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6.11* The Geometry of Orthogonal Operators

By Theorem 6.22 (p. 383), any rigid motion on a finite-dimensional real inner product space is the composite of an orthogonal operator and a translation. Thus, to understand the geometry of rigid motions thoroughly, we must analyze the structure of orthogonal operators. In this section, we show that any orthogonal operator on a finite-dimensional real inner product space can be described in terms of rotations and reflections.

This material assumes familiarity with the results about direct sums developed at the end of Section 5.2 and with the definition of the determinant of a linear operator given in Section 5.1 as well as elementary properties of the determinant in Exercise 8 of Section 5.1.

We now extend our earlier definitions of rotation and reflection on $R^{2}$ $R^{2}$ to all 2-dimensional real inner product spaces.

Definitions.

Let T be a linear operator on a two-dimensional real inner product space V.

We call T a rotation if there exists an orthonormal basis $β = {x_{1}, x_{2}}$ $β = {x_{1}, x_{2}}$ for V and a real number $θ$ $θ$ such that

T (x_{1}) = (\cos θ) x_{1} + (\sin θ) x_{2} and T (x_{2}) = (- \sin θ) x_{1} + (\cos θ) x_{2} .

$T (x_{1}) = (\cos θ) x_{1} + (\sin θ) x_{2} and T (x_{2}) = (- \sin θ) x_{1} + (\cos θ) x_{2} .$

We call T a reflection if there exists a one-dimensional subspace W of V such that $T (x) = - x$ $T (x) = - x$ for all $x \in W$ $x \in W$ and $T (y) = y$ $T (y) = y$ for all $y \in W^{⊥}$ $y \in W^{⊥}$ . In this context, T is called a reflection of V about $W^{⊥}$ $W^{⊥}$ .

As a convenience, we define rotations and reflections on a 1-dimensional inner product space.

Definitions.

A linear operator T on a 1-dimensional inner product space V is called a rotation if T is the identity and a reflection if $T (x) = - x$ $T (x) = - x$ for all $x \in V$ $x \in V$ .

Trivially, rotations and reflections on 1-dimensional inner product spaces are orthogonal operators. It should be noted that rotations and reflections on 2-dimensional real inner product spaces (or composites of these) are orthogonal operators (see Exercise 2).

Example 1

Some Typical Reflections

(a) Define $T : R^{2} \to R^{2}$ $T : R^{2} \to R^{2}$ by $T (a, b) = (- a, b)$ $T (a, b) = (- a, b)$ , and let $W = span ({e_{1}})$ $W = span ({e_{1}})$ . Then $T (x) = - x$ $T (x) = - x$ for all $x \in W$ $x \in W$ , and $T (y) = y$ $T (y) = y$ for all $y \in W^{⊥}$ $y \in W^{⊥}$ . Thus T is a reflection of $R^{2}$ $R^{2}$ about $W^{⊥} = span ({e_{2}})$ $W^{⊥} = span ({e_{2}})$ , the y-axis.
(b) Define $T : R^{2} \to R^{2}$ $T : R^{2} \to R^{2}$ by $T (a, b) = (b, a)$ $T (a, b) = (b, a)$ , and let $W = span ({1, 1})$ $W = span ({1, 1})$ . Clearly $T (w) = w$ $T (w) = w$ for all $w \in W$ $w \in W$ . Let $(a, b) \in W^{⊥}$ $(a, b) \in W^{⊥}$ . Then (a, b) is orthogonal to (1, 1) and hence, $a + b = 0$ $a + b = 0$ . So $b = - a$ $b = - a$ . Thus $W^{⊥} = span ({(1, - 1)})$ $W^{⊥} = span ({(1, - 1)})$ . It follows that $T (a, b) = (a, - a) = - (- a, a) = - (b, a)$ $T (a, b) = (a, - a) = - (- a, a) = - (b, a)$ . Hence T is the reflection of $R^{2}$ $R^{2}$ about W.

