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Suppose we have a quadratic eigenvalue problem $\lambda^2 M + \lambda C + K$. Under what conditions is the following statement true: If $\lambda$ is an eigenvalue, so is $1/\lambda$?

Here, $M$, $C$, and $K$ are square matrices (not necessarily full rank). This is of interest to me since I have such systems for which I know (based on physical arguments) that the eigenvalues must come in reciprocal pairs, but I don't know what this necessarily implies about the matrix properties. From a cursory look through Google, it seems that palindromic QEPs have this property, but I'm wondering if this property is more general.

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It might help to elaborate a little. Why is this interesting to you? Where does this problem arise? Also, it help to pinpoint what you are actually asking to define the notation. What are M,C,K? Where does the polynmial you wrote down actually live? etc... –  José Figueroa-O'Farrill Nov 26 '09 at 11:35
    
Odds are that your pencil is palindromic or can be reduced to a palindromic one using some tricks. Can you tell us more on your physical problem? –  Federico Poloni Sep 25 '10 at 14:59

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Consider the linearization $$ L(\lambda)=\lambda\begin{pmatrix}M&0\cr 0&1 \end{pmatrix}+\begin{pmatrix}C&K\cr -1&0\end{pmatrix}=\lambda A+B. $$ The eigenvalues of the original problem coincide with those for the linearization.

Now, the eigenvalues of the linearized problem are the roots of the polynomial $\det(\lambda A+B)$, while their reciprocals are roots of the polynomial $\lambda^{2n}\det(\lambda^{-1}A+B)=\det(\lambda B+A)$. If we require that these coincide with multiplicities, then the two polynomials must be linearly dependent. (In fact, they will be $\pm 1$ times each other.) If $A$ is nondegenerate, this implies that $A^{-1}B$ is similar to its inverse matrix. The latter condition does not give us much immediately, but at least we know that $\det A=\pm\det B$; that is, $\det M=\pm\det K$. We also get that the traces of $A^{-1}B$ and $B^{-1}A$ coincide, which can be rewritten as a condition on $M$, $C$, and $K$ (hopefully).

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