Solving Eigenvalues of a system time-varying which is 5x5 matrix

Hi @Ahmed Salem ,

However, I was curious and wanted to solve for the eigenvalues of the system matrix A(t) by implement the stability analysis in MATLAB by defining the variables (L, λ, φ(t)), construct the matrices (L, A(t)), and solve the characteristic equation. Below is a sample MATLAB code snippet to achieve this,

lambda1 = 1; % Example value for lambda1

lambda2 = 2; % Example value for lambda2

lambda3 = 3; % Example value for lambda3

lambda = 0.5; % Example value for lambda

phi_t = rand(2, 3); % Example time-varying matrix

After defining the variables, construct the A_t matrix using the defined variables:

L = diag([lambda1, lambda2, lambda3]);

A_t = [L, zeros(3, 2); lambda * phi_t, zeros(2, 2)];

Then, proceed with solving the characteristic equation for stability:

syms s

eqn = det(s*eye(5) - A_t) == 0;

lambda_values = solve(eqn, 'MaxDegree', 5);

disp(lambda_values); % Display the eigenvalues for stability

Please see attached results.

This will help you determine the eigenvalues that will ensure system stability.

Now, after going through attached pdf , this is what I understood by providing brief summary below.

The system dynamics are described by the matrix equation ε ̇ = A(t)ε, where A(t) is a matrix involving gains and system information. To simplify, assuming η ̇ = 0 reduces the system to e ̇ = −Le. Determining the matrix L involves choosing desired eigenvalues and forming L accordingly. The characteristic equation det(sI − L) = 0 helps in selecting stable eigenvalues. Substituting L back into A(t) yields the full system matrix. Stability is ensured by ensuring the eigenvalues of the matrix A(t) have negative real parts. This stability condition is met by solving det(sI − A(t)) = 0 with the given structure of A(t) to find the appropriate lambda values. I will start by assuming η ̇ = 0 to simplify the system to e ̇ = −Le, where L is a known matrix based on desired eigenvalues λ1, λ2, and λ3. So, I will define the matrix L using the desired eigenvalues:

L = diag([lambda1, lambda2, lambda3]);

Calculate the characteristic equation for L to ensure stability:

syms s

det(s*eye(3) - L) = 0;

Substitute the obtained L back into the original A(t) matrix to form a 5x5 matrix A(t) with λ as the variable:

syms lambda

A = [lambda, 1, 0, 0, 0; 0, lambda, 0, 0, 0; 0, 0, lambda, 0, 0; 0, 0, 0, lambda, 0;

phi(1,1), phi(1,2), 0, 0, lambda];

Solve for λ such that the eigenvalues of A(t) have negative real parts to ensure stability.

syms s

det(s*eye(5) - A) = 0;

Hope this helps resolve your problem. Please let me know if you have any questions.