Where does the eigenvalues of 3-by-3 matrix meet?

I have a 3x3 Hermitian matrix dependent on 10 variables. Each combination of these variables yields three eigenvalues: E1, E2, and E3. My objective is to identify the set of variables for which either two or all three eigenvalues are identical.

While I could developed a code utilizing 10 nested FOR loops to achieve a numerical solution, I am curious if there exists a more efficient approach. (Also, an analytical solution to this problem would be highly advantageous.)

Presented below is the matrix in question:

with

here J = S = a = 1. And I will have 10 variables as following and their limits for which I want to know that eigenvalues will be same:

• K_A (-2 to +2)
• K_B (-2 to +2)
• K_C (-2 to +2)
• J_A (-2 to +2)
• J_B (-2 to +2)
• J_C (-2 to +2)
• D_z (-2 to +2)
• k_x ( to )
• k_y ( to )
• k_z (-pi to +pi)

Relevant code (the matrix is written here in simplified form):

clear; clc;

%constants:
S = 1;
J = 1;
a = 1;


%limits of variables:

KAmin = -2;
KBmin = KAmin;
KCmin = KAmin;

KAmax = 2;
KBmax = KAmax;
KCmax = KAmax;

JAmin = -2;
JBmin = JAmin;
JCmin = JAmin;

JAmax = 2;
JBmax = JAmax;
JCmax = JAmax;

Dzmin = -2;
Dzmaxx = 2;

kxmin = -4*pi/(3); 
kxmax = 4*pi/(3);
kymin = -2*pi/(sqrt(3)); 
kymax = 2*pi/(sqrt(3));
kzmin = -pi; 
kzmax = pi;



%variable:
KA = 
KB =
KC =

JA = 
JB = 
JC = 

Dz =

kx = 
ky = 
kz =



% matrix:

TA = 2*KA*S - 2*JA*S*(cos(kz*c) - 1);
TB = 2*KB*S - 2*JB*S*(cos(kz*c) - 1);
TC = 2*KC*S - 2*JC*S*(cos(kz*c) - 1);

H = [                                                4*J*S+TA, -(S*cos((a*kx)/4 + (3^(1/2)*a*ky)/4)*(Dz*4i + 4*J))/2, (S*cos((a*kx)/4 - (3^(1/2)*a*ky)/4)*(Dz*4i - 4*J))/2
    (S*cos((a*kx)/4 + (3^(1/2)*a*ky)/4)*(Dz*4i - 4*J))/2,                                                 4*J*S+TB,                   -(S*cos((a*kx)/2)*(Dz*4i + 4*J))/2
    -(S*cos((a*kx)/4 - (3^(1/2)*a*ky)/4)*(Dz*4i + 4*J))/2,                     (S*cos((a*kx)/2)*(Dz*4i - 4*J))/2,                                                4*J*S+TC];

[Matt J]

Well, just taking the case where all 3 eigenvalues are equal, that would imply that the Hermitian matrix is a scalar multiple of eye(3). That condition gives 10 equations in 10 unknowns, which means you might have a finite set of solutions. However, for 2 equal eigenvalues, you will have to remove one of the equations. With more unknowns than equations, it is unlikely you will have a finite (or even a countable) set of solutions.