Rank one decomposition of a positive semi-definite matrix with inequality trace constraints
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Suppose there is a square matrix A and a positive semi-definite matrix
, such that
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528394/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528399/image.png)
Is there any ways I could do the rank one decomposition of matrix X, such that for
,
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528404/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528409/image.png)
and keep the inquality constraints
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528414/image.png)
Or at least hold for the most significant (largest eigenvalue)
?
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528419/image.png)
Many thanks!
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Matt J
2021년 2월 23일
편집: Matt J
2021년 2월 23일
Is there any ways I could do the rank one decomposition of matrix X, such that
The obvious answer seems to be to test each k to see which satisfies
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528434/image.png)
and choose any subset of them.
Or at least hold for the most significant (largest eigenvalue) ?
I don't know why you think this is a special case if your first requirement. This is not possible in general, as can be seen from the example A=diag([1,-4]) and X=diag(4,1). In this case, you can only satisfy the requirement with the least significant eigenvalue,
x1 =
2
0
x2 =
0
1
>> x1.'*A*x1, x2.'*A*x2
ans =
4
ans =
-4
댓글 수: 2
Matt J
2021년 2월 23일
If trace(A*X)<=0, There will always be some
satisfying the constraint. Once you have the
, you can check each one, as I mentioned.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528714/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/528719/image.png)
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