The next theorem characterizes all orthogonal operators on a two-dimensional real inner product space V. The proof follows from Theorem 6.23 (p. 384) since all two-dimensional real inner product spaces are structurally identical. For a rigorous justification, apply Theorem 2.21 (p. 105), where $β$ $β$ is an orthonormal basis for V. By Exercise 15 of Section 6.2, the resulting isomorphism $ϕ_{β} : V \to R^{2}$ $ϕ_{β} : V \to R^{2}$ preserves inner products. (See Exercise 8.)

Theorem 6.46.

Let T be an orthogonal operator on a two-dimensional real inner product space V. Then T is either a rotation or a reflection. Furthermore, T is a rotation if and only if $\det (T) = 1$ $\det (T) = 1$ , and T is a reflection if and only if $\det (T) = - 1$ $\det (T) = - 1$ .

A complete description of the reflections of $R^{2}$ $R^{2}$ is given in Section 6.5.

Corollary.

Let V be a two-dimensional real inner product space.

(a) The composite of a reflection and a rotation on V is a reflection on V.
(b) The composite of two reflections on V is a rotation on V.
(c) The product of two rotations on V is a rotation on V.

Proof.

If $T_{1}$ $T_{1}$ is a reflection on V and $T_{2}$ $T_{2}$ is a rotation on V, then by Theorem 6.46, $\det (T_{1}) = 1$ $\det (T_{1}) = 1$ and $\det (T_{2}) = - 1$ $\det (T_{2}) = - 1$ . Let $T = T_{2} T_{1}$ $T = T_{2} T_{1}$ be the composite. Since $T_{2}$ $T_{2}$ and $T_{1}$ $T_{1}$ are orthogonal, so is T. Moreover, $\det (T) = \det (T_{2}) \cdot \det (T_{1}) = - 1$ $\det (T) = \det (T_{2}) \cdot \det (T_{1}) = - 1$ . Thus, by Theorem 6.46, T is a reflection. The proof for $T_{1} T_{2}$ $T_{1} T_{2}$ is similar.

The proofs of (b) and (c) are similar to that of (a).

We now study orthogonal operators on spaces of higher dimension.

Lemma. If T is a linear operator on a nonzero finite-dimensional real vector space v, then there exists a T-invariant subspace W of V such that $1 \leq \dim (W) \leq 2$ $1 \leq \dim (W) \leq 2$ .

Proof.

Fix an ordered basis $β = {y_{1}, y_{2}, \dots, y_{n}}$ $β = {y_{1}, y_{2}, \dots, y_{n}}$ for V, and let $A = {[T]}_{β}$ $A = {[T]}_{β}$ . Let $ϕ_{β} : V \to R^{n}$ $ϕ_{β} : V \to R^{n}$ be the linear transformation defined by $ϕ_{β} (y_{i}) = e_{i}$ $ϕ_{β} (y_{i}) = e_{i}$ for $i = 1, 2, \dots, n$ $i = 1, 2, \dots, n$ . Then $ϕ_{β}$ $ϕ_{β}$ is an isomorphism, and, as we have seen in Section 2.4, the diagram in Figure 6.10 commutes, that is, $L_{A} ϕ_{β} = ϕ_{β} T$ $L_{A} ϕ_{β} = ϕ_{β} T$ . As a consequence, it suffices to show that there exists an $L_{A}$ $L_{A}$ -invariant subspace Z of $R^{n}$ $R^{n}$ such that $1 \leq \dim (Z) \leq 2$ $1 \leq \dim (Z) \leq 2$ . If we then define $W = ϕ_{β}^{- 1} (Z)$ $W = ϕ_{β}^{- 1} (Z)$ , it follows that W satisfies the conclusions of the lemma (see Exercise 12).

