Python v / s MATLAB中的SVD命令

Question

I have m = 10, n = 5, A=randn(m,n);[U,S,V]=svd(A); This returns a full 10x5 S matrix in MATLAB whereas Python only returns S as a 5x1 array. How do I recover the complete S matrix in Python? I have tried looking up several StackOverflow posts online but surprisingly doesn't shed light on this.

Also, how much does a Python IDE matter? I use Spyder but have been told that Vim is perhaps the most common.

Thanks a lot.

Answer 1

The SVD of a matrix can be written as

A = U S V^H

Where the ^H signifies the conjugate transpose . Matlab's svd command returns U, S and V, while numpy.linalg.svd returns U, the diagonal of S, and V^H. Thus, to get the same S and V as in Matlab you need to reconstruct the S and also get the V:

import numpy

m = 10
n = 5
A = numpy.random.randn(m, n)
U, sdiag, VH = numpy.linalg.svd(A)
S = numpy.zeros((m, n))
numpy.fill_diagonal(S, sdiag)
V = VH.T.conj()  # if you know you have real values only you can leave out the .conj()

Answer 2

To recover the Complete matrix you can do as follow :

import numpy as np
m = 10
n = 5
A=np.random.randn(m,n)
U,S,V =np.linalg.svd(A)

It's right that S.shape = (5,) .

You want something similar to https://www.mathworks.com/help/matlab/ref/svd.html with A = 4x2 where final S = 4×2 too.

To do that you define a matrix B = np.zeros(A.shape) . And you fill its diagonal with the element of S. By diagonal I mean where i==j as follow :

B = np.zeros(A.shape)
for i in range(m) :
    for j in range(n) :
        if i == j : B[i,j] = S[j]

Now B.shape = (10,5) as expected Or in a more compact form :

C = np.array([[S[j] if i==j else 0 for j in range(n)] for i in range(m)])   

I hope it helps

For the second question, I use gedit (standard text editor) running the code in ipython shell.

You can have a look to jupyter too