Lehmer's conjecture

In mathematics, the quadratic eigenvalue problem (QEP), is to find scalar eigenvalues $\lambda$, left eigenvectors $y$ and right eigenvectors $x$ such that

$Q(\lambda )x=0\text{ and }{y}^{\ast }Q(\lambda )=0,$

where $Q(\lambda )={\lambda }^2{A}_2+\lambda A_1+A_0$, with matrix coefficients $A_2,A_1,A_0\in {\mathbb{ℂ}}^{n\times n}$ and we require that $A_2\ne 0$, (so that we have a nonzero leading coefficient). There are $2n$ eigenvalues that may be infinite or finite, and possibly zero. This is a special case of a nonlinear eigenproblem. $Q(\lambda )$ is also known as a quadratic matrix polynomial.

Applications

A QEP can result in part of the dynamic analysis of structures discretized by the finite element method. In this case the quadratic, $Q(\lambda )$ has the form $Q(\lambda )={\lambda }^{2}M+\lambda C+K$, where $M$ is the mass matrix, $C$ is the damping matrix and $K$ is the stiffness matrix. Other applications include vibro-acoustics and fluid dynamics.

Methods of Solution

Direct methods for solving the standard or generalized eigenvalue problems $Ax=\lambda x$ and $Ax=\lambda Bx$ are based on transforming the problem to Schur or Generalized Schur form. However, there is no analogous form for quadratic matrix polynomials. One approach is to transform the quadratic matrix polynomial to a linear matrix pencil ($A-\lambda B$), and solve a generalized eigenvalue problem. Once eigenvalues and eigenvectors of the linear problem have been determined, eigenvectors and eigenvalues of the quadratic can be determined.

The most common linearization is the first companion linearization

$L(\lambda )=\lambda \left[\begin{array}{cc}M& 0\\ 0& I_n\end{array}\right]+\left[\begin{array}{cc}C& K\\ -I_n& 0\end{array}\right],$

where $I_n$ is the $n$-by-$n$ identity matrix, with corresponding eigenvector

$z=\left[\begin{array}{c}\lambda x\\ x\end{array}\right].$

We solve $L(\lambda )z=0$ for $\lambda$ and $z$, for example by computing the Generalized Schur form. We can then take the first $n$ components of $z$ as the eigenvector $x$ of the original quadratic $Q(\lambda )$.