# Recitation 9B

Denote a 2\times 2 orthogonal matrix by \begin{bmatrix}a & c\\ b & d\end{bmatrix}. Since it is an orthogonal matrix, the entries satisfy a^2+b^2=1, ac+bd=0, c^2+d^2=1. Since a^2+b^2=1, it is standard to parametrize a and b by a=\cos\theta and b=\sin\theta. Using the last two equations, we can get c=\mp \sin\theta, d=\pm\cos\theta. In other words, we have two types of orthogonal matrix A = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}, B = \begin{bmatrix}\cos\theta & \sin\theta \\ \sin\theta & -\cos\theta\end{bmatrix}.

Problem 1: What does the linear transformation S that sends x to Ax do? What about the linear transformation T that sends x to A^2x? Use these results to recover the double angle formulas for trigonometric functions.

Solution: Since the linear transformation S sends \begin{bmatrix}1\\0\end{bmatrix} to \begin{bmatrix}\cos\theta \\ \sin\theta\end{bmatrix} and \begin{bmatrix}0\\1\end{bmatrix} to \begin{bmatrix}-\sin\theta \\ \cos\theta\end{bmatrix}S is a rotation about the origin by an angle of \theta. Since T which is a linear transformation that sends x to Ax and then to A(Ax), T is a composition of S with S itself. Hence T is a rotation about the origin by an angle of 2\theta. On one hand, the matrix associated to T is \begin{bmatrix}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta\end{bmatrix}. On the other hand, the matrix associated to T is A^2=\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}^2=\begin{bmatrix}\cos^2\theta-\sin^2\theta & -2\sin\theta\cos\theta \\2\sin\theta\cos\theta & \cos^2\theta-\sin^2\theta\end{bmatrix}. Comparing the entries, we obtain \cos 2\theta = \cos^2\theta - \sin^2\theta, \sin 2\theta = 2\sin\theta\cos\theta.

Problem 2: What are the eigenvalues and correspondent eigenvectors of the matrix B? How are those two eigenvectors related to each other? What does the linear transformation S that sends x to Bx do? What about the linear transformation T that sends x to B^2x?

Solution: To get the eigenvalues of the matrix B, consider the characteristic equation det\left(B-\lambda I\right)=\begin{vmatrix}\cos\theta-\lambda & \sin\theta \\ \sin\theta & -\cos\theta-\lambda\end{vmatrix}=(\cos\theta-\lambda)(-\cos\theta-\lambda)-\sin^2\theta=\lambda^2-1=0. This gives us two eigenvalues \lambda_1 = 1, \lambda_2=-1. For \lambda_1=1, the eigenvector can be \mathbf{u}=(\sin\theta, 1-\cos\theta). For \lambda_2=-1, the eigenvector can be \mathbf{v}=(-\sin\theta, 1+\cos\theta). These two vectors \mathbf{u}, \mathbf{v} are perpendicular to each other because their inner product is 0. Now we have two perpendicular directions \mathbf{u} and \mathbf{v}. On one hand, because the vector \mathbf{u} is correspondent to the eigenvalue 1, S preserves all the vectors who have the same direction as \mathbf{u}. On the other hand, because the vector \mathbf{v} is correspondent to the eigenvalue -1, S reverses the direction of all the vectors who have the same direction as \mathbf{v}. Therefore S is a reflection across the line through the origin with direction \mathbf{u}. Since the composition of two same reflections does nothing, the linear transformation T that sends x to B^2x is an identity map. Also it is easy to check B^2 is indeed an identity matrix.