How to create random orthonormal matrix in python numpy

Question

Is there a method that I can call to create a random orthonormal matrix in python? Possibly using numpy? Or is there a way to create a orthonormal matrix using multiple numpy methods? Thanks.

Answer 1

Version 0.18 of scipy has scipy.stats.ortho_group and scipy.stats.special_ortho_group. The pull request where it was added is https://github.com/scipy/scipy/pull/5622

For example,

In [24]: from scipy.stats import ortho_group  # Requires version 0.18 of scipy

In [25]: m = ortho_group.rvs(dim=3)

In [26]: m
Out[26]: 
array([[-0.23939017,  0.58743526, -0.77305379],
       [ 0.81921268, -0.30515101, -0.48556508],
       [-0.52113619, -0.74953498, -0.40818426]])

In [27]: np.set_printoptions(suppress=True)

In [28]: m.dot(m.T)
Out[28]: 
array([[ 1.,  0., -0.],
       [ 0.,  1.,  0.],
       [-0.,  0.,  1.]])

Answer 2

You can obtain a random n x n orthogonal matrix Q, (uniformly distributed over the manifold of n x n orthogonal matrices) by performing a QR factorization of an n x n matrix with elements i.i.d. Gaussian random variables of mean 0 and variance 1. Here is an example:

import numpy as np
from scipy.linalg import qr

n = 3
H = np.random.randn(n, n)
Q, R = qr(H)

print (Q.dot(Q.T))

[[  1.00000000e+00  -2.77555756e-17   2.49800181e-16]
 [ -2.77555756e-17   1.00000000e+00  -1.38777878e-17]
 [  2.49800181e-16  -1.38777878e-17   1.00000000e+00]]

EDIT: (Revisiting this answer after the comment by @g g.) The claim above on the QR decomposition of a Gaussian matrix providing a uniformly distributed (over the, so called, Stiefel manifold) orthogonal matrix is suggested by Theorems 2.3.18-19 of this reference. Note that the statement of the result suggests a "QR-like" decomposition, however, with the triangular matrix R having positive elements.

Apparently, the qr function of scipy (numpy) function does not guarantee positive diagonal elements for R and the corresponding Q is actually not uniformly distributed. This has been observed in this monograph, Sec. 4.6 (the discussion refers to MATLAB, but I guess both MATLAB and scipy use the same LAPACK routines). It is suggested there that the matrix Q provided by qr is modified by post multiplying it with a random unitary diagonal matrix.

Below I reproduce the experiment in the above reference, plotting the empirical distribution (histogram) of phases of eigenvalues of the "direct" Q matrix provided by qr, as well as the "modified" version, where it is seen that the modified version does indeed have a uniform eigenvalue phase, as would be expected from a uniformly distributed orthogonal matrix.

from scipy.linalg import qr, eigvals
from seaborn import distplot

n = 50
repeats = 10000

angles = []
angles_modified = []
for rp in range(repeats):
    H = np.random.randn(n, n)
    Q, R = qr(H)
    angles.append(np.angle(eigvals(Q)))
    Q_modified = Q @ np.diag(np.exp(1j * np.pi * 2 * np.random.rand(n)))
    angles_modified.append(np.angle(eigvals(Q_modified))) 

fig, ax = plt.subplots(1,2, figsize = (10,3))
distplot(np.asarray(angles).flatten(),kde = False, hist_kws=dict(edgecolor="k", linewidth=2), ax= ax[0])
ax[0].set(xlabel='phase', title='direct')
distplot(np.asarray(angles_modified).flatten(),kde = False, hist_kws=dict(edgecolor="k", linewidth=2), ax= ax[1])
ax[1].set(xlabel='phase', title='modified');

Answer 3

This is the rvs method pulled from the https://github.com/scipy/scipy/pull/5622/files, with minimal change - just enough to run as a stand alone numpy function.

import numpy as np    

def rvs(dim=3):
     random_state = np.random
     H = np.eye(dim)
     D = np.ones((dim,))
     for n in range(1, dim):
         x = random_state.normal(size=(dim-n+1,))
         D[n-1] = np.sign(x[0])
         x[0] -= D[n-1]*np.sqrt((x*x).sum())
         # Householder transformation
         Hx = (np.eye(dim-n+1) - 2.*np.outer(x, x)/(x*x).sum())
         mat = np.eye(dim)
         mat[n-1:, n-1:] = Hx
         H = np.dot(H, mat)
         # Fix the last sign such that the determinant is 1
     D[-1] = (-1)**(1-(dim % 2))*D.prod()
     # Equivalent to np.dot(np.diag(D), H) but faster, apparently
     H = (D*H.T).T
     return H

It matches Warren's test, https://stackoverflow.com/a/38426572/901925

Answer 4

An easy way to create any shape (n x m) orthogonal matrix:

import numpy as np

n, m = 3, 5

H = np.random.rand(n, m)
u, s, vh = np.linalg.svd(H, full_matrices=False)
mat = u @ vh

print(mat @ mat.T) # -> eye(n)

Note that if n > m, it would obtain mat.T @ mat = eye(m).

Answer 5

from scipy.stats import special_ortho_group
num_dim=3
x = special_ortho_group.rvs(num_dim)

Documentation

Answer 6

if you want a none Square Matrix with orthonormal column vectors you could create a square one with any of the mentioned method and drop some columns.

Answer 7

Numpy also has qr factorization. https://numpy.org/doc/stable/reference/generated/numpy.linalg.qr.html

import numpy as np

a = np.random.rand(3, 3)
q, r = np.linalg.qr(a)

q @ q.T
# array([[ 1.00000000e+00,  8.83206468e-17,  2.69154044e-16],
#        [ 8.83206468e-17,  1.00000000e+00, -1.30466244e-16],
#        [ 2.69154044e-16, -1.30466244e-16,  1.00000000e+00]])