Dear Fahreddın, I think, the norm of the eigenvectors corresponds to some generic amplitude. But that is something you cannot extract from the solution of the eigenvalue problem but it depends on the initial deflection or velocities. So I think you should be able to use the normalized values just as well as the non-, un- or not normalized ones. Octave seems to normalize that way that, transpose(Z).B.Z=I, where Z is the matrix of eigenvectors, B is matrix B of the generalized eigenvalue problem and I is the identity. It uses lapack functions. But that's only true if A,B are symmetric. If not it normalizes the magnitude of largest element of each eigenvector to 1. I believe you can get it like that. If U is a Matrix with normalization factors it is diagonal and Z.A contains the normalized column vectors. then it is: transpose(Z.A).B.Z.A =transpose(A).transpose(Z).B.Z.A =A.transpose(Z).B.Z.A=I and thus invert(A).invert(A)=transpose(Z).B.Z As A is diagonal invert(A) has the reciprocal elements on the diagonal. So you can easily extract them A=diag(1/sqrt(diag(transpose(Z).B.Z))) I hope that's correct. Best Regards Lennart