数学 - Python - 線形代数学 - 線形写像と行列 – 線形写像の合成(正弦と余弦の加法定理、ノルムの保存、基底と恒等写像に対応する行列)

ラング線形代数学(上)
原著

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax
• 参考書籍
  • Pythonからはじめる数学入門
  • JavaScriptリファレンス 第6版
  • インタラクティブ・データビジュアライゼーション

ラング線形代数学(上)(S.ラング (著)、芹沢 正三 (翻訳)、ちくま学芸文庫)の5章(線形写像と行列)、3(線形写像の合成)、練習問題8.を取り組んでみる。

1. 
   $\begin{array}{l}{F}_{\theta }{F}_{\theta }\text{'}\\ =\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}\cos \theta \text{'}& -\sin \theta \text{'}\\ \sin \theta \text{'}& \cos \theta \text{'}\end{array}\right)\\ =\left(\begin{array}{cc}\cos \theta \cos {\theta }^{\prime }-\sin \theta \sin \theta \text{'}& -\cos \theta \sin \theta \text{'}-\sin \theta \cos {\theta }^{\prime }\\ \sin \theta \cos \theta \text{'}+\cos \theta \sin \theta \text{'}& -\sin \theta \sin \theta \text{'}+\cos \theta \cos {\theta }^{\prime }\end{array}\right)\\ =\left(\begin{array}{cc}\cos \left(\theta +{\theta }^{\prime }\right)& -\sin \left(\theta +\theta \text{'}\right)\\ \sin \left(\theta +{\theta }^{\prime }\right)& \cos \left(\theta +{\theta }^{\prime }\right)\end{array}\right)\\ =F_{\theta +{\theta }^{\prime }}\end{array}$
   
