Null after solving for unknowns by setting the matrix det as zero

Question

Lingling Fan 2019-5-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/463001-null-after-solving-for-unknowns-by-setting-the-matrix-det-as-zero

编辑： Lingling Fan 2019-5-20

eigen_vec_null_puzz.m

Hi community,

I attach my code here. I try to solve kz by setting the matrix T_mat det as zero. val_arr_sol is the solution. But later when I wanted to get the null of matrix T_mat, it fails.

The reason is related to the low accuracy? How could I improve my method so that I can get the null of a matrix with higher accuracy towards zero determinant, such as 1e^-15?

Thanks in advance

clear;
epsxx = -1.7778;
epsyy = -2.1250;
epsyz = 0.0000 + 1.0417i;
kx = 0;
ky = -5;
k0 =  0.6000;
val_arr_sol = [
    0.0000 - 5.0636i
    0.0000 - 5.0578i
    0.0000 + 5.0578i
    0.0000 + 5.0636i];
for i = 1:4
    kz = val_arr_sol(i);
    eps_mat = [epsxx, 0, 0;
        0, epsyy, epsyz;
        0, -epsyz, epsyy];
    eta_mat = inv(eps_mat);
    nabla_matrix = 1i*...
        [0, -kz, ky;...
        kz, 0, -kx;...
        -ky, kx, 0];
    rhs_matrix = [1, 0; 0, 1; -kx/kz, -ky/kz];
    Tot_matrix = (nabla_matrix*eta_mat*nabla_matrix-k0^2*eye(3))*rhs_matrix;
    T_mat = Tot_matrix(1:2, :);
    
    
    det(T_mat)
    null(T_mat)
    
end
%% The result is: 
 eigen_vec_null_puzz
ans =
   7.5827e-07
ans =
  2×0 empty double matrix
ans =
  -3.6005e-06
ans =
  2×0 empty double matrix
ans =
  -3.6005e-06
ans =
  2×0 empty double matrix
ans =
   7.5827e-07
ans =
  2×0 empty double matrix

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Answer 1

Lingling Fan 2019-5-20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/463001-null-after-solving-for-unknowns-by-setting-the-matrix-det-as-zero#answer_375789

编辑：Lingling Fan 2019-5-20

在 MATLAB Online 中打开

follow up:

I improved the code by putting the analytical solution function of the determinant inside the eigenvector null solver function. The det is closer to zero, about 1e^-15. But the null of the matrix is still empty.

%% test the eigen vector
clear;
epsxx = -1.7778;
epsyy = -2.1250;
epsyz = 0.0000 + 1.0417i;
kx = 0;
ky = -5;
k0 =  0.6000;
% val_arr_sol = [
%     0.0000 - 5.0636i
%     0.0000 - 5.0578i
%     0.0000 + 5.0578i
%     0.0000 + 5.0636i];
%% what if I solve for the problem once again and without any another transfer
val_arr_sol = eig_val_MO_x(epsxx, epsyy, epsyz, kx, ky, k0);
for i = 1:4
    kz = val_arr_sol(i);
    eps_mat = [epsxx, 0, 0;
        0, epsyy, epsyz;
        0, -epsyz, epsyy];
    eta_mat = inv(eps_mat);
    nabla_matrix = 1i*...
        [0, -kz, ky;...
        kz, 0, -kx;...
        -ky, kx, 0];
    rhs_matrix = [1, 0; 0, 1; -kx/kz, -ky/kz];
    Tot_matrix = (nabla_matrix*eta_mat*nabla_matrix-k0^2*eye(3))*rhs_matrix;
    T_mat = Tot_matrix(1:2, :);
    
    
    det(T_mat)
    null(T_mat)
    
end
    
%% results
%% test the eigen vector
ans =
  -1.9425e-16
ans =
  2×0 empty double matrix
ans =
  -1.9425e-16
ans =
  2×0 empty double matrix