# Generic eigenstructures of Hermitian pencils

Fernando De Terán Departamento de Matemáticas, Universidad Carlos III de
Madrid, Avda. Universidad 30, 28911 Leganés, Spain<EMAIL_ADDRESS>Andrii
Dmytryshyn School of Science and Technology, Örebro University, 701 82,
Örebro, Sweden<EMAIL_ADDRESS>Froilán M. Dopico Departamento de
Matemáticas, Universidad Carlos III de Madrid, Avda. Universidad 30, 28911
Leganés, Spain<EMAIL_ADDRESS>
###### Abstract

We obtain the generic complete eigenstructures of complex Hermitian $n\times
n$ matrix pencils with rank at most $r$ (with $r\leq n$). To do this, we prove
that the set of such pencils is the union of a finite number of bundle
closures, where each bundle is the set of complex Hermitian $n\times n$
pencils with the same complete eigenstructure (up to the specific values of
the finite eigenvalues). We also obtain the explicit number of such bundles
and their codimension. The cases $r=n$, corresponding to general Hermitian
pencils, and $r<n$ exhibit surprising differences, since for $r<n$ the generic
complete eigenstructures can contain only real eigenvalues, while for $r=n$
they can contain real and non-real eigenvalues. Moreover, we will see that the
sign characteristic of the real eigenvalues plays a relevant role for
determining the generic eigenstructures of Hermitian pencils.

###### keywords:

Matrix pencil, rank, strict equivalence, congruence, Hermitian matrix pencil,
orbit, bundle, closure, sign characteristic.

###### AMS:

15A22, 15A18, 15A21, 15A54.

##  1 Introduction

The complete eigenstructure of a matrix pencil (or just a “pencil”, for short)
is an intrinsic information of the pencil that is relevant in many of the
applied problems where matrix pencils (or, more in general, matrix polynomials
and rational matrices) arise (see, for instance, [11, 12, 13, 32, 40, 41] and
the references therein). More precisely, the complete eigenstructure of a
pencil is the set of invariants of the pencil under strict equivalence, and it
is encoded in the Kronecker canonical form (see [24, Ch. XII] or [27, §3] for
a more recent reference). In many applications where matrix pencils arise
(either by themselves or by means of linearizations of matrix polynomials and
rational matrices) the coefficient matrices have some particular symmetries,
which lead to structured matrix pencils. These include (skew-)symmetric,
(skew-)Hermitian, (anti-)palindromic, or alternating matrix pencils (see, for
instance, [21, 31, 32]).

The problem addressed in the present work is an instance of the general
problem of determining the most likely complete eigenstructure of matrix
pencils within some particular set, ${\cal S}$. We use the word generic for
the most likely complete eigenstructure, to mean that all pencils within the
set ${\cal S}$ are in the closure of the set of pencils having the generic
eigenstructure. In other words, in every neighborhood of any particular pencil
in ${\cal S}$ there is at least one pencil having the generic eigenstructure.
The generic complete eigenstructure of general $n\times n$ pencils consist, as
it is well-known, of $n$ different simple eigenvalues. However, when some
restrictions are imposed to the pencils, so that we restrict ourselves to
pencils in some particular set $\bf{\cal S}$, then it is not, in general, so
easy to identify the generic complete eigenstructure (for instance, it is not
trivial to obtain the generic complete eigenstructure of general $m\times n$
pencils when $m\neq n$, see below). As a consequence, the problem of
describing the generic complete eigenstructure of matrix pencils within a
particular set has attracted the attention of researchers for several decades.
The research on this problem has allowed to describe the generic
eigenstructure for the following sets of matrix pencils:

  * •

Singular $n\times n$ (namely, square) pencils [42].

  * •

Rectangular pencils of a fixed size [14] (though the credit of the result, as
mentioned in [14, p. 85], goes back to, at least, [39]).

  * •

General $m\times n$ pencils with rank at most $r$ (smaller than
$\min\\{m,n\\}$) [6] (revisited in [9]).

  * •

Palindromic and alternating $n\times n$ pencils with rank at most $r$ (smaller
than $n$) [4].

  * •

Complex symmetric $n\times n$ pencils with rank at most $r$ (smaller than $n$)
[5].

  * •

Complex skew-symmetric $n\times n$ pencils with rank at most $2r$ (smaller
than $n$) [16].

If we remove the restriction for the pencils being of bounded rank, then the
generic complete eigenstructure of structured $n\times n$ pencils is also
known for the following structures (though, up to our knowledge, some of them
are not explicitly provided in the literature):

  * •

Complex symmetric $n\times n$ pencils: it is the same as for general (non-
structured) matrix pencils, namely $n$ different eigenvalues. A way to see
this is the following. Consider the set of $n\times n$ symmetric pencils as a
manifold depending on $n(n+1)$ complex variables, encoded in a vector $X$
(these variables come from the upper triangular part, including the main
diagonal, of the leading and the trailing coefficient matrices of the pencil).
Assume that $f(X,\lambda)=\det{\cal
P}(X,\lambda)=\sum_{i=0}^{n}p_{i}(X)\lambda^{i}$ is the determinant of a
general $n\times n$ symmetric pencil, ${\cal P}(X,\lambda)$. Then, the subset
of singular symmetric pencils is a proper algebraic set of
$\mathbb{C}^{n(n+1)}$, defined by the polynomial equations
$p_{0}(X)=\cdots=p_{n}(X)=0$. Second, the subset of pencils with a multiple
eigenvalue is also an algebraic set, namely
Res($f(X,\lambda),f^{\prime}(X,\lambda))=0$, where $f^{\prime}(X,\lambda)$ is
the derivative of $f(X,\lambda)$ with respect to the variable $\lambda$, and
Res denotes the resultant (which is a polynomial in $X$) [30, Ch. I, §3].
Therefore, the set of $n\times n$ symmetric pencils with $n$ different
eigenvalues is the complementary of the union of two algebraic sets. Since it
is also nonempty (for instance, $\mathop{\operator@font
diag}\nolimits(\lambda-1,\lambda-2,\ldots,\lambda-n)$ is such a pencil), we
conclude that it is a generic set.

  * •

Complex skew-symmetric $n\times n$ pencils: we have been unable to find an
explicit expression for the generic complete eigenstructures for this
structure. However, it can be deduced from the canonical form under congruence
of skew-symmetric pencils in [38] and the developments in [16]. More
precisely, it depends on whether $n$ is even or odd. If $n$ is even, then the
generic canonical form consists of $n/2$ skew-symmetric Jordan blocks
associated with different eigenvalues of multiplicity $2$, whereas if $n$ is
odd we add to the previous structure a $0$ diagonal block. In other words, the
generic complete eigenstructure consists of $n/2$ distinct eigenvalues of
multiplicity exactly $2$ (if $n$ is even), together with one left and one
right minimal indices equal to $0$ (when $n$ is odd).

  * •

$\top$-palindromic pencils: the generic complete eigenstructure is also
different depending on whether $n$ is even or odd [8, Th. 6]. More precisely,
it consists of $n/2$ pairs of different simple complex values of the form
$(\mu,1/\mu)$ (if $n$ is even), together with a simple eigenvalue $-1$ (when
$n$ is odd). For $\top$-anti-palindromic pencils the eigenvalue $-1$ is
replaced by $1$. For $*$-palindromic pencils the generic complete
eigenstructure can be found in [7, Th. 5.4] and it consists of $n/2$ pairs of
different simple complex values of the form $(\mu,1/\overline{\mu})$ with
$|\mu|>1$ (if $n$ is even), together with a simple eigenvalue which is an
unspecified complex number of modulus $1$ (when $n$ is odd).

  * •

$\top$-alternating pencils: in this case, the generic complete eigenstructure
(that can be obtained from the one of $\top$-palindromic pencils by means of a
Cayley transformation, see [4]) consists of $n/2$ pairs of different simple
complex values of the form $(\mu,-\mu)$, if $n$ is even, together with a
simple eigenvalue $\infty$ (for $\top$-even pencils) or $0$ (for $\top$-odd
pencils), when $n$ is odd. For $*$-alternating pencils, the pairs $(\mu,-\mu)$
are replaced by $(\mu,-\overline{\mu})$.

In the present work, we describe the generic complete eigenstructure of
Hermitian $n\times n$ matrix pencils with rank at most $r$, for any $0\leq
r\leq n$. When $r=n$, what we obtain is the generic complete eigenstructure of
general $n\times n$ Hermitian pencils (without restrictions), and for this
reason this case is addressed separately. We will prove (in Theorems 10 and
12) that the number of generic complete eigenstructures in the set of $n\times
n$ Hermitian pencils with rank at most $r$ is equal to
$\left(\left\lfloor\frac{r}{2}\right\rfloor+1\right)\left\lfloor\frac{r+3}{2}\right\rfloor$.
However, there are relevant differences between the case $r=n$ and $r<n$,
namely:

  * •

When $r=n$, all generic eigenstructures correspond to regular pencils having
$n$ simple eigenvalues. Some of these eigenvalues are real, and the other ones
are pairs of non-real complex conjugate numbers. Only in one of these
eigenstructures (namely, when all eigenvalues are real) there are no non-real
eigenvalues.

  * •

When $r<n$, however, none of the generic eigenstructures have non-real
eigenvalues.

In both cases, each of the generic eigenstructures differs from the others in
the number of real eigenvalues, and in the sign characteristics of these
eigenvalues. This emphasizes the relevance of the sign characteristic for
Hermitian pencils, which is a quantity that does not arise in the other
structures mentioned above.

It is worth to emphasize also that the number of generic complete
eigenstructures of $n\times n$ Hermitian pencils with rank at most $r$ is
always greater than $1$, so there is no a unique generic eigenstructure. In
fact, there are many when $r$ is large. This should not be surprising when
$r<n$ since it happens also for general square singular pencils [14] and for
singular symmetric pencils [5]. However, it may seem surprising in the case of
general $n\times n$ Hermitian pencils (namely, when $r=n$). We will see that
the lack of uniqueness when $r=n$ is a consequence of the different forms to
distribute the eigenvalues between real and non-real ones and also of the
different possible sign characteristics of the real eigenvalues.

The rest of the paper is organized as follows: in Section 2 we introduce the
notation and some basic notions that are used throughout the manuscript.
Section 3 presents some technical results that are needed to prove the main
results of the paper. The main results are introduced in Sections 4 and 5.
More precisely, Theorem 10 describes the generic complete eigenstructures of
Hermitian $n\times n$ matrix pencils, and in Theorem 12 we provide the generic
complete eigenstructures of Hermitian $n\times n$ matrix pencils with rank at
most $r$, with $r<n$. The codimension of these generic complete
eigenstructures are computed in Section 6, whereas in Section 7 we provide
some numerical experiments to show that all the generic complete
eigenstructures of general Hermitian pencils arise in numerical computations,
and that non-real eigenvalues do not typically appear in singular Hermitian
pencils. Finally, in Section 8 we present a summary of the contributions of
the manuscript.

##  2 Basic definitions and notation

By $\mathbb{R}$ and $\mathbb{C}$ we denote the fields of real and complex
numbers, respectively. We also follow the standard notation ${\rm re}(\mu)$
and ${\rm im}(\mu)$ for, respectively, the real and imaginary parts of the
complex number $\mu$, and $\rm i$ for the imaginary unit (namely, ${\rm
i}=\sqrt{-1}$).

A matrix pencil is of the form ${\cal P}(\lambda)=A+\lambda B$, with
$A,B\in\mathbb{C}^{m\times n}$, and $\lambda$ being a scalar variable (matrix
pencils can also be seen as pairs of $m\times n$ complex matrices $(A,B)$,
see, for instance, [27]). We use calligraphic letters, as above, to denote
matrix pencils. Sometimes, and for the sake of brevity, we will remove the
variable $\lambda$ and just write ${\cal P}$. The pencil ${\cal P}(\lambda)$
is called regular if $m=n$ and $\det{\cal P}(\lambda)$ is not identically zero
(as a polynomial in $\lambda$) and it is called singular otherwise. For a
matrix pencil ${\cal P}(\lambda)$ as above, we set ${\cal
P}(\lambda)^{*}=(A+\lambda B)^{*}=A^{*}+\lambda B^{*}$, where $*$ denotes the
conjugate transpose. It is important to note that the complex conjugation does
not affect the variable $\lambda$.

In this paper, we are interested in complex Hermitian matrix pencils, namely
those with $A^{*}=A$ and $B^{*}=B$. An important part of this work focuses on
Hermitian matrix pencils with bounded rank, where the rank of the pencil
${\cal P}$, denoted $\mathop{\operator@font rank}\nolimits{\cal P}$, is the
size of the largest non-identically zero minor of ${\cal P}$ (namely, the rank
of ${\cal P}$ when viewed as a matrix with entries in the field of rational
functions in $\lambda$). The set of complex Hermitian $n\times n$ pencils is
denoted by $\operatorname{PENCIL}_{n\times n}^{H}$, and
$\operatorname{PENCIL}_{n\times n}^{H}(r)$ denotes the set of complex
Hermitian $n\times n$ pencils with rank at most $r$.

The signature of a Hermitian constant matrix $A\in\mathbb{C}^{n\times n}$ is
the tuple $(\sigma_{+},\sigma_{-},\sigma_{0})$, where $\sigma_{+}$ is the
number of positive eigenvalues, $\sigma_{-}$ is the number of negative
eigenvalues, and $\sigma_{0}$ is the multiplicity of the $0$ eigenvalue of
$A$.

Two $n\times n$ pencils ${\cal H}_{1}(\lambda)$ and ${\cal H}_{2}(\lambda)$
are $*$-congruent if there exists a nonsingular matrix
$Q\in\mathbb{C}^{n\times n}$ such that ${\cal H}_{1}(\lambda)=Q^{*}\,{\cal
H}_{2}(\lambda)\,Q$. Note that, if ${\cal H}_{1}$ and ${\cal H}_{2}$ are
$*$-congruent, then ${\cal H}_{1}$ is Hermitian if and only if ${\cal H}_{2}$
is Hermitian. Since in the rest of this paper we only use the relation of
“$*$-congruence”, we will often refer to it simply as “congruence”.

The closure of a subset of $n\times n$ complex matrix pencils ${\cal S}$,
denoted by $\overline{\cal S},$ is considered in the Euclidean topology of the
space $\mathbb{C}^{2n^{2}}\simeq\mathbb{C}^{n\times
n}\times\mathbb{C}^{n\times n}$, which is identified with the set of $n\times
n$ matrix pencils, when considered as pairs of $n\times n$ matrices. Also,
open sets and open neighborhoods, as well as the notion of convergence, are
considered in this topology. Through the identification above,
$\operatorname{PENCIL}_{n\times n}^{H}(r)$ becomes a subset of
$\mathbb{C}^{2n^{2}}$ and we can consider in $\operatorname{PENCIL}_{n\times
n}^{H}(r)$ the subspace topology induced by the Euclidean topology of
$\mathbb{C}^{2n^{2}}$.

The direct sum of the pencils ${\cal P}_{1},\ldots,{\cal P}_{k}$ is a block
diagonal pencil whose diagonal blocks are ${\cal P}_{1},\ldots,{\cal P}_{k}$,
in this order. We will denote it by either $\mathop{\operator@font
diag}\nolimits({\cal P}_{1},\ldots,{\cal P}_{k})$ or $\bigoplus_{i=1}^{k}{\cal
P}_{i}$.

Following [5, p. 909], let ${\cal L}_{d}(\lambda):=\lambda G_{d}+F_{d}$, where

$F_{d}:=\begin{bmatrix}0&1&&\\\ &\ddots&\ddots&\\\ &&0&1\\\
\end{bmatrix}_{d\times(d+1)}\qquad\mbox{\rm and}\qquad
G_{d}:=\begin{bmatrix}1&0&&\\\ &\ddots&\ddots&\\\ &&1&0\\\
\end{bmatrix}_{d\times(d+1)},$

and define the Hermitian (actually, real symmetric) pencil

${\cal M}_{d}(\lambda):=\begin{bmatrix}0&{\cal L}_{d}(\lambda)^{\top}\\\ {\cal
L}_{d}(\lambda)&0\end{bmatrix}_{(2d+1)\times(2d+1)}.$

The pencil ${\cal M}_{0}$ is a $1\times 1$ null matrix, and it is a degenerate
case of ${\cal M}_{d}$ obtained after joining ${\cal L}_{0}$ and ${\cal
L}_{0}^{\top}$, namely a null column and a null row, respectively.

We are also going to use the following pencils:

  * •

Symmetric Jordan-like blocks associated with a finite eigenvalue:

${\cal J}_{k}^{H}(\mu):=\left[\begin{array}[]{cccc}&&1&\lambda-\mu\\\
&\iddots&\iddots&\\\ 1&\lambda-\mu&&\\\
\lambda-\mu&&&\end{array}\right]_{k\times k}\quad(\mu\in\mathbb{C}).$

  * •

Hermitian Jordan-like blocks associated with the infinite eigenvalue:

${\cal J}_{k}^{H}(\infty):=\left[\begin{array}[]{cccc}&&\lambda&1\\\
&\iddots&\iddots&\\\ \lambda&1&&\\\ 1&&&\end{array}\right]_{k\times k}.$

  * •

Hermitian Jordan-like blocks associated with a pair of complex conjugate
eigenvalues:

${\cal J}_{k}^{H}(\mu,\overline{\mu})=\begin{bmatrix}0&{\cal
J}_{k}^{H}(\overline{\mu})\\\ {\cal J}_{k}^{H}(\mu)&0\end{bmatrix}_{2k\times
2k}.$

Note that

$\begin{array}[]{lc}{\cal J}_{1}^{H}(\mu)={\cal
J}_{1}(\mu)=\lambda-\mu,&\mbox{for $\mu\in\mathbb{C}$, and}\\\ {\cal
J}_{1}^{H}(\infty)={\cal J}_{1}(\infty)=1.\end{array}$

The last two blocks are standard Jordan $1\times 1$ blocks, and for this
reason we will just write ${\cal J}_{1}(\mu)$ and ${\cal J}_{1}(\infty)$,
respectively, omitting the superscript $H$. Note also that the Jordan-like
block ${\cal J}_{k}^{H}(\mu)$ is Hermitian if and only if $\mu\in\mathbb{R}$.

We also warn the reader that ${\cal J}_{k}^{H}(\mu,\overline{\mu})$ has size
$2k\times 2k$, instead of $k\times k$.

The following result, which provides a canonical form for Hermitian pencils
under $*$-congruence, can be found in [27, Th. 6.1], but here we present it as
in [10, Th. 1].

