Problem 25 - Algebra: Pure and Applied

Question

For \(\theta \in \mathbb{R}\), let \(A(\theta) \in S L(2, \mathbb{R})\) be the matrix representing a rotation of \(\theta\) radians: $$ A(\theta)=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] $$ (a) Show that \(H=\\{A(\theta) \mid \theta \in \mathbb{R}\\}\) is a subgroup of the special linear group \(S L(2, \mathbb{R})\) (b) Find the inverse of \(A(2 \pi / 3)\). (c) Find the order of \(A(2 \pi / 3)\).

Step-by-Step Solution

Answer

(a) H is a subgroup. (b) Inverse is \( \left[ \begin{array}{cc} -1/2 & \sqrt{3}/2 \\ -\sqrt{3}/2 & -1/2 \end{array} \right] \). (c) Order is 3.

Step 1: Verify Identity Element

To confirm that \( H \) is a subgroup of \( SL(2, \mathbb{R}) \), check for the identity element. The identity matrix \( I \) is given by \( A(0) = \left[\begin{array}{cc} 1 & 0 \ 0 & 1 \end{array}\right] \). Since \( \cos(0) = 1 \) and \( \sin(0) = 0 \), \( I \) is in \( H \). Thus, the identity element is present.

Step 2: Verify Closure Under Multiplication

For closure, multiply two arbitrary elements \( A(\theta) \) and \( A(\phi) \) in \( H \): \[A(\theta)A(\phi) = \left[\begin{array}{cc} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \phi & -\sin \phi \ \sin \phi & \cos \phi \end{array}\right]\]After performing matrix multiplication, \[A(\theta)A(\phi) = \left[\begin{array}{cc} \cos(\theta+\phi) & -\sin(\theta+\phi) \ \sin(\theta+\phi) & \cos(\theta+\phi) \end{array}\right] = A(\theta + \phi)\]Since \( A(\theta + \phi) \in H \), closure is satisfied.

Step 3: Verify Existence of Inverses

The inverse of \( A(\theta) \) in \( H \) is \( A(-\theta) \), because \[A(\theta)A(-\theta) = \left[\begin{array}{cc} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos(-\theta) & -\sin(-\theta) \ \sin(-\theta) & \cos(-\theta) \end{array}\right] = I\]Hence, each element has an inverse in \( H \), verifying it is a subgroup.

Step 4: Calculate the Inverse of \(A(2\pi/3)\)

The inverse of a rotation matrix \( A(\theta) \) is \( A(-\theta) \). Thus, \[A(2\pi/3)^{-1} = A(-2\pi/3) = \left[\begin{array}{cc} \cos(-2\pi/3) & -\sin(-2\pi/3) \ \sin(-2\pi/3) & \cos(-2\pi/3) \end{array}\right] = \left[\begin{array}{cc} -1/2 & \sqrt{3}/2 \ -\sqrt{3}/2 & -1/2 \end{array}\right]\]

Step 5: Find the Order of \(A(2\pi/3)\)

The order of an element \( A(\theta) \) in \( H \) is the smallest positive integer \( n \) such that \( A(n\theta) = I \). This means \( n\theta = 2\pi k \) for some integer \( k \). Solving for \( n \) with \( \theta = 2\pi/3 \), we find \[ n\cdot2\pi/3 = 2\pi \implies n = 3 \]Thus, \( A(2\pi/3) \) has order 3.

Key Concepts

Rotation Matrix

A rotation matrix is a fundamental concept in linear algebra. It's a matrix that represents a rotation of points in the Cartesian coordinate system. For a rotation by an angle \( \theta \), the associated rotation matrix \( A(\theta) \) is structured as follows:

• The first row is \([ \cos \theta, -\sin \theta ]\).
• The second row is \([ \sin \theta, \cos \theta ]\).

This matrix operates in the plane and rotates points around the origin. Specifically, for counterclockwise rotation in two dimensions, this matrix form succinctly encodes the rotation transformation.
When \( \theta = 0 \), it represents the identity matrix, ensuring no change to the points it's applied to. It's important to appreciate that the determinant of a rotation matrix is always 1, making it part of the special linear group \( SL(2, \mathbb{R}) \). The components \( \cos \theta \) and \( \sin \theta \) adhere to trigonometric identities, supporting the integrity and functionality of such transformations.

Subgroups

A subgroup is a smaller group formed within a larger algebraic structure—in this case, within the special linear group \( SL(2, \mathbb{R}) \). To qualify as a subgroup, certain properties must hold:

• Identity Element: The subgroup must contain an identity element. Here, the identity matrix \( A(0) \) serves this role.
• Closure: The group is closed under the group operation. This means multiplying two matrices, like \( A(\theta) \) and \( A(\phi) \), results in another matrix within the subgroup, namely \( A(\theta + \phi) \).
• Inverse Element: For every element in the group, there should be another one inside the group that when multiplied together yield the identity element. For any \( \theta \), \( A(-\theta) \) is the inverse of \( A(\theta) \).

By confirming these criteria, we can confidently say that \( H = \{A(\theta) \mid \theta \in \mathbb{R}\} \) constitutes a subgroup of \( SL(2, \mathbb{R}) \). This structure is particularly useful in mathematical contexts that involve rotative symmetries and any algebraic operations abiding by rotation principles.

Inverse Matrix

The concept of an inverse matrix is pivotal in solving systems of linear equations and understanding transformations. For a rotation matrix \( A(\theta) \), the inverse can be found using the property that rotating by \( -\theta \) undoes the rotation by \( \theta \). Specifically, the inverse matrix is:\[A(\theta)^{-1} = A(-\theta) = \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \ \sin(-\theta) & \cos(-\theta) \end{bmatrix}\]When \( \theta = 2\pi/3 \), you can calculate the inverse matrix by using trigonometric values:

• \( \cos(-2\pi/3) = -1/2 \) and \( \sin(-2\pi/3) = -\sqrt{3}/2 \)
• The inverse matrix \( A(2\pi/3)^{-1} \) becomes:\[\begin{bmatrix} -1/2 & \sqrt{3}/2 \ -\sqrt{3}/2 & -1/2 \end{bmatrix}\]

Understanding how to derive this inverse helps in areas such as computer graphics, where reversing the direction of a rotation is often required.

Matrix Order

Matrix order in the context of groups is a concept related to determining the smallest integer \( n \) such that applying the operation of the matrix \( A(\theta) \) repeatedly \( n \) times results in the identity matrix. This is also akin to determining the number of steps needed to return to the starting point in rotations.
For \( A(\theta) \) where \( \theta = 2\pi/3 \), the order is found by ensuring \( A(n \cdot 2\pi/3) = I \) for some integer \( n \). Equating \( n \cdot 2\pi/3 \) to full rotations (multiples of \( 2\pi \)), we solve
\[n \cdot \frac{2\pi}{3} = 2\pi k \Rightarrow n = 3n\]This result implies that three applications of \( A(2\pi/3) \) cycle the transformation back to the identity matrix. Such an understanding of matrix order is useful in cyclical processes and symmetry operations where detecting repetition and periodicity is essential.