I am rewriting a program from Matlab to Python. I realised a difference in a multiplication between arrays. Here is an example:
A = [-1822.87977846-4375.93518777j
3675.88618351+3824.34290883j
971.68964707-2393.36758923j]
In Matlab:
A*A'= 5.7282e+07
In Python:
np.dot(A,A) = -21723405.178+39418085.0343j
How to obtain the same result of A'*A in Numpy?
Thank you.
First of all remember, in MATLAB, ' is different than .'.
' does complex conjugate transpose
.' does non-conjugate transpose
On a real value vector or matrix both operators obtain similar result. However, on complex vectors or matrices they obtain different results. Check the links to find matlab examples for both.
In MATLAB, you can do the following:
A.'*A
ans =
-2.172340517799748e+07 + 3.941808503424492e+07i
.' and ' in Matlab. Most would just use '. Well done +1
On the Python side
In [488]: A=np.array( [-1822.87977846-4375.93518777j ,
...: 3675.88618351+3824.34290883j,
...: 971.68964707-2393.36758923j])
In [489]: A
Out[489]:
array([-1822.87977846-4375.93518777j, 3675.88618351+3824.34290883j,
971.68964707-2393.36758923j])
In [490]: A.conj()
Out[490]:
array([-1822.87977846+4375.93518777j, 3675.88618351-3824.34290883j,
971.68964707+2393.36758923j])
In [491]: A.dot(A.conj())
Out[491]: (57281826.560119703+0j)
In [492]: A.dot(A)
Out[492]: (-21723405.177997477+39418085.034244925j)
In [497]: np.vdot(A,A)
Out[497]: (57281826.560119703+0j)
In Octave, as pointed out in the other answer
>> A'*A
ans = 5.7282e+07
>> A.'*A
ans = -2.1723e+07 + 3.9418e+07i
>>
.real to my answers to remove the 0j part.To obtain the A*A' Matlab operation in Numpy, you can do:
np.dot(A,np.conj(A))
Thank you!