Under what conditions do eigenvalues of a quadratic eigenvalue problem come in reciprocal pairs?

Question

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.

Answer 1

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).