数学 - Python - 線形代数学 - 線形写像と行列 – 線形写像に対応する行列(恒等写像、時計方向の基底の回転、反時計周りの基底の回転)

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(線形写像に対応する行列)、練習問題3、4、5.を取り組んでみる。

3. 
   $\begin{array}{l}{E}_1\text{'}=\left(\cos \theta \right)E_1+\left(\sin \theta \right)E_2\\ E_2\text{'}=\left(-\sin \theta \right)E_1+\left(\cos \theta \right)E_2\\ \\ \left(\sin \theta \right)E_1\text{'}=\left(\sin \theta \cos \theta \right)E_1+\left({\sin }^2\theta \right)E_2\\ \left(\cos \theta \right)E_2\text{'}=\left(-\sin \theta \cos \theta \right)E_1+\left({\cos }^2\theta \right)E_2\\ \\ \left(\sin \theta \right)E_1\text{'}+\left(\cos \theta \right)E_2\text{'}=E_2\\ \\ \left(\cos \theta \right)E_1\text{'}=\left({\cos }^2\theta \right)E_1+\left(\sin \theta \cos \theta \right)E_2\\ \left(\sin \theta \right)E_2\text{'}=\left(-{\sin }^2\theta \right)E_1+\left(\sin \theta \cos \theta \right)E_2\\ \\ \left(\cos \theta \right)E_1\text{'}-\left(\sin \theta \right)E_2\text{'}=E_1\\ \\ id\left(E_1\right)=E_1=\left(\cos \theta \right)E_1\text{'}+\left(-\sin \theta \right)E_2\text{'}\\ id\left(E_2\right)=E_2=\left(\sin \theta \right)E_1\text{'}+\left(\cos \theta \right)E_2\text{'}\\ \\ \left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\\ \\ \left(\begin{array}{cc}\cos \left(-\theta \right)& \sin \left(-\theta \right)\\ -\sin \left(-\theta \right)& \cos \left(-\theta \right)\end{array}\right)=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\end{array}$
4. 
   $\begin{array}{l}{E}_1\text{'}=\left(\cos \frac{\pi }{4}\right)E_1+\left(\sin \frac{\pi }{4}\right)E_2\\ =\frac{1}{\sqrt{2}}{E}_1+\frac{1}{\sqrt{2}}{E}_2\\ E_{2^{\prime }}\text{'}=\left(-\sin \frac{\pi }{4}\right)E_1+\left(\cos \frac{\pi }{4}\right)E_2\\ =-\frac{1}{\sqrt{2}}{E}_1+\frac{1}{\sqrt{2}}{E}_2\\ \\ M=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\\ F\left(\left(\begin{array}{c}1\\ 2\end{array}\right)\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{c}1\\ 2\end{array}\right)\\ =\left(\begin{array}{c}-\frac{1}{\sqrt{2}}\\ \frac{3}{\sqrt{2}}\end{array}\right)\end{array}$
5. 
   $\begin{array}{l}{E}_1\text{'}=\left(\cos \frac{\pi }{2}\right)E_1+\left(\sin \frac{\pi }{2}\right)E_2\\ =0E_1+E_2\\ \\ E_{2^{\prime }}\text{'}=\left(-\sin \frac{\pi }{2}\right)E_1+\left(\cos \frac{\pi }{2}\right)E_2\\ =-E_1+0E_2\\ \\ F\left(\left(\begin{array}{c}-1\\ 3\end{array}\right)\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}-1\\ 3\end{array}\right)\\ =\left(\begin{array}{c}-3\\ -1\end{array}\right)\end{array}$

コード(Emacs)

Python 3

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

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

Θ = symbols('Θ')
M = Matrix([[cos(Θ), -sin(Θ)],
            [sin(Θ), cos(Θ)]])

pprint(M)

t = [(pi / 4, Matrix([1, 2])),
     (pi / 2, Matrix([-1, 3]))]

for i, (Θ0, X) in enumerate(t, 4):
    print('{0}.'.format(i))
    pprint(Θ0)
    M0 = M.subs({Θ: Θ0})
    pprint(M0)
    pprint(X)
    pprint(M0 * X)
    print()

入出力結果(Terminal, IPython)

$ ./sample3.py
⎡cos(Θ)  -sin(Θ)⎤
⎢               ⎥
⎣sin(Θ)  cos(Θ) ⎦
4.
π
─
4
⎡√2  -√2 ⎤
⎢──  ────⎥
⎢2    2  ⎥
⎢        ⎥
⎢√2   √2 ⎥
⎢──   ── ⎥
⎣2    2  ⎦
⎡1⎤
⎢ ⎥
⎣2⎦
⎡-√2 ⎤
⎢────⎥
⎢ 2  ⎥
⎢    ⎥
⎢3⋅√2⎥
⎢────⎥
⎣ 2  ⎦

5.
π
─
2
⎡0  -1⎤
⎢     ⎥
⎣1  0 ⎦
⎡-1⎤
⎢  ⎥
⎣3 ⎦
⎡-3⎤
⎢  ⎥
⎣-1⎦

$