Exercise 6.11.1

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1.: No. It may be the composition of two or more rotations. For example, $T (x, y) = (R_{𝜃} (x), R_{𝜃} (y))$ is orthogonal but not a rotation or a reflection, where $x, y \in ℝ^{2}$ and $R_{𝜃}$ is the rotation transformation about the angle $𝜃$ .
2.: Yes. See the Corollary after Theorem 6.45.
3.: Yes. See Exercise 6.11.6.
4.: No. For example, $U (x, y) = (R_{𝜃} (x), y)$ and $T (x, y) = (x, R_{𝜃} (y))$ are two rotations, where $x, y \in ℝ^{2}$ and $R_{𝜃}$ is the rotation transformation about the angle $𝜃$ . But $UT$ is not a rotation.
5.: Yes. It comes from the definition of rotation.
6.: No. In two-dimensional real space, the composite of two reflections is a rotation.
7.: No. It may contain one reflection. For example, the mapping $T (x) = - x$ could not be the composite of rotations since the only rotation in $ℝ$ is the identity.
8.: No. It may be the composite of some rotations and a reflection. For example, $T (x, y) = (- x, R_{𝜃} (y))$ has $\det (T) = - 1$ but it’s not a reflection, where $x \in ℝ$ , $y \in ℝ^{2}$ , and $R_{𝜃}$ is the rotation about the angle $𝜃$ .
9.: Yes. Let $T$ be the reflection about $W$ . We have that $W \oplus W^{⊥} = V$ . So one of $W$ and $W^{⊥}$ could not be zero. But every nonzero vector in $W$ is an eigenvector with eigenvalue $1$ while that in $W^{⊥}$ is an eigenvector with eigenvalue $- 1$ . So $T$ must have eigenvalues.
10.: No. The rotation on two-dimensional space has no eigenvector unless it’s the identity mapping.

jephianlin

2011-06-27 00:00

Exercise 6.11.1

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