Using Numpy (np.linalg.svd) for Singular Value Decomposition
Im reading Abdi & Williams (2010) "Principal Component Analysis", and I'm trying to redo the SVD to attain values for further PCA.
The article states that following SVD:
X = P D Q^t
I load my data in a np.array X.
X = np.array(data)
P, D, Q = np.linalg.svd(X, full_matrices=False)
D = np.diag(D)
But i do not get the above equality when checking with
X_a = np.dot(np.dot(P, D), Q.T)
X_a and X are the same dimensions, but the values are not the same. Am I missing something, or is the functionality of the np.linalg.svd function not compatible somehow with the equation in the paper?
Answer
TL;DR: numpy's SVD computes X = PDQ, so the Q is already transposed.
SVD decomposes the matrix X effectively into rotations P and Q and the diagonal matrix D. The version of linalg.svd() I have returns forward rotations for P and Q. You don't want to transform Q when you calculate X_a.
import numpy as np
X = np.random.normal(size=[20,18])
P, D, Q = np.linalg.svd(X, full_matrices=False)
X_a = np.matmul(np.matmul(P, np.diag(D)), Q)
print(np.std(X), np.std(X_a), np.std(X - X_a))
I get: 1.02, 1.02, 1.8e-15, showing that X_a very accurately reconstructs X.
If you are using Python 3, the @ operator implements matrix multiplication and makes the code easier to follow:
import numpy as np
X = np.random.normal(size=[20,18])
P, D, Q = np.linalg.svd(X, full_matrices=False)
X_a = P @ diag(D) @ Q
print(np.std(X), np.std(X_a), np.std(X - X_a))
print('Is X close to X_a?', np.isclose(X, X_a).all())