Quadratic eigenvalue problem: Difference between revisions

Revision as of 21:39, 10 March 2023

Revision as of 00:33, 28 March 2023

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\end{bmatrix}.

</math>

For convenience, one often takes <math>N</math> theto be the <math>n\times n</math> [[identity matrix]]. We solve <math> L(\lambda) z = 0 </math> for <math> \lambda </math> and <math>z</math>, for example by computing the Generalized Schur form. We can then

take the first <math>n</math> components of <math>z</math> as the eigenvector <math>x</math> of the original quadratic <math>Q(\lambda)</math>.