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Suppose we have a quadratic eigenvalue problem $(A_{0}+\lambda A_{1}+ \lambda^{2} A_{2})x=0$. I'd to know if there are conditions under which the problem is known to have a small number of distinct eigenvalues (say, 3 or 4). If it helps, I can assume that $A_{0}=I,A_{1}=I-J$ and that $A_{2}$ is negative semidefinite.

P.S. $J$ is the all-ones matrix.

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What is the matrix $J$? – Federico Poloni Jan 15 '13 at 14:30
@FedericoPoloni: the all-ones matrix. – Felix Goldberg Jan 15 '13 at 15:18

The following paper may help, Algorithms for the Nonlinear Eigenvalue Problem, SIAM Journal of Numer. Anal., vol. 10, No. 4, P.674.

Your equation may be cast into a linear eigenvalue problem, as $A_0$ is non-singular (see the matrix $B$ after eq.(1.5) in the same paper).

I think probably the problem becomes $\lambda I - B = 0$.

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Assume that $A_0$ is invertible. The problem reduces to obtain conditions so that the pseudo Hamiltonian $M=\begin{pmatrix}0&-I\\{A_0}^{-1}A_2&{A_0}^{-1}A_1\end{pmatrix}$ has a small number of eigenvalues (change $\lambda$ with $-1/\lambda$). Here we want that $M=\begin{pmatrix}0&-I\\A_2&I-J\end{pmatrix}$ admits at most $4$ distinct eigenvalues. It suffices to say that $M$ admits an annulator polynomial of degree $4$.

Let $P$ be a polynomial of degree $4$. Then $P(M)=0$ can be written as a system of $4$ matrix equations of degree $2$ in the unknown $A_2$. Of course, it is possible that, for some unlucky choices of $P$, there are no solutions in $A_2$.

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