######  Theorem 1.

(Hermitian Kronecker canonical form). Every $n\times n$ Hermitian matrix
pencil, ${\cal H}(\lambda)$, is $*$-congruent to a direct sum of blocks of the
form

  * (i)

blocks $\sigma{\cal J}_{k}^{H}(a)$, with $a\in\mathbb{R}$ and
$\sigma\in\\{+1,-1\\}$;

  * (ii)

blocks $\sigma{\cal J}_{k}^{H}(\infty)$, with $\sigma\in\\{+1,-1\\}$;

  * (iii)

blocks ${\cal J}_{k}^{H}(\mu,\overline{\mu})$, with $\mu\in\mathbb{C}$ having
positive imaginary part;

  * (iv)

blocks ${\cal M}_{k}(\lambda)$.

The parameters $a,k,\sigma$, and $\mu$ may be distinct in different blocks.
These parameters, as well as the number of blocks of each type, are uniquely
determined by ${\cal H}$, and they are the invariants of ${\cal H}$ under
$*$-congruence. Furthermore, the direct sum is unique up to permutation of
blocks. We will refer to this direct sum as the Hermitian Kronecker canonical
form of ${\cal H}$, and we denote it by HKCF$({\cal H})$.

The values $a$ associated with the blocks in part (i) of Theorem 1 are the
real eigenvalues of ${\cal H}$, whereas the values $\mu,\overline{\mu}$
associated with blocks in part (iii) are the pairs of (non-real) complex
conjugate eigenvalues of ${\cal H}$. Both the real and the complex conjugate
eigenvalues conform the set of finite eigenvalues of ${\cal H}$. Moreover, if
at least one block like the ones in part (ii) appears in HKCF(${\cal H}$),
then ${\cal H}$ has an infinite eigenvalue. The list of signs $\sigma$
appearing in the blocks $\sigma{\cal J}_{k}^{H}(a)$ and $\sigma{\cal
J}_{k}^{H}(\infty)$, given in a certain order, is known as the sign
characteristic of the pencil ${\cal H}$ [27]. We emphasize that the sign
characteristic of Hermitian matrix pencils and polynomials has been defined in
several equivalent ways in the literature (see, for instance, [28, 29, 34]).
Each block ${\cal M}_{k}$ in part (iv) is associated with a couple of left and
right minimal indices equal to $k$ [24]. The set of eigenvalues together with
the number, sign characteristics, and sizes of the blocks associated to them
in the HKCF(${\cal H}$) in Theorem 1, and the number and sizes of the blocks
${\cal M}_{k}(\lambda)$ associated to the minimal indices, constitute the
complete eigenstructure of ${\cal H}$.

Note that ${\cal H}$ is regular if and only if HCKF(${\cal H}$) does not
contain blocks ${\cal M}_{k}$.

The Hermitian orbit of the $n\times n$ Hermitian pencil ${\cal H}$, denoted by
${\cal O}^{H}(\cal H)$, is the set of matrix pencils which are $*$-congruent
with ${\cal H}$, namely

${\cal O}^{H}({\cal H}):=\\{Q^{*}{\cal H}(\lambda)Q:\ \
Q\in\mathbb{C}^{n\times n}\ \mbox{is invertible}\\}.$

Note that all pencils in ${\cal O}^{H}({\cal H})$ are Hermitian.

The Hermitian bundle of ${\cal H}$, denoted by ${\cal B}^{H}({\cal H})$, is
the set of all Hermitian pencils having the same HKCF as ${\cal H}$ except
maybe for the specific values of their distinct finite eigenvalues. Thus, all
the pencils in ${\cal B}^{H}({\cal H})$ have the same number of distinct
finite eigenvalues and, moreover, there exists an ordering of such distinct
finite eigenvalues for which each eigenvalue has the same number and sizes of
associated Hermitian canonical blocks (with the same signs associated with the
blocks of real eigenvalues).

######  Remark 2.

In our definition of Hermitian bundle we allow the finite eigenvalues to vary
from one pencil to another in the same bundle. However, the blocks (with their
signs) of the infinite eigenvalue are equal for all pencils in the bundle, in
contrast with the standard approach for nonstructured pencils [22, 23]. The
reason for introducing this restriction on the infinite eigenvalue is related
to the sign characteristic and to the fact that we expect the Hermitian
bundles to have the following property: if ${\cal H}_{1}\in\overline{{\cal
B}^{H}}({\cal H}_{2})$ then ${\cal B}^{H}({\cal
H}_{1})\subseteq\overline{{\cal B}^{H}}({\cal H}_{2})$. This property is
necessary for considering the set $\operatorname{PENCIL}_{n\times n}^{H}$ a
stratified manifold whose strata are the bundles, since the closure of a
strata must be the finite union of itself with strata of smaller dimensions.
This, however, does not hold if we allow finite eigenvalues to become the
infinite one in a bundle or vice versa. Let us illustrate this situation in
the simple case of $\operatorname{PENCIL}_{1\times 1}^{H}=\\{a+\lambda
b\,:\,a,b\in\mathbb{R}\\}$. The possible canonical forms of these $1\times 1$
Hermitian pencils are $+{\cal J}_{1}(\alpha)=\lambda-\alpha,-{\cal
J}_{1}(\alpha)=-(\lambda-\alpha)$, with $\alpha\in\mathbb{R}$ and finite,
$+{\cal J}_{1}(\infty)=1,-{\cal J}_{1}(\infty)=-1$, and $\mathcal{M}_{0}=0$
(the only singular $1\times 1$ pencil). If we include $\lambda-\alpha=+{\cal
J}_{1}(\alpha)$ for all finite $\alpha\in\mathbb{R}$ and $1=+{\cal
J}_{1}(\infty)$ in the same bundle, as might seem natural taking into account
the definition of bundles for unstructured pencils, then $+{\cal
J}_{1}(\infty)=1\in\overline{{\cal B}^{H}}(-{\cal J}_{1}(\beta))$, where
$\beta\in\mathbb{R}$ is finite, since $-(\frac{\lambda}{m}-1)$ converges to
$1$ as the natural number $m$ tends to infinity, and
$-(\frac{\lambda}{m}-1)\in{\cal B}^{H}(-{\cal J}_{1}(\beta))$. However,
$+{\cal J}_{1}(\alpha)=\lambda-\alpha\not\in\overline{{\cal B}^{H}}(-{\cal
J}_{1}(\beta))$ for any $\alpha$. This means that the previous desired
property of bundles does not hold, since $\overline{{\cal B}^{H}}(-{\cal
J}_{1}(\beta))$ would not include the whole bundle to which
$\lambda-\alpha=+{\cal J}_{1}(\alpha)$ and $1=+{\cal J}_{1}(\infty)$ belong.
Note that the problem remains if $1=+{\cal J}_{1}(\infty)$ is included in the
same bundle as $-(\lambda-\beta)=-{\cal J}_{1}(\beta)$ for all finite
$\beta\in\mathbb{R}$, since taking the sequence
$\\{\frac{\lambda}{m}+1\\}\subset{\cal B}^{H}(+{\cal J}_{1}(\alpha))$, that
tends to $1$ as well, we see that $1=+{\cal J}_{1}(\infty)\in\overline{{\cal
B}^{H}}(+{\cal J}_{1}(\alpha))$, but $-(\lambda-\beta)\not\in\overline{{\cal
B}^{H}}(+{\cal J}_{1}(\alpha))$ for any $\beta$. For these reasons, we only
allow the finite eigenvalues to vary in the pencils of a given bundle. Using
the ideas above, it is possible to construct higher dimensional examples with
similar difficulties.

Next, we introduce a notation that allows us to express some arguments
concisely and we state without proof a few very simple properties of Hermitian
bundles that are often used. If ${\cal H}\in\operatorname{PENCIL}_{n\times
n}^{H}$ and ${\cal H}_{1},{\cal H}_{2}\in{\cal B}^{H}({\cal H})$, then we will
write

$\mathrm{HKCF}({\cal H}_{1})\simeq\mathrm{HKCF}({\cal H}_{2})$

to mean that the $\mathrm{HKCF}$s of ${\cal H}_{1}$ and ${\cal H}_{2}$ are the
same up to the values of their distinct finite eigenvalues.

######  Lemma 3.

Let ${\cal H}_{1},{\cal H}_{2}\in\operatorname{PENCIL}_{n\times n}^{H}$ and
$Q\in\mathbb{C}^{n\times n}$ be nonsingular. Then

  * (a)

${\cal H}_{1}\in{\cal B}^{H}({\cal H}_{2})$ if and only if ${\cal B}^{H}({\cal
H}_{1})={\cal B}^{H}({\cal H}_{2})$,

  * (b)

${\cal H}_{1}\in{\cal B}^{H}({\cal H}_{2})$ if and only if
$\mathrm{HKCF}({\cal H}_{1})\simeq\mathrm{HKCF}({\cal H}_{2})$,

  * (c)

${\cal H}_{1}\in{\cal B}^{H}({\cal H}_{2})$ if and only if $Q^{*}{\cal
H}_{1}Q\in{\cal B}^{H}({\cal H}_{2})$,

  * (d)

${\cal H}_{1}\in\overline{{\cal B}^{H}}({\cal H}_{2})$ if and only if
$Q^{*}{\cal H}_{1}Q\in\overline{{\cal B}^{H}}({\cal H}_{2})$.

##  3 Some technical results

In this section, we present some results that are needed to prove the main
theorems of the paper (in Sections 4 and 5). We first provide a block anti-
triangular decomposition of Hermitian matrix pencils. This result can be
proven in the same way as the analogous result for symmetric matrix pencils
[5, Theorem 2] but using the factorization $W^{*}=U^{*}R$ (equivalently,
$W=R^{*}U$), where $U^{*}$ is unitary and $R$ is upper-triangular, see e.g.,
[26, p. 89, Theorem 2.1.14].

######  Theorem 4.

(Block anti-triangular form of Hermitian pencils). Let ${\cal H}(\lambda)$ be
a Hermitian pencil. Then, there is a unitary matrix $U$ such that

(1) ${\cal H}(\lambda)=U^{*}\left[\begin{array}[]{ccc}{\cal A}(\lambda)&{\cal
B}(\lambda)&{\cal H}_{\mbox{\rm\tiny right}}(\lambda)\\\ {\cal
B}(\lambda)^{*}&{\cal H}_{\mbox{\rm\tiny reg}}(\lambda)&0\\\ {\cal
H}_{\mbox{\rm\tiny right}}(\lambda)^{*}&0&0\end{array}\right]U,$

where:

  * (i)

${\cal A}(\lambda)$ is a Hermitian pencil.

  * (ii)

${\cal H}_{\mbox{\rm\tiny reg}}(\lambda)$ is a regular Hermitian pencil whose
elementary divisors are exactly those of ${\cal H}(\lambda)$.

  * (iii)

${\cal H}_{\mbox{\rm\tiny right}}(\lambda)$ is a pencil whose complete
eigenstructure consists only of the right minimal indices of ${\cal
H}(\lambda)$.

As a consequence, ${\cal H}_{\mbox{\rm\tiny right}}(\lambda)^{*}$ is a pencil
whose complete eigenstructure consists only of the left minimal indices of
${\cal H}(\lambda)$.

In the proofs of the main Theorems 10 and 12 we make use of the following
results.

######  Lemma 5.

Let ${\cal A}_{1},\ldots,{\cal A}_{s}$ and ${\cal H}_{1},\ldots,{\cal H}_{s}$
be Hermitian pencils of different sizes such that, for $i,j=1,\ldots,s$,

  * (a)

the sizes of ${\cal A}_{i}$ and ${\cal H}_{i}$ are equal, ${\cal
A}_{i}\in{\cal B}^{H}\left({\cal H}_{i}\right)$, and

  * (b)

${\cal H}_{i}$ and ${\cal H}_{j}$ have no finite eigenvalues in common for
$i\neq j$.

Then ${\cal A}_{1}\oplus\cdots\oplus{\cal A}_{s}\in\overline{{\cal
B}^{H}}\left({\cal H}_{1}\oplus\cdots\oplus{\cal H}_{s}\right)$. If, in
addition, ${\cal A}_{i}$ and ${\cal A}_{j}$ have no finite eigenvalues in
common for $i\neq j$, then ${\cal A}_{1}\oplus\cdots\oplus{\cal A}_{s}\in{\cal
B}^{H}\left({\cal H}_{1}\oplus\cdots\oplus{\cal H}_{s}\right)$.

###### Proof.

Case 1. Let us assume first that ${\cal A}_{i}$ and ${\cal A}_{j}$ have no
finite eigenvalues in common for $i\neq j$. Then

$\displaystyle\mathrm{HKCF}({\cal A}_{1}\oplus\cdots\oplus{\cal A}_{s})$
$\displaystyle=\mathrm{HKCF}({\cal
A}_{1})\oplus\cdots\oplus\mathrm{HKCF}({\cal A}_{s})$
$\displaystyle\simeq\mathrm{HKCF}({\cal
H}_{1})\oplus\cdots\oplus\mathrm{HKCF}({\cal H}_{s})=\mathrm{HKCF}({\cal
H}_{1}\oplus\cdots\oplus{\cal H}_{s}),$

and Lemma 3-(b) implies ${\cal A}_{1}\oplus\cdots\oplus{\cal A}_{s}\in{\cal
B}^{H}\left({\cal H}_{1}\oplus\cdots\oplus{\cal H}_{s}\right)$.

Case 2. Next, assume that ${\cal A}_{i}$ and ${\cal A}_{j}$ have finite
eigenvalues in common for some $i\neq j$. Let
$\\{\lambda_{1},\ldots,\lambda_{t}\\}:=\\{\lambda\in\mathbb{C}\,:\,\mbox{$\lambda$
is a finite eigenvalue of ${\cal A}_{i}$ and of ${\cal A}_{j}$ for $i\neq
j$}\\}$, where if $\lambda$ is an eigenvalue of exactly $\ell$ pencils ${\cal
A}_{i}$ (with $\ell>1$) then it is repeated $\ell$ times in the previous set.
Since there are infinitely many different sequences of real numbers
$\\{a_{n}\\}_{n\in\mathbb{N}}$ with all their terms distinct whose limit is
zero (as $m$ tends to infinity), we can choose $t$ of these sequences
$\\{c_{1}^{(m)}\\},\ldots,\\{c_{t}^{(m)}\\}$ and replace in
$\mathrm{HKCF}({\cal A}_{1}),\ldots,\mathrm{HKCF}({\cal A}_{s})$ the common
finite eigenvalues $\lambda_{1},\lambda_{2},\ldots,\lambda_{t}$ by
$\lambda_{1}+c_{1}^{(m)},\ldots,\lambda_{t}+c_{t}^{(m)}$ to get sequences
$\\{\mathrm{HKCF}({\cal
A}_{1}^{(m)})\\},\ldots,\allowbreak\\{\mathrm{HKCF}({\cal A}_{s}^{(m)})\\}$ of
pencils in $\mathrm{HKCF}$s such that, for all $m\in\mathbb{N}$,

  * (p1)

$\mathrm{HKCF}({\cal A}_{i}^{(m)})\simeq\mathrm{HKCF}({\cal A}_{i})\in{\cal
B}^{H}\left({\cal H}_{i}\right)$ for $i=1\ldots,s$,

  * (p2)

$\mathrm{HKCF}({\cal A}_{i}^{(m)})$ and $\mathrm{HKCF}({\cal A}_{j}^{(m)})$
have no finite eigenvalues in common for $i\neq j$, and

  * (p3)

$\lim_{m\rightarrow\infty}\mathrm{HKCF}({\cal
A}_{i}^{(m)})=\mathrm{HKCF}({\cal A}_{i})$, for $i=1,\ldots,s$.

Then, the result in Case 1 implies

$\mathrm{HKCF}({\cal A}_{1}^{(m)}\oplus\cdots\oplus{\cal
A}_{s}^{(m)})=\mathrm{HKCF}({\cal
A}_{1}^{(m)})\oplus\cdots\oplus\mathrm{HKCF}({\cal A}_{s}^{(m)})\in{\cal
B}^{H}\left({\cal H}_{1}\oplus\cdots\oplus{\cal H}_{s}\right)$

for all $m$. Then, from (p3) above, $\mathrm{HKCF}({\cal
A}_{1}\oplus\cdots\oplus{\cal A}_{s})\in\overline{{\cal B}^{H}}\left({\cal
H}_{1}\oplus\cdots\oplus{\cal H}_{s}\right)$, and the result follows from
Lemma 3-(d). ∎

######  Remark 6.

We emphasize that, in Lemma 5, the closure in the condition ${\cal
A}_{1}\oplus\cdots\oplus{\cal A}_{s}\in\overline{{\cal B}^{H}}\left({\cal
H}_{1}\oplus\cdots\oplus{\cal H}_{s}\right)$ cannot be removed. Consider the
following $1\times 1$ Hermitian pencils: ${\cal A}_{1}={\cal
A}_{2}=\lambda-1$, ${\cal H}_{1}=\lambda+1$, and ${\cal H}_{2}=\lambda$. Thus,
${\cal A}_{1}\in{\cal B}^{H}\left({\cal H}_{1}\right)$, ${\cal A}_{2}\in{\cal
B}^{H}\left({\cal H}_{2}\right)$, and ${\cal A}_{1}$ and ${\cal A}_{2}$ have
finite eigenvalues in common. Obviously ${\cal A}_{1}\oplus{\cal
A}_{2}=(\lambda-1)\oplus(\lambda-1)\notin{\cal B}^{H}\left({\cal
H}_{1}\oplus{\cal H}_{2}\right)$. However, ${\cal A}_{1}\oplus{\cal
A}_{2}=(\lambda-1)\oplus(\lambda-1)\in\overline{{\cal B}^{H}}\left({\cal
H}_{1}\oplus{\cal H}_{2}\right)$, since, for each $m=1,2,\ldots$,
$(\lambda-(1-1/m))\oplus(\lambda-(1+1/m))\in{\cal B}^{H}\left({\cal
H}_{1}\oplus{\cal H}_{2}\right)$. Observe also that in this example, ${\cal
B}^{H}\left({\cal H}_{1}\right)={\cal B}^{H}\left({\cal H}_{2}\right)$, but
the hypotheses of Lemma 5 require to take different pencils representing this
bundle.

Lemma 5 allows us to prove the following lemma, which is the one we will
actually use to prove our main results.

######  Lemma 7.

Let ${\cal H}_{1},\ldots,{\cal H}_{s}$ and $\widetilde{\cal
H}_{1},\ldots,\widetilde{\cal H}_{s}$ be Hermitian pencils of different sizes
such that, for $i,j=1,\ldots,s$,

  * (a)

the sizes of ${\cal H}_{i}$ and $\widetilde{\cal H}_{i}$ are equal and
$\overline{{\cal B}^{H}}\left({\cal H}_{i}\right)\subseteq\overline{{\cal
B}^{H}}\left(\widetilde{\cal H}_{i}\right)$, and

  * (b)

$\widetilde{\cal H}_{i}$ and $\widetilde{\cal H}_{j}$ have no finite
eigenvalues in common for $i\neq j$.

