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Given $A\in\mathbb{C}^{n\times n}$, I want to use LMI or SDP to find feasibility of $x>0$ in the following inequality:

$$(I-A^*A)+x(\frac{A+A^*}{2})\prec0,$$

where $D\prec0$ means that $D$ is negative definite matrix.


Let $A_1=I-A^*A$ and $A_2=\frac{A+A^*}{2}$, then we get an LMI: $A_1+xA_2\prec0.$

Example 1:

Let $A=\begin{bmatrix}-0.2511+i0.9327&i0.03&0\\0&0.2511+i1.0673&0.01\\0&0&-0.45+i0.7794\end{bmatrix}$

I have tried to use the following cvx code on Matlab to check the feasibility for Example 1:

i=sqrt(-1);
A = [-0.2511 + 0.9327i,   0.0000 + 0.0300i,   0.0000 + 0.0000i;
   0.0000 + 0.0000i,   0.2511 + 1.0673i,   0.0100 + 0.0000i;
   0.0000 + 0.0000i,   0.0000 + 0.0000i,  -0.4500 + 0.7794i];
n=length(A);
I=eye(n);

L0=(I-A'*A);
L1=0.5.*(A+A');

cvx_begin
variable x semidefinite;
minimize 0
subject to
-(L0+x*L1) == semidefinite(n);
cvx_end

Output:

Status: Infeasible
Optimal value (cvx_optval): +Inf

However, $x=0.6165$ satisfies the above inequality.

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  • $\begingroup$ Does this answer your question? $\endgroup$
    – nicoguaro
    Sep 26 at 13:26
  • $\begingroup$ @nicoguaro yes, looks like it is easier to solve LMI as SDP in Matlab $\endgroup$
    – Lee
    Sep 27 at 4:18
  • $\begingroup$ In that case, I suggest that you revert your post to the original state and answer your own question. $\endgroup$
    – nicoguaro
    Sep 27 at 11:25
  • 1
    $\begingroup$ @nicoguaro I have done it! thanks $\endgroup$
    – Lee
    Sep 30 at 15:06
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Here is the correct answer:

   i=sqrt(-1);
    A = [-0.2511 + 0.9327i,   0.0000 + 0.0300i,   0.0000 + 0.0000i;
       0.0000 + 0.0000i,   0.2511 + 1.0673i,   0.0100 + 0.0000i;
       0.0000 + 0.0000i,   0.0000 + 0.0000i,  -0.4500 + 0.7794i];
    n=length(A);
    I=eye(n);
    
    L0=(A'*A-I);
    L1=0.5.*(A+A');
    
    cvx_begin
    variable x;
    minimize 0
    subject to
    (L0-x*L1) == hermitian_semidefinite(n);
    cvx_end

Output: x=0.7376

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