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
Is there any ways I could do the rank one decomposition of matrix X, such that for 
, 

and keep the inquality constraints
Or at least hold for the most significant (largest eigenvalue) 
?
Many thanks!
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  Matt J
      
      
 2021 年 2 月 23 日
        
      編集済み: Matt J
      
      
 2021 年 2 月 23 日
  
      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
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
2 件のコメント
  Matt J
      
      
 2021 年 2 月 23 日
				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.
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