Then $\overline{{\cal B}^{H}}\left({\cal H}_{1}\oplus\cdots\oplus{\cal
H}_{s}\right)\subseteq\overline{{\cal B}^{H}}\left(\widetilde{\cal
H}_{1}\oplus\cdots\oplus\widetilde{\cal H}_{s}\right).$

###### Proof.

By definition of closure, we only need to prove ${\cal B}^{H}\left({\cal
H}_{1}\oplus\cdots\oplus{\cal H}_{s}\right)\subseteq\overline{{\cal
B}^{H}}\left(\widetilde{\cal H}_{1}\oplus\cdots\oplus\widetilde{\cal
H}_{s}\right).$ Let ${\cal D}\in{\cal B}^{H}\left({\cal
H}_{1}\oplus\cdots\oplus{\cal H}_{s}\right)$. This implies
$\mathrm{HKCF}({\cal D})\simeq\mathrm{HKCF}({\cal
H}_{1}\oplus\cdots\oplus{\cal H}_{s})=\mathrm{HKCF}({\cal
H}_{1})\oplus\cdots\oplus\mathrm{HKCF}({\cal H}_{s}).$ Therefore,
$\mathrm{HKCF}({\cal D})=\mathrm{HKCF}({\cal
D}_{1})\oplus\cdots\oplus\mathrm{HKCF}({\cal D}_{s})$, with
$\mathrm{HKCF}({\cal D}_{i})\simeq\mathrm{HKCF}({\cal H}_{i})$, which,
according to Lemma 3 (b), implies $\mathrm{HKCF}({\cal D}_{i})\in{\cal
B}^{H}\left({\cal H}_{i}\right)\subseteq\overline{{\cal
B}^{H}}\left(\widetilde{\cal H}_{i}\right)$, for $i=1,\ldots s$. Thus, there
exist sequences of Hermitian pencils, $\\{{\cal D}_{i}^{(m)}\\}\subseteq{\cal
B}^{H}\left(\widetilde{\cal H}_{i}\right)$, such that
$\lim_{m\rightarrow\infty}{\cal D}_{i}^{(m)}=\mathrm{HKCF}({\cal D}_{i})$, for
$i=1,\ldots s$. In addition, from Lemma 5,

${\cal D}_{1}^{(m)}\oplus\cdots\oplus{\cal D}_{s}^{(m)}\in\overline{{\cal
B}^{H}}\left(\widetilde{\cal H}_{1}\oplus\cdots\oplus\widetilde{\cal
H}_{s}\right)$

for all $m$. Since, $\lim_{m\rightarrow\infty}({\cal
D}_{1}^{(m)}\oplus\cdots\oplus{\cal D}_{s}^{(m)})=\mathrm{HKCF}({\cal
D}_{1})\oplus\cdots\oplus\mathrm{HKCF}({\cal D}_{s})=\mathrm{HKCF}({\cal D})$,
we get $\mathrm{HKCF}({\cal D})\in\overline{{\cal B}^{H}}\left(\widetilde{\cal
H}_{1}\oplus\cdots\oplus\widetilde{\cal H}_{s}\right)$, which combined with
Lemma 3-(d) implies ${\cal D}\in\overline{{\cal B}^{H}}\left(\widetilde{\cal
H}_{1}\oplus\cdots\oplus\widetilde{\cal H}_{s}\right)$. ∎

Lemma 7 will be combined in the proofs of the main results with the next
proposition dealing with the canonical blocks appearing in Theorem 1.

######  Proposition 8.

Let $\sigma\in\\{+1,-1\\}$ and $k>0$, $d,d_{1},\ldots,d_{t}\geq 0$ be integer
numbers.

  * (a1)

If $k$ is even and $a\in\mathbb{R}$, then

$\overline{{\cal B}^{H}}\left(\sigma{\cal
J}_{k}^{H}(a)\right)\subseteq\,\overline{{\cal
B}^{H}}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k}{2}-1},\overline{\mu_{\frac{k}{2}-1}}),{\cal
J}_{1}(a_{1}),-{\cal J}_{1}(a_{2}))\right),$

and

$\overline{{\cal B}^{H}}\left(\sigma{\cal
J}_{k}^{H}(\infty)\right)\subseteq\,\overline{{\cal
B}^{H}}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k}{2}-1},\overline{\mu_{\frac{k}{2}-1}}),\sigma{\cal
J}_{1}(a_{1}),\sigma{\cal J}_{1}(a_{2}))\right),$

with $a_{1},a_{2}\in\mathbb{R}$,
$\mu_{1},\ldots,\mu_{\frac{k}{2}-1}\in\mathbb{C}$ having positive imaginary
part, and $a_{1},a_{2},\mu_{1},\allowbreak\ldots,\mu_{\frac{k}{2}-1}$
different to each other.

  * (a2)

If $k$ is odd and $a\in\mathbb{R}\cup\\{\infty\\}$, then

$\overline{{\cal B}^{H}}\left(\sigma{\cal
J}_{k}^{H}(a)\right)\subseteq\,\overline{{\cal
B}^{H}}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k-1}{2}},\overline{\mu_{\frac{k-1}{2}}}),\sigma{\cal
J}_{1}(\widetilde{a}))\right),$

with $\widetilde{a}\in\mathbb{R}$,
$\mu_{1},\ldots,\mu_{\frac{k-1}{2}}\in\mathbb{C}$ having positive imaginary
part, and $\mu_{1},\ldots,\mu_{\frac{k-1}{2}}$ different to each other.

  * (b)

If $\mu\in\mathbb{C}$ has positive imaginary part, then

$\overline{{\cal B}^{H}}\left({\cal
J}_{k}^{H}(\mu,\overline{\mu})\right)\subseteq\,\overline{{\cal
B}^{H}}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{k},\overline{\mu_{k}}))\right),$

with $\mu_{1},\ldots,\mu_{k}\in\mathbb{C}$ being different to each other and
having positive imaginary part.

  * (c)

If $\mu\in\mathbb{C}$ has positive imaginary part, then $\overline{{\cal
B}^{H}}\left({\cal J}_{k}^{H}(\mu,\overline{\mu})\oplus{\cal
M}_{d}\right)\subseteq\overline{{\cal O}^{H}}\left({\cal M}_{d+k}\right)$.

  * (d)

$\overline{{\cal O}^{H}}\left(\mathop{\operator@font diag}\nolimits({\cal
M}_{d_{1}},\ldots,{\cal M}_{d_{t}})\right)\subseteq\overline{{\cal
O}^{H}}(\mathop{\operator@font diag}\nolimits(\underbrace{{\cal
M}_{\alpha+1},\ldots,{\cal M}_{\alpha+1}}_{\mbox{\tiny$s$}},\underbrace{{\cal
M}_{\alpha},\ldots,{\cal M}_{\alpha})}_{\mbox{\tiny$t-s$}}),$ with
$\sum_{i=1}^{t}d_{i}=t\alpha+s$ being the Euclidean division of
$\sum_{i=1}^{t}d_{i}$ by $t$.

###### Proof.

Part (a1). Assume that $k$ is even and $a\in\mathbb{R}$, and note that
$\sigma{\cal J}^{H}_{k}(a)$ is the limit of the following sequence of
Hermitian pencils (as $m\in\mathbb{N}$ tends to infinity):

$\sigma{\cal
S}_{(a,k)}^{(\varepsilon,m)}(\lambda):=\sigma\left[\begin{array}[]{cccccc}&&&1&\lambda-a\\\
&&1&\lambda-a\\\ &\iddots&\iddots&&\\\ 1&\lambda-a&\\\
\lambda-a&&&&\varepsilon/m\end{array}\right]_{k\times k},$

with $\varepsilon>0$. By a direct calculation, we get $\det\sigma{\cal
S}_{(a,k)}^{(\varepsilon,m)}(\lambda)=(-1)^{\frac{k}{2}}\left((\lambda-a)^{k}-\frac{\varepsilon}{m}\right)$,
so the eigenvalues of $\sigma{\cal S}_{(a,k)}^{(\varepsilon,m)}(\lambda)$ are
$a+(\frac{\varepsilon}{m})^{1/k}$, where $(\frac{\varepsilon}{m})^{1/k}$
denote the $k$ different $k$th roots of $\frac{\varepsilon}{m}$. Since $k$ is
even and $\frac{\varepsilon}{m}>0$, these consist of $2$ real values and
$\frac{k}{2}-1$ pairs of complex conjugate non-real distinct numbers.
Therefore,

${\rm HKCF}\left(\sigma{\cal
S}_{(a,k)}^{(\varepsilon,m)}\right)=\mathop{\operator@font
diag}\nolimits\left({\cal J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k}{2}-1},\overline{\mu_{\frac{k}{2}-1}}),\pm{\cal
J}_{1}(a_{1}),\pm{\cal J}_{1}(a_{2})\right),$

where
$\mu_{1},\ldots,\mu_{\frac{k}{2}-1},\overline{\mu_{1}},\ldots,\overline{\mu_{\frac{k}{2}-1}}$
are the non-real eigenvalues of $\sigma{\cal
S}_{(a,k)}^{(\varepsilon,m)}(\lambda)$, $a_{1},a_{2}$ are the real
eigenvalues, and we take $\mu_{1},\ldots,\mu_{\frac{k}{2}-1}$ with positive
imaginary part. In order to fix the signs in $\left(\pm{\cal
J}_{1}(a_{1}),\pm{\cal J}_{1}(a_{2})\right)$, we will use that the matrix
leading terms of $\sigma{\cal S}_{(a,k)}^{(\varepsilon,m)}(\lambda)$ and of
the pencil in the right-hand side of the previous equality are $*$-congruent
and so have the same signature. Since the leading term of $\sigma{\cal
S}_{(a,k)}^{(\varepsilon,m)}$ is a symmetric $k\times k$ matrix with trace $0$
and eigenvalues $\pm 1$, its signature is $(k/2,k/2,0)$. Now, since the
signature of the leading term of each block ${\cal
J}_{1}^{H}(\mu_{i},\overline{\mu_{i}})$ is $(1,1,0)$, for
$i=1,\ldots,\frac{k}{2}-1$, then the signature of the leading coefficient of
the remaining block $\mathop{\operator@font diag}\nolimits(\pm{\cal
J}_{1}(a_{1}),\pm{\cal J}_{1}(a_{2}))$ must be $(1,1,0)$ as well. As a
consequence,

$\sigma{\cal S}_{(a,k)}^{(\varepsilon,m)}(\lambda)\in{\cal
B}^{H}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k}{2}-1},\overline{\mu_{\frac{k}{2}-1}}),{\cal
J}_{1}(a_{1}),-{\cal J}_{1}(a_{2}))\right),$

with $a_{1},a_{2}\in\mathbb{R}$, and
$a_{1},a_{2},\mu_{1},\ldots,\mu_{\frac{k}{2}-1}$ being different to each
other, as wanted. Since the arguments above are independent of the specific
value of $a\in\mathbb{R}$, using Lemma 3-(d) we conclude that

${\cal B}^{H}\left(\sigma{\cal J}_{k}^{H}(a)\right)\subseteq\overline{{\cal
B}^{H}}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k}{2}-1},\overline{\mu_{\frac{k}{2}-1}}),{\cal
J}_{1}(a_{1}),-{\cal J}_{1}(a_{2}))\right),$

and the result follows by the definition of closure.

Now, let us prove the claim for $\sigma{\cal J}_{k}^{H}(\infty)$. In this
case, we consider the Hermitian perturbation

(2) $\sigma{\cal
S}_{(\infty,k)}^{(\varepsilon,m)}(\lambda):=\sigma\begin{bmatrix}&&\lambda&1\\\
&\iddots&\iddots\\\ \lambda&1&&\\\
1&&&\frac{\varepsilon}{m}\lambda\end{bmatrix},$

with $\varepsilon>0$. Note that $\sigma{\cal
S}_{(\infty,k)}^{(\varepsilon,m)}(\lambda)$ tends to $\sigma{\cal
J}_{k}^{H}(\infty)$ as $m$ tends to infinity. A direct calculation gives
$\det(\sigma{\cal
S}_{(\infty,k)}^{(\varepsilon,m)}(\lambda))=(-1)^{k/2}(1-\frac{\varepsilon}{m}\lambda^{k})$,
so the eigenvalues of $\sigma{\cal S}_{(\infty,k)}^{(\varepsilon,m)}(\lambda)$
are the $k$th roots of $\frac{m}{\varepsilon}$. Since $\varepsilon>0$, these
are $\frac{k}{2}-1$ pairs of complex conjugate (non-real) numbers and two real
numbers. Hence,

${\rm HKCF}\left(\sigma{\cal
S}_{(\infty,k)}^{(\varepsilon,m)}\right)=\mathop{\operator@font
diag}\nolimits\left({\cal J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k}{2}-1},\overline{\mu_{\frac{k}{2}-1}}),\pm{\cal
J}_{1}(a_{1}),\pm{\cal J}_{1}(a_{2})\right),$

with $a_{1},a_{2}\in\mathbb{R}$ and $\mu_{1},\ldots,\mu_{\frac{k}{2}-1}$
having positive imaginary part, and all them being different to each other.
The signs of $(\pm{\cal J}_{1}(a_{1}),\pm{\cal J}_{1}(a_{2}))$ are determined
again by the signatures of the leading terms. The leading term of $\sigma{\cal
S}_{(\infty,k)}^{(\varepsilon,m)}$ is

$T^{(\varepsilon,m)}:=\sigma\begin{bmatrix}&&1&0\\\ &\iddots&\iddots&\\\
1&0&\\\ 0&&&\frac{\varepsilon}{m}\end{bmatrix}.$

The eigenvalues of $\sigma T^{(\varepsilon,m)}$ are the following: $(k-2)/2$
of them are equal to $+1$, $(k-2)/2$ are equal to $-1$, another one is equal
to $\sigma$, and the last one is equal to $\sigma\frac{\varepsilon}{m}$. Since
the signature of the leading term of ${\cal J}_{1}^{H}(\mu,\overline{\mu})$ is
$(1,1,0)$, the signature of the leading term of $\mathop{\operator@font
diag}\nolimits(\pm{\cal J}_{1}(a_{1}),\pm{\cal J}_{1}(a_{2}))$ must be the one
of the leading term of $\mathop{\operator@font diag}\nolimits(\sigma{\cal
J}_{1}(a_{1}),\allowbreak\mbox{sign}(\sigma\frac{\varepsilon}{m})\,{\cal
J}_{1}(a_{2}))$. Since $\varepsilon>0$ and all the pencils in ${\cal
B}^{H}\left(\sigma{\cal J}_{k}^{H}(\infty)\right)$ are congruent to each
other, the result follows.

Part (a2). The argument when $a\in\mathbb{R}$ is similar to the one in part
(a1). Thus, we only sketch the main ideas. In this case, $\det(\sigma{\cal
S}_{(a,k)}^{(\varepsilon,m)}(\lambda))=\sigma^{k}(-1)^{\frac{k-1}{2}}\left((\lambda-a)^{k}+\frac{\varepsilon}{m}\right)$,
so $\sigma{\cal S}_{(a,k)}^{(\varepsilon,m)}$ has only one real eigenvalue,
and the remaining $k-1$ eigenvalues are couples of complex conjugate non-real
distinct numbers, namely $\sigma{\cal S}_{(a,k)}^{(\varepsilon,m)}$ is
$*$-congruent to $\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k-1}{2}},\overline{\mu_{\frac{k-1}{2}}}),\pm{\cal
J}_{1}(\widetilde{a}))$, for some $\mu_{1},\ldots,\mu_{\frac{k-1}{2}}$ being
different to each other and having positive imaginary part, and
$\widetilde{a}\in\mathbb{R}$. The signature of the leading coefficient of
$\sigma{\cal S}_{(a,k)}^{(\varepsilon,m)}$ is $((k+1)/2,(k-1)/2,0)$ if
$\sigma=1$ and $((k-1)/2,(k+1)/2,0)$ if $\sigma=-1$. Since, again, the
signature of the leading coefficient of ${\cal J}_{1}^{H}(\mu,\overline{\mu})$
is $(1,1,0)$, we conclude that the sign of the block ${\cal J}_{1}(a)$ must be
equal to $\sigma$. The fact that the arguments above are again independent of
the specific value of $a\in\mathbb{R}$ allows us to get the desired inclusion
of the bundle closures.

For $\sigma{\cal J}_{k}^{H}(\infty)$ in (a2) we can use a similar argument as
for the case $k$ even. In this case the eigenvalues of (2) are the $k$th roots
of $-\frac{m}{\varepsilon}$, so there is just one real root and $(k-1)/2$
pairs of complex conjugate (non-real) roots. Therefore,

${\rm HKCF}\left(\sigma{\cal
S}_{(\infty,k)}^{(\varepsilon,m)}\right)=\mathop{\operator@font
diag}\nolimits\left({\cal J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{\frac{k-1}{2}},\overline{\mu_{\frac{k-1}{2}}}),\pm{\cal
J}_{1}(\widetilde{a})\right),$

with $\widetilde{a}\in\mathbb{R}$ and $\mu_{1},\ldots,\mu_{\frac{k-1}{2}}$
being different to each other and having positive imaginary part. Reasoning as
for the $k$ even case with the signatures of the leading terms of $\sigma{\cal
S}_{(\infty,k)}^{(\varepsilon,m)}$ and of ${\rm HKCF}\left(\sigma{\cal
S}_{(\infty,k)}^{(\varepsilon,m)}\right)$, which are equal to each other, we
conclude that the sign of the block $\pm{\cal J}_{1}(\widetilde{a})$ must be
equal to $\sigma$.