A diagram of a square linear map T. — Figure 6.10

Figure 6.10 Full Alternative Text

The matrix A can be considered as an $n \times n$ $n \times n$ matrix over C and, as such, can be used to define a linear operator U on $C^{n}$ $C^{n}$ by $U (v) = A v$ $U (v) = A v$ . Since U is a linear operator on a finite-dimensional vector space over C, it has an eigenvalue $λ \in C$ $λ \in C$ . Let $x \in C^{n}$ $x \in C^{n}$ be an eigenvector corresponding to $λ$ $λ$ . We may write $λ = λ_{1} + i λ_{2}$ $λ = λ_{1} + i λ_{2}$ , where $λ_{1}$ $λ_{1}$ and $λ_{2}$ $λ_{2}$ are real, and

x = (\begin{array}{c} a_{1} + i b_{1} \\ a_{2} + i b_{2} \\ ⋮ \\ a_{n} + i b_{n} \end{array}),

$x = (\begin{array}{c} a_{1} + i b_{1} \\ a_{2} + i b_{2} \\ ⋮ \\ a_{n} + i b_{n} \end{array}),$

where the $a_{i}$ $a_{i}$ ’s and $b_{i}$ $b_{i}$ ’s are real. Thus, setting

x_{1} = (\begin{array}{c} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{array}) and x_{2} = (\begin{array}{c} b_{1} \\ b_{2} \\ ⋮ \\ b_{n} \end{array}),

$x_{1} = (\begin{array}{c} a_{1} \\ a_{2} \\ ⋮ \\ a_{n} \end{array}) and x_{2} = (\begin{array}{c} b_{1} \\ b_{2} \\ ⋮ \\ b_{n} \end{array}),$

we have $x = x_{1} + i x_{2}$ $x = x_{1} + i x_{2}$ , where $x_{1}$ $x_{1}$ and $x_{2}$ $x_{2}$ have real entries. Note that at least one of $x_{1}$ $x_{1}$ or $x_{2}$ $x_{2}$ is nonzero since $x \neq 0$ $x \neq 0$ . Hence

U (x) = λ x = (λ_{1} + i λ_{2}) (x_{1} + i x_{2}) = (λ_{1} x_{1} - λ_{2} x_{2}) + i (λ_{1} x_{2} + λ_{2} x_{1}) .

$U (x) = λ x = (λ_{1} + i λ_{2}) (x_{1} + i x_{2}) = (λ_{1} x_{1} - λ_{2} x_{2}) + i (λ_{1} x_{2} + λ_{2} x_{1}) .$

Similarly,

U (x) = A (x_{1} + i x_{2}) = A x_{1} + i A x_{2} .

$U (x) = A (x_{1} + i x_{2}) = A x_{1} + i A x_{2} .$

Comparing the real and imaginary parts of these two expressions for U(x), we conclude that

A x_{1} = λ_{1} x_{1} - λ_{2} x_{2} and A x_{2} = λ_{1} x_{2} + λ_{2} x_{1} .

$A x_{1} = λ_{1} x_{1} - λ_{2} x_{2} and A x_{2} = λ_{1} x_{2} + λ_{2} x_{1} .$

Finally, let $Z = span ({x_{1}, x_{2}})$ $Z = span ({x_{1}, x_{2}})$ , the span being taken as a subspace of $R^{n}$ $R^{n}$ . Since $x_{1} \neq 0$ $x_{1} \neq 0$ or $x_{2} \neq 0$ $x_{2} \neq 0$ , Z is a nonzero subspace. Thus $1 \leq \dim (Z) \leq 2$ $1 \leq \dim (Z) \leq 2$ , and the preceding pair of equations shows that Z is $L_{A}$ $L_{A}$ -invariant.

Theorem 6.47.

Let T be an orthogonal operator on a nonzero finite-dimensional real inner product space V. Then there exists a collection of pairwise orthogonal T-invariant subspaces ${W_{1}, W_{2}, \dots, W_{m}}$ ${W_{1}, W_{2}, \dots, W_{m}}$ of V such that

(a) $1 \leq \dim (W_{i}) \leq 2$ $1 \leq \dim (W_{i}) \leq 2$ for $i = 1, 2, \dots, m .$ $i = 1, 2, \dots, m .$
(b) $V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{m} .$ $V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{m} .$

Proof.

The proof is by mathematical induction on dim(V). If $\dim (V) = 1$ $\dim (V) = 1$ , the result is obvious. So assume that the result is true whenever $\dim (V) < n$ $\dim (V) < n$ for some fixed integer $n > 1$ $n > 1$ .

