Using Numpy (np.linalg.svd) for Singular Value Decomposition

Question

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 1

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())