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 ( θ ) = [ cos θ − sin θ sin θ cos θ ] R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} R ( θ ) = [ cos θ sin θ − sin θ cos θ ]
(1) R ( θ ) R(\theta) R ( θ )
(2) R ( θ ) R(\theta) R ( θ ) U U U
(3) 対角化の結果を用いて、R ( θ ) n R(\theta)^n R ( θ ) n n n n
(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 ( θ ) = [ cos θ − sin θ sin θ cos θ ] R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} R ( θ ) = [ cos θ sin θ − sin θ cos θ ]
(1) Find all the eigenvalues of the matrix R ( θ ) R(\theta) R ( θ )
(2) Diagonalize the matrix R ( θ ) R(\theta) R ( θ ) U U U
(3) Find the R ( θ ) n R(\theta)^n R ( θ ) n n n n
(4) Derive the sum formulas for the matrix R ( θ ) R(\theta) R ( θ )
题目描述
给定二维旋转矩阵
R ( θ ) = [ cos θ − sin θ sin θ cos θ ] . R(\theta)=
\begin{bmatrix}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta
\end{bmatrix}. R ( θ ) = [ cos θ sin θ − sin θ cos θ ] .
求 R ( θ ) R(\theta) R ( θ )
使用酉矩阵 U U U R ( θ ) R(\theta) R ( θ )
利用对角化结果求 R ( θ ) n R(\theta)^n R ( θ ) n n n n
利用对角化结果推导正弦与余弦的加法公式。
Kai Let A A A R ( θ ) R(\theta) R ( θ ) E E E
(1)
Eigenvalues:
∣ A − λ E ∣ = 0 ⇒ ∣ cos θ − λ − sin θ sin θ cos θ − λ ∣ = 0 ⇒ λ 2 − 2 λ cos θ + 1 = 0 ⇒ { λ 1 = cos θ − i sin θ λ 2 = cos θ + i sin θ \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} ∣ A − λ E ∣ = 0 ⇒ cos θ − λ sin θ − sin θ cos θ − λ = 0 ⇒ λ 2 − 2 λ cos θ + 1 = 0 ⇒ { λ 1 = cos θ − i sin θ λ 2 = cos θ + i sin θ Corresponding eigenvectors:
{ v 1 = ( − i , 1 ) v 2 = ( i , 1 ) \begin{cases}
v_1 = (-i, 1) \\
v_2 = (i, 1)
\end{cases} { v 1 = ( − i , 1 ) v 2 = ( i , 1 ) (2)
When sin θ ≠ 0 \sin\theta\neq0 sin θ = 0 v 1 v_1 v 1 v 2 v_2 v 2 θ = k π \theta=k\pi θ = kπ ( − 1 ) k (-1)^k ( − 1 ) k C 2 \mathbb C^2 C 2 U U U
U = 1 2 [ − i i 1 1 ] U = \frac{1}{\sqrt{2}} \begin{bmatrix}
-i & i \\ 1 & 1
\end{bmatrix} U = 2 1 [ − i 1 i 1 ] U − 1 = U ∗ = U ‾ T = 1 2 [ i 1 − i 1 ] U^{-1} = U^*=\overline{U}^{\,T} = \frac{1}{\sqrt{2}} \begin{bmatrix}
i & 1 \\ -i & 1
\end{bmatrix} U − 1 = U ∗ = U T = 2 1 [ i − i 1 1 ] U − 1 A U = [ cos θ − i sin θ 0 0 cos θ + i sin θ ] U^{-1}AU = \begin{bmatrix}
\cos \theta - i \sin \theta & 0 \\
0 & \cos \theta + i \sin \theta
\end{bmatrix} U − 1 A U = [ cos θ − i sin θ 0 0 cos θ + i sin θ ] (3)
Let D D D [ cos θ − i sin θ 0 0 cos θ + i sin θ ] \begin{bmatrix} \cos \theta - i \sin \theta & 0 \\ 0 & \cos \theta + i \sin \theta \end{bmatrix} [ cos θ − i sin θ 0 0 cos θ + i sin θ ]
D n = [ ( cos θ − i sin θ ) n 0 0 ( cos θ + i sin θ ) n ] = [ ( e − i θ ) n 0 0 ( e i θ ) n ] = [ e − i n θ 0 0 e i n θ ] = [ cos ( − n θ ) + i sin ( − n θ ) 0 0 cos ( n θ ) + i sin ( n θ ) ] = [ cos ( n θ ) − i sin ( n θ ) 0 0 cos ( n θ ) + i sin ( n θ ) ] \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} D n = [ ( cos θ − i sin θ ) n 0 0 ( cos θ + i sin θ ) n ] = [ ( e − i θ ) n 0 0 ( e i θ ) n ] = [ e − in θ 0 0 e in θ ] = [ cos ( − n θ ) + i sin ( − n θ ) 0 0 cos ( n θ ) + i sin ( n θ ) ] = [ cos ( n θ ) − i sin ( n θ ) 0 0 cos ( n θ ) + i sin ( n θ ) ] Hence
A n = U D n U − 1 = [ cos ( n θ ) − sin ( n θ ) sin ( n θ ) cos ( n θ ) ] \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} A n = U D n U − 1 = [ cos ( n θ ) sin ( n θ ) − sin ( n θ ) cos ( n θ ) ] (4)
R ( α ) R ( β ) = U [ cos α − i sin α 0 0 cos α + i sin α ] [ cos β − i sin β 0 0 cos β + i sin β ] U − 1 = U [ e − i α e − i β 0 0 e i α e i β ] U − 1 = U [ e − i ( α + β ) 0 0 e i ( α + β ) ] U − 1 = U [ cos ( α + β ) − i sin ( α + β ) 0 0 cos ( α + β ) + i sin ( α + β ) ] U − 1 = R ( α + β ) \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} R ( α ) R ( β ) = U [ cos α − i sin α 0 0 cos α + i sin α ] [ cos β − i sin β 0 0 cos β + i sin β ] U − 1 = U [ e − i α e − i β 0 0 e i α e i β ] U − 1 = U [ e − i ( α + β ) 0 0 e i ( α + β ) ] U − 1 = U [ cos ( α + β ) − i sin ( α + β ) 0 0 cos ( α + β ) + i sin ( α + β ) ] U − 1 = R ( α + β ) Hence, we have
R ( α ) R ( β ) = [ cos α cos β − sin α sin β − sin α cos β − cos α sin β sin α cos β + cos α sin β cos α cos β − sin α sin β ] = R ( α + β ) = [ cos ( α + β ) − sin ( α + β ) sin ( α + β ) cos ( α + β ) ] \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} R ( α ) R ( β ) = [ cos α cos β − sin α sin β sin α cos β + cos α sin β − sin α cos β − cos α sin β cos α cos β − sin α sin β ] = R ( α + β ) = [ cos ( α + β ) sin ( α + β ) − sin ( α + β ) cos ( α + β ) ] which implies
sin ( α + β ) = sin α cos β + cos α sin β cos ( α + β ) = cos α cos β − sin α sin β \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} sin ( α + β ) cos ( α + β ) = sin α cos β + cos α sin β = cos α cos β − sin α sin β