Suppose $\dim (V) = n$ $\dim (V) = n$ . By the lemma, there exists a T-invariant subspace $W_{1}$ $W_{1}$ of V such that $1 \leq \dim (W) \leq 2$ $1 \leq \dim (W) \leq 2$ . If $W_{1} = V$ $W_{1} = V$ , the result is established. Otherwise, $W_{1}^{⊥} \neq {0}$ $W_{1}^{⊥} \neq {0}$ . By Exercise 13, $W_{1}^{⊥}$ $W_{1}^{⊥}$ is T-invariant and the restriction of T to $W_{1}^{⊥}$ $W_{1}^{⊥}$ is orthogonal. Since $\dim (W_{1}^{⊥}) < n$ $\dim (W_{1}^{⊥}) < n$ , we may apply the induction hypothesis to $T_{W_{1}^{⊥}}$ $T_{W_{1}^{⊥}}$ and conclude that there exists a collection of pair-wise orthogonal T-invariant subspaces ${W_{1}, W_{2}, \dots, W_{m}}$ ${W_{1}, W_{2}, \dots, W_{m}}$ of $W_{1}^{⊥}$ $W_{1}^{⊥}$ such that $1 \leq \dim (W_{i}) \leq 2$ $1 \leq \dim (W_{i}) \leq 2$ for $i = 2, 3, \dots, m$ $i = 2, 3, \dots, m$ and $W_{1}^{⊥} = W_{2} \oplus W_{3} \oplus \dots \oplus W_{m}$ $W_{1}^{⊥} = W_{2} \oplus W_{3} \oplus \dots \oplus W_{m}$ . Thus ${W_{1}, W_{2}, \dots, W_{m}}$ ${W_{1}, W_{2}, \dots, W_{m}}$ is pairwise orthogonal, and by Exercise 13(d) of Section 6.2,

V = W_{1} \oplus W_{1}^{⊥} = W_{1} \oplus W_{2} \oplus \dots \oplus W_{m} .

$V = W_{1} \oplus W_{1}^{⊥} = W_{1} \oplus W_{2} \oplus \dots \oplus W_{m} .$

Applying Theorem 6.46 in the context of Theorem 6.47, we conclude that the restriction of T to $W_{i}$ $W_{i}$ is either a rotation or a reflection for each $i = 1, 2, \dots, m .$ $i = 1, 2, \dots, m .$ Thus, in some sense, T is composed of rotations and reflections. Unfortunately, very little can be said about the uniqueness of the decomposition of V in Theorem 6.47. For example, the $W_{i}$ $W_{i}$ ’s, the number m of $W_{i}$ $W_{i}$ ’s, and the number of $W_{i}$ $W_{i}$ ’s for which $T_{W_{i}}$ $T_{W_{i}}$ is a reflection are not unique. Although the number of $W_{i}$ $W_{i}$ ’s for which $T_{W_{i}}$ $T_{W_{i}}$ is a reflection is not unique, whether this number is even or odd is an intrinsic property of T. Moreover, we can always decompose V so that $T_{W_{i}}$ $T_{W_{i}}$ is a reflection for at most one $W_{i}$ $W_{i}$ . These facts are established in the following result.

Theorem 6.48.

Let $T, V, W_{1}, \dots, W_{m}$ $T, V, W_{1}, \dots, W_{m}$ be as in Theorem 6.47.

(a) The number of $W_{i}$ $W_{i}$ ’s for which $T_{W_{i}}$ $T_{W_{i}}$ is a reflection is even or odd according to whether $\det (T) = 1$ $\det (T) = 1$ or $\det (T) = - 1$ $\det (T) = - 1$ .
(b) It is always possible to decompose V as in Theorem 6.47 so that the number of $W_{i}$ $W_{i}$ ’s for which $T_{W_{i}}$ $T_{W_{i}}$ is a reflection is zero or one according to whether $\det (T) = 1$ $\det (T) = 1$ or $\det (T) = - 1$ $\det (T) = - 1$ . Furthermore, if $T_{W_{i}}$ $T_{W_{i}}$ is a reflection, then $\dim (W_{i}) = 1$ $\dim (W_{i}) = 1$ .

