数学 - Python - 線形代数学 - 線形写像と行列 - 線形写像の合成 - ベクトル空間、基底、恒等写像に対応する行列、連立方程式、転置行列

学習環境

ラング線形代数学(上) (ちくま学現文庫)(S.ラング (著)、芹沢正三 (翻訳)、筑摩書房)の5章(線形写像と行列)、3(線形写像の合成)、練習問題3の解答を求めてみる。

1. $\begin{array}{l} (1, 1, 0) = a_{11} (2, 1, 1) + a_{21} (0, 0, 1) + a_{31} (- 1, 1, 1) \\ (- 1, 1, 1) = a_{12} (2, 1, 1) + a_{22} (0, 0, 1) + a_{32} (- 1, 1, 1) \\ (0, 1, 2) = a_{13} (2, 1, 1) + a_{23} (0, 0, 1) + a_{33} (- 1, 1, 1) \\ {\begin{cases} 2 a_{11} - a_{31} = 1 \\ a_{11} + a_{31} = 1 \\ a_{11} + a_{21} + a_{31} = 0 \end{cases} \\ 3 a_{11} = 2 \\ a_{11} = \frac{2}{3} \\ a_{31} = \frac{1}{3} \\ a_{21} = - 1 \\ {\begin{cases} 2 a_{12} - a_{32} = - 1 \\ a_{12} + a_{32} = 1 \\ a_{12} + a_{22} + a_{32} = 1 \end{cases} \\ a_{12} = 0 \\ a_{32} = 1 \\ a_{22} = 0 \\ {\begin{cases} 2 a_{13} - a_{33} = 0 \\ a_{13} + a_{33} = 1 \\ a_{13} + a_{23} + a_{33} = 2 \end{cases} \\ a_{13} = \frac{1}{3} \\ a_{33} = \frac{2}{3} \\ a_{23} = 1 \end{array}$
  
  よって、問題の基底に関する恒等線形写像に対応する行列は、
  
  $[\begin{matrix} \frac{2}{3} & 0 & \frac{1}{3} \\ - 1 & 0 & 1 \\ \frac{1}{3} & 1 & \frac{2}{3} \end{matrix}]$
2. $\begin{array}{l} (3, 2, 1) = a_{11} (1, 1, 0) + a_{21} (- 1, 2, 4) + a_{31} (2, - 1, 1) \\ (0, - 2, 5) = a_{12} (1, 1, 0) + a_{22} (- 1, 2, 4) + a_{32} (2, - 1, 1) \\ (1, 1, 2) = a_{13} (1, 1, 0) + a_{23} (- 1, 2, 4) + a_{33} (2, - 1, 1) \\ {\begin{cases} a_{11} - a_{21} + 2 a_{31} = 3 \\ a_{11} + 2 a_{21} - a_{31} = 2 \\ 4 a_{21} + a_{31} = 1 \end{cases} \\ a_{31} = 1 - 4 a_{21} \\ a_{11} - 9 a_{21} = 1 \\ a_{11} + 6 a_{21} = 3 \\ 15 a_{21} = 2 \\ a_{21} = \frac{2}{15} \\ a_{31} = \frac{7}{15} \\ a_{11} = \frac{33}{15} \\ {\begin{cases} a_{12} - a_{22} + 2 a_{32} = 0 \\ a_{12} + 2 a_{22} - a_{32} = - 2 \\ 4 a_{22} + a_{32} = 5 \end{cases} \\ a_{32} = 5 - 4 a_{22} \\ a_{12} - 9 a_{22} = - 10 \\ a_{12} + 6 a_{22} = 3 \\ a_{22} = \frac{13}{15} \\ a_{32} = \frac{23}{15} \\ a_{12} = \frac{45}{15} - \frac{78}{15} = - \frac{33}{15} \\ {\begin{cases} a_{13} - a_{23} + 2 a_{33} = 1 \\ a_{13} + 2 a_{23} - a_{33} = 1 \\ 4 a_{23} + a_{33} = 2 \end{cases} \\ a_{33} = 2 - 4 a_{23} \\ a_{13} - 9 a_{23} = - 3 \\ a_{13} + 6 a_{23} = 3 \\ a_{23} = \frac{6}{15} \\ a_{33} = \frac{6}{15} \\ a_{13} = \frac{9}{15} \\ \frac{1}{15} [\begin{matrix} 33 & - 33 & 9 \\ 2 & 13 & 6 \\ 7 & 23 & 6 \end{matrix}] \end{array}$

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import symbols, Matrix, solve, Rational print('3.') class Test(TestCase): def test_a(self): A = Matrix(symbols('a:9')).reshape(3, 3) X = Matrix([[1, 1, 0], [-1, 1, 1], [0, 1, 2]]) s = solve(A * Matrix([[2, 1, 1], [0, 0, 1], [-1, 1, 1]]) - X) self.assertEqual(Matrix(list(s.values())).reshape(3, 3).T, Matrix([[Rational(2, 3), 0, Rational(1, 3)], [-1, 0, 1], [Rational(1, 3), 1, Rational(2, 3)]])) def test_b(self): A = Matrix(symbols('a:9')).reshape(3, 3) X = Matrix([[3, 2, 1], [0, -2, 5], [1, 1, 2]]) s = solve(A * Matrix([[1, 1, 0], [-1, 2, 4], [2, -1, 1]]) - X) self.assertEqual(Matrix(list(s.values())).reshape(3, 3).T, Rational(1, 15) * Matrix([[33, -33, 9], [2, 13, 6], [7, 23, 6]])) if __name__ == "__main__": main()

Kamimura's blog

2020年5月24日日曜日

数学 - Python - 線形代数学 - 線形写像と行列 - 線形写像の合成 - ベクトル空間、基底、恒等写像に対応する行列、連立方程式、転置行列

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