Lanczos法的目的：将实对称矩阵转换成为三对角矩阵（稀疏矩阵），从而便于计算机储存和后续的计算。
在三对角矩阵矩阵上，采用QR分解，得到矩阵的特征值。
# -*- coding: utf-8 -*-"""Created on Tue Nov 20 12:41:40 2018@author: yujin.wang"""import numpy as npimport scipy as sciimport matplotlib.pyplot as pltimport timeimport sysdef lanczos(A,b,nmax):m = np.size(A,1)alpha = []beta = [0]qprev = np.zeros(m)q = b / np.linalg.norm(b)for n in range(nmax):v = np.dot(q,A) #*sci.linalg.inv(q)temp = np.dot(v,q.T)alpha.append(temp[0][0,0])v = v - np.dot(beta[-1],qprev)-np.dot(alpha[-1],q)beta.append(sci.linalg.norm(v))qprev = qq = v/beta[-1]beta = beta[1:-1]T = np.diag(alpha) + np.diag(beta,1) +np.diag(beta,-1)return Tdef qreigen(A,num=100):m = np.size(A,1)p = np.eye(m)for i in range(num):v = np.diag(A)q,r = sci.linalg.qr(A)A = np.dot(r,q)p = np.dot(p,q)tol = max(np.diag(A))-max(v)print 'TOL:',tol,'Max. Eig:',max(v)if np.abs(tol) <1e-6:break#s = np.diag(np.diag(r))return p,np.diag(A)if __name__ == '__main__':A = np.matrix([[5,1,3],[1,2,0],[3,0,11]])print np.linalg.eig(A)[0]print '$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$'p,s = qreigen(A)print sprint '$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$'b = np.matrix([1,1,1])tridiag = lanczos(A,b,23)p,s = qreigen(tridiag,num=200)print max(s),min(s)参考：https://baike.baidu.com/item/Lanczos%E7%AE%97%E6%B3%95/9849921?fr=aladdin
http://www.cnblogs.com/qxred/p/dcalgorithm.html
http://www.doc88.com/p-1446190703774.html