Invert QEP.

For most iterative methods for solving a generalized eigenvalue problem, the formulation (9.4), with either (9.5) or with (9.10), is suitable if one wants to determine a few of the exterior eigenvalues and eigenvectors. If one wants to compute some of the smallest (in modulus) eigenvalues and eigenvectors, then the obvious transformation is $\mu=1/\lambda$, and, after multiplying the QEP (9.1) with $\mu^2$, we obtain the invert QEP:

$$\left( M + \mu C + \mu^2 K \right) x = 0 .$$ (256)

Here it is assumed that $0$ is not an eigenvalue of the original QEP (9.1), i.e., that $K$ is nonsingular.

The QEP for the triplet $\{K,C,M\}$ can be linearized as discussed in §9.2.2, for instance, as (9.4) with (9.5), where $M$ interchanged with $K$. We can reformulate this generalized linearized eigenproblem in terms of $\lambda$, instead of $\mu$, which leads to

$$A z = \frac{1}{\lambda} B z,$$ (257)

where

$$A = \twobytwo{-C}{-M}{I}{0}, \quad B = \twobytwo{K}{0}{0}{I}, \quad z = \twobyone{x}{\lambda x}.$$ (258)

Note that from the factorization

$$B - \lambda A = \twobytwo{I}{\lambda M}{0}{I} \twobytwo{\lambda^2 M + \lambda C + K}{0}{0}{I} \twobytwo{I}{0}{-\lambda I}{I}$$

we know that the pencil $B - \lambda A$ is equivalent to

$$\twobytwo{\lambda^2 M + \lambda C + K}{0}{0}{I} .$$

Since $\det(B - \lambda A) = \det( \lambda^2 M + \lambda C + K )$, we conclude that the matrix pencil $B - \lambda A$ is regular if and only if the quadratic matrix polynomial $\lambda^2 M + \lambda C + K$ is regular and the eigenvalues of the original QEP (9.1) coincide with the eigenvalues of the matrix pencil $B - \lambda A$.

For the special case (9.2), we may formulate the generalized eigenvalue problem $Az = \frac{1}{\lambda}Bz$, with

$$A = \twobytwo{C}{M}{M}{0}, \quad B = \twobytwo{-K}{0}{0}{M}.$$ (259)

In this case, both matrices are Hermitian, but indefinite. Linearization with (9.15) results after left multiplication of (9.14) with a block-diagonal matrix $\mbox{diag}(-I,-M)$. Therefore, if $\mbox{det}(M) \neq 0$, then the pencil $B - \lambda A$ is regular if and only if the quadratic matrix polynomial $\lambda^2 M + \lambda C + K$ is regular.