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Next: Spectral Transformations for QEP Up: Quadratic Eigenvalue Problems Z. Bai, Previous: Introduction   Contents   Index


Transformation to Linear Form

It is easy to see that the QEP in (9.1) is equivalent to the following generalized ``linear'' eigenvalue problem:[*]
\begin{displaymath}
A z = \lambda B z \quad \mbox{and} \quad
w^{\ast} A = \lambda w^{\ast} B ,
\end{displaymath} (248)

where
\begin{displaymath}
A = \twobytwo{0}{I}{-K}{-C}, \quad
B = \twobytwo{I}{0}{0}{M}
\end{displaymath} (249)

and
\begin{displaymath}
z = \twobyone{x}{\lambda x}, \quad
w = \twobyone{(\lambda M + C)^{\ast} y }{y}.
\end{displaymath} (250)

The generalized eigenvalue problem (9.4) is commonly called a linearization of the QEP (9.1). It can be shown that for any matrices $A$ and $B$ of the above forms, the right and left eigenvectors $z$ and $w $ have the structures described in (9.6).

Note that from the factorization

\begin{displaymath}
A - \lambda B = \twobytwo{0}{I}{-I}{-\lambda M - C}
\twobyt...
... M + \lambda C + K}{0}{0}{I}
\twobytwo{I}{0}{-\lambda I }{I},
\end{displaymath} (251)

we can conclude that the pencil $A - \lambda B$ is equivalent[*] to the matrix
\begin{displaymath}
\twobytwo{\lambda^2 M + \lambda C + K }{0}{0}{I}
\end{displaymath} (252)

and

\begin{displaymath}
{\rm det}(A - \lambda B ) = {\rm det}(\lambda^2 M + \lambda C + K).
\end{displaymath}

This means that the eigenvalues of the original QEP (9.1) coincide with the eigenvalues of the generalized eigenvalue problem (9.4). Furthermore, we have that For the theory on regular pencils $(A,B)$, see, for instance, [425, Chap. VI]. We will assume that at least $M$ is nonsingular throughout this section.

A disadvantage of the above reduction to linear form is that if the matrices $M$, $C$, and $K$ are all Hermitian, then this is not reflected in the reduced form (9.5), where $A$ is non-Hermitian. This can be repaired as follows.

In fact, the matrix pair $(A,B)$ in (9.4) can be chosen in a more general form

\begin{displaymath}
A = \twobytwo{0}{W}{- K}{ - C}, \quad
B = \twobytwo{W}{0}{0}{ M},
\end{displaymath}

where $W$ can be any arbitrary nonsingular matrix. Note that now the matrix pencil $A - \lambda B$ is equivalent to the matrix polynomial (9.8) if and only if $W$ is nonsingular, and because of (9.7),

\begin{displaymath}
\det( A - \lambda B ) = \det(W)\cdot \det( \lambda^2 M + \lambda C + K ).
\end{displaymath}

For example, if the matrices $M$, $K$, and $C$ are all symmetric, as in the special case (9.2), and $K$ is nonsingular, then we may choose $W = - K$, which leads to the following symmetric generalized ``linear'' eigenvalue problem
\begin{displaymath}
A z = \lambda B z \quad \mbox{and} \quad
w^{\ast} A = \lambda w^{\ast} B ,
\end{displaymath} (253)

where
\begin{displaymath}
A = \twobytwo{0}{-K}{-K}{ -C}, \quad
B = \twobytwo{-K}{0}{0}{M}, \quad
\end{displaymath} (254)

and
\begin{displaymath}
z = \twobyone{x}{\lambda x}, \quad
w = \twobyone{y}{\bar\lambda y}.
\end{displaymath} (255)

Both $A$ and $B$ are symmetric, but may be indefinite.


next up previous contents index
Next: Spectral Transformations for QEP Up: Quadratic Eigenvalue Problems Z. Bai, Previous: Introduction   Contents   Index
Susan Blackford 2000-11-20