概述

向量是对数的拓展，一个向量表示一组数；而矩阵则可以视为对向量的拓展，一个矩阵表示一组向量。

看待一个矩阵有两个视角，行向量视角和列向量视角。

当行数和列数相等时候，称为方阵，方阵有很多特殊的性质。有很多特殊的性质的矩阵，是方阵。

实现矩阵类

from vector import Vector


class Matrix:
	def __init__(self, list2d):
		self._value = list2d.copy()

	def __repr__(self):
		return "Matrix({})".format(self._value)

	# 此处简单设置交互式与print模式打印的内容相同
	__str__ = __repr__

	def shape(self):
		return len(self._value), len(self._value[0])

	def row_num(self):
		"""返回矩阵行数"""
		return self.shape()[0]

	def col_num(self):
		"""返回矩阵列数"""
		return self.shape()[1]

	def size(self):
		"""矩阵元素个数"""
		r, c = self.shape()
		return r * c

	def __getitem__(self, pos):
		"""返回指定未知数的元素，pos的形式是元组"""
		assert pos[0] < self.row_num() and pos[1] < self.col_num
		r, c = pos
		return self._value[r][c]

	def row_vector(self, idx):
		"""返回第idx个行向量"""
		return Vector(self._value[idx])

	def col_vector(self, idx):
		"""返回第idx个列向量"""
		return Vector([row[idx] for row in self._value)
		

矩阵基本运算和性质

矩阵加法

两个同形矩阵加法，
$$A=\left(\begin{array}{cccc}{a}_{11}& a_{12}& ...& a_{1c}\\ a_{21}& a_{22}& ...& a_{2c}\\ ...& ...& ...& ...\\ a_{r1}& a_{r2}& ...& a_{rc}\end{array}\right)B=\left(\begin{array}{cccc}{b}_{11}& b_{12}& ...& b_{1c}\\ b_{21}& b_{22}& ...& b_{2c}\\ ...& ...& ...& ...\\ b_{r1}& b_{r2}& ...& b_{rc}\end{array}\right)$$
$$A+B=\left(\begin{array}{cccc}{a}_{11}+b_{11}& a_{12}+b_{12}& ...& a_{1c}+b_{1c}\\ a_{21}+b_{21}& a_{22}+b_{22}& ...& a_{2c}+b_{2c}\\ ...& ...& ...& ...\\ a_{r1}+b_{r1}& a_{r2}+b_{r2}& ...& a_{rc}+b_{rc}\end{array}\right)$$

矩阵数乘

一个实数与一个矩阵的乘法运算，
$$A=\left(\begin{array}{cccc}{a}_{11}& a_{12}& ...& a_{1c}\\ a_{21}& a_{22}& ...& a_{2c}\\ ...& ...& ...& ...\\ a_{r1}& a_{r2}& ...& a_{rc}\end{array}\right)$$
$$k\cdot A=\left(\begin{array}{cccc}k\cdot a_{11}& k\cdot a_{12}& ...& k\cdot a_{1c}\\ k\cdot a_{21}& k\cdot a_{22}& ...& k\cdot a_{2c}\\ ...& ...& ...& ...\\ k\cdot a_{r1}& k\cdot a_{r2}& ...& k\cdot a_{rc}\end{array}\right)$$

矩阵运算性质

1. 交换律
   $A+B=B+A$

2. 结合律
   $(A+B)+C=A+(B+C)$
   $(c\cdot k)\cdot A=c\cdot (k\cdot A)$ 。其中c和k是实数
   $k\cdot (A+B)=k\cdot A+k\cdot B$
   $(c+k)\cdot A=c\cdot A+k\cdot A$

3. 任何一个矩阵$A$，都存在一个相同形状的矩阵$O$，满足
   $A+O=A$

矩阵基本运算代码实现

接之前Matrix类代码，

	def __add__(self, another):
		assert self.shape() == another.shape()
		return Matrix([[a+b for a, b in zip(self.row_vector(i), another.row_vector(i))] for i in range(self.row_num()])

	def __sub__(self, another):
		assert self.shape() == another.shape()
		return Matrix([[a-b for a, b in zip(self.row_vector(i), another.row_vector(i))] for i in range(self.row_num()])

	def __mul__(self, k):
		return Matrix([[k*a for a in self.row_vector(i)] for i in range(self.row_num()])

	def __rmul__(self, k):
		return Matrix([[k*a for a in self.row_vector(i)] for i in range(self.row_num()])

	def __truediv__(self, k):
		return Matrix([[a/k for a in self.row_vector(i)] for i in range(self.row_num()])

	def __pos__(self):
		return self
	
	def __neg__(self):
		return -1 * self
	
	@classmethod
	def __zero__(cls, r, c):
		return cls([[0] * c] for _ in range(r))