【详细】使用正定分解矩阵拟合线性一次方程

1.如何对方程 $AX=b$求解，A=[x1,x2,x3,…xn], b=[y1,y2,y3…yn]都为列向量。
输入A,b的散点图如下：

数学推导如下：

$令C=A^{T}*A$ $变换为：A^{T}*AX=A^{T}*b$ $即：CX=A^{T}*b=B$

$C为正定矩阵可以表示为一个下三角矩阵L和其转置L^{T}的积：$
$L*L^{T}X=B$
$令：L^{T}X=Y 、 先通过 L*Y=B求得Y，再通过 L^{T}X=Y求得X$

import matplotlib.pyplot as plt
import numpy as np
import tensorflow as tf

sess = tf.Session()
A = np.linspace(0, 10, 100)
b = A + np.random.normal(0, 1, 100)
x_vals_column = np.transpose(np.matrix(A))
ones_column = np.transpose(np.matrix(np.repeat(1, 100)))
A = np.column_stack((x_vals_column, ones_column))
b = np.transpose(np.matrix(b))
A_tensor = tf.constant(A)   #输入矩阵A
b_tensor = tf.constant(b)   #输出矩阵B

tA_A = tf.matmul(tf.transpose(A_tensor), A_tensor)   #求得正定矩阵C
L = tf.cholesky(tA_A)  #三角矩阵L
tA_b = tf.matmul(tf.transpose(A_tensor), b)
sol1 = tf.matrix_solve(L, tA_b) #求得Y
sol2 = tf.matrix_solve(tf.transpose(L), sol1) #求得X

solution_eval = sess.run(sol2)
slope = solution_eval[0][0]
y_intercept = solution_eval[1][0]
print('slope: ' + str(slope))
print('y'_intercept: ' + str(y_intercept))

best_fit = []
for i in x_vals:
	best_fit.append(slope*i+y_intercept)
plt.plot(x_vals, y_vals, 'o', label='Data')
plt.plot(x_vals, best_fit, 'r-', label='Best' fit line',
linewidth=3)
plt.legend(loc='upper left')
plt.show()

拟合一次方程图像如下：