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This section is divided into a number of subsections, links to which are:

# Rotations

An in-depth study of rotations can become a lifelong project so we restrict our self to rotations in ℝ² and ℝ³. Unlike two dimensional case, rotations do not commute. For instance, Ry(45°)∘Rx(90°) ≠ Rx(90°)∘Ry(45°), where Rx(θ) is rotation around the x-axis by θ in ℝ³. (As usual, B∘A means the composition of action A followed by action B.) By a rotation, we always mean a linear transformation in either ℝ² or ℝ³. It is a customary to define rotation to be positive if rotation occurs in counterclockwise direction.

The general linear group associated to 𝔽n is the set of invertible matrices in 𝔽n×n under matrix multiplication. This groups is usually denoted as GLn(𝔽) or just GLn.

Matrices that correspond to rotation operations provide a very valuable example of special class of matrices.

The orthogonal group associated to 𝔽n is $O_n \left( \mathbb{F} \right) = \left\{ \mathbf{A} \in \mbox{GL}_n (\mathbb{F}) \ : \ \mathbf{A}^{\mathrm T} = \mathbf{A}^{-1} \right\} .$

Example 1: Let us consider an arbitrary 2-by-2 matrix from O2(ℝ) $\mathbf{A} = \begin{bmatrix} a&b \\ c&d \end{bmatrix} ,$ where 𝑎, b, c, and d are some real numbers. Then we have $\mathbf{A}\,\mathbf{A}^{\mathrm T} = \begin{bmatrix} a^2 + b^2 & ac + bd \\ ac + bd & c^2 + d^2 \end{bmatrix}$ Since this product of these two matrices must be equal the identity matrix, we get the conditions: \begin{align*} a^2 + b^2 &= 1 , \\ c^2 + d^2 &= 1, \\ ac + bd &= 0. \end{align*} From first two equations, we derive $a = \pm \cos\varphi , \quad b = \pm \sin\varphi . \quad c = \pm \cos \psi , \quad d= \pm\sin\psi .$ Eﬀectively, there are two possibilities for the sign choices in the latter in order to satisfy 𝑎c + bd = 0. One is $\cos\varphi \,\cos\psi + \sin\varphi \,\sin\psi = \cos \left( \varphi - \psi \right) = 0 .$ The second is $\cos\varphi \,\cos\psi - \sin\varphi \,\sin\psi = \cos \left( \varphi + \psi \right) = 0 .$ In the ﬁrst case, we can take φ − ψ = π/2 and use trigonometric identities to ﬁnd that cos ψ = sin φ and sin ψ = − cos φ. We then have $\mathbf{A} = \begin{bmatrix} \cos\varphi & \sin\varphi \\ \sin\varphi & -\cos\varphi \end{bmatrix} , \qquad \det\mathbf{A} = -1. .$ The same approach applies in the second case. We can take φ + ψ = π/2 and use identities to get cos ψ = − sin φ and sin ψ = cos φ. This gives us $\mathbf{A} = \begin{bmatrix} \cos\varphi & \sin\varphi \\ -\sin\varphi & \cos\varphi \end{bmatrix} , \qquad \det\mathbf{A} = 1. .$ Some examples of matrices in O₂(ℝ), aside from the identity matrix I₂, are $\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} , \qquad \begin{bmatrix} 0&1 \\ -1& 0 \end{bmatrix} , \qquad \begin{bmatrix} 1&0 \\ 0& -1 \end{bmatrix} , \qquad \begin{bmatrix} -1&0 \\ 0& 1 \end{bmatrix} .$ Later, we will see that each element in O₂(ℝ) is either a rotation or an orthogonal reﬂection.    ■
End of Example 1
The previous example shows that rotations on ℝ² are commutative. We treat the identity matrix as a rotation through 0 radians. With that, we can say that the rotations on ℝ² are the mappings in O₂(ℝ) with determinant 1. Since det(A B) = detA detB for matrices A and B when the product A B is deﬁned, and since det(A−1 = 1/det(A when det(A) ≠ 0, we see that the rotations on ℝ² form a subgroup of O₂(ℝ). That subgroup is denoted SO₂(ℝ) and is called the special orthogonal group on ℝ².

In general, the orthogonal group in dimension n has two connected components. The one that contains the identity element is a normal subgroup, called the special orthogonal group, and denoted SO₂(ℝ) or SO(n). It consists of all orthogonal matrices of determinant 1. This group is also called the rotation group, generalizing the fact that in dimensions 2 and 3, its elements are the usual rotations around a point (in dimension 2) or a line (in dimension 3). In low dimension, these groups have been widely studied, see SO(2), SO(3) and SO(4). The other component consists of all orthogonal matrices of determinant −1. This component does not form a group, as the product of any two of its elements is of determinant 1, and therefore not an element of the component.

1. Show that GLn(𝔽) is not a vector space.
2. Let $\mathbf{A} = \begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} , \quad \mathbf{v} = \begin{bmatrix} 1 + \cos\theta \\ \sin\theta \end{bmatrix} , \quad \mathbf{x} = \begin{bmatrix} \sin\theta \\ 1 - \cos\theta \end{bmatrix} .$ Show that A v is a scalar multiple of v and that A x is a scalar multiple of x.

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