Proof.

(a) Let r denote the number of $W_{i}$ $W_{i}$ ’s in the decomposition for which $T_{W_{i}}$ $T_{W_{i}}$ is a reflection. Then, by Exercise 14,

\det (T) = \det (T_{W_{i}}) \cdot \det (T_{W_{2}}) \cdot \dots \cdot \det (T_{w_{m}}) = {(- 1)}^{r},

$\det (T) = \det (T_{W_{i}}) \cdot \det (T_{W_{2}}) \cdot \dots \cdot \det (T_{w_{m}}) = {(- 1)}^{r},$

proving (a).

(b) Let $E = {x \in V : T (x) = - x};$ $E = {x \in V : T (x) = - x};$ then E is a T-invariant subspace of V. Let $W = E^{⊥};$ $W = E^{⊥};$ then W is T-invariant. So by applying Theorem 6.47 to $T_{W}$ $T_{W}$ , we obtain a collection of pairwise orthogonal T-invariant subspaces ${W_{1}, W_{2}, \dots, W_{k}}$ ${W_{1}, W_{2}, \dots, W_{k}}$ of W such that $W = W_{1} \oplus W_{2} \oplus \dots \oplus W_{k}$ $W = W_{1} \oplus W_{2} \oplus \dots \oplus W_{k}$ , and for $1 \leq i \leq k$ $1 \leq i \leq k$ , the dimension of each $W_{i}$ $W_{i}$ is either 1 or 2. Observe that, for each $i = 1, 2, \dots, k, T_{W_{i}}$ $i = 1, 2, \dots, k, T_{W_{i}}$ is a rotation. For otherwise, if $T_{W_{i}}$ $T_{W_{i}}$ is a reflection, there exists a nonzero $x \in W_{i}$ $x \in W_{i}$ for which $T (x) = - x$ $T (x) = - x$ . But then, $x \in W_{i} \cap E \subseteq E^{⊥} \cap E = {0}$ $x \in W_{i} \cap E \subseteq E^{⊥} \cap E = {0}$ , a contradiction. If $E = {0}$ $E = {0}$ , the result follows. Otherwise, choose an orthonormal basis $β$ $β$ for E containing p vectors $(p > 0)$ $(p > 0)$ . It is possible to decompose $β$ $β$ into a pairwise disjoint union $β = β_{1} \cup β_{2} \cup \dots \cup β_{r}$ $β = β_{1} \cup β_{2} \cup \dots \cup β_{r}$ such that each $β_{i}$ $β_{i}$ contains exactly two vectors for $i < r$ $i < r$ , and $β_{r}$ $β_{r}$ contains two vectors if p is even and one vector if p is odd. For each $i = 1, 2, \dots, r$ $i = 1, 2, \dots, r$ , let $W_{k + i} = span (β_{i})$ $W_{k + i} = span (β_{i})$ . Then, clearly, ${W_{1}, W_{2}, \dots, W_{k}, \dots, W_{k + r}}$ ${W_{1}, W_{2}, \dots, W_{k}, \dots, W_{k + r}}$ is pairwise orthogonal, and

V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{k} \oplus \dots \oplus W_{k + r} .

$V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{k} \oplus \dots \oplus W_{k + r} .$ (27)

Moreover, if any $β_{i}$ $β_{i}$ contains two vectors, then

\det (T_{W_{k + i}}) = \det ({[T_{W_{k + i}}]}_{β_{i}}) = \det (\begin{array}{r} - 1 & 0 \\ 0 & - 1 \end{array}) = 1.

$\det (T_{W_{k + i}}) = \det ({[T_{W_{k + i}}]}_{β_{i}}) = \det (\begin{array}{r} - 1 & 0 \\ 0 & - 1 \end{array}) = 1.$

So $T_{W_{k + r}}$ $T_{W_{k + r}}$ is a rotation, and hence $T_{W_{j}}$ $T_{W_{j}}$ is a rotation for $j < k + r$ $j < k + r$ . If $β_{r}$ $β_{r}$ consists of one vector, then $\dim (W_{k + r}) = 1$ $\dim (W_{k + r}) = 1$ and

\det (T_{W_{k + r}}) = \det ({[T_{W_{k + r}}]}_{β_{r}}) = \det (- 1) = - 1.