Part (b). Note that, given an arbitrary parameter $\varepsilon>0$, the
Hermitian pencil

$\begin{array}[]{ccl}{\cal
J}_{k}^{H}(\mu,\overline{\mu})+T^{(k,m,\varepsilon)}&:=&\left[\begin{array}[]{cccc|cccc}&&&&&&1&\lambda-\overline{\mu}\\\
&&&&&\iddots&\iddots&\\\ &&&&1&\lambda-\overline{\mu}&&\\\
&&&&\lambda-\overline{\mu}&&&\\\ \hline\cr&&1&\lambda-\mu&&&&\\\
&\iddots&\iddots&&&&&\\\ 1&\lambda-\mu&&&&&&\\\
\lambda-\mu&&&&&&&\end{array}\right]\\\
&&+\left[\begin{array}[]{cccc|cccc}&&&&&&&\frac{\varepsilon}{m}\\\
&&&&&&\iddots&\\\ &&&&&\frac{(k-1)\varepsilon}{m}&&\\\
&&&&\frac{k\varepsilon}{m}&&&\\\ \hline\cr&&&\frac{k\varepsilon}{m}&&&&\\\
&&\frac{(k-1)\varepsilon}{m}&&&&&\\\ &\iddots&&&&&&\\\
\frac{\varepsilon}{m}&&&&&&&\end{array}\right]\end{array}$

has eigenvalues
$\mu-\frac{\varepsilon}{m},\overline{\mu-\frac{\varepsilon}{m}},\ldots,\mu-k\frac{\varepsilon}{m},\overline{\mu-k\frac{\varepsilon}{m}}$,
which are all different to each other, so

${\cal J}_{k}^{H}(\mu,\overline{\mu})+T^{(k,m,\varepsilon)}\in{\cal
B}^{H}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{k},\overline{\mu_{k}})\right),$

with $\mu_{1},\ldots,\mu_{k}$ being different to each other and having
positive imaginary part. Then $\lim_{m\rightarrow\infty}{\cal
J}_{k}^{H}(\mu,\overline{\mu})+T^{(k,m,\varepsilon)}={\cal
J}_{k}^{H}(\mu,\overline{\mu})$, which implies that ${\cal
J}_{k}^{H}(\mu,\overline{\mu})\in\overline{{\cal
B}^{H}}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{k},\overline{\mu_{k}})\right)$. Since this is valid for any
$\mu\in\mathbb{C}$ with positive imaginary part, we get that ${\cal
B}^{H}({\cal J}_{k}^{H}(\mu,\overline{\mu}))\subseteq\overline{{\cal
B}^{H}}\left(\mathop{\operator@font diag}\nolimits({\cal
J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\allowbreak\ldots,\allowbreak{\cal
J}_{1}^{H}(\mu_{k},\overline{\mu_{k}})\right),$ using Lemma 3-(d), which
implies the result because of the definition of closure.

Part (c). First note that, if $P$ is the block permutation matrix

$P=\begin{bmatrix}I_{k}&0&0&0\\\ 0&0&I_{k}&0\\\ 0&I_{d+1}&0&0\\\
0&0&0&I_{d}\\\ \end{bmatrix},$

then

$\displaystyle P^{*}\Big{(}$ $\displaystyle{\cal
J}_{k}^{H}(\mu,\overline{\mu})\oplus{\cal M}_{d}(\lambda)\Big{)}P$
$\displaystyle=\small\begin{bmatrix}I_{k}&0&0&0\\\ 0&0&I_{d+1}&0\\\
0&I_{k}&0&0\\\ 0&0&0&I_{d}\\\ \end{bmatrix}\begin{bmatrix}0&{\cal
J}_{k}^{H}(\overline{\mu})&0&0\\\ {\cal J}_{k}^{H}(\mu)&0&0&0\\\ 0&0&0&{\cal
L}_{d}(\lambda)^{\top}\\\ 0&0&{\cal
L}_{d}(\lambda)&0\end{bmatrix}\begin{bmatrix}I_{k}&0&0&0\\\ 0&0&I_{k}&0\\\
0&I_{d+1}&0&0\\\ 0&0&0&I_{d}\\\ \end{bmatrix}$
$\displaystyle=\begin{bmatrix}0&0&{\cal J}_{k}^{H}(\overline{\mu})&0\\\
0&0&0&{\cal L}_{d}(\lambda)^{\top}\\\ {\cal J}_{k}^{H}(\mu)&0&0&0\\\ 0&{\cal
L}_{d}(\lambda)&0&0\end{bmatrix}$ $\displaystyle=\begin{bmatrix}0&{\cal
J}_{k}^{H}(\overline{\mu})\oplus{\cal L}_{d}(\lambda)^{\top}\\\ {\cal
J}_{k}^{H}(\mu)\oplus{\cal L}_{d}(\lambda)&0\\\ \end{bmatrix}.$

Therefore, ${\cal J}_{k}^{H}(\mu,\overline{\mu})\oplus{\cal M}_{d}(\lambda)$
is $*$-congruent to $\left[\begin{smallmatrix}0&{\cal
J}_{k}^{H}(\overline{\mu})\oplus{\cal L}_{d}(\lambda)^{\top}\\\ {\cal
J}_{k}^{H}(\mu)\oplus{\cal L}_{d}(\lambda)&0\\\ \end{smallmatrix}\right]$.

By [2, Section 5.1] (or [36, Lemma 5]), there exist two invertible matrices
$R$ and $Q$ and an arbitrarily small (entry-wise for each coefficient) matrix
pencil ${\cal E}$ such that

$R\left({\cal J}_{k}^{H}(\mu)\oplus{\cal L}_{d}(\lambda)+{\cal
E}\right)Q={\cal L}_{d+k}(\lambda).$

Applying the $*$ operator to both sides of the previous equality, we obtain

$Q^{*}\left({\cal J}_{k}^{H}(\overline{\mu})\oplus{\cal
L}_{d}(\lambda)^{\top}+{\cal E}^{*}\right)R^{*}={\cal
L}_{d+k}(\lambda)^{\top}.$

Combining these two identities we obtain

$\displaystyle\begin{bmatrix}Q^{*}&0\\\ 0&R\\\
\end{bmatrix}\left(\begin{bmatrix}0&{\cal
J}_{k}^{H}(\overline{\mu})\oplus{\cal L}_{d}(\lambda)^{\top}\\\ {\cal
J}_{k}^{H}(\mu)\oplus{\cal L}_{d}(\lambda)&0\\\
\end{bmatrix}+\begin{bmatrix}0&{\cal E}^{*}\\\ {\cal E}&0\\\
\end{bmatrix}\right)\begin{bmatrix}Q&0\\\ 0&R^{*}\\\ \end{bmatrix}$
$\displaystyle=\begin{bmatrix}0&{\cal L}_{d+k}(\lambda)^{\top}\\\ {\cal
L}_{d+k}(\lambda)&0\\\ \end{bmatrix}.$

Since $\cal E$ is arbitrarily small, this implies that
$\left[\begin{smallmatrix}0&{\cal J}_{k}^{H}(\overline{\mu})\oplus{\cal
L}_{d}(\lambda)^{\top}\\\ {\cal J}_{k}^{H}(\mu)\oplus{\cal
L}_{d}(\lambda)&0\\\ \end{smallmatrix}\right]\in\overline{{\cal O}^{H}}({\cal
M}_{d+k)}$, so ${\cal J}_{k}^{H}(\mu,\overline{\mu})\oplus{\cal
M}_{d}(\lambda)\in\overline{{\cal O}^{H}}({\cal M}_{d+k})$ as well. Again, the
fact that this is valid for any $\mu\in\mathbb{C}$ with positive imaginary
part allows us to get the inclusion of the corresponding bundle-orbit
closures.

Part (d). Note that $\mathop{\operator@font diag}\nolimits({\cal
M}_{d_{1}},\ldots,{\cal M}_{d_{t}})\in\overline{{\cal
O}^{H}}(\mathop{\operator@font diag}\nolimits({\cal M}_{\alpha+1},\ldots,{\cal
M}_{\alpha+1},\allowbreak{\cal M}_{\alpha},\ldots,{\cal M}_{\alpha}))$ (with
the number of blocks in the statement) was already proven in the proof of
Theorem 3 in [5]. This result immediately implies the inclusion of the
corresponding orbit closures. ∎

The final technical result of this section is Lemma 9, which will allow us to
simplify the proofs of our main results (Theorems 10 and 12).

######  Lemma 9.

Let $\widetilde{r},r$, and $n$ be nonnegative integers such that
$0\leq\widetilde{r}<r\leq n$. Let $\widetilde{\cal
H}\in\operatorname{PENCIL}_{n\times n}^{H}$ be a Hermitian pencil with
$\mathop{\operator@font rank}\nolimits\widetilde{\cal H}=\widetilde{r}$. Then,
there exists ${\cal H}\in\operatorname{PENCIL}_{n\times n}^{H}$ with
$\mathop{\operator@font rank}\nolimits{\cal H}=r$ such that $\overline{{\cal
B}^{H}}(\widetilde{\cal H})\subseteq\overline{{\cal B}^{H}}({\cal H})$.

###### Proof.

Note that the $\mathrm{HKCF}(\widetilde{\cal H})$ has at least one block
${\cal M}_{d}$. Thus, in the first part of the proof, we consider the sequence
of perturbed blocks ${\cal M}_{d}^{(m)}:={\cal M}_{d}+\frac{1}{m}E_{d}$, where
$m=1,2,\ldots$, and $E_{d}$ is a $(2d+1)\times(2d+1)$ constant matrix that has
$1$ in the $(1,1)$-entry and zeroes elsewhere. It is immediate to prove that
$\det{\cal M}_{d}^{(m)}=\pm\frac{1}{m}$, which implies that ${\cal
M}_{d}^{(m)}$ is regular with all its eigenvalues equal to $\infty$. Moreover
the rank of the leading coefficient of ${\cal M}_{d}^{(m)}$ is $2d$, which
implies that $\mathrm{HKCF}({\cal M}_{d}^{(m)})=\pm{\cal
J}_{2d+1}^{H}(\infty)$. In order to determine the correct sign $\sigma$ note
that the signatures of the zero degree coefficients of ${\cal M}_{d}^{(m)}$
and $\sigma{\cal J}_{2d+1}^{H}(\infty)$ must be the same. The signature of the
zero degree coefficient of ${\cal M}_{d}^{(m)}$ is $(d+1,d,0)$, while the one
of the zero degree coefficient of ${\cal J}_{2d+1}^{H}(\infty)$ is also
$(d+1,d,0)$. Thus, the correct sign is $\sigma=+1$ and we have proved that,
for all $m=1,2,\ldots$,

(3) $\mathrm{HKCF}({\cal M}_{d}^{(m)})=\mathrm{HKCF}({\cal
M}_{d}+\frac{1}{m}E_{d})={\cal J}_{2d+1}^{H}(\infty).$

Next, note that it is enough to find a Hermitian pencil ${\cal H}$ with
$\mathop{\operator@font rank}\nolimits{\cal H}=r$ such that ${\cal
B}^{H}(\widetilde{\cal H})\subseteq\overline{{\cal B}^{H}}({\cal H})$ due to
the definition of closure. Let $\widehat{\cal H}\in{\cal
B}^{H}(\widetilde{\cal H})$. Then

$\mathrm{HKCF}(\widehat{\cal H})=\left(\bigoplus_{i=1}^{n-r}{\cal
M}_{d_{i}}\right)\oplus\left(\bigoplus_{i=n-r+1}^{n-\widetilde{r}}{\cal
M}_{d_{i}}\right)\oplus\mathrm{HKCF}_{reg}(\widehat{\cal H}),$

where $\mathrm{HKCF}_{reg}(\widehat{\cal H})$ includes all the blocks of types
(i), (ii), and (iii) in Theorem 1 of the $\mathrm{HKCF}$ of $\widehat{\cal
H}$. Observe that, for all the pencils in ${\cal B}^{H}(\widetilde{\cal H})$,
the parameters $d_{1},\ldots,d_{n-\widetilde{r}}$ are the same and
$\mathrm{HKCF}_{reg}(\widehat{\cal
H})\simeq\mathrm{HKCF}_{reg}(\widetilde{\cal H})$. Then, let us define the
sequence of pencils

${\cal H}^{(m)}=\left(\bigoplus_{i=1}^{n-r}{\cal
M}_{d_{i}}\right)\oplus\left(\bigoplus_{i=n-r+1}^{n-\widetilde{r}}\left({\cal
M}_{d_{i}}+\frac{1}{m}E_{d_{i}}\right)\right)\oplus\mathrm{HKCF}_{reg}(\widehat{\cal
H}),$

that satisfies $\lim_{m\rightarrow\infty}{\cal
H}^{(m)}=\mathrm{HKCF}(\widehat{\cal H})$. Moreover (3) implies that, for all
$m\in\mathbb{N}$, $\mathop{\operator@font rank}\nolimits{\cal H}^{(m)}=r$ and

(4) $\mathrm{HKCF}({\cal H}^{(m)})=\left(\bigoplus_{i=1}^{n-r}{\cal
M}_{d_{i}}\right)\oplus\left(\bigoplus_{i=n-r+1}^{n-\widetilde{r}}{\cal
J}_{2d_{i}+1}^{H}(\infty)\right)\oplus\mathrm{HKCF}_{reg}(\widehat{\cal H}),$

which is independent of $m$. From the right-hand side of (4), we define

${\cal H}:=\left(\bigoplus_{i=1}^{n-r}{\cal
M}_{d_{i}}\right)\oplus\left(\bigoplus_{i=n-r+1}^{n-\widetilde{r}}{\cal
J}_{2d_{i}+1}^{H}(\infty)\right)\oplus\mathrm{HKCF}_{reg}(\widetilde{\cal
H}),$

which is independent of $m$ and of the particular $\widehat{\cal H}$ in ${\cal
B}^{H}(\widetilde{\cal H})$ we are considering. Observe that
$\mathop{\operator@font rank}\nolimits{\cal H}=r$ and that, since
$\mathrm{HKCF}_{reg}(\widehat{\cal
H})\simeq\mathrm{HKCF}_{reg}(\widetilde{\cal H})$, then ${\cal
H}^{(m)}\in{\cal B}^{H}({\cal H})$ for all $m$. Since
$\lim_{m\rightarrow\infty}{\cal H}^{(m)}=\mathrm{HKCF}(\widehat{\cal H})$, we
conclude that $\mathrm{HKCF}(\widehat{\cal H})\in\overline{{\cal B}^{H}}({\cal
H})$, which combined with Lemma 3-(d) implies $\widehat{\cal
H}\in\overline{{\cal B}^{H}}({\cal H})$, and this proves the result. ∎

##  4 The case of general Hermitian pencils

In Theorem 10 we present the generic eigenstructures of complex Hermitian
$n\times n$ pencils, and we describe the set $\operatorname{PENCIL}_{n\times
n}^{H}$ as a finite union of closed sets, which are the closures of the
bundles corresponding to these generic eigenstructures. Theorem 10 is the
first main result of this paper.

######  Theorem 10.

(Generic complete eigenstructures of Hermitian matrix pencils). Let $n\geq 2$,
$0\leq d\leq\lfloor\frac{n}{2}\rfloor$, and $0\leq c\leq n-2d$. Let us define
the following complex Hermitian $n\times n$ regular matrix pencils:

(5) $\begin{array}[]{ccl}{\cal R}_{c,d}(\lambda)&:=&\mathop{\operator@font
diag}\nolimits({\cal J}_{1}^{H}(\mu_{1},\overline{\mu_{1}}),\ldots,{\cal
J}_{1}^{H}(\mu_{d},\overline{\mu_{d}}),\\\ &&{\cal J}_{1}(a_{1}),\dots,{\cal
J}_{1}(a_{c}),-{\cal J}_{1}(a_{c+1}),\ldots,-{\cal
J}_{1}(a_{n-2d})),\end{array}$

where $a_{1},\ldots,a_{n-2d}\in\mathbb{R}$,
$\mu_{1},\ldots,\mu_{d}\in\mathbb{C}\setminus\mathbb{R}$ have positive
imaginary part, $a_{i}\neq a_{j}$, and $\mu_{i}\neq\mu_{j}$, for $i\neq j$.
Then:

  1. (i)

For every complex Hermitian $n\times n$ matrix pencil ${\cal H}(\lambda)$
there exist integers $c$ and $d$ such that $\overline{{\cal B}^{H}}({\cal
H})\subseteq\,\overline{{\cal B}^{H}}({\cal R}_{c,d})$.

  2. (ii)

${\cal B}^{H}({\cal R}_{c^{\prime},d^{\prime}})\cap\overline{{\cal
B}^{H}}({\cal R}_{c,d})=\emptyset$ whenever $d\neq d^{\prime}$ or $c\neq
c^{\prime}$.

  3. (iii)

The set $\operatorname{PENCIL}_{n\times n}^{H}$ is equal to
$\displaystyle\bigcup_{\begin{array}[]{c}\mbox{\tiny$0\leq
d\leq\lfloor\frac{n}{2}\rfloor$}\\\ \mbox{\tiny$0\leq c\leq
n-2d$}\end{array}}\overline{{\cal B}^{H}}({\cal R}_{c,d})$.

###### Proof.

Claim (iii) is an immediate consequence of claim (i).

Let us prove (i). As a consequence of Lemma 9, with $r=n$, we only need to
prove it when ${\cal H}$ is a regular pencil, i.e., when ${\cal H}$ has rank
exactly $r$. The HKCF of any Hermitian $n\times n$ regular pencil ${\cal H}$
is a direct sum of blocks of the form ${\cal J}_{k}^{H}(\mu,\overline{\mu})$
and $\sigma{\cal J}_{\ell}^{H}(a)$, with $\mu$ having positive imaginary part
and $a\in\mathbb{R}\cup\\{\infty\\}$. By combining claims (a1)–(a2)–(b) in
Proposition 8 with Lemma 7, the closure of the Hermitian bundle corresponding
to ${\cal H}$ is included in the closure of the Hermitian bundle of a direct
sum of blocks ${\cal J}_{1}^{H}(\mu_{i},\overline{\mu_{i}})$ and $\pm{\cal
J}_{1}(a_{i})$, where $\mu_{i}$ has positive imaginary part and
$a_{i}\in\mathbb{R}$, as long as we take all the eigenvalues in this last
direct sum to be distinct. This is always allowed by Proposition 8 since the
parameters appearing in the bundles in that result are arbitrary (subjected to
their defining conditions).

