数学 - Python - 線形代数学 - 行列 – 行列の乗法(非可換、対角行列、転置行列の乗法)

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3章(行列)、3(行列の乗法)、練習問題5、6、7、8、9.を取り組んでみる。

5. 
   $\begin{array}{l}AB=(\begin{array}{cc}4& 2\\ 5& -1\end{array})\hfill \\ BA=(\begin{array}{cc}2& 4\\ 4& 1\end{array})\hfill \end{array}$
6. 
   $\begin{array}{l}CA=\left(\begin{array}{cc}7& 14\\ 21& -7\end{array}\right)\\ AC=\left(\begin{array}{cc}7& 14\\ 21& -7\end{array}\right)\\ CB=\left(\begin{array}{cc}14& 0\\ 7& 7\end{array}\right)\\ BC=\left(\begin{array}{cc}14& 0\\ 7& 7\end{array}\right)\\ \\ C=a{I}_n\\ DC=CD=aD\end{array}$
7. 
   $XA=\left(\begin{array}{ccc}3& 1& 5\end{array}\right)$
8. 
   $\begin{array}{l}(0,1,0)\hfill \\ A_2\hfill \\ (0,0,1)\hfill \\ A_3\hfill \\ n\times n\hfill \\ (0,\cdots ,a_i,\cdots ,0)\hfill \\ a_i=1\hfill \\ A_i\hfill \end{array}$
9. 
   $\begin{array}{l}{}^t\left(AB\right)=\left(\begin{array}{cc}-1& -2\\ 2& 3\end{array}\right)\\ {}^{t}B=\left(\begin{array}{cc}-1& 1\\ 1& 0\end{array}\right)\\ {}^{t}A=\left(\begin{array}{cc}2& 3\\ 1& 1\end{array}\right)\\ {}^{t}B{}^{t}A=\left(\begin{array}{cc}-1& -2\\ 2& 3\end{array}\right)\\ \\ \\ {}^t\left(AB\right)=\left(\begin{array}{cc}1& 11\\ 3& 1\end{array}\right)\\ {}^{t}B=\left(\begin{array}{ccc}1& 2& 3\\ 1& 0& -1\end{array}\right)\\ {}^{t}A=\left(\begin{array}{cc}2& 3\\ 1& 1\\ -1& 2\end{array}\right)\\ {}^{t}B{}^{t}A=\left(\begin{array}{cc}1& 11\\ 3& 1\end{array}\right)\\ \\ \\ {}^t\left(AB\right)=\left(\begin{array}{cc}13& 0\\ 7& 2\\ 1& -5\end{array}\right)\\ {}^{t}B=\left(\begin{array}{ccc}1& 2& 3\\ 1& 1& 1\\ 0& -1& 5\end{array}\right)\\ {}^{t}A=\left(\begin{array}{cc}2& 3\\ 4& 0\\ 1& -1\end{array}\right)\\ {}^{t}B{}^{t}A=\left(\begin{array}{cc}13& 0\\ 7& 2\\ 1& -5\end{array}\right)\\ \\ \\ {}^t\left(ABC\right)\\ ={}^t\left(\left(AB\right)C\right)\\ ={}^{t}C{}^t\left(AB\right)\\ ={}^{t}C{}^{t}B{}^{t}A\end{array}$

コード(Emacs)

Python 3

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

from sympy import pprint, Matrix, randMatrix, symbols
import random
import functools

print('5.')
a = Matrix([[1, 2],
            [3, -1]])
b = Matrix([[2, 0],
            [1, 1]])

pprint(a * b)
pprint(b * a)

print('6.')
c = Matrix([[7, 0],
            [0, 7]])
for m in [c * a, a * c, c * b, b * c]:
    pprint(m)

n = 10
d = randMatrix(n, n)

a = symbols('a')
i = Matrix([[1 if i == j else 0 for j in range(n)]
            for i in range(n)])

c = a * i
print(d * c == c * d == a * d)

print('7.')
x = Matrix([[1, 0, 0]])
a = Matrix([[3, 1, 5],
            [2, 0, 1],
            [1, 1, 7]])
pprint(x * a)

print('8.')
x = Matrix([[0, 1, 0]])
a = randMatrix(3, 3)
pprint(x * a == a[1, :])

x = Matrix([[0, 0, 1]])
pprint(x * a == a[2, :])

print('9.')
m = random.randrange(1, 10)
n = random.randrange(1, 10)
l = random.randrange(1, 10)
a = randMatrix(m, n)
b = randMatrix(n, l)
c = randMatrix(l, random.randrange(1, 10))

print((a * b * c).transpose() == functools.reduce(lambda x, y: x * y,
                                                  map(lambda x: x.transpose(),
                                                      [c, b, a])))

入出力結果(Terminal, IPython)

$ ./sample5.py
5.
⎡4  2 ⎤
⎢     ⎥
⎣5  -1⎦
⎡2  4⎤
⎢    ⎥
⎣4  1⎦
6.
⎡7   14⎤
⎢      ⎥
⎣21  -7⎦
⎡7   14⎤
⎢      ⎥
⎣21  -7⎦
⎡14  0⎤
⎢     ⎥
⎣7   7⎦
⎡14  0⎤
⎢     ⎥
⎣7   7⎦
True
7.
[3  1  5]
8.
True
True
9.
True
$