## short cut COV applied with complex numbers

Yogan Sganzerla

### Yogan Sganzerla (view profile)

さんによって質問されました 2019 年 8 月 13 日

さんによって 回答されました 2019 年 8 月 20 日
I have the same troublem using the short cut cov() in MATLAB.
Well, my matrix A is defined as
A = 5.0000 + 2.0000i 3.0000 + 2.0000i
8.0000 + 5.0000i 7.0000 + 3.0000i
And what I am looking for is the best way to apply this equation:
I solved by hand and got 4.5 + 7.5i for the element COV(1,2). However, when I used the function cov I got as shown below.
And using the equation: 0.5*conj((A-mean(A))') * (A-mean(A)) I received as result:
So I would like to know, how should I procede with this situation? Which method fits better?

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on 20 Aug 2019

From your description, it's a bit unclear on how you want to implement these formulae on matrix A. cov(A) returns the covariance matrix with the corresponding column variances along the diagonal. Please refer the following link on description of covariance(cov) function in MATLAB, to understand how it works :-
Note : In Matlab when you use the transpose operator on a matrix with complex numbers, it gives you a Hermitian (complex-transpose) matrix. So, for getting only the transpose, you must apply the conj function after applying the transpose to the matrix, it seems like that is what you have done in the other equation. Also, in the second equation, it’s a bit unclear why you’re using the multiplication factor of 0.5.
Hope this helps!

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