Now, let us prove (ii). We are going to prove it first when $d\neq
d^{\prime}$. By contradiction, assume that ${\cal H}(\lambda)$ is any pencil
in ${\cal B}^{H}({\cal R}_{c^{\prime},d^{\prime}})$ but ${\cal
H}\in\overline{{\cal B}^{H}}({\cal R}_{c,d})$, with ${\cal
R}_{c^{\prime},d^{\prime}}(\lambda)$ as in the statement. Then, there is a
sequence of pencils in ${\cal B}^{H}({\cal R}_{c,d})$, say $\\{{\cal
H}_{m}(\lambda)\\}$, converging to ${\cal H}(\lambda)$. Since ${\cal
H}_{m}\in{\cal B}^{H}({\cal R}_{c,d})$, the eigenvalues of ${\cal H}_{m}$ are
of the form
$\mu_{1,m},\overline{\mu_{1,m}},\ldots,\mu_{d,m},\overline{\mu_{d,m}}$, and
$a_{1,m},\ldots,a_{n-2d,m}$, with $\mu_{1,m},\ldots,\mu_{d,m}$ having positive
imaginary part, $a_{1,m},\ldots,a_{n-2d,m}\in\mathbb{R}$, and
$\mu_{i,m}\neq\mu_{j,m}$, $a_{i,m}\neq a_{j,m}$, for $i\neq j$. Analogously,
the eigenvalues of ${\cal H}$ are
$\mu_{1}^{\prime},\overline{\mu_{1}^{\prime}},\ldots,\mu_{d^{\prime}}^{\prime},\overline{\mu_{d^{\prime}}^{\prime}}$,
and $a^{\prime}_{1},\ldots,a^{\prime}_{n-2d^{\prime}}\in\mathbb{R}$, with
$\mu_{1}^{\prime},\ldots,\mu_{d^{\prime}}^{\prime}$ having positive imaginary
part, and all these numbers being different to each other.

Let us assume first that $d^{\prime}>d$. By the continuity of the eigenvalues
of regular pencils (see, for instance, Theorem 2.1 in [37, Ch. 6]), at least
one of the sequences $\\{a_{i,m}\\}$ converges to some $\mu_{j}^{\prime}$. But
this is not possible, since $a_{i,m}\in\mathbb{R}$ and
$\mu_{j}^{\prime}\in\mathbb{C}\setminus\mathbb{R}$.

Now, assume that $d>d^{\prime}$. By the continuity of the eigenvalues again,
there is at least one sequence $\\{\mu_{i,m}\\}$ converging to some
$a_{j}^{\prime}$. Since $a_{j}^{\prime}\in\mathbb{R}$, this implies that ${\rm
im}(\mu_{i,m})$ converges to $0$, and this in turn implies that
$\overline{\mu_{i,m}}$ converges to $a_{j}^{\prime}$ as well. Therefore, the
limit of $\\{{\cal H}_{m}\\}$ (namely, a pencil with HKCF equal to ${\cal
R}_{c^{\prime},d^{\prime}}$) has $a_{j}^{\prime}$ as an eigenvalue with
algebraic multiplicity at least $2$. This is in contradiction with the fact
that the real eigenvalues of ${\cal R}_{c^{\prime},d^{\prime}}$ are different
to each other.

It remains to prove the statement when $d=d^{\prime}$ but $c\neq c^{\prime}$.
Note that the signature of the leading coefficient of ${\cal
J}_{1}(\mu,\overline{\mu})$ (namely, $\left[\begin{smallmatrix}0&1\\\
1&0\end{smallmatrix}\right]$) is $(1,1,0)$. Therefore, the signature of the
leading coefficient of ${\cal R}_{c^{\prime},d}$ is
$(c^{\prime}+d,n-c^{\prime}-d,0)$, which is equal to the signature of the
leading coefficient of ${\cal H}(\lambda)$ since the signature is invariant
under $*$-congruence. On the other hand, using again that the signature is
invariant under $*$-congruence, the signature of the leading coefficient of
${\cal H}_{m}$ is the same as the signature of the leading coefficient of
${\cal R}_{c,d}$, namely $(c+d,n-c-d,0)$. Then, $c^{\prime}\neq c$ implies
that either $c+d<c^{\prime}+d$ or $n-c-d<n-c^{\prime}-d$, which contradicts
Theorem 4.3 in [27]. ∎

The following result is an immediate consequence of Theorem 10.

######  Corollary 11.

The number of different generic eigenstructures in
$\operatorname{PENCIL}_{n\times n}^{H}$ is equal to
$\displaystyle\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)\left\lfloor\frac{n+3}{2}\right\rfloor$.

###### Proof.

The number of different pairs $(c,d)$ in Theorem 10 is

$\begin{array}[]{ccl}\displaystyle\sum_{d=0}^{\lfloor
n/2\rfloor}(n-2d+1)&=&(n+1)\left(\lfloor\frac{n}{2}\rfloor+1\right)-\lfloor\frac{n}{2}\rfloor\left(\lfloor\frac{n}{2}\rfloor+1\right)\\\
&=&\left(\lfloor\frac{n}{2}\rfloor+1\right)\left(n+1-\lfloor\frac{n}{2}\rfloor\right)=\left(\lfloor\frac{n}{2}\rfloor+1\right)\lfloor\frac{n+3}{2}\rfloor.\end{array}$

∎

Note that the number of generic eigenstructures in
$\operatorname{PENCIL}_{n\times n}^{H}$, according to Corollary 11, is larger
than $1$, for all $n\in\mathbb{N}$. This is a consequence of the different
forms to distribute in regular Hermitian pencils the eigenvalues between real
and non-real ones and also of the different possible sign characteristics of
the real eigenvalues. The fact that there are more than one (in fact, many for
large $n$) generic eigenstructures of regular Hermitian pencils is in stark
contrast with the number of generic eigenstructures of regular unstructured
pencils and of regular complex symmetric pencils, which have only one generic
eigenstructure corresponding to all the eigenvalues being different and
simple.

##  5 The case of Hermitian pencils with bounded rank

In Theorem 12, which is the second main result of this paper, we prove that
$\operatorname{PENCIL}_{n\times n}^{H}(r)$ is the union of a finite number of
closures of Hermitian bundles, and we explicitly provide the HKCF of each of
these bundles. As a consequence, these HKCFs are the generic canonical forms
of complex Hermitian $n\times n$ matrix pencils with rank at most $r$. In
other words, they provide the generic complete eigenstructures of these
pencils. Surprisingly, none of these eigenstructures contain non-real
eigenvalues, unlike what happens with the generic regular complete
eigenstructures for general Hermitian pencils provided in Theorem 10.

######  Theorem 12.

(Generic complete eigenstructures of Hermitian matrix pencils with bounded
rank). Let $n$ and $r$ be integers such that $n\geq 2$ and $1\leq r\leq n-1$.
Set $d=0,1,\ldots,\lfloor\frac{r}{2}\rfloor\,$ and let $d=(n-r)\alpha+s$ be
the Euclidean division of $d$ by $n-r$. Let us define the following complex
Hermitian $n\times n$ matrix pencils with rank $r$:

(6) $\begin{array}[]{ccl}{\cal K}_{c,d}(\lambda)&:=&\mathop{\operator@font
diag}\nolimits(\overbrace{{\cal M}_{\alpha+1},\ldots,{\cal
M}_{\alpha+1}}^{s},\overbrace{{\cal M}_{\alpha},\ldots,{\cal
M}_{\alpha}}^{n-r-s},\\\ &&{\cal J}_{1}(a_{1}),\dots,{\cal
J}_{1}(a_{c}),-{\cal J}_{1}(a_{c+1}),\ldots,-{\cal
J}_{1}(a_{r-2d})),\end{array}$

where $a_{1},\ldots,a_{r-2d}\in\mathbb{R},a_{i}\neq a_{j}$ for $i\neq j$, and
$c=0,1,\ldots,r-2d$. Then:

  1. (i)

For every complex Hermitian $n\times n$ matrix pencil ${\cal H}(\lambda)$ with
rank at most $r$, there exist nonnegative integers $c$ and $d$, with $0\leq
d\leq\lfloor\frac{r}{2}\rfloor$ and $0\leq c\leq r-2d$, such that
$\overline{{\cal B}^{H}}({\cal H})\subseteq\overline{{\cal B}^{H}}({\cal
K}_{c,d})$.

  2. (ii)

${\cal B}^{H}({\cal K}_{c^{\prime},d^{\prime}})\cap\overline{{\cal
B}^{H}}({\cal K}_{c,d})=\emptyset$ whenever $d\neq d^{\prime}$ or $c\neq
c^{\prime}$.

  3. (iii)

The set $\operatorname{PENCIL}_{n\times n}^{H}(r)$ is a closed subset of
$\operatorname{PENCIL}^{H}_{n\times n}$, and it is equal to
$\displaystyle\bigcup_{\begin{array}[]{c}\mbox{\tiny$0\leq
d\leq\lfloor\frac{r}{2}\rfloor$}\\\ \mbox{ \tiny$0\leq c\leq
r-2d$}\end{array}}\overline{{\cal B}^{H}}({\cal K}_{c,d})$.

###### Proof.

The proof has the same structure as the one for Theorem 3 in [5], adapted to
Hermitian pencils instead of symmetric ones. However, we emphasize that some
interesting differences also appear related to the role played by the sign
characteristic and by the blocks ${\cal J}_{k}^{H}(\mu,\overline{\mu})$
corresponding to pairs of non-real complex conjugate eigenvalues.

The set $\operatorname{PENCIL}_{n\times n}^{H}(r)$ is a closed subset of
$\operatorname{PENCIL}^{H}_{n\times n}$ because it is the intersection of
$\operatorname{PENCIL}^{H}_{n\times n}$ with the set of complex $n\times n$
matrix pencils with rank at most $r$, which is a closed set. Therefore, claim
(iii) is an immediate consequence of (i), so we only need to prove (i) and
(ii).

Let us start proving (i). Because of Lemma 9, we can restrict ourselves to
Hermitian $n\times n$ pencils with rank exactly $r$. So let ${\cal
H}(\lambda)$ be a Hermitian pencil with $\mathop{\operator@font
rank}\nolimits{\cal H}=r$. By Theorem 1, we may assume that

${\rm HKCF}({\cal H})=\mathop{\operator@font
diag}\nolimits\left(\bigoplus_{t}\sigma_{t}{\cal
J}_{k_{t}}^{H}(b_{t}),\bigoplus_{j}{\cal
J}_{k_{j}}^{H}(\lambda_{j},\overline{\lambda_{j}}),\bigoplus_{\ell}{\cal
M}_{d_{\ell}}\right),$

with $b_{t}\in\mathbb{R}\cup\\{\infty\\}$ and $\lambda_{j}\in\mathbb{C}$ with
positive imaginary part (the number of blocks ${\cal J}_{k_{t}}^{H}(b_{t})$,
${\cal J}_{k_{j}}^{H}(\lambda_{j},\overline{\lambda_{j}})$ is not relevant,
and the number of blocks ${\cal M}_{d_{\ell}}(\lambda)$ is $n-r$, by the rank-
nullity Theorem). Then, by (a1), (a2), and (b) in Proposition 8, together with
Lemma 7 and Lemma 3, $\overline{{\cal B}^{H}}({\cal H})$ is included in the
closure of the Hermitian bundle of

$\widehat{\cal H}(\lambda):=\mathop{\operator@font
diag}\nolimits\left(\bigoplus_{i=1}^{c}{\cal
J}_{1}(a_{i}),\bigoplus_{i=c+1}^{r-2d}(-{\cal
J}_{1}(a_{i})),\bigoplus_{p}{\cal
J}_{1}^{H}(\mu_{p},\overline{\mu_{p}}),\bigoplus_{\ell}{\cal
M}_{d_{\ell}}(\lambda)\right),$

for some $a_{i}\in\mathbb{R}$ and $\mu_{p}$ having positive imaginary part,
and for some $1\leq c\leq r-2d$, with $0\leq d\leq\lfloor\frac{r}{2}\rfloor$.
Moreover, since there are infinitely many possible choices for the distinct
eigenvalues in the bundles of the right-hand side of each of the inclusions in
parts (a1), (a2), and (b) in Proposition 8, the values $a_{i}$ can be taken to
be different to each other, for $i=1,\ldots,r-2d$, and the same happens for
the values $\mu_{p}$. Since $r\leq n-1$, there is at least one block ${\cal
M}_{d_{1}}(\lambda)$ in the previous direct sum defining $\widehat{\cal
H}(\lambda)$. Then, by repeatedly applying Proposition 8–(c) and Lemma 7 (as
many times as the number of ${\cal J}^{H}_{1}(\mu_{p},\overline{\mu_{p}})$
blocks in $\widehat{\cal H}$), the closure of the Hermitian bundle of
$\widehat{\cal H}$ is included in the closure of the Hermitian bundle of

(7) $\widetilde{\cal H}(\lambda):=\mathop{\operator@font
diag}\nolimits\left(\bigoplus_{i=1}^{c}{\cal
J}_{1}(a_{i}),\bigoplus_{i=c+1}^{r-2d}(-{\cal J}_{1}(a_{i})),{\cal
M}_{d_{1}+q}(\lambda),\bigoplus_{\ell>1}{\cal M}_{d_{\ell}}(\lambda)\right),$

for some $q\geq 0$ (which is equal to the number of all blocks in
$\bigoplus_{p}{\cal J}_{1}^{H}(\mu_{p},\overline{\mu_{p}})$). To get this
inclusion we first split (modulo permutation of direct summands)
$\widehat{\cal H}$ into ${\cal H}_{1}$ and ${\cal H}_{2}$, where ${\cal
H}_{2}$ is the direct sum of ${\cal M}_{d_{1}}$ with ${\cal
J}^{H}_{1}(\mu_{p},\overline{\mu_{p}})$ which, for simplicity, is the last
block of this kind in $\widehat{\cal H}$, and ${\cal H}_{1}$ is the direct sum
of the remaining blocks. Then, by part (c) in Proposition 8 and Lemma 7,
$\overline{{\cal B}^{H}}(\widehat{\cal H})=\overline{{\cal B}^{H}}({\cal
H}_{1}\oplus{\cal H}_{2})\subseteq\overline{{\cal B}^{H}}(\widetilde{\cal
H}_{1}\oplus\widetilde{\cal H}_{2})$, with $\widetilde{\cal H}_{1}={\cal
H}_{1}$ and $\widetilde{\cal H}_{2}={\cal M}_{d_{1}+1}$, since
$\widetilde{\cal H}_{1}$ and $\widetilde{\cal H}_{2}$ do not have common
eigenvalues. Now, we proceed in the same way with the new pencil
$\widetilde{\cal H}_{1}\oplus\widetilde{\cal H}_{2}$, and so on until we get
$\widetilde{\cal H}(\lambda)$ in (7). Applying claim (d) in Proposition 8,
together with Lemma 7 again, to the ${\cal M}_{d}$ blocks of $\widetilde{\cal
H}(\lambda)$, we conclude that the closure of the Hermitian bundle of
$\widetilde{\cal H}$ is in turn included in the closure of the Hermitian
bundle of

$\mathop{\operator@font diag}\nolimits\left(\bigoplus_{i=1}^{c}{\cal
J}_{1}(a_{i}),\bigoplus_{i=c+1}^{r-2d}(-{\cal J}_{1}(a_{i})),\bigoplus{\cal
M}_{\alpha+1}(\lambda),\bigoplus{\cal M}_{\alpha}\right),$

for some fixed $\alpha$, and where the total number of blocks ${\cal
M}_{\alpha}$ and ${\cal M}_{\alpha+1}$ in the previous direct sum is $n-r$.
Taking this into account, if $s$ is the number of blocks ${\cal
M}_{\alpha+1}$, then the number of blocks ${\cal M}_{\alpha}$ is $n-r-s$. The
value of $\alpha$ can be obtained by adding up the number of rows (or columns)
in the previous pencil and equating to $n$, namely:

$n=r-2d+s(2\alpha+3)+(n-r-s)(2\alpha+1)=n+2(\alpha(n-r)+s-d),$

which implies $d=\alpha(n-r)+s$, as claimed.

Summarizing, we have proved that $\overline{{\cal B}^{H}}({\cal
H})\subseteq\overline{{\cal B}^{H}}({\cal K}_{c,d})$, for some $c,d$ as in the
statement. This proves (i).

Now, let us prove (ii). First, we need to see that if $d^{\prime}>d$, then for
any ${\cal H}\in{\cal B}^{H}({\cal K}_{c^{\prime},d^{\prime}})$, ${\cal
H}\notin\overline{{\cal B}^{H}}({\cal K}_{c,d})$, for any $c$ and
$c^{\prime}$. By Lemma 3-(d) and the definition of bundle, we can take ${\cal
H}={\cal K}_{c^{\prime},d^{\prime}}$ for certain distinct real numbers
$a_{1},\ldots,a_{r-2d^{\prime}}$. Then, the same argument as the one in the
proof of Theorem 3 in [5] is still valid, i.e., ${\cal
K}_{c^{\prime},d^{\prime}}\in\overline{{\cal B}^{H}}({\cal K}_{c,d})$ would be
against the majorization of the Weyr sequence of right minimal indices, see,
for instance, [3, Lemma 1.2].

It remains to prove that if $d^{\prime}<d$ or $d^{\prime}=d$ but $c\neq
c^{\prime}$, then ${\cal B}^{H}({\cal
K}_{c^{\prime},d^{\prime}})\cap\overline{{\cal B}^{H}}({\cal
K}_{c,d})=\emptyset$ too.

The first part of the rest of the proof is almost the same as the proof of
Theorem 3 in [5], replacing $\top$ in that proof by ∗. However, there are
several relevant differences, so we include here the proof for completeness
and for the sake of reader.

By contradiction, if ${\cal B}^{H}({\cal
K}_{c^{\prime},d^{\prime}})\cap\overline{{\cal B}^{H}}({\cal
K}_{c,d})\neq\emptyset$, then at least one pencil $S(\lambda)$ congruent to
${\cal K}_{c^{\prime},d^{\prime}}(\lambda)$ as in (6), with $a_{i}\neq a_{j}$,
for $i\neq j$, and $a_{i}\in\mathbb{R}$, must be the limit of a sequence of
pencils in ${\cal B}^{H}({\cal K}_{c,d})$. Let $\\{{\cal
S}_{m}(\lambda)\\}_{m\in\mathbb{N}}$ be a sequence of pencils with ${\cal
S}_{m}(\lambda)\in{\cal B}^{H}({\cal K}_{c,d})$, for all $m\in\mathbb{N}$.
Then, by Theorem 4

(8) ${\cal S}_{m}(\lambda)=Q_{m}^{*}\left[\begin{array}[]{ccc}{\cal
A}_{m}(\lambda)&{\cal B}_{m}(\lambda)&{\cal S}_{\tiny\mbox{\rm
right}}^{(m)}(\lambda)\\\ {\cal B}_{m}(\lambda)^{*}&S_{\tiny\mbox{\rm
reg}}^{(m)}(\lambda)&0\\\ {\cal S}_{\tiny\mbox{\rm
right}}^{(m)}(\lambda)^{*}&0&0\end{array}\right]Q_{m},$

with $Q_{m}\in\mathbb{C}^{n\times n}$ being a unitary matrix, for all
$m\in\mathbb{N}$, and

  * •

${\cal S}_{\tiny\mbox{\rm right}}^{(m)}(\lambda)$ has size $d\times(n-r+d)$,
and complete eigenstructure consisting of the right minimal indices of ${\cal
K}_{c,d}(\lambda)$,

  * •

${\cal S}_{\tiny\mbox{\rm right}}^{(m)}(\lambda)^{*}$ has size $(n-r+d)\times
d$, and complete eigenstructure consisting of the left minimal indices of
${\cal K}_{c,d}(\lambda)$,

  * •

${\cal S}^{(m)}_{\tiny\mbox{\rm reg}}(\lambda)$ is a regular (Hermitian)
pencil of size $(r-2d)\times(r-2d)$, with $r-2d$ distinct real eigenvalues and
$c$ of them having positive sign characteristic.

