シュミットの直交化を自分で書く

import numpy as np
A = np.array([[1,1,0],[0,1,1],[1,0,1]],dtype=float)
Q = np.zeros_like(A)
m = shape(Q)[0]
n = shape(Q)[1]
cnt = 0
for a in A.T:
    u = np.copy(a)
    for i in range(0, cnt):
        u -= np.dot(np.dot(Q[:, i].T, a), Q[:, i]) #                
    e = u / np.linalg.norm(u)  #    
    Q[:, cnt] = e
    cnt += 1
print Q

from scipy import linalg
A = np.array([[1,1,0],[0,1,1],[1,0,1]])
a = linalg.orth(A)
print array(linalg.orth(A),dtype=float)
print dot(A,A.T)-1