求矩阵全部特征值和特征向量的QR方法

import numpy as np
import math


def sgn1(x):
    if x > 0:
        return 1
    elif x == 0:
        return 0
    elif x < 0:
        return -1


ai2 = np.mat([[-1, 2, 1],
            [2, -4, 1],
            [1, 1, -6]], dtype=float)
n = ai2.shape[0]
for i in range(n-2):
    c = -1*sgn1(ai2[i+1, i])*np.sqrt(np.sum(np.multiply(ai2[i+1:n, i], ai2[i+1:n, i])))
    lou = np.sqrt(2*c*(c-ai2[i+1, i]))
    l = ai2[i+1:n, i]
    l1 =l.copy()
    l1[0] = l1[0]-c
    b = l1/lou
    a = np.mat(np.zeros((i+1, 1)))
    u = np.vstack((a, b))
    I = np.mat(np.eye(n))
    H = I-2*u*u.T
    ai2 = H*ai2*H.T
err = 1
ai3 = ai2.copy()
for t in range(88):
    i = 0
    sita = math.atan(ai2[i+1, i]/ai2[i, i])
    c = math.cos(sita)
    s = math.sin(sita)
    V1 = np.mat(np.eye(n))
    V1[i,i] = c
    V1[i+1,i] = -s
    V1[i,i+1] = s
    V1[i+1, i+1] = c
    ai2 = V1*ai2
    i = 1
    sita = math.atan(ai2[i+1, i]/ai2[i, i])
    c = math.cos(sita)
    s = math.sin(sita)
    V2 = np.mat(np.eye(n))
    V2[i,i] = c
    V2[i+1,i] = -s
    V2[i,i+1] = s
    V2[i+1, i+1] = c
    ai2 = V2*ai2
    ai2 = ai2*V1.T*V2.T
print('迭代88次后得：')
print(ai2)
print('矩阵的特征值为{:.7},{:.7}，{:.7}'.format(ai2[0, 0],ai2[1, 1],ai2[2, 2]))