Now, let us assume that ${\cal S}_{m}(\lambda)$ converges to some pencil
${\cal S}(\lambda)\in{\cal B}^{H}({\cal K}_{c^{\prime},d^{\prime}})$. Since
the set of unitary $n\times n$ matrices is a compact subset of the metric
space $(\mathbb{C}^{n\times n},\|\cdot\|_{2})$, the sequence
$\\{Q_{m}\\}_{m\in\mathbb{N}}$ has a convergent subsequence, say
$\\{Q_{m_{j}}\\}_{j\in\mathbb{N}}$, whose limit is a unitary matrix (see, for
instance, [26, Lemma 2.1.8]). Set

${\cal H}_{m}(\lambda):=\left[\begin{array}[]{ccc}{\cal A}_{m}(\lambda)&{\cal
B}_{m}(\lambda)&{\cal S}_{\tiny\mbox{\rm right}}^{(m)}(\lambda)\\\ {\cal
B}_{m}(\lambda)^{*}&{\cal S}_{\tiny\mbox{\rm reg}}^{(m)}(\lambda)&0\\\ {\cal
S}_{\tiny\mbox{\rm right}}^{(m)}(\lambda)^{*}&0&0\end{array}\right]$

for the matrix in the middle of the right-hand side in (8). Then the sequence
$\\{{\cal H}_{m_{j}}\\}_{j\in\mathbb{N}}$ is convergent as well, since ${\cal
H}_{m_{j}}(\lambda)=Q_{m_{j}}{\cal S}_{m_{j}}(\lambda)Q_{m_{j}}^{*}$, and both
$\\{Q_{m_{j}}\\}_{j\in\mathbb{N}}$ (and, as a consequence,
$\\{Q_{m_{j}}\\}_{j\in\mathbb{N}}$ and $\\{Q_{m_{j}}^{*}\\}_{j\in\mathbb{N}}$)
and $\\{{\cal S}_{m_{j}}\\}_{j\in\mathbb{N}}$ are convergent, because any
subsequence of ${\cal S}_{m}$ converges to its limit. Moreover, by continuity
of the zero blocks in the block-structure, $\\{{\cal
H}_{m_{j}}\\}_{j\in\mathbb{N}}$ converges to a matrix pencil of the form

(9) ${\cal H}(\lambda)=\left[\begin{array}[]{ccc}{\cal A}(\lambda)&{\cal
B}(\lambda)&{\cal C}(\lambda)\\\ {\cal B}(\lambda)^{*}&{\cal R}(\lambda)&0\\\
{\cal C}(\lambda)^{*}&0&0\end{array}\right],$

with

  * •

${\cal C}(\lambda)$ being of size $d\times(n-r+d)$,

  * •

${\cal C}(\lambda)^{*}$ being of size $(n-r+d)\times d$, and

  * •

${\cal R}(\lambda)$ being of size $(r-2d)\times(r-2d)$.

Therefore, the sequence $\\{{\cal S}_{m_{j}}\\}_{j\in\mathbb{N}}$ converges to
$Q^{*}{\cal H}(\lambda)Q$, where
$Q:=\displaystyle\lim_{j\rightarrow\infty}Q_{m_{j}}$ is unitary. Since
$\\{{\cal S}_{m}\\}_{m\in\mathbb{N}}$ is convergent, any subsequence must
converge to its limit, so $\displaystyle\lim_{m\rightarrow\infty}{\cal
S}_{m}={\cal S}=Q^{*}{\cal H}(\lambda)Q$. In the rest of the proof, it is
important to bear in mind that ${\cal S}(\lambda)$ (resp. ${\cal
S}(\lambda_{0})$ for any particular $\lambda_{0}\in\mathbb{C}$) has rank at
most $r$, and has rank exactly $r$ if and only if $\mathop{\operator@font
rank}\nolimits{\cal C}(\lambda)=\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda)^{*}=d$ (resp. $\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda_{0})=\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda_{0})^{*}=d$) and $\mathop{\operator@font rank}\nolimits{\cal
R}(\lambda)=r-2d$ (resp. $\mathop{\operator@font rank}\nolimits{\cal
R}(\lambda_{0})=r-2d$). This follows easily from the block structure of ${\cal
H}(\lambda)$ and the fact that $\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda)\leq d$ and $\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda)^{*}\leq d$.

Let us first assume that $d^{\prime}<d$ and $\lim_{m\rightarrow\infty}{\cal
S}_{m}={\cal S}\in{\cal B}^{H}({\cal K}_{c^{\prime},d^{\prime}})$. Then,

(10) $\displaystyle{\cal S}\in{\cal B}^{H}({\cal
K}_{c^{\prime},d^{\prime}}),\qquad\mbox{\rm with
$d^{\prime}<d$},\qquad\mbox{\rm and}$ (11) $\displaystyle\mbox{\rm${\cal S}$
has exactly $r-2d^{\prime}$ (real) simple eigenvalues}.$

Note that conditions (10)–(11) mean that ${\cal S}$ is congruent to ${\cal
K}_{c^{\prime},d^{\prime}}$, for some real eigenvalues
$a_{1},\ldots,a_{r-2d^{\prime}}$ in (6), different from each other. Moreover,
(10) implies that

(12) $\mathop{\operator@font rank}\nolimits{\cal S}=r,$

which is equivalent to $\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda)=\mathop{\operator@font rank}\nolimits{\cal C}(\lambda)^{*}=d$ and
$\mathop{\operator@font rank}\nolimits{\cal R}(\lambda)=r-2d$. Then, ${\cal
R}$ in (9) is a regular pencil with $r-2d$ eigenvalues (counting
multiplicities). Let us denote these eigenvalues by
$\widetilde{a}_{1},\ldots,\widetilde{a}_{r-2d}$ (in principle some of them
might be infinite). By (10), the pencil ${\cal S}$ has more than $r-2d$
eigenvalues, which are all real and distinct. If $\mathop{\operator@font
rank}\nolimits{\cal C}(\lambda_{0})=\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda_{0})^{*}=d$ for all $\lambda_{0}\in\mathbb{R}$, then
$\mathop{\operator@font rank}\nolimits{\cal S}(\mu)=\mathop{\operator@font
rank}\nolimits{\cal S}=r$ for all $\mu\in\mathbb{R}$ such that
$\mu\neq\widetilde{a}_{i}$ ($i=1,\ldots,r-2d$), which means that ${\cal S}$
has at most $r-2d$ real eigenvalues, which is a contradiction. Therefore,
there must be some $\lambda_{0}\in\mathbb{R}$ such that
$\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda_{0})=\mathop{\operator@font rank}\nolimits{\cal
C}(\lambda_{0})^{*}<d$. In particular, such $\lambda_{0}$ is an eigenvalue of
${\cal S}$, since the number of linearly independent rows of ${\cal
S}(\lambda_{0})$ is less than $r$. Now, we are going to see that, in this
case, $\lambda_{0}$ is an eigenvalue of ${\cal S}$ with algebraic multiplicity
at least 2, which is in contradiction with (11) as well. For this, we will
prove that all $r\times r$ non-identically zero minors of ${\cal H}$ have
$(\lambda-\lambda_{0})^{2}$ as a factor. In order for an $r\times r$ submatrix
of ${\cal H}$ to have non-identically zero determinant, it must contain fewer
than $d+1$ columns among the last $n-r+d$ columns of ${\cal H}$ (namely, those
corresponding to ${\cal C}$), and fewer than $d+1$ rows among the last $n-r+d$
rows of ${\cal H}$ (namely, those corresponding to ${\cal C}^{*}$). This is
because any set of $d+1$ columns among the last $n-r+d$ columns of ${\cal H}$
is linearly dependent, and the same for the last $n-r+d$ rows. As a
consequence, any $r\times r$ non-identically zero minor, $M(\lambda)$, of
${\cal H}$ is of the form:

$M(\lambda):=\det\left[\begin{array}[]{ccc}{\cal A}(\lambda)&{\cal
B}(\lambda)&{\cal C}_{1}(\lambda)\\\ {\cal B}(\lambda)^{*}&{\cal
R}(\lambda)&0\\\ {\cal C}_{2}(\lambda)^{*}&0&0\end{array}\right],$

where ${\cal C}_{1}$ (respectively, ${\cal C}_{2}^{*}$) is a submatrix of
${\cal C}$ (resp., ${\cal C}^{*}$) with size $d\times d$. Therefore,
$M(\lambda)=\pm\det{\cal R}\cdot\det{\cal C}_{1}\cdot\det{\cal C}_{2}^{*}$.
Since $\mathop{\operator@font rank}\nolimits{\cal C}_{1}(\lambda_{0})<d$ and
$\mathop{\operator@font rank}\nolimits{\cal C}_{2}(\lambda_{0})^{*}<d$, the
binomial $(\lambda-\lambda_{0})=(\lambda-\overline{\lambda_{0}})$ is a factor
of both $\det{\cal C}_{1}$ and $\det{\cal C}_{2}^{*}$, so
$(\lambda-\lambda_{0})^{2}$ is a factor of $M(\lambda)$. Note that
$\lambda_{0}\in\mathbb{R}$ is key in this conclusion, in order to guarantee
$(\lambda-\lambda_{0})=(\lambda-\overline{\lambda_{0}})$.

Therefore, the gcd of all $r\times r$ non-identically zero minors of ${\cal
H}$ is a multiple of $(\lambda-\lambda_{0})^{2}$. This implies (see, for
instance, [24, p. 141]) that the algebraic multiplicity of $\lambda_{0}$ as an
eigenvalue of ${\cal H}$, and so of ${\cal S}$, is at least $2$.

Now, assume that $d^{\prime}=d$ but $c^{\prime}\neq c$. Then,
$\lim_{m\rightarrow\infty}{\cal S}_{m}={\cal S}\in{\cal B}^{H}({\cal
K}_{c^{\prime},d})$, which implies that ${\cal S}(\lambda)$ is congruent to
${\cal K}_{c^{\prime},d}(\lambda)$ as in (6), for some real distinct
eigenvalues $a_{1},\ldots,a_{r-2d}$. Thus, if we express ${\cal
S}(\lambda)=\lambda X+Y$ and ${\cal K}_{c^{\prime},d}(\lambda)=\lambda
X_{c^{\prime},d}+Y_{c^{\prime},d}$, with $X,Y,X_{c^{\prime},d}$ and
$Y_{c^{\prime},d}$ being constant Hermitian matrices, then $X$ and
$X_{c^{\prime},d}$ are $*$-congruent and both have the same signature. This
signature is

(13) $\mathrm{signature}(X)=(c^{\prime}+m_{+},r-2d-c^{\prime}+m_{-},m_{0}),$

where $(m_{+},m_{-},m_{0})$ is the signature of the blocks
$\mathop{\operator@font diag}\nolimits(\overbrace{{\cal
M}_{\alpha+1},\ldots,{\cal M}_{\alpha+1}}^{s},\overbrace{{\cal
M}_{\alpha},\ldots,{\cal M}_{\alpha}}^{n-r-s})$ in ${\cal
K}_{c^{\prime},d}(\lambda)$. On the other hand, if we express ${\cal
S}_{m}(\lambda)=\lambda X_{m}+Y_{m}\in{\cal B}^{H}({\cal K}_{c,d})$ and ${\cal
K}_{c,d}(\lambda)=\lambda X_{c,d}+Y_{c,d}$, with $X_{m},Y_{m},X_{c,d}$ and
$Y_{c,d}$ being constant Hermitian matrices, then $X_{m}$ and $X_{c,d}$ are
$*$-congruent and both have the same signature. This signature is

(14) $\mathrm{signature}(X_{m})=(c+m_{+},r-2d-c+m_{-},m_{0}).$

Thus, $c^{\prime}\neq c$ implies that $c+m_{+}<c^{\prime}+m_{+}$ or
$r-2d-c+m_{-}<r-2d-c^{\prime}+m_{-}$, which together with
$\lim_{m\rightarrow\infty}X_{m}=X$, (13), and (14) contradict Theorem 4.3 in
[27].

∎

It is interesting to compare Theorem 12 with Theorem 3 in [5], which gives the
generic complete eigenstructures of symmetric $n\times n$ pencils with bounded
rank. In particular, the generic singular symmetric pencils contain complex
eigenvalues, that may be non-real. However, this is not the case of generic
Hermitian pencils, that can only contain real eigenvalues. Moreover, the
number of generic eigenstructures for Hermitian pencils, provided in Theorem
12, is larger than the one for symmetric pencils, and this is due to the
presence of the sign characteristic. The following result, which is an
immediate consequence of Theorem 12, provides the number of different generic
complete eigenstructures in the Hermitian case.

######  Corollary 13.

The number of different generic bundles provided in Theorem 12 is equal to
$\displaystyle\left(\left\lfloor\frac{r}{2}\right\rfloor+1\right)\left\lfloor\frac{r+3}{2}\right\rfloor$.

###### Proof.

The result is a consequence of the following identities:

$\begin{array}[]{ccl}\displaystyle\sum_{d=0}^{\lfloor
r/2\rfloor}(r-2d+1)&=&\displaystyle(r+1)\sum_{d=0}^{\lfloor
r/2\rfloor}1-2\sum_{d=0}^{\lfloor
r/2\rfloor}d=(r+1)\left(\left\lfloor\frac{r}{2}\right\rfloor+1\right)-\left\lfloor\frac{r}{2}\right\rfloor\left(\left\lfloor\frac{r}{2}\right\rfloor+1\right)\\\
&=&\displaystyle\left\lfloor\frac{r+3}{2}\right\rfloor\left(\left\lfloor\frac{r}{2}\right\rfloor+1\right).\end{array}$

∎

In particular, when $r=n-1$, there are $(\lfloor
n/2\rfloor+1)(\lfloor(n-1)/2\rfloor+1)$ different generic bundles. This
implies that the set of complex singular Hermitian $n\times n$ pencils is the
union of $(\lfloor n/2\rfloor+1)(\lfloor(n-1)/2\rfloor+1)$ different bundle
closures. This number is greater than $1$, provided that $n\geq 2$. This is in
contrast with Theorem 4 in [42], where it is claimed that the set of singular
complex Hermitian $n\times n$ pencils is an irreducible closed set. Then,
according to this result, the union of all closures of the generic Hermitian
bundles of rank $n-1$ would be an irreducible closed (in the Zariski topology)
set. Related to this, it is interesting to compare with the case of symmetric
pencils, considered in [5]. It is proved in that reference (Theorem 3) that
the set of symmetric singular pencils is the union of a finite number of
bundle closures, a result which is the counterpart of Theorem 12 for symmetric
pencils. Moreover, in Theorem 4 of [5] it is also proved that each of these
bundle closures is a closed (in the Zariski topology) irreducible set.
However, the arguments in the proof of that result are not valid anymore for
Hermitian pencils, since the map $\Phi$ in that proof, which is essentially a
congruence of matrix pencils, in the case of Hermitian pencils should be a
$*$-congruence instead, which is not a polynomial map.

######  Remark 14 (Skew-Hermitian matrix pencils).

A matrix pencil ${\cal N}(\lambda)=A+\lambda B$, with
$A,B\in\mathbb{C}^{n\times n}$, is called skew Hermitian when $A^{*}=-A$ and
$B^{*}=-B$. Notably, ${\cal N}$ is skew Hermitian if and only if ${\rm i}{\cal
N}$ is Hermitian. Therefore the generic complete eigenstructures of skew-
Hermitian matrix pencils can be obtained from Theorems 10 and 12, multiplying
(5) and (6) by ${\rm i}$.

##  6 Codimension computations

The Hermitian orbit of a Hermitian pencil ${\cal H}(\lambda)$ is a
differentiable manifold over $\mathbb{R}$, whose tangent space at the point
${\cal H}$ is the following vector subspace of $\operatorname{PENCIL}_{n\times
n}^{H}$ (over $\mathbb{R}$)

(15) $T^{H}({\cal H}):=\\{P^{*}{\cal H}(\lambda)+{\cal H}(\lambda)P:\,\
P\in\mathbb{C}^{n\times n}\\}.$

A way to see that this is the tangent space follows similar arguments to the
ones used in [14, p. 74] for orbits under strict equivalence. More precisely,
consider a small perturbation of ${\cal H}$ in ${\cal O}^{H}({\cal H})$,
namely $(I+\delta P)^{*}{\cal H}(\lambda)(I+\delta P)={\cal
H}(\lambda)+\delta\cdot\left(P^{*}{\cal H}(\lambda)+{\cal
H}(\lambda)P\right)+O(\delta^{2})$, for some “small” real quantity $\delta$
and $P\in\mathbb{C}^{n\times n}$, and then take the first-order term of this
perturbation (in $\delta$), namely $P^{*}{\cal H}(\lambda)+{\cal H}(\lambda)P$
(see also [18, p. 1432] for the congruence orbit). Note also that, since
${\cal H}$ is Hermitian, then all points in $T^{H}({\cal H})$ belong to
$\operatorname{PENCIL}_{n\times n}^{H}$, so $T^{H}({\cal H})$ is a (real)
vector subspace of $\operatorname{PENCIL}_{n\times n}^{H}$.

The dimension of $T^{H}({\cal H})$ is the _dimension of the Hermitian orbit_
of ${\cal H}$ and the dimension of the normal space to the orbit at the point
${\cal H}$ is the _codimension of the Hermitian orbit_ of ${\cal H}$ (denoted
by $\operatorname{codim}_{\mathbb{R}}{\cal O}^{H}({\cal H})$), where the
orthogonality is defined with respect to the Frobenius inner product $\langle
A+\lambda B,C+\lambda D\rangle=\operatorname{trace}(AC^{*}+BD^{*}),$ and
$\operatorname{trace}(X)$ denotes the trace of the matrix $X$. We also
emphasize that the normal space to $T^{H}({\cal H})$ is considered in the
vector space $\operatorname{PENCIL}_{n\times n}^{H}$ (and not in
$\operatorname{PENCIL}_{n\times n}$).

In order to get the codimension of ${\cal O}^{H}({\cal H})$ we need the
following result.

