Hiroshima-University
広島大学 先進理工系科学研究科 情報科学プログラム 2021年1月実施 専門科目I 問題1
Author samparker, 祭音Myyura
Description 2次元の回転行列 \(R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\) について以下の問いに答えよ。
(1) \(R(\theta)\) のすべての固有値と対応する固有ベクトルを求めよ。
(2) \(R(\theta)\) をユニタリ行列 \(U\) を用いて対角化せよ。
(3) 対角化の結果を用いて、\(R(\theta)^n\) を求めよ。ただし、\(n\) は正の整数とする。
(4) 対角化の結果を用いて、正弦および余弦の加法定理を導出せよ。
Let \(R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\) be the rotation matrix in the two-dimensional Euclidean space.
(1) Find all the eigenvalues of the matrix \(R(\theta)\) and the corresponding eigenvectors.
(2) Diagonalize the matrix \(R(\theta)\) by using a unitary matrix \(U\) .
(3) Find the \(R(\theta)^n\) for the natural number \(n\) by using the result of the diagonalization.
(4) Derive the sum formulas for the matrix \(R(\theta)\) and cosine by using the result of the diagonalization.
Kai Let \(A\) denote matrix \(R(\theta)\) and \(E\) denote the identity matrix.
(1)
Eigenvalues:
\[
\begin{aligned}
&|A - \lambda E| = 0 \\
&\Rightarrow \begin{vmatrix}
\cos \theta - \lambda & -\sin \theta \\
\sin \theta & \cos \theta - \lambda
\end{vmatrix} = 0 \\
&\Rightarrow \lambda^2 - 2 \lambda \cos \theta + 1 = 0 \\
&\Rightarrow \begin{cases}
\lambda_1 = \cos \theta - i\sin \theta \\
\lambda_2 = \cos \theta + i\sin \theta
\end{cases}
\end{aligned}
\]
Corresponding eigenvectors:
\[
\begin{cases}
v_1 = (-i, 1) \\
v_2 = (i, 1)
\end{cases}
\]
(2)
\[
U = \frac{1}{\sqrt{2}} \begin{bmatrix}
-i & i \\ 1 & 1
\end{bmatrix}
\]
\[
U^{-1} = (U^{*})^T = \frac{1}{\sqrt{2}} \begin{bmatrix}
i & 1 \\ -i & 1
\end{bmatrix}
\]
\[
U^{-1}AU = \begin{bmatrix}
\cos \theta - i \sin \theta & 0 \\
0 & \cos \theta + i \sin \theta
\end{bmatrix}
\]
(3)
Let \(D\) denote \(\begin{bmatrix} \cos \theta - i \sin \theta & 0 \\ 0 & \cos \theta + i \sin \theta \end{bmatrix}\) . Then
\[
\begin{aligned}
D^n &= \begin{bmatrix}
(\cos \theta - i \sin \theta)^n & 0 \\
0 & (\cos \theta + i \sin \theta)^n
\end{bmatrix} \\
&= \begin{bmatrix}
(e^{-i\theta})^n & 0 \\
0 & (e^{i\theta})^n
\end{bmatrix}
=
\begin{bmatrix}
e^{-in\theta} & 0 \\
0 & e^{in\theta}
\end{bmatrix} \\
&= \begin{bmatrix}
\cos (-n\theta) + i \sin (-n\theta) & 0 \\
0 & \cos (n\theta) + i \sin (n\theta)
\end{bmatrix} \\
&= \begin{bmatrix}
\cos (n\theta) - i \sin (n\theta) & 0 \\
0 & \cos (n\theta) + i \sin (n\theta)
\end{bmatrix}
\end{aligned}
\]
Hence
\[
\begin{aligned}
A^n = UD^nU^{-1} = \begin{bmatrix}
\cos (n\theta) & -\sin (n\theta) \\
\sin (n\theta) & \cos (n\theta)
\end{bmatrix}
\end{aligned}
\]
(4)
\[
\begin{aligned}
R(\alpha) R(\beta) &= U \begin{bmatrix}
\cos \alpha - i \sin \alpha & 0 \\
0 & \cos \alpha + i \sin \alpha
\end{bmatrix}
\begin{bmatrix}
\cos \beta - i \sin \beta & 0 \\
0 & \cos \beta + i \sin \beta
\end{bmatrix} U^{-1} \\
&= U \begin{bmatrix}
e^{-i \alpha} e^{-i \beta} & 0 \\
0 & e^{i \alpha} e^{i \beta}
\end{bmatrix} U^{-1} \\
&= U \begin{bmatrix}
e^{-i (\alpha + \beta)} & 0 \\
0 & e^{i (\alpha + \beta)}
\end{bmatrix} U^{-1} \\
&= U \begin{bmatrix}
\cos (\alpha + \beta) - i \sin (\alpha + \beta) & 0 \\
0 & \cos (\alpha + \beta) + i \sin (\alpha + \beta)
\end{bmatrix} U^{-1} \\
&= R(\alpha + \beta)
\end{aligned}
\]
Hence, we have
\[
\begin{aligned}
R(\alpha) R(\beta) &= \begin{bmatrix}
\cos \alpha \cos \beta - \sin \alpha \sin \beta & -\sin \alpha \cos \beta - \cos \alpha \sin \beta \\
\sin \alpha \cos \beta + \cos \alpha \sin \beta & \cos \alpha \cos \beta - \sin \alpha \sin \beta
\end{bmatrix} \\
&= R(\alpha + \beta) \\
&= \begin{bmatrix}
\cos (\alpha + \beta) & -\sin (\alpha + \beta) \\
\sin (\alpha + \beta) & \cos (\alpha + \beta)
\end{bmatrix}
\end{aligned}
\]
which implies
\[
\begin{aligned}
\sin (\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\
\cos (\alpha + \beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta
\end{aligned}
\]