$\det (T_{W_{k + r}}) = \det ({[T_{W_{k + r}}]}_{β_{r}}) = \det (- 1) = - 1.$

Thus $T_{W_{k + r}}$ $T_{W_{k + r}}$ is a reflection by Theorem 6.47, and we conclude that the decomposition in (27) satisfies (b).

Example 2

Orthogonal Operators on a Three-Dimensional Real Inner Product Space

Let T be an orthogonal operator on a three-dimensional real inner product space V. Then, by Theorem 6.48(b), V can be decomposed into a direct sum of T-invariant orthogonal subspaces so that the restriction of T to each is either a rotation or a reflection, with at most one reflection. Let

V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{m}

$V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{m}$

be such a decomposition. Clearly, $m = 2$ $m = 2$ or $m = 3$ $m = 3$ .

If $m = 2$ $m = 2$ , then $V = W_{1} \oplus W_{2}$ $V = W_{1} \oplus W_{2}$ . Without loss of generality, suppose that $\dim (W_{1}) = 1$ $\dim (W_{1}) = 1$ and $\dim (W_{2}) = 2$ $\dim (W_{2}) = 2$ . Thus $T_{W_{1}}$ $T_{W_{1}}$ is a reflection or the identity on $W_{1}$ $W_{1}$ , and $W_{2}$ $W_{2}$ is a rotation.

If $m = 3$ $m = 3$ , then $V = W_{1} \oplus W_{2} \oplus W_{3}$ $V = W_{1} \oplus W_{2} \oplus W_{3}$ and $\dim (W_{i}) = 1$ $\dim (W_{i}) = 1$ for all i. If $T_{W_{i}}$ $T_{W_{i}}$ is not a reflection, then it is the identity on $W_{i}$ $W_{i}$ . If no $T_{W}$ $T_{W}$ is a reflection, then T is the identity operator.

Exercises

Label the following statements as true or false. Assume that the underlying vector spaces are one or two-dimensional real inner product spaces.
1. (a) Any orthogonal operator is either a rotation or a reflection.
2. (b) The composite of any two rotations is a rotation.
3. (c) The identity operator is a rotation.
4. (d) The composite of two reflections is a reflection.
5. (e) Any orthogonal operator is a composite of rotations.
6. (f) For any orthogonal operator T, if $\det (T) = - 1$ $\det (T) = - 1$ , then T is a reflection.
7. (g) Reflections always have eigenvalues.
8. (h) Rotations always have eigenvalues.
9. (i) If T is an operator on a 2-dimensional space V and W is a subspace of dimension 1 such that T is a reflection of V about $W^{⊥}$ $W^{⊥}$ , then W is the eigenspace of T corresponding to the eigenvalue $λ = - 1$ $λ = - 1$ .
10. (j) The composite of an orthogonal operator and a translation is an orthogonal operator.
Prove that rotations, reflections, and composites of rotations and reflections are orthogonal operators.
Let

$A = (\begin{array}{r} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & - \frac{1}{2} \end{array}) and B = (\begin{array}{r} 1 & 0 \\ 0 & - 1 \end{array}) .$ $A = (\begin{array}{r} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & - \frac{1}{2} \end{array}) and B = (\begin{array}{r} 1 & 0 \\ 0 & - 1 \end{array}) .$
1. (a) Prove that $L_{A}$ $L_{A}$ is a reflection.
2. (b) Find the subspace of $R^{2}$ $R^{2}$ on which $L_{A}$ $L_{A}$ acts as the identity.
3. (c) Prove that $L_{A B}$ $L_{A B}$ and $L_{B A}$ $L_{B A}$ are rotations.
For any real number $ϕ$ $ϕ$ , let

