \frac{n-1}{2}\rfloor =\frac{1}{2}(n-2)$$

rotations.