Exercise 6.11.8

If $T$ is an orthogonal operator, we know that the determinant of $T$ should be $\pm 1$. Now pick an orthonormal basis $\beta$. If $\det <!--\; FUNCTION\; APPLICATION\; -->(T)=1$, we have $\det <!--\; FUNCTION\; APPLICATION\; -->({[T]}_{\beta })=1$ and hence ${[T]}_{\beta }$ is a rotation matrix by Theorem 6.23. By Exercise 6.2.15 we know that the mapping ${\varphi }_{\mathit{beta}}$, who maps $x\in V$ into its coordinates with respect to $\beta$, preserve the inner product. Hence $T$ is a rotation when ${[T]}_{\beta }$ is a rotation. On the other hand, if $\det <!--\; FUNCTION\; APPLICATION\; -->(T)=\det <!--\; FUNCTION\; APPLICATION\; -->([T])=-1$, we know that ${[T]}_{\beta }$ is a reflection matrix by Theorem 6.23. Again, $T$ is a reflection since ${[T]}_{\beta }$ is a reflection matrix.