How to solve the bivariate LMI problem?

[yunjiandaiju]

Recently have been reading an article where I need to solve for a bivariate LMI. i have not been able to get the right answer, hopefully this can help me. Article name: From Noisy Data to Feedback Controllers:Nonconservative Design via a Matrix S-Lemma.Here is the code I used.
clear, clc;
% oo=1;
% EE=cell(100,100);
% cc=zeros(100,100);
% while(1)
%状态空间
A = [0.850 -0.038 -0.380;
0.735 0.815 1.594;
-0.664 0.697 -0.064];
B = [1.431 0.705;
1.620 -1.129;
0.913 0.369];
u = randn(2,20);
x0 = randn(3, 1);
%% 生成噪音
n = 3; % 噪声向量的长度
num_samples = 100; % 要生成的噪声向量数量
desired_norm_square = 0.5; %限制范数界
scaled_noises = zeros(n, num_samples);
for i = 1:num_samples
noise = randn(n, 1); % 生成高斯分布的随机噪声
while norm(noise, 2)^2 > desired_norm_square
noise = randn(n, 1); % 如果范数平方超过了阈值，重新生成噪声
end
scaled_noises(:, i) = noise;
end
% 从生成的噪声中随机抽取20列
num_selected = 20;
selected_indices = randperm(num_samples, num_selected);
w= scaled_noises(:, selected_indices);
%% 生成响应
t=20;x=[];x(:,1)=x0;
for i=1:t
x(:,i+1)=Ax(:,i)+Bu(:,i)+w(:,i);
end
x_i0=x(:,1:20);
x_i1=x(:,2:21);
U_i=u;
%% 求解LMI
%定义给定矩阵
I_3=eye(3);
I_0=zeros(3,3);
I_01=zeros(2,3);
I_02=zeros(1,20);
I_03=zeros(3,20);
I_04=zeros(20,3);
I_05=zeros(3,2);
I_06=zeros(2,3);
I_07=zeros(2,2);
a=randn;
b=randn;
fai_11=t0.5eye(3);
fai_22=-eye(20);
%% 定义N
A1=[I_3,x_i1;I_0,-x_i0;I_01,-U_i;I_0,I_03];
A2=[fai_11 I_03;I_04 fai_22];
N=aA1A2A1’;
[F2,G2]=eig(N);
G2;
N11=N(1:3,1:3);
N12=N(1:3,4:6);
N13=N(1:3,7:8);
N14=N(1:3,9:11);
N21=N(4:6,1:3);
N22=N(4:6,4:6);
N23=N(4:6,7:8);
N24=N(4:6,9:11);
N31=N(7:8,1:3);
N32=N(7:8,4:6);
N33=N(7:8,7:8);
N34=N(7:8,9:11);
N41=N(9:11,1:3);
N42=N(9:11,4:6);
N43=N(9:11,7:8);
N44=N(9:11,9:11);
% n = 23; % 方阵的大小
% N1 = zeros(n);
% % 设置左上角为3x3单位阵
% N1(1:3, 1:3) = 200.5eye(3);
% % 设置右下角为3x3单位阵
% N1(n-2:n, n-2:n) = -eye(3);
% N=aA1N1A1’;
%% 定义决策变量
% %CVX工具箱
cvx_begin
cvx_solver sdpt3
variable P(3, 3) semidefinite
variable L(2, 3)
minimize trace(P)
subject to
% [P-(bI_3)-N11,I_0-N12,I_05-N13,I_0-N14;I_0-N21,-P-N22,-(L’)-N23,I_0-N24;I_06-N31,-L-N32,I_07-N33,L-N34; I_0-N41,I_0-N42,(L’)-N43,P-N44]==semidefinite(11);% 半正定约束
[[P-(bI_3),I_0,I_05,I_0;I_0,-P,-(L’),I_0;I_06,-L,I_07,L;I_01,I_0,(L’),P]-N]==semidefinite(11);% 半正定约束
cvx_end
%定义优化问题
k=Linv(P);
U=A+(Bk);
[F1,E]=eig(U);
E
% EE{oo,oo}=E;
% oo=oo+1;
% if oo==100
% break;
% end
% end
% for i=1:100
% if any(max(EE{i,i})>=1)
% cc(i,i)=1;
% end
% end
% a=sum(cc(:));
% disp(a);

The picture shows the LMI to be solved, where P and L are variables. p is a real symmetric semipositive definite matrix and L is unbounded.I can work out the solution every time, but I think he's wrong. I brought the solution back into the system equations and could not verify that the system was stable. I hope I can get some help with this.

[johanlofberg]

This is a YALMIP forum, not a CVX forum.
https://yalmip.github.io/tutorial/basics/

Not clear why you are talking about bivariate, as it appears linear since your only decision variables are P and L

[yunjiandaiju]

I see what you mean, but mathematically shouldn't N be symmetric? Is this due to the fact that I'm giving too large a value? The difference in Frobenius norms between the matrix N and its transpose N' was calculated and the result is 5.8604e-09. this is very close to zero, but not exactly zero.

[johanlofberg]

As I said, you have horrible numerics, and thus have to fix that to ensure you actually have symmetric data

[yunjiandaiju]

So it's caused by the presence of very large numbers in N. If I fix this, he should be symmetric.

[johanlofberg]

Large, small, trickle.flow or what ever which is causing numerical issues.

I hope this isn't a surprise to you that floating-point numbers have errors

>> x = sqrt(2);x^2-2

ans =

   4.4409e-16

Since you know the matrix should be symmetric and it is just numerical noise, just symmetrize it afterwards (although best is of course if you can understand why you have crap data, since crap data in an SDP model will most likely cause numerical problems once the solver tries to solve it. crap in crap out)

[yunjiandaiju]

Okay, thank you Professor. I have figured out what my problem was, thank you very much.