1.使用QR分解获取特征值和特征向量
将矩阵A进行QR分解，得到正规正交矩阵Q与上三角形矩阵R。由上可知Ak为相似矩阵，当k增加时，Ak收敛到上三角矩阵，特征值为对角项。
2.奇异值分解(SVD)
其中U是m×m阶酉矩阵；Σ是半正定m×n阶对角矩阵；而V*，即V的共轭转置，是n×n阶酉矩阵。
将矩阵A乘它的转置，得到的方阵可用于求特征向量v，进而求出奇异值σ和左奇异向量u。
#coding:utf8
import numpy as np
np.set_printoptions(precision=4, suppress=True) def householder_reflection(A):
"""Householder变换"""
(r, c) = np.shape(A)
Q = np.identity(r)
R = np.copy(A)
for cnt in range(r - 1):
x = R[cnt:, cnt]
e = np.zeros_like(x)
e[0] = np.linalg.norm(x)
u = x - e
v = u / np.linalg.norm(u)
Q_cnt = np.identity(r)
Q_cnt[cnt:, cnt:] -= 2.0 * np.outer(v, v)
R = np.dot(Q_cnt, R) # R=H(n-1)*...*H(2)*H(1)*A
Q = np.dot(Q, Q_cnt) # Q=H(n-1)*...*H(2)*H(1) H为自逆矩阵
return (Q, R) def eig(A, epsilon=1e-10):
'''采用QR分解法计算特征值和特征向量 '''
(q,r_)=householder_reflection(A)
h = np.identity(A.shape[0])
for i in range(50):
B=np.dot(r_,q)
h=h.dot(q)
(q,r)=gram_schmidt(B)
if abs(r.trace()-r_.trace())< epsilon:
print("Converged in {} iterations!".format(i))
break
r_=r
return r,h def svd(A):
'''奇异值分解'''
n, m = A.shape
svd_ = []
k = min(n, m)
v_=eig(np.dot(A.T, A))[1] #np.linalg.eig(np.dot(A.T, A))[1]
for i in range(k):
v=v_.T[i]
u_ = np.dot(A, v)
s = np.linalg.norm(u_)
u = u_ / s
svd_.append((s, u, v))
ss, us, vs = [np.array(x) for x in zip(*svd_)]
return us.T,ss, vs if __name__ == "__main__": mat = np.array([
[2, 5, 3],
[1, 2, 1],
[4, 1, 1],
[3, 5, 2],
[5, 3, 1],
[4, 5, 5],
[2, 4, 2],
[2, 2, 5],
], dtype='float64')
u,s,v = svd(mat)
print u
print s
print v
print np.dot(np.dot(u,np.diag(s)),v)