CODEWITHSUNDEEP
pythonmatlab
Gze
Gze

Reputation: 408

On the division and multiplication of matrices in python

Here the question with details and I think it's clearer,

suppose I have a matrix h of size 4 x 4 , and a vector of x of size 4 x 1, if we have y is the output of multiplication between h and x which means y = h * x; whose size is 1 x 4. So when I multiply again the inverse of every column in h by vector y, I should be able to get a vector equivalent of vector x which means $x = h^{-1} * y $. But unfortunately, I can't get that in python.

for example, let's first do that in MATLAB: 

clear all 
clc 

h = (randn(4,4) + 1j*randn(4,4));  %any matrix of 4 x 4 
x = [1 + 1j ; 0; 0 ; 0];           % a vector of 4 x 1 

y = h * x ;                       % y is the output of multiplication 
x2 = [];
for ii = 1 : 4 
    x1 = pinv(h(:,ii))*y;        %multiply every column of h^(-1) with y  
    x2 = [x2 x1];                % the output
end

in that case, the output x2 is as expected, a vector 1 x 4 as below: 

x2 =

   1.0000 + 1.0000i   0.7249 + 0.5054i  -0.0202 + 0.0104i   0.2429 + 0.0482i

In MATLAB, that's ok.

Now let's do that in python: 

import numpy as np

h = np.random.randn(4,4) + 1j*np.random.randn(4,4)
x = [[1+1j],[0+0j],[0+0j],[0+0j]]
y = h.dot(x)

x2 = []
for ii in range(4):
    x1 = np.divide(y, h[:,ii])
    x2.append(x1)
    print(x2)

Although x2 is supposed to be a vector of dimension 1 x 4 similar as in output of above MATLAB code, but in that case, I get x2 a matrix of size 4 x 4 !!

please any help.

Upvotes: 2

Views: 909

Answers (1)

norok2
norok2

Reputation: 26896

There are two issues here:

  • np.divide() is for element-wise division, you may be looking for np.linalg.pinv() instead.
  • MATLAB is col major (FORTRAN-style), while NumPy is row major (C-style) so getting a list as a NumPy array will get you to a shape (n,) with n the length of the list and not an object of size (1, n) as MATLAB would.

The Python code equivalent (sort of, I'll do preallocation) to your MATLAB one, would be:

import numpy as np

h = np.random.randn(4, 4) + 1j * np.random.randn(4, 4)
x = np.array([[1 + 1j], [0 + 0j], [0 + 0j], [0 + 0j]])
# y = h.dot(x)  <-- now NumPy supports also `@` in place of `np.dot()`
y = h @ x

x2 = np.zeros((1, 4), dtype=np.complex)
for i in range(4):
    x2[0, i] = np.linalg.pinv(h[:, i:i + 1]) @ y

as you can see, the shape of the output is enforced right away.

Upvotes: 1

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