BUG: orthogonal_procrustes returns the transpose/inverse

Describe your issue.

The scipy.linalg.orthogonal_procrustes function will return the transpose (or inverse) of the desired orthogonal matrix instead of the correct one on some occasions.

If the code below is run a few times, it will print True sometimes and False other times. At the times when it prints False, I find that R_hat is the transpose of R and not R itself. However, if I uncomment the line

# R = np.linalg.qr(np.random.randn(2,2))[0]

and comment the line

R = scipy.stats.ortho_group.rvs(2)

I will always get True as expected. It seems that the way that the random orthogonal matrix R is sampled affects the result of orthogonal_procrustes. Sampling using scipy.stats.ortho_group.rvs(2) seems to be causing the problem,

Reproducing Code Example

import numpy as np
import scipy.linalg
import scipy.stats

X = np.random.randn(2,3)

# sample random orthogonal matrix
# R = np.linalg.qr(np.random.randn(2,2))[0]
R = scipy.stats.ortho_group.rvs(2)

# rotate/reflect each column vector in X using R
Y = R @ X

# estimate R using orthogonal procrustes
R_hat = scipy.linalg.orthogonal_procrustes(X.T,Y.T)[0]

print(np.allclose(R,R_hat))

Error message

No error message.

SciPy/NumPy/Python version information

1.5.4 1.19.2 sys.version_info(major=3, minor=8, micro=11, releaselevel=‘final’, serial=0)

[mhdadk]

Here is the solution to my problem above in case anyone needs it. According to the documentation for the scipy.linalg.orthogonal_procrustes function:

Returns R : (N, N) ndarray The matrix solution of the orthogonal Procrustes problem. Minimizes the Frobenius norm of (A @ R) - B, subject to R.T @ R = I.

I initially thought that this function solves the problem (R @ A) - B and not (A @ R) - B, which was the source of my confusion. The following code works fine

import numpy as np
import scipy.linalg
import scipy.stats

# each row is a 2D vector
X = np.random.randn(3,2)

# sample random orthogonal matrix
R = scipy.stats.ortho_group.rvs(2)

# rotate/reflect each row vector in X using R
Y = X @ R

# estimate R using orthogonal procrustes. NOTE: this function solves
# the problem A * R - B and NOT R * A - B
R_hat = scipy.linalg.orthogonal_procrustes(X,Y)[0]

print(np.allclose(R,R_hat))

Note that each row of the matrix X is a separate vector and not each column as before.

[ilayn]

Ah what @melissawm said. My apologies, I missed the page refresh