######  Theorem 15.

The codimension of the Hermitian orbit of an $n\times n$ Hermitian matrix
pencil $A+\lambda B$ is equal to the dimension of the solution space of

(16) $\begin{split}X^{*}A+AX=0,\\\ X^{*}B+BX=0.\end{split}$

###### Proof.

Define the mapping

$f:\mathbb{C}^{n\times n}\rightarrow T^{H}(A+\lambda B),\quad X\mapsto
X^{*}(A+\lambda B)+(A+\lambda B)X,$

where $T^{H}(A+\lambda B)$ is the tangent space of ${\cal O}^{H}(A+\lambda B)$
at the point $A+\lambda B$. The mapping $f$ is a surjective homomorphism of
vector spaces over $\mathbb{R}$. Therefore
$\dim_{\mathbb{R}}\mathbb{C}^{n\times n}=\dim_{\mathbb{R}}T^{H}(A+\lambda
B)+\dim_{\mathbb{R}}V(A+\lambda B),$ where $V(A+\lambda
B):=\\{X\in\mathbb{C}^{n\times n}:\ X^{*}(A+\lambda B)+(A+\lambda
B)X=0\\}=\\{X\in\mathbb{C}^{n\times n}:\ X^{*}A+AX=0=X^{*}B+BX\\}$. At every
point $A+\lambda B$ there is an isomorphism

$\operatorname{PENCIL}_{n\times n}^{H}\ \simeq T^{H}(A+\lambda B)\oplus
N(A+\lambda B),$

in which $N(A+\lambda B)$ is the normal space to $T^{H}(A+\lambda B)$ at the
point $A+\lambda B$ with respect to the inner product. Therefore,

$\displaystyle\operatorname{codim}_{\mathbb{R}}{\cal O}^{H}(A+\lambda B)$
$\displaystyle=\dim_{\mathbb{R}}N(A+\lambda
B)=\dim_{\mathbb{R}}(\operatorname{PENCIL}_{n\times
n}^{H})-\dim_{\mathbb{R}}T^{H}(A+\lambda B)$
$\displaystyle=\dim_{\mathbb{R}}(\operatorname{PENCIL}_{n\times
n}^{H})-\dim_{\mathbb{R}}\mathbb{C}^{n\times n}+\dim_{\mathbb{R}}V(A,B)$
$\displaystyle=2n^{2}-2n^{2}+\dim_{\mathbb{R}}V(A,B)=\dim_{\mathbb{R}}V(A,B).$

∎

By Theorem 15, for obtaining the codimension of ${\cal O}^{H}(A+\lambda B)$,
when $A+\lambda B$ is Hermitian, it is enough to obtain the dimension over
$\mathbb{R}$ of the solution space of the system of matrix equations (16). Now
we focus on computing the latter.

Let $A+\lambda B=(A_{1}+\lambda B_{1})\oplus(A_{2}+\lambda B_{2})$ be a
Hermitian matrix pencil. Consider the system of matrix equations (16)
associated with $A+\lambda B$. Partitioning the unknown matrix $X$ we rewrite
the system (16) as follows

$\displaystyle\begin{bmatrix}X_{11}^{*}&X_{21}^{*}\\\
X_{12}^{*}&X_{22}^{*}\end{bmatrix}\begin{bmatrix}A_{1}&0\\\
0&A_{2}\end{bmatrix}+\begin{bmatrix}A_{1}&0\\\
0&A_{2}\end{bmatrix}\begin{bmatrix}X_{11}&X_{12}\\\
X_{21}&X_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},$
$\displaystyle\begin{bmatrix}X_{11}^{*}&X_{21}^{*}\\\
X_{12}^{*}&X_{22}^{*}\end{bmatrix}\begin{bmatrix}B_{1}&0\\\
0&B_{2}\end{bmatrix}+\begin{bmatrix}B_{1}&0\\\
0&B_{2}\end{bmatrix}\begin{bmatrix}X_{11}&X_{12}\\\
X_{21}&X_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix}.$

Operating in the left-hand side of the previous identities we obtain

(17)
$\displaystyle\begin{bmatrix}X_{11}^{*}A_{1}+A_{1}X_{11}&X_{21}^{*}A_{2}+A_{1}X_{12}\\\
X_{12}^{*}A_{1}+A_{2}X_{21}&X_{22}^{*}A_{2}+A_{2}X_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\
0&0\end{bmatrix},$
$\displaystyle\begin{bmatrix}X_{11}^{*}B_{1}+B_{1}X_{11}&X_{21}^{*}B_{2}+B_{1}X_{12}\\\
X_{12}^{*}B_{1}+B_{2}X_{21}&X_{22}^{*}B_{2}+B_{2}X_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\
0&0\end{bmatrix},$

where in each equation the off-diagonal blocks are the conjugate transposed of
each other. In the previous system of equations there are two types of
equations, namely: (a) equations of the form $X^{*}A+AX=0$ and (b) equations
of the form $YA+BZ=0$. The coefficients $A$ and $B$ are the coefficients of
the pencils in the diagonal blocks of the direct sum. Then we define, for the
Hermitian matrix pencils $A_{i}+\lambda B_{i}$ and $A_{j}+\lambda B_{j}$, the
following systems of matrix equations:

$\displaystyle\operatorname{syst}(A_{i}+\lambda B_{i}):$
$\displaystyle\qquad\qquad X^{*}A_{i}+A_{i}X=0,$ $\displaystyle
X^{*}B_{i}+B_{i}X=0;$ $\displaystyle\operatorname{syst}(A_{i}+\lambda
B_{i},\,A_{j}+\lambda B_{j}):$ $\displaystyle\qquad\qquad ZA_{j}+A_{i}Y=0,$
$\displaystyle ZB_{j}+B_{i}Y=0.$

In the system (16), one can assume via a $*$-congruence and a change of
variable that $A+\lambda B$ is given in HKCF. Then, we can decouple the system
into a set of systems like (17) (partitioned according to the number of blocks
in the HKCF, which obviously may be larger than $2$) where in each system
$A_{1}+\lambda B_{1}$ and $A_{2}+\lambda B_{2}$ are canonical blocks. To
obtain the dimension of the solution space of (16) it is thus enough to sum up
the dimensions of the solution spaces of all systems (17).

Following Arnold [1, §5.5], and taking into account that the non-real
eigenvalues of Hermitian pencils appear in conjugate pairs, given a Hermitian
pencil ${\cal H}(\lambda)$, the codimension of ${\cal B}^{H}({\cal H})$ over
$\mathbb{R}$, denoted by $\operatorname{codim}_{\mathbb{R}}{\cal B}^{H}({\cal
H})$, is equal to the codimension of ${\cal O}^{H}({\cal H})$ minus the number
of different eigenvalues of ${\cal H}$ (see also [18, p. 1441] for congruence
bundles of general pencils). From this definition and the codimension of
${\cal O}^{H}({\cal T}_{c,d})$ we can obtain the codimension of ${\cal
B}^{H}({\cal T}_{c,d})$, with ${\cal T}_{c,d}(\lambda)$ being the generic
Hermitian pencils in either Theorem 10 or 12.

###  6.1 Codimension of generic regular bundles

In this section, we compute the codimension of the generic bundles in Theorem
10.

We start by computing the dimension of the solution space of
$\operatorname{syst}(A_{i}+\lambda B_{i})$ for $A_{i}+\lambda
B_{i}\in\\{\sigma{\cal J}_{1}(a),{\cal J}_{1}^{H}(\mu,\overline{\mu})\\}$,
with $a\in\mathbb{R}$ and $\mu\in\mathbb{C}$ having positive imaginary part.
First consider $\sigma=1$ and $a\in\mathbb{R}$, resulting in

$\operatorname{syst}({\cal
J}_{1}(a)):\qquad\overline{x}a+ax=0,\quad\overline{x}+x=0,$

whose solution is $x={\rm i}b$, with $b\in\mathbb{R}$. The dimension of this
solution space is equal to $1$. The solution of $\operatorname{syst}(-{\cal
J}_{1}(a))$ (i.e., $\sigma=-1$) remains the same.

Consider the system $\operatorname{syst}({\cal
J}_{1}^{H}(\mu,\overline{\mu}))$:

(18) $\displaystyle\begin{bmatrix}\overline{x_{11}}&\overline{x_{21}}\\\
\overline{x_{12}}&\overline{x_{22}}\end{bmatrix}\begin{bmatrix}0&1\\\
1&0\end{bmatrix}+\begin{bmatrix}0&1\\\
1&0\end{bmatrix}\begin{bmatrix}x_{11}&x_{12}\\\
x_{21}&x_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},$
$\displaystyle\begin{bmatrix}\overline{x_{11}}&\overline{x_{21}}\\\
\overline{x_{12}}&\overline{x_{22}}\end{bmatrix}\begin{bmatrix}0&\overline{\mu}\\\
\mu&0\end{bmatrix}+\begin{bmatrix}0&\overline{\mu}\\\
\mu&0\end{bmatrix}\begin{bmatrix}x_{11}&x_{12}\\\
x_{21}&x_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix}.$

Multiplying the matrices in the first equation in (18) we have

$\displaystyle\begin{bmatrix}\overline{x_{21}}+x_{21}&\overline{x_{11}}+x_{22}\\\
\overline{x_{22}}+x_{11}&\overline{x_{12}}+x_{12}\end{bmatrix}=\begin{bmatrix}0&0\\\
0&0\end{bmatrix},$

whose solution is then replaced into the second equation:

$\displaystyle\begin{bmatrix}\overline{x_{11}}&-{\rm i}b_{21}\\\ -{\rm
i}b_{12}&-{x}_{11}\end{bmatrix}\begin{bmatrix}0&\overline{\mu}\\\
\mu&0\end{bmatrix}+\begin{bmatrix}0&\overline{\mu}\\\
\mu&0\end{bmatrix}\begin{bmatrix}x_{11}&{\rm i}b_{12}\\\ {\rm
i}b_{21}&-\overline{x_{11}}\end{bmatrix}=\begin{bmatrix}0&0\\\
0&0\end{bmatrix},$

where $b_{12},b_{21}\in\mathbb{R}$, or, equivalently,

$\displaystyle\begin{bmatrix}{\rm
i}b_{21}(\overline{\mu}-\mu)&\overline{x_{11}}\overline{\mu}-\overline{x_{11}}\overline{\mu}\\\
{x}_{11}\mu-{x}_{11}\mu&{\rm
i}b_{12}(\mu-\overline{\mu})\end{bmatrix}=\begin{bmatrix}0&0\\\
0&0\end{bmatrix}.$

Therefore $b_{12}=b_{21}=0,x_{11}\in\mathbb{C}$, and the dimension over
$\mathbb{R}$ of the solution space of $\operatorname{syst}({\cal
J}_{1}^{H}(\mu,\overline{\mu}))$ is $2$.

Now we compute the dimension of the solution space of
$\operatorname{syst}(A_{i}+\lambda B_{i},\,A_{j}+\lambda B_{j})$ for
$A_{i}+\lambda B_{i},A_{j}+\lambda B_{j}\in\\{\sigma{\cal J}_{1}(a),{\cal
J}_{1}^{H}(\mu,\overline{\mu})\\}$, with $a\in\mathbb{R},\mu\in\mathbb{C}$ and
$\operatorname{im}\mu>0$.

The system $\operatorname{syst}(\sigma_{i}{\cal
J}_{1}(a_{i}),\,\sigma_{j}{\cal J}_{1}(a_{j}))$ reads $z+y=0=a_{j}z+a_{i}y$
when $\sigma_{i}=\sigma_{j}$ and $z-y=0=a_{j}z-a_{i}y$ when
$\sigma_{i}=-\sigma_{j}$. Since $a_{i}\neq a_{j}$, in both cases the only
solution is $y=z=0$, so the dimension of the solution space of
$\operatorname{syst}(\sigma_{i}{\cal J}_{1}(a_{i}),\,\sigma_{j}{\cal
J}_{1}(a_{j}))$ is $0$.

Next type of systems we need to consider is $\operatorname{syst}({\cal
J}_{1}^{H}(\mu,\overline{\mu}),{\cal J}_{1}^{H}(\eta,\overline{\eta}))$, where
$\mu,\eta$ have positive imaginary parts and $\mu\neq\eta$:

$\displaystyle\begin{bmatrix}z_{11}&z_{12}\\\
z_{21}&z_{22}\end{bmatrix}\begin{bmatrix}0&1\\\
1&0\end{bmatrix}+\begin{bmatrix}0&1\\\
1&0\end{bmatrix}\begin{bmatrix}y_{11}&y_{12}\\\
y_{21}&y_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},$
$\displaystyle\begin{bmatrix}z_{11}&z_{12}\\\
z_{21}&z_{22}\end{bmatrix}\begin{bmatrix}0&\overline{\mu}\\\
\mu&0\end{bmatrix}+\begin{bmatrix}0&\overline{\eta}\\\
\eta&0\end{bmatrix}\begin{bmatrix}y_{11}&y_{12}\\\
y_{21}&y_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},$

or, equivalently,

$\displaystyle\begin{bmatrix}z_{11}&z_{12}\\\
z_{21}&z_{22}\end{bmatrix}\begin{bmatrix}0&\overline{\mu}\\\
\mu&0\end{bmatrix}-\begin{bmatrix}0&\overline{\eta}\\\
\eta&0\end{bmatrix}\begin{bmatrix}z_{22}&z_{21}\\\
z_{12}&z_{11}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix}.$

After performing the matrix multiplications we obtain:

$\displaystyle\begin{bmatrix}z_{12}(\mu-\overline{\eta})&z_{11}(\overline{\mu}-\overline{\eta})\\\
z_{22}(\mu-\eta)&z_{21}(\overline{\mu}-\eta)\end{bmatrix}=\begin{bmatrix}0&0\\\
0&0\end{bmatrix}.$

If ${\rm re}(\mu)\neq{\rm re}(\eta)$ then $z_{ij}=0$. If ${\rm re}(\mu)={\rm
re}(\eta)$, taking into account that ${\rm im}(\mu),{\rm im}(\eta)>0$, and
$\mu\neq\eta$, we have ${\rm im}(\mu)\neq\pm{\rm im}(\eta)$, which in turn
result in $z_{ij}=0$. Thus, the dimension of the solution space of
$\operatorname{syst}({\cal J}_{1}^{H}(\mu,\overline{\mu}),{\cal
J}_{1}^{H}(\eta,\overline{\eta}))$ is zero.

Finally, we consider the systems $\operatorname{syst}({\cal
J}_{1}^{H}(\mu,\overline{\mu}),\sigma{\cal J}_{1}(a))$, where $\mu$ has a
positive imaginary part, $\sigma=\pm 1$, and $a\in\mathbb{R}$. For $\sigma=1$,
the system is

$\displaystyle\begin{bmatrix}z_{11}&z_{12}\end{bmatrix}\begin{bmatrix}0&1\\\
1&0\end{bmatrix}+\begin{bmatrix}1\end{bmatrix}\begin{bmatrix}y_{11}&y_{12}\\\
\end{bmatrix}=\begin{bmatrix}0&0\end{bmatrix},$
$\displaystyle\begin{bmatrix}z_{11}&z_{12}\end{bmatrix}\begin{bmatrix}0&\overline{\mu}\\\
\mu&0\end{bmatrix}+\begin{bmatrix}a\end{bmatrix}\begin{bmatrix}y_{11}&y_{12}\end{bmatrix}=\begin{bmatrix}0&0\end{bmatrix},$

which is equivalent to a single equation

$\displaystyle\begin{bmatrix}z_{11}&z_{12}\end{bmatrix}\begin{bmatrix}0&\overline{\mu}\\\
\mu&0\end{bmatrix}+\begin{bmatrix}a\end{bmatrix}\begin{bmatrix}-z_{12}&-z_{11}\end{bmatrix}=\begin{bmatrix}0&0\end{bmatrix}.$

Multiplying the matrices we get

$\displaystyle\begin{bmatrix}z_{12}\mu-
az_{12}&z_{11}\overline{\mu}-az_{11}\end{bmatrix}=\begin{bmatrix}0&0\end{bmatrix}.$

Since ${\rm im}(\mu)>0$, we must have $z_{ij}=0$ and thus $y_{ij}=0$. For
$\sigma=-1$, the solution is exactly the same. Thus, the dimension of the
solution space of $\operatorname{syst}({\cal
J}_{1}^{H}(\mu,\overline{\mu}),\sigma{\cal J}_{1}(a))$ is always $0$.

We note that only the dimensions of the solution spaces for
$\operatorname{syst}({\cal J}_{1}^{H}(\mu,\overline{\mu}))$ and
$\operatorname{syst}(\sigma{\cal J}_{1}(a))$ (2 and 1, respectively)
contribute to the dimension of the solution spaces for $A+\lambda B$ being
equal to the generic Hermitian canonical form ${\cal R}_{c,d}$. Summing up all
these dimensions of the solution spaces of matrix equations we obtain
$\operatorname{codim}_{\mathbb{R}}{\cal O}^{H}({\cal R}_{c,d})=n$, which
implies the following theorem:

######  Theorem 16.

The codimension in $\operatorname{PENCIL}_{n\times n}^{H}$ of the Hermitian
bundle of ${\cal R}_{c,d}(\lambda)$ of generic Hermitian pencils in Theorem 10
is

$\operatorname{codim}_{\mathbb{R}}{\cal B}^{H}({\cal R}_{c,d})=0.$

In contrast with the singular case, considered in Section 6.2 below,
$\operatorname{codim}_{\mathbb{R}}{\cal O}^{H}({\cal R}_{c,d})$ and
$\operatorname{codim}_{\mathbb{R}}{\cal B}^{H}({\cal R}_{c,d})$ do not depend
on the values of $c$ or $d$.

###  6.2 Codimension of generic singular bundles with bounded rank

In this subsection we obtain the codimensions of the generic bundles ${\cal
B}^{H}({\cal K}_{c,d})$ in Theorem 12. We want to emphasize that, as a
consequence of Theorem 17, the generic bundles in Theorem 12 have different
codimension whenever $d\neq d^{\prime}$ (but those for which $d=d^{\prime}$
and $c\neq c^{\prime}$ have the same codimension). Thus, the codimension of
the generic bundles does not depend on the sign characteristic.

