Skip to main content

Recitation 9B

Denote a 2×22\times 2 orthogonal matrix by [acbd]\begin{bmatrix}a & c\\ b & d\end{bmatrix}. Since it is an orthogonal matrix, the entries satisfy a2+b2=1,ac+bd=0,c2+d2=1.a^2+b^2=1, ac+bd=0, c^2+d^2=1. Since a2+b2=1a^2+b^2=1, it is standard to parametrize aa and bb by a=cos⁡θa=\cos\theta and b=sin⁡θb=\sin\theta. Using the last two equations, we can get c=∓sin⁡θ,d=±cos⁡θc=\mp \sin\theta, d=\pm\cos\theta. In other words, we have two types of orthogonal matrix A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ],B=[cos⁡θsin⁡θsin⁡θ−cos⁡θ].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 SS that sends xx to AxAx do? What about the linear transformation TT that sends xx to A2xA^2x? Use these results to recover the double angle formulas for trigonometric functions.

Solution: Since the linear transformation SS sends [10]\begin{bmatrix}1\\0\end{bmatrix} to [cos⁡θsin⁡θ]\begin{bmatrix}\cos\theta \\ \sin\theta\end{bmatrix} and [01]\begin{bmatrix}0\\1\end{bmatrix} to [−sin⁡θcos⁡θ]\begin{bmatrix}-\sin\theta \\ \cos\theta\end{bmatrix}, SS is a rotation about the origin by an angle of θ\theta. Since TT which is a linear transformation that sends xx to AxAx and then to A(Ax)A(Ax), TT is a composition of SS with SS itself. Hence TT is a rotation about the origin by an angle of 2θ2\theta. On one hand, the matrix associated to TT is [cos⁡2θ−sin⁡2θsin⁡2θcos⁡2θ]\begin{bmatrix}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta\end{bmatrix}. On the other hand, the matrix associated to TT is A2=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]2=[cos⁡2θ−sin⁡2θ−2sin⁡θcos⁡θ2sin⁡θcos⁡θcos⁡2θ−sin⁡2θ]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⁡2θ=cos⁡2θ−sin⁡2θ,sin⁡2θ=2sin⁡θcos⁡θ.\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 BB? How are those two eigenvectors related to each other? What does the linear transformation SS that sends xx to BxBx do? What about the linear transformation TT that sends xx to B2xB^2x?

Solution: To get the eigenvalues of the matrix BB, consider the characteristic equation det(B−λI)=∣cos⁡θ−λsin⁡θsin⁡θ−cos⁡θ−λ∣=(cos⁡θ−λ)(−cos⁡θ−λ)−sin⁡2θ=λ2−1=0det\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 λ1=1,λ2=−1\lambda_1 = 1, \lambda_2=-1. For λ1=1\lambda_1=1, the eigenvector can be u=(sin⁡θ,1−cos⁡θ)\mathbf{u}=(\sin\theta, 1-\cos\theta). For λ2=−1\lambda_2=-1, the eigenvector can be v=(−sin⁡θ,1+cos⁡θ)\mathbf{v}=(-\sin\theta, 1+\cos\theta). These two vectors u,v\mathbf{u}, \mathbf{v} are perpendicular to each other because their inner product is $$. Now we have two perpendicular directions u\mathbf{u} and v\mathbf{v}. On one hand, because the vector u\mathbf{u} is correspondent to the eigenvalue 11, SS preserves all the vectors who have the same direction as u\mathbf{u}. On the other hand, because the vector v\mathbf{v} is correspondent to the eigenvalue −1-1, SS reverses the direction of all the vectors who have the same direction as v\mathbf{v}. Therefore SS is a reflection across the line through the origin with direction u\mathbf{u}. Since the composition of two same reflections does nothing, the linear transformation TT that sends xx to B2xB^2x is an identity map. Also it is easy to check B2B^2 is indeed an identity matrix.