Introduction

In this section, we consider the quadratic eigenvalue problem (QEP) of the form

$$(\lambda^2 M + \lambda C + K)x = 0 \quad\mbox{and}\quad y^{\ast} (\lambda^2 M + \lambda C + K ) = 0,$$ (245)

where $M, K$, and $C$ are given matrices of size $n \times n$. The nontrivial $n$-vectors $x$, $y$, and the corresponding scalars $\lambda$ are the right, left eigenvectors, and eigenvalues, respectively.

The matrix function $L(\lambda) \equiv \lambda^2 M + \lambda C + K$ is a special case of a matrix polynomial, or a $\lambda$-matrix; see, for example, [187,284,194]. In this case, it is a $\lambda$-matrix of degree 2. The matrix function $L(\lambda)$ is said to be regular if $\det(L(\lambda))$ is not identical to zero for all $\lambda$. Otherwise, it is called singular.

An important special case of the quadratic eigenvalue problem is when

$$M^{\ast} = M > 0, \quad C^{\ast} = C, \quad \mbox{and} \quad K^{\ast} = K >0.$$ (246)

These matrices are sometimes called mass, damping, and stiffness matrices, respectively, referring to their origin in mechanical engineering models; see, for instance, [145]. In some problems, the stiffness matrix $K$ is only semi-positive definite. In this case, we may consider a shifted QEP to be discussed in §9.2.3.

One of the factors that makes the QEP different from standard eigenproblems $Ax = \lambda x$, or generalized eigenproblems $Ax = \lambda Bx$, is that there are $2n$ eigenvalues for QEP, with at most $2n$ right (and left) eigenvectors. Of course, in an $n$-dimensional space the right (and left) eigenvectors no longer form an independent set. This is illustrated by the following simple example. The triplet

$$M=\left[ \begin{array}{cc} 5 & 2 \\ 1 & 4 \end{array} \ri... ...ft[ \begin{array}{cc} 2 & 0 \\ 0 & 12 \end{array} \right]$$

has $4$ different (but pairwise conjugate) eigenvalues (rounded to five decimals):

$$\begin{array}{ll} \lambda_1=-0.9396+ 1.5749i, & \lambda_2=-0... ...-0.0049+ 0.6296i, & \lambda_4=-0.0049- 0.6296i. \\ \end{array}$$

The associated eigenvectors (normalized so that the first coordinate is equal to 1) are:

$$\begin{array}{ll} x_1= (1, -2.4756-0.9779i)^T, & x_2= (1, -2... ...0326-0.0132i)^T, & x_4= (1, ~0.0326+0.0132i)^T. \\ \end{array}$$

The four eigenvectors are obviously dependent, but, in actual problems, each of them may represent a relevant state of the system.

One has to be careful with Rayleigh quotients for quadratic eigenproblems. Indeed, given $x$ as a right eigenvector for the QEP (9.1), i.e.,

$$( \lambda^2 M + \lambda C + K ) x = 0,$$

one can form a quadratic Rayleigh quotient:

$$\lambda^2 ( x^{\ast} M x ) + \lambda ( x^{\ast} C x ) + ( x^{\ast} K x) = 0.$$ (247)

However, this equation has two roots; one of the roots is an eigenvalue, the other root may be a spurious one. For instance, if we compute the quadratic Rayleigh quotient for our example, with $(\lambda_1,x_1)$, then clearly, the pair $(\lambda_1,x_1)$ satisfies equation (9.3). If we solve equation (9.3), then we find the two roots $\mu_1=-0.9396+1.5749i$, $\mu_2=-0.8776-1.6057i$. We see that $\lambda_1$ is recovered (by $\mu_1$) and the other root has no meaning for the given QEP.

In an effort to decide which of the two is the desired one and which is the spurious one, one could compute the residual vector

$$r_{\mu}\equiv (\mu^2 M +\mu C + K) x_1 ,$$

and this leads to $\Vert r_{\mu_1}\Vert _2\approx 8.4 \times 10^{-14}$, $\Vert r_{\mu_2}\Vert _2\approx 12.5$, which, in this case, clearly points out that $\mu_2$ is not an eigenvalue. We cannot exclude the possibility that in contrived examples, one might make a wrong choice, which may lead to a delay in a specific iterative solution method.

For more general matrices, we can have defectiveness, as for the standard eigenproblems, which means that there is not necessarily a complete set of eigenvectors. In the next section, we will relate the QEP to a generalized standard problem, which helps to shed more light on this matter.