non linear eigen value problem

I want to find the eigen values of

k0=magic(2);
k1=1e-3*k0;
c1=[1 0;0 1] 

c1 = 2×2

1 0 0 1

c2=[0 1;1 0]

c2 = 2×2

0 1 1 0

B=kron(c1,k0)+kron(c2,k1)

B = 4×4

1.0000 3.0000 0.0010 0.0030 4.0000 2.0000 0.0040 0.0020 0.0010 0.0030 1.0000 3.0000 0.0040 0.0020 4.0000 2.0000

syms lambda1a lambda1b lambda2a lambda2b
x = sym('x',[size(B,2),1])

x = 

l1=[lambda1a;lambda1b ]

l1 = 

l2=[lambda2a;lambda2b ]

l2 = 

% lambda1 is 2 by 1 matrix and lambda2 is 2 by 1 matrix 
% I want to solve the problem

B*x

ans = 

(kron(diag(l1),c1)+kron(diag(l2),c2))*x

ans = 

%I ant to find l1 and l2

[Torsten]

Could you correct your code ? There seems to be a dimension mismatch.

[rakesh kumar]

(kron(diag(l1),c1)+kron(diag(l2),c2))

[Walter Roberson]

l1 and l2 are each 2 x 1 matrices. You cannot take the eigenvalues of non-square matrices.

If you take eig(diag(l1)) and eig(diag(l2)) so that you are making them into 2 x 2 diagonal matrices, then the eigenvalues are just the contents of l1 and l2

Answer 1

[Christine Tobler]

This isn't the standard definition of a nonlinear eigenvalue problem, where you would have only one scalar lambda.

Am I understanding correctly that you are looking for four scalar values lambda... for which there exists an non-zero vector x which satisfies

B*x == (kron(diag(l1),c1)+kron(diag(l2),c2))*x

In that case, I would take this to the symbolic toolbox, by trying to solve for values of lambda... for which

det(B == (kron(diag(l1),c1)+kron(diag(l2),c2))) == 0

(using DET is fine for symbolic calculations, we only run into trouble using it when subjected to round-off error).