Diagonalization in eigs with Generalized Eigenvalue Problem with Positive Semidefinitive matrix

After I execute an eigs command in Matlab 2020b, using as input matrix A and B, i.e. a generalized eigenvalue problem, and 'SM' as sigma, it appears that unstable eigenvectors are obtained when A is a positive semidefinitive matrix, eventhougth the output eigenvalues are fine. The function I use is:

[V, D] = eigs(A, B, ArbitraryNumberOfEigenvalues, 'SM');

In short, the mathematical problem I'm coding is to model the response of a Finite Element Method Vibro-Acoustic problem, hence if we normalize the eigenvectors in respect to the B matrix and diagonalize the A matrix, we should be obtaining again the eigenvalues.

%% Normalization

nm = size(D, 1);
    
for j = 1 : nm

    fm = V(:, j).' * B * V(:, j);
    V(:, j) = V(:, j) / sqrt(fm);

end

%% Diagonalization

D = V.' * A * V;

But, as I said, the curve becomes really unstable:

Now, if I impose a boundary condition to the matrices, as example, excluding the rows and collums from 49th to 72th, A matrix becomes Positive Definite and the curve congerve smoothly:

I believe both curves should converge smoothly. Unfortunalety, I can't just use the output eigenvalues matrix, because I will use the eigenvectors to multiply with other matrices. Is this instability expected ? Is there any workaround ?

Thanks.

[Christine Tobler]

EIGS with the 'sm' option is based on solving linear systems with the A matrix. When A is semidefinite, this is going to be badly conditioned (A is singular strictly speaking). Usually EIGS should give a warning about this, saying that the matrix is badly conditioned. In a singular case, the results can be wrong in these cases.

You can probably fix the issue by looking for all eigenvalues closest to some moderately small number, for example

sigma = -100;
[V, D] = eigs(A, B, ArbitraryNumberOfEigenvalues, sigma);

The "moderately small" should be with respect to an estimated norm of the matrix A, the goal being to get to a point where A - sigma*B is a well-conditioned matrix.

[Bruno Luong]

Just wonder: Is your B matrix strictly definite positive? I see you divide by square root of

fm = V(:, j).' * B * V(:, j)

It can causes problem if B is not coercive operator.

[Thiago Morhy]

Bruno Luong, yes, B is positive definite.

I normalize the eigenvectors so that when I diagonalize the B matrix, I obtain an identity matrix. (And consequently, obtain the Eigenvalue matrix, if diagonalize the A matrix instead).

[Bohan]

What is the output created by eig or eigs in Matlab if the B matrix is not strictly positive definite? I tried some examples and there are output but I am not sure what this means.

[Torsten]

@Bohan

Could you be more explicit ? Please include the examples and explain what you don't understand.

Answer 1

[Thiago Morhy]

Cristine Tobler's comment answered my problem.

Yes, eigs was giving me an alert of singularity of matrix A. Using, now, the command:

sigma = -100;
[V, D] = eigs(A, B, ArbitraryNumberOfEigenvalues, sigma);

I obtain the following curve of eigenvalues:

It converges flawlessly. And as the lower eigenvalue is approximately zero, a -1 sigma works as well. Thanks for the help.