   $\begin{array}{l}{F}_{\theta }{F}_{-\theta }\\ =\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}\cos \left(-\theta \right)& -\sin \left(-\theta \right)\\ \sin \left(-\theta \right)& \cos \left(-\theta \right)\end{array}\right)\\ =\left(\begin{array}{cc}\cos \theta \cos \left(-\theta \right)-\sin \theta \sin \left(-\theta \right)& -\cos \theta \sin \left(-\theta \right)-\sin \theta \cos \left(-\theta \right)\\ \sin \theta \cos \left(-\theta \right)+\cos \theta \sin \left(-\theta \right)& -\sin \theta \sin \left(-\theta \right)+\cos \theta \cos \left(-\theta \right)\end{array}\right)\\ =\left(\begin{array}{cc}\cos \theta \cos \theta +\sin \theta \sin \theta & \sin \theta \cos \theta -\cos \theta \sin \theta \\ \sin \theta \cos \theta -\cos \theta \sin \theta & \sin \theta \sin \theta +\cos \theta \cos \theta \end{array}\right)\\ =\left(\begin{array}{cc}\cos \left(\theta -\theta \right)& \sin \left(\theta -\theta \right)\\ \sin \left(\theta -\theta \right)& \cos \left(\theta -\theta \right)\end{array}\right)\\ =\left(\begin{array}{cc}\cos 0& \sin 0\\ \sin 0& \cos 0\end{array}\right)\\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\end{array}$
2. 
   $\begin{array}{l}X=\left(x,y\right)\\ F_{\theta }\left(X\right)=\left(\begin{array}{c}x\cos \theta -y\sin \theta \\ x\sin \theta +y\cos \theta \end{array}\right)\\ \Vert X\Vert =\sqrt{x^2+y^2}\\ \Vert F_{\theta }\left(X\right)\Vert \\ =\sqrt{{\left(x\cos \theta -y\sin \theta \right)}^2+{\left(x\sin \theta +y\cos \theta \right)}^2}\\ =\sqrt{x^2{\cos }^2\theta +y^2{\sin }^2\theta -2xy{\sin }^2\theta +x^2{\sin }^2\theta +y^2{\cos }^2\theta +2xy\sin \theta \cos \theta }\\ =\sqrt{x^2\left({\sin }^2\theta +{\cos }^2\theta \right)+y^2\left({\sin }^2\theta +{\cos }^2\theta \right)}\\ =\sqrt{x^2+y^2}\end{array}$
3. 1. 
      $\begin{array}{l}id\left(1,1,0\right)={}^t\left(1,1,0\right)=a_{11}\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)+a_{21}\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)+a_{31}\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)\\ id\left(-1,1,1\right)={}^t\left(-1,1,1\right)=a_{12}\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)+a_{22}\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)+a_{32}\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)\\ id\left(0,1,2\right)={}^t\left(0,1,2\right)=a_{13}\left(\begin{array}{c}2\\ 1\\ 1\end{array}\right)+a_{23}\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)+a_{33}\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)\\ 2a_{11}-a_{31}=1\\ a_{31}=2a_{11}-1\\ \\ a_{11}+2a_{11}-1=1\\ a_{11}=\frac{2}{3}\\ a_{31}=\frac{1}{3}\\ \\ \frac{2}{3}+a_{21}+\frac{1}{3}=0\\ a_{21}=-1\\ \\ 2a_{12}-a_{32}=-1\\ a_{32}=2a_{12}+1\\ \\ a_{12}+2a_{12}+1=1\\ a_{12}=0\\ a_{32}=1\\ \\ a_{22}+1=1\\ a_{22}=0\\ \\ 2a_{13}-a_{33}=0\\ a_{33}=2a_{13}\\ \\ a_{13}+2a_{13}=1\\ a_{13}=\frac{1}{3}\\ a_{33}=\frac{2}{3}\\ \\ \frac{1}{3}+a_{23}+\frac{2}{3}=2\\ a_{23}=1\\ \\ \left(\begin{array}{ccc}\frac{2}{3}& 0& \frac{1}{3}\\ -1& 0& 1\\ \frac{1}{3}& 1& \frac{2}{3}\end{array}\right)\end{array}$
   2. 
      $\begin{array}{l}id\left(3,2,1\right)={}^t\left(3,2,1\right)=a_{11}\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)+a_{21}\left(\begin{array}{c}-1\\ 2\\ 4\end{array}\right)+a_{31}\left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)\\ id\left(0,-2,5\right)={}^t\left(0,-2,5\right)=a_{12}\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)+a_{22}\left(\begin{array}{c}-1\\ 2\\ 4\end{array}\right)+a_{32}\left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)\\ id\left(1,1,2\right)={}^t\left(1,1,2\right)=a_{13}\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)+a_{23}\left(\begin{array}{c}-1\\ 2\\ 4\end{array}\right)+a_{33}\left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)\\ a_{11}-a_{21}+2a_{31}=3\\ a_{11}=a_{21}-2a_{31}+3\\ \\ a_{21}-2a_{31}+3+2a_{21}-a_{31}=2\\ 3a_{21}-3a_{31}=-1\\ a_{31}=a_{21}+\frac{1}{3}\\ a_{11}=a_{21}-2a_{21}-\frac{2}{3}+3=-a_{21}+\frac{7}{3}\\ \\ 4a_{21}+a_{21}+\frac{1}{3}=1\\ 5a_{21}=\frac{2}{3}\\ a_{21}=\frac{2}{15}\\ a_{11}=-\frac{2}{15}+\frac{7}{3}=\frac{33}{15}\\ a_{31}=\frac{2}{15}+\frac{1}{3}=\frac{7}{15}\\ \\ a_{12}-a_{22}+2a_{32}=0\\ a_{12}=a_{22}-2a_{32}\\ \\ a_{22}-2a_{32}+2a_{22}-a_{32}=-2\\ 3a_{22}-3a_{32}=-2\\ a_{22}=a_{32}-\frac{2}{3}\\ a_{12}=a_{32}-\frac{2}{3}-2a_{32}=-a_{32}-\frac{2}{3}\\ \\ 4a_{32}-\frac{8}{3}+a_{32}=5\\ a_{32}=1+\frac{8}{15}=\frac{23}{15}\\ a_{12}=-\frac{23}{15}-\frac{2}{3}=-\frac{-33}{15}\\ a_{22}=\frac{23}{15}-\frac{2}{3}=\frac{13}{15}\\ \\ a_{13}-a_{23}+2a_{33}=1\\ a_{13}=a_{23}-2a_{33}+1\\ \\ a_{23}-2a_{33}+1+2a_{23}-a_{33}=1\\ 3a_{23}-3a_{33}=0\\ a_{33}=a_{23}\\ a_{13}=a_{23}-2a_{23}+1=-a_{23}+1\\ \\ 4a_{23}+a_{23}=2\\ a_{23}=\frac{2}{5}\\ a_{33}=\frac{2}{5}\\ a_{13}=\frac{3}{5}\\ \\ \left(\begin{array}{ccc}\frac{33}{15}& -\frac{33}{15}& \frac{3}{5}\\ \frac{2}{15}& \frac{13}{15}& \frac{2}{5}\\ \frac{7}{15}& \frac{23}{15}& \frac{2}{5}\end{array}\right)\end{array}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

from sympy import pprint, symbols, Matrix, sin, cos, Rational

print('1.')
Θ, Θ1 = symbols('Θ Θ1')

M = Matrix([[cos(Θ), -sin(Θ)],
            [sin(Θ), cos(Θ)]])
M1 = M.subs({Θ: Θ1})

A = M * M1
B = M.subs({Θ: Θ + Θ1})
pprint(A)
pprint(B)

print('2.')
x, y = symbols('x y')
X = Matrix([x, y])
a = (M * X).norm()
b = X.norm()
pprint(a)
pprint(b)

入出力結果(Terminal, IPython)

$ ./sample1.py
1.
⎡-sin(Θ)⋅sin(Θ1) + cos(Θ)⋅cos(Θ1)  -sin(Θ)⋅cos(Θ1) - sin(Θ1)⋅cos(Θ)⎤
⎢                                                                  ⎥
⎣sin(Θ)⋅cos(Θ1) + sin(Θ1)⋅cos(Θ)   -sin(Θ)⋅sin(Θ1) + cos(Θ)⋅cos(Θ1)⎦
⎡cos(Θ + Θ1)  -sin(Θ + Θ1)⎤
⎢                         ⎥
⎣sin(Θ + Θ1)  cos(Θ + Θ1) ⎦
2.
   _________________________________________________
  ╱                      2                        2 
╲╱  │x⋅sin(Θ) + y⋅cos(Θ)│  + │x⋅cos(Θ) - y⋅sin(Θ)│  
   _____________
  ╱    2      2 
╲╱  │x│  + │y│  
$