$A = (\begin{array}{r} \cos ϕ & \sin ϕ \\ \sin ϕ & - \cos ϕ \end{array}) .$ $A = (\begin{array}{r} \cos ϕ & \sin ϕ \\ \sin ϕ & - \cos ϕ \end{array}) .$
1. (a) Prove that $L_{A}$ $L_{A}$ is a reflection.
2. (b) Find the axis in $R^{2}$ $R^{2}$ about which $L_{A}$ $L_{A}$ reflects.
For any real number $ϕ$ $ϕ$ , define $T_{ϕ} = L_{A}$ $T_{ϕ} = L_{A}$ , where

$A = (\begin{array}{r} \cos ϕ & - \sin ϕ \\ \sin ϕ & \cos ϕ \end{array}) .$ $A = (\begin{array}{r} \cos ϕ & - \sin ϕ \\ \sin ϕ & \cos ϕ \end{array}) .$
1. (a) Prove that any rotation on $R^{2}$ $R^{2}$ is of the form $T_{ϕ}$ $T_{ϕ}$ for some $ϕ$ $ϕ$ .
2. (b) Prove that $T_{ϕ} T_{ψ} = T_{(ϕ + ψ)}$ $T_{ϕ} T_{ψ} = T_{(ϕ + ψ)}$ for any $ϕ, ψ \in R$ $ϕ, ψ \in R$ .
3. (c) Deduce that any two rotations on $R^{2}$ $R^{2}$ commute.
Prove that if T is a rotation on a 2-dimensional inner product space, then —T is also a rotation.
Prove that if T is a reflection on a 2-dimensional inner product space, then $T^{2}$ $T^{2}$ is the identity operator.
Prove Theorem 6.46 using the hints preceding the statement of the theorem.
Prove that no orthogonal operator can be both a rotation and a reflection.
Prove that if V is a two-dimensional real inner product space, then the composite of two reflections on V is a rotation of V.
Let V be a one- or a two-dimensional real inner product space. Define $T : V \to V$ $T : V \to V$ by $T (x) = - x$ $T (x) = - x$ . Prove that T is a rotation if and only if $\dim (V) = 2$ $\dim (V) = 2$ .
Complete the proof of the lemma to Theorem 6.47 by showing that $W = ϕ_{β}^{- 1} (Z)$ $W = ϕ_{β}^{- 1} (Z)$ satisfies the required conditions.
Let T be an orthogonal [unitary] operator on a finite-dimensional real [complex] inner product space V. If W is a T-invariant subspace of V, prove the following results.
1. (a) $T_{W}$ $T_{W}$ is an orthogonal [unitary] operator on W.
2. (b) $W^{⊥}$ $W^{⊥}$ is a T-invariant subspace of V. Hint: Use the fact that $T_{W}$ $T_{W}$ is one-to-one and onto to conclude that, for any $y \in W, T* (y) = T^{- 1} (y) \in W$ $y \in W, T* (y) = T^{- 1} (y) \in W$ .
3. (c) $T_{W^{⊥}}$ $T_{W^{⊥}}$ is an orthogonal [unitary] operator on W.
Let T be a linear operator on a finite-dimensional vector space V, where V is a direct sum of T-invariant subspaces, say, $V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{k}$ $V = W_{1} \oplus W_{2} \oplus \dots \oplus W_{k}$ . Prove that $\det (T) = \det (T_{W_{1}}) \cdot \det (T_{W_{2}}) \cdot \dots \cdot \det (T_{W_{k}})$ $\det (T) = \det (T_{W_{1}}) \cdot \det (T_{W_{2}}) \cdot \dots \cdot \det (T_{W_{k}})$ .
Complete the proof of the corollary to Theorem 6.48.
Let V be a real inner product space of dimension 2. For any $x, y \in V$ $x, y \in V$ such that $x \neq y$ $x \neq y$ and $| | x | | = | | y | | = 1$ $| | x | | = | | y | | = 1$ , show that there exists a unique rotation T on V such that $T (x) = y$ $T (x) = y$ . Visit goo.gl/ahQT67 for a solution.

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Table of Contents for 6.11* The Geometry of Orthogonal Operators

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Table of Contents for
6.11* The Geometry of Orthogonal Operators