# Find $x$ that satisfy $(I-A^*A)+x(\frac{A+A^*}{2})\prec0$ using LMI or SDP on Matlab

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.

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

   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