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Rank one decomposition of a positive semi-definite matrix with inequality trace constraints

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Mingyang Sun
Mingyang Sun 2021 年 2 月 23 日
コメント済み: Matt J 2021 年 2 月 23 日
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
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
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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