# Nonorthogonal eigenvectors for general eigenvalue problem with eig() and eigs()

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William el 5 de Jun. de 2024
Respondida: Christine Tobler el 5 de Jun. de 2024
I am working with a finite element model and want to decompose my system using the eigenvectors of the system. The eigenvalue problem is of the form: K*V = M*V*D, where V is the eigenvector matrix and D is the eigenvalue matrix. Both M and K are block diagonal symmetric matrices. For the generalized eigenvalue problem it should be the case that V'*M*V = a matrix whose only nonzero terms are along the diagonal, but I am getting nonzero terms on either side of these. Using the code included below for something like N =5, I get (after rounding to nearest 3rd decimal to remove near zero terms)
abs( Orthgonal_Check_eig) = [ 78.8050 0 0 0 0
0 43.4540 0 0 0
0 0 0 34.9310 0
0 0 34.9310 0 0
0 0 0 0 29.6430]
Where I am taking the absolute value because this case has complex eigenvectors (though I believe I've seen it with real eigenvector matrices as well). If I rerun the code it does not always happen, but I am trying to understand why this can occur for both eig() and eigs() so that I can determine if it is the conditioning of the matrices or if it is inherent to the approximations inside of eig() and eigs(). Is there a better one of the two to use to avoid this issue?
N = 5;
M = zeros(N,N);
K = zeros(N,N);
for ii = 1:N
% Diagonal Terms
M(ii,ii) = rand*50;
K(ii,ii) = rand*500;
if ii ~=N
% Off Diagonal Terms
off_diagonal = rand*50;
M(ii,ii+1) = off_diagonal;
K(ii,ii+1) = off_diagonal*10;
M(ii+1,ii) = off_diagonal;
K(ii+1,ii) = off_diagonal*10;
end
end
M= round(M);
K= round(K);
[Veig,~] = eig(K,M);
[Veigs,~] = eigs(K,M,3);
Orthgonal_Check_eig = Veig'*M*Veig;
Orthgonal_Check_eigs =Veigs'*M*Veigs;
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Christine Tobler el 5 de Jun. de 2024
For simple eigenvalue problem A*x = lambda*x, the eigenvalues are real and the eigenvectors can form an orthogonal basis only if A is symmetric. In the generalized case, this condition becomes that A must be symmetric and B must be symmetric positive definite.
So what you're expecting is true if K is symmetric (Hermitian if K is complex) and M is symmetric positive definite (Hermitian if M is complex). However, in this example M is not positive definite:
>> eig(M)
ans =
-32.1828
30.7556
37.5342
50.4526
110.4404
When M is not positive definite, there is no guarantee that the eigenvalues are all real, which you can see by computing them:
eig(K, M)
ans =
11.0502 + 0.0000i
11.1566 + 5.5626i
11.1566 - 5.5626i
6.8143 + 0.0000i
6.4945 + 0.0000i
This is also why your proof in the comment above doesn't work out: It's assuming that the eigenvectors and eigenvalues of (K, M) are real (and therefore a transpose instead of a conjugate transpose should be used). If you use conjugate transpose there, the terms don't cancel anymore there.
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### Más respuestas (1)

Torsten el 5 de Jun. de 2024
Editada: Torsten el 5 de Jun. de 2024
From the documentation:
[V,D] = eig(A,B) returns diagonal matrix D of generalized eigenvalues and full matrix V whose columns are the corresponding right eigenvectors, so that A*V = B*V*D.
I don't see where it is guaranteed that V'*B*V is a diagonal matrix. I also don't see that V should be a unitary matrix ( V' = inv(V) ) (maybe it is if B^(-1)*A is Hermitian).
If the inverse exists,
inv(B*V)*A*V
should be diagonal and equal D.
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William el 5 de Jun. de 2024
Editada: William el 5 de Jun. de 2024
I have noticed that sometimes the eig()/eigs() functions will output matrcies such that V'*B*V = eye(N), at least when M is a diagonal matrix, but I haven't seen this happen with these block diagonal cases. The proof I found for V'*B*V being diagonal is that if = V(:,i), and = D(i,i) then
eqn1 :
eqn2 :
Subtracting *(eqn1) from *(eqn2),
And so (likewise for B), then this equation becomes
=0
So the off diagonal terms have to be zero when for this condition to hold, and thus V'*B*V is a diagonal matrix. I am not sure if this must be true for all generalized eigenvalue problems but I know it is true for the model I am using.
Edit: I realize now that this condition is still satisfied if there is a case where but . But I am not sure what this means about A and B if they have repeated real eigenvalues.
Torsten el 5 de Jun. de 2024
I already wrote the V' does not need to be inv(V).
From the equation
A*V = B*V*D
that MATLAB says to be true it follows that
inv(B*V)*A*V = D
if B*V is invertible - nothing less and nothing more.

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