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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}](img3312.png) |
(248) |
where
![\begin{displaymath}
A = \twobytwo{0}{I}{-K}{-C}, \quad
B = \twobytwo{I}{0}{0}{M}
\end{displaymath}](img3313.png) |
(249) |
and
![\begin{displaymath}
z = \twobyone{x}{\lambda x}, \quad
w = \twobyone{(\lambda M + C)^{\ast} y }{y}.
\end{displaymath}](img3314.png) |
(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
and
of the above forms,
the right and left eigenvectors
and
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}](img3315.png) |
(251) |
we can conclude that the pencil
is equivalent
to the matrix
![\begin{displaymath}
\twobytwo{\lambda^2 M + \lambda C + K }{0}{0}{I}
\end{displaymath}](img3321.png) |
(252) |
and
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
is regular if and only if
is regular;
- if
(hence
) is nonsingular, then
is regular;
- if
(hence
) is nonsingular, then
is regular.
For the theory on regular pencils
, see, for instance,
[425, Chap. VI].
We will assume that at least
is nonsingular throughout this section.
A disadvantage of the above reduction to linear form is that if
the matrices
,
, and
are all Hermitian, then this is not reflected in
the reduced form (9.5), where
is
non-Hermitian. This can be repaired as follows.
In fact, the matrix pair
in (9.4) can be
chosen in a more general form
where
can be any arbitrary nonsingular matrix.
Note that now the matrix pencil
is equivalent to the
matrix polynomial (9.8) if and only if
is nonsingular,
and because of (9.7),
For example, if the matrices
,
, and
are all symmetric, as
in the special case (9.2), and
is nonsingular, then we may choose
, 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}](img3312.png) |
(253) |
where
![\begin{displaymath}
A = \twobytwo{0}{-K}{-K}{ -C}, \quad
B = \twobytwo{-K}{0}{0}{M}, \quad
\end{displaymath}](img3327.png) |
(254) |
and
![\begin{displaymath}
z = \twobyone{x}{\lambda x}, \quad
w = \twobyone{y}{\bar\lambda y}.
\end{displaymath}](img3328.png) |
(255) |
Both
and
are symmetric, but may be indefinite.
Next: Spectral Transformations for QEP
Up: Quadratic Eigenvalue Problems Z. Bai,
Previous: Introduction
  Contents
  Index
Susan Blackford
2000-11-20