######  Theorem 17.

The codimension in $\operatorname{PENCIL}_{n\times n}^{H}$ of the Hermitian
$n\times n$ bundle of generic Hermitian matrix pencils in Theorem 12 is

$\displaystyle\operatorname{codim}_{\mathbb{R}}{\cal B}^{H}({\cal K}_{c,d})$
$\displaystyle=2(n-d)(n-r).$

###### Proof.

We are first going to compute the codimension of ${\cal O}^{H}({\cal
K}_{c,d})$. By Theorem 15, this codimension is equal to the dimension of the
solution space of (16), with ${\cal K}_{c,d}(\lambda)=A+\lambda B$. By the
arguments after the proof of Theorem 15, we need to obtain the dimension of
the solution spaces of $\operatorname{syst}(A_{i}+\lambda B_{i})$ and
$\operatorname{syst}(A_{i}+\lambda B_{i},A_{j}+\lambda B_{j})$, where
$A_{i}+\lambda B_{i}$ and $A_{j}+\lambda B_{j}$ are the canonical blocks
appearing in ${\cal K}_{c,d}(\lambda)$.

We recall that the dimension of the solution space of
$\operatorname{syst}(\sigma{\cal J}_{1}(a))$ for $a\in\mathbb{R}$ is computed
in Section 6.1 and is equal to $1$.

Now we consider the system $\operatorname{syst}({\cal M}_{k})$:

$\displaystyle\begin{bmatrix}X_{11}^{*}&X_{21}^{*}\\\
X_{12}^{*}&X_{22}^{*}\end{bmatrix}\begin{bmatrix}0&F_{k}^{\top}\\\
F_{k}&0\end{bmatrix}+\begin{bmatrix}0&F_{k}^{\top}\\\
F_{k}&0\end{bmatrix}\begin{bmatrix}X_{11}&X_{12}\\\
X_{21}&X_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},$
$\displaystyle\begin{bmatrix}X_{11}^{*}&X_{21}^{*}\\\
X_{12}^{*}&X_{22}^{*}\end{bmatrix}\begin{bmatrix}0&G_{k}^{\top}\\\
G_{k}&0\end{bmatrix}+\begin{bmatrix}0&G_{k}^{\top}\\\
G_{k}&0\end{bmatrix}\begin{bmatrix}X_{11}&X_{12}\\\
X_{21}&X_{22}\end{bmatrix}=\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},$

where $X$ is partitioned conformally with the $2\times 2$ block structure of
${\cal M}_{k}.$ Note that the conjugation of $X$ is the only difference
compared to the case described in [20, Section 3.2]. Multiplying the matrices
we have

(19)
$\displaystyle\begin{bmatrix}X_{21}^{*}F_{k}+F_{k}^{\top}X_{21}&X_{11}^{*}F_{k}^{\top}+F_{k}^{\top}X_{22}\\\
X_{22}^{*}F_{k}+F_{k}X_{11}&X_{12}^{*}F_{k}^{\top}+F_{k}X_{12}\end{bmatrix}=\begin{bmatrix}0&0\\\
0&0\end{bmatrix},$
$\displaystyle\begin{bmatrix}X_{21}^{*}G_{k}+G_{k}^{\top}X_{21}&X_{11}^{*}G_{k}^{\top}+G_{k}^{\top}X_{22}\\\
X_{22}^{*}G_{k}+G_{k}X_{11}&X_{12}^{*}G_{k}^{\top}+G_{k}X_{12}\end{bmatrix}=\begin{bmatrix}0&0\\\
0&0\end{bmatrix}.$

Since the pairs of blocks at positions $(1,2)$ and $(2,1)$ are equal to each
other up to transposition and conjugation, equation (19) decomposes into three
independent subsystems. Let us first consider the subsystem corresponding to
the $(1,1)$-blocks:

(20) $\displaystyle X_{21}^{*}F_{k}+F_{k}^{\top}X_{21}=0,$ $\displaystyle
X_{21}^{*}G_{k}+G_{k}^{\top}X_{21}=0.$

In order to satisfy the second equation of (20), $X_{21}$ must have the form

$X_{21}=\begin{bmatrix}{\rm i}b_{11}&x_{12}&x_{13}&\ldots&x_{1k}&0\\\
-\overline{x_{12}}&{\rm i}b_{22}&x_{23}&\ldots&x_{2k}&0\\\
-\overline{x_{13}}&-\overline{x_{23}}&{\rm i}b_{33}&\ldots&x_{3k}&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
-\overline{x_{1k}}&-\overline{x_{2k}}&-\overline{x_{3k}}&\ldots&{\rm
i}b_{kk}&0\\\ \end{bmatrix},$

with $b_{11},\ldots,b_{kk}\in\mathbb{R}$. Substituting $X_{21}$ in the first
equation of (20), we obtain

(21) $\begin{bmatrix}0&-{\rm i}b_{11}&-x_{12}&\ldots&-x_{1,k-1}&-x_{1k}\\\
{\rm i}b_{11}&\overline{x_{12}}+x_{12}&-{\rm
i}b_{22}-x_{13}&\ldots&x_{1k}-x_{2,k-1}&-x_{2k}\\\ -\overline{x_{12}}&{\rm
i}b_{22}+\overline{x_{13}}&\overline{x_{23}}+x_{23}&\ldots&x_{2k}-x_{3,k-1}&-x_{3k}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
-\overline{x_{1,k-1}}&\overline{x_{1k}}-\overline{x_{2,k-1}}&\overline{x_{2k}}-\overline{x_{3,k-1}}&\ldots&\overline{x_{k-1,k}}+x_{k-1,k}&-{\rm
i}b_{kk}\\\
-\overline{x_{1k}}&-\overline{x_{2k}}&-\overline{x_{3k}}&\ldots&{\rm
i}b_{kk}&0\end{bmatrix}=0.$

The first row of (21) gives $b_{11}=0=x_{12}=\cdots=x_{1k}$. Replacing this
into the equations obtained from the upper diagonal entries in the second row
of (21) gives $b_{22}=0=x_{23}=\cdots=x_{2k}$. Proceeding recursively in this
way, we conclude that $X_{21}=0$.

Now consider the subsystem corresponding to the $(2,2)$-blocks:

(22) $\displaystyle X_{12}^{*}F_{k}^{\top}+F_{k}X_{12}=0,$ $\displaystyle
X_{12}^{*}G_{k}^{\top}+G_{k}X_{12}=0.$

In order to satisfy the second equation of (22), $X_{12}^{*}$ must have the
form

$X_{12}^{*}=\begin{bmatrix}{\rm
i}b_{11}&x_{12}&x_{13}&\ldots&x_{1k}&x_{1,k+1}\\\ -\overline{x_{12}}&{\rm
i}b_{22}&x_{23}&\ldots&x_{2k}&x_{2,k+1}\\\
-\overline{x_{13}}&-\overline{x_{23}}&{\rm i}b_{33}&\ldots&x_{3k}&x_{3,k+1}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
-\overline{x_{1k}}&-\overline{x_{2k}}&-\overline{x_{3k}}&\ldots&{\rm
i}b_{kk}&x_{k,k+1}\\\ \end{bmatrix},$

with $b_{11},\ldots,b_{kk}\in\mathbb{R}$. Replacing it in the first equation
of (22), we obtain

$\begin{bmatrix}\overline{x_{12}}+x_{12}&x_{13}-{\rm
i}b_{22}&x_{14}-x_{23}&\ldots&x_{1,k+1}-x_{2k}\\\ \overline{x_{13}}+{\rm
i}b_{22}&x_{23}+\overline{x_{23}}&x_{24}-{\rm
i}b_{33}&\ldots&x_{2,k+1}-x_{3k}\\\ \vdots&\ddots&\ddots&\ddots&\vdots\\\
\overline{x_{1k}}-\overline{x_{2,k-1}}&\overline{x_{2k}}-\overline{x_{3,k-1}}&\ldots&x_{k-1,k}+\overline{x_{k-1,k}}&x_{k-1,k+1}-{\rm
i}b_{kk}\\\
\overline{x_{1,k+1}}-\overline{x_{2k}}&\overline{x_{2,k+1}}-\overline{x_{3k}}&\ldots&\overline{x_{k-1,k+1}}+{\rm
i}b_{kk}&x_{k,k+1}+\overline{x_{k,k+1}}\end{bmatrix}=0.$

Equating the diagonal entries of this identity gives
$x_{i,i+1}+\overline{x_{i,i+1}}=0$, for $i=1,\ldots,k$, which implies that
$x_{i,i+1}={\rm i}b_{i,i+1}$, with $b_{i,i+1}\in\mathbb{R}$. Equating the
upper diagonal entries in turn gives $x_{i,j}=x_{i+1,j-1}$, for $j\geq i+2$.
Therefore, $X_{12}^{*}={\rm i}[b_{ij}]$ is a Hankel matrix (namely, each skew
diagonal is constant) and $b_{ij}\in\mathbb{R}$.

Finally, using similar arguments to the ones in [19, Section 3.2] (replacing
$\top$ by $*$ in that reference leads to the same solution), the solution of
the off-diagonal subsystem

$\displaystyle X_{22}^{*}F_{k}+F_{k}X_{11}=0,$ $\displaystyle
X_{22}^{*}G_{k}+G_{k}X_{11}=0,$

is $X_{11}=\alpha I_{k+1}$ and $X_{22}=-\overline{\alpha}I_{k}$, with
$\alpha\in\mathbb{C}$.

Summing up, the solution of system (19) is

$X=\begin{bmatrix}\alpha I_{k+1}&X_{12}\\\
0_{k,k+1}&-\overline{\alpha}I_{k}\end{bmatrix},$

where, as we have seen, $X_{12}^{*}={\rm i}[b_{ij}]$ is a Hankel matrix and
$b_{ij}\in\mathbb{R}$. The number of independent real parameters of the matrix
$X$ above is $2k+2$, namely $2k$ coming from $X_{12}$ and $2$ coming from
$\alpha\in\mathbb{C}$. Hence the dimension over $\mathbb{R}$ of the solution
space of $\operatorname{syst}({\cal M}_{k})$ is $2k+2$.

Now we compute the dimension of the solution space of
$\operatorname{syst}(A_{i}+\lambda B_{i},\,A_{j}+\lambda B_{j})$ for
$A_{i}+\lambda B_{i},A_{j}+\lambda B_{j}\in\\{\sigma{\cal J}_{1}(a),{\cal
M}_{k}\\}$, with $a\in\mathbb{R}$. The system
$\operatorname{syst}(\sigma_{i}{\cal J}_{1}(a_{i}),\,\ \sigma_{j}{\cal
J}_{1}(a_{j}))$ was handled in Section 6.1 and the dimension of its solution
space is equal to $0$. The dimension of the solution of
$\operatorname{syst}(\sigma{\cal J}_{1}(a),\,{\cal M}_{k})$, and
$\operatorname{syst}({\cal M}_{m_{i}},\,{\cal M}_{m_{j}})$ follows directly
from the dimension of the corresponding systems in [20, Corollary 2.2], see
also [15, Corollary 2.1] and [17, Theorem 2.7]. Namely, the dimension of
$\operatorname{syst}({\cal J}_{1}(a),\,{\cal M}_{k})$ is equal to $2$ (from
here, we can conclude that the dimension of the solution space of
$\operatorname{syst}(-{\cal J}_{1}(a),\,{\cal M}_{k})$ is also equal to $2$,
using the change of variables $Y^{\prime}=-Y$); and as for
$\operatorname{syst}({\cal M}_{m_{i}},\,{\cal M}_{m_{j}})$ the dimension is
equal to $2\cdot(2\max\\{m_{i},m_{j}\\}+\varepsilon_{ij})$, where
$\varepsilon_{ij}=2$ if $m_{i}=m_{j}$ and $\varepsilon_{ij}=1$ otherwise. Note
that, for the generic Hermitian pencils in Theorem 12,
$m_{i},m_{j}\in\\{\alpha,\alpha+1\\}$.

We summarize in Table 1 the dimension of the solution space of
$\operatorname{syst}({\cal H})$ and in Table 2 the dimension of the solution
space of the systems $\operatorname{syst}({\cal H}_{1},{\cal H}_{2})$, with
${\cal H}$, ${\cal H}_{1}$, and ${\cal H}_{2}$ being all possible pairs of
blocks in ${\cal K}_{c,d}(\lambda)$. Each entry of the table contains the
dimension of the solution spaces of all systems obtained from the
corresponding blocks, so it is the product of the dimension obtained with the
precedent arguments for each system multiplied by the number of blocks of each
kind in ${\cal K}_{c,d}(\lambda)$. The lower diagonal entries in Table 2 are
not considered to avoid repetitions.

${\cal H}$ | $\sum\dim(\operatorname{syst}({\cal H}))$
---|---
$\sigma{\cal J}_{1}(a)$ | $r-2d$
${\cal M}_{\alpha}$ | $(2\alpha+2)(n-r-s)$
${\cal M}_{\alpha+1}$ | $(2\alpha+4)s$
Table 1: Sum of dimensions of the solution spaces of $\operatorname{syst}({\cal H})$, for all blocks of each of the forms ${\cal H}$ in ${\cal K}_{c,d}(\lambda)$ ${\cal H}_{1}$ ${\cal H}_{2}$ | $\sigma{\cal J}_{1}(a_{i})$ | ${\cal M}_{\alpha}$ | ${\cal M}_{\alpha+1}$
---|---|---|---
$\sigma{\cal J}_{1}(a_{i})$ | $0$ | $2(r-2d)(n-r-s)$ | $2(r-2d)s$
${\cal M}_{\alpha}$ | – | $2(2\alpha+2)\binom{n-r-s}{2}$ | $2(2\alpha+3)(n-r-s)s$
${\cal M}_{\alpha+1}$ | – | – | $2(2\alpha+4)\binom{s}{2}$
Table 2: Sum of dimensions of the solution spaces of
$\operatorname{syst}({\cal H}_{1},{\cal H}_{2})$, for all blocks of each of
the forms ${\cal H}_{1},{\cal H}_{2}$ in ${\cal K}_{c,d}(\lambda)$

Summing up the dimensions of the solution spaces for all the subsystems we
obtain the following codimension of ${\cal O}^{H}({\cal K}_{c,d})$:

$\displaystyle\operatorname{codim}_{\mathbb{R}}{\cal O}^{H}({\cal K}_{c,d})$
$\displaystyle\phantom{a}=r-2d+2(\alpha+2)s+2(\alpha+1)(n-r-s)+2(n-r)(r-2d)$
$\displaystyle\phantom{a}+2(n-r-s)(n-r-s-1)(2\alpha+2)/2+2s(s-1)(2\alpha+4)/2$
$\displaystyle\phantom{a}+2s(n-r-s)(2(\alpha+1)+1)$
$\displaystyle\phantom{a}=r-2d+2(\alpha(n-r)+n-r+s)+2(n-r)(r-2d)$
$\displaystyle\phantom{a}+2(n-r-s)(n-r-s-1)(\alpha+1)+2s(s-1)(\alpha+1)+2s(s-1)$
$\displaystyle\phantom{a}+4s(n-r-s)(\alpha+1)+2s(n-r-s)$
$\displaystyle\phantom{a}=r-2d+2(d+n-r)+2(n-r)(r-2d)$
$\displaystyle\phantom{a}+2(\alpha+1)((n-r-s)(n-r-1)+s(n-r-1))+2s(n-r-1)$
$\displaystyle\phantom{a}=r-2d+2(d+n-r)+2(n-r)(r-2d)$
$\displaystyle\phantom{a}+2(\alpha+1)(n-r)(n-r-1)+2s(n-r-1)$
$\displaystyle\phantom{a}=r-2d+2(d+n-r)+2(n-r)(r-2d)+2(d+n-r)(n-r-1)$
$\displaystyle\phantom{a}=r-2d+2(n-d)(n-r),$

where in the third and fifth identities we have used that $\alpha(n-r)+s=d$.
As a consequence, the codimensions of the generic bundles are:

$\displaystyle\operatorname{codim}_{\mathbb{R}}{\cal B}^{H}({\cal K}_{c,d})$
$\displaystyle=r-2d+2(n-d)(n-r)-r+2d=2(n-d)(n-r).$

∎

We want to emphasize that the generic bundle with smallest codimension
(namely, with largest dimension) is the one with largest $d$, namely the one
having the smallest number of eigenvalues (this number is equal to $0$ if $r$
is even and to $1$ if $r$ is odd).

##  7 Numerical illustration of the theoretical results

Despite this paper is of a theoretical nature, we provide a couple of
numerical experiments to illustrate and support the main results (namely,
Theorems 10 and 12). The Matlab code for these experiments is available on
GitHub.111https://github.com/dmand/generic_herm_experiments.git

######  Example 1.

The purpose of this experiment is to show that all the generic complete
eigenstructures of regular Hermitian matrix pencils in Theorem 10 arise
numerically when computing the eigenvalues of a simple family of randomly
generated regular Hermitian matrix pencils.

Using [33] we generate the matrix coefficients, i.e., $A$ and $B$, of a
Hermitian matrix pencil $A+\lambda B$, and shift these matrix coefficients, by
adding to each of the matrices a diagonal matrix with the same value on the
diagonal: $A+w_{j}I$ and $B+w_{j}I$. Then we compute the eigenvalues of
$(A+w_{j}I)+\lambda(B+w_{j}I)$ with the Matlab function eig($A$,$B$) and we
count the number of real eigenvalues in the output. In Figure 1 we show the
outcome after repeating this computation $350$ times for $20\times 20$
Hermitian matrix pencils with the shifts $w_{j}=(j\log j)/100$. We see that
the number of the real eigenvalues varies from $0$ to $20$ (the size of the
pencils), and that all possible numbers (namely, all even numbers between $0$
and $20$) are attained. However, as $j$ increases, there is a larger number of
real eigenvalues. This is the expected output, since the diagonal entries of
the coefficients matrices are increasing, while the size of all non-diagonal
entries remains the same.

Fig. 1: Number of real eigenvalues for $20\times 20$ regular random Hermitian
matrix pencils $(A+w_{j}I)+\lambda(B+w_{j}I)$, where $w_{j}=(j\log j)/100,\
j=1,\dots,350$.

######  Example 2.

The purpose of this experiment is to show that singular Hermitian matrix
pencils generically do not have pairs of complex conjugate eigenvalues. We
generate Hermitian matrix pencils of a given rank $r$ using the result of [10,
Theorem 2] and compute their eigenvalues using the MultiParEig Toolbox for
Matlab [35], see also [25]. In extensive set of experiments we have never seen
a pencil with a pair of complex conjugate eigenvalues. For example, after
running $50000$ experiments with $17\times 17$ Hermitian matrix pencils of
rank $9$, we get $9,7,5,3$, or $1$ eigenvalues (all real) and no non-real
eigenvalues.

##  8 Conclusions

We have proved that the set of complex Hermitian $n\times n$ matrix pencils
with rank at most $r$ (with $r\leq n$) is the union of a finite number of
closed sets, which are the closures of the bundles of certain pencils. These
pencils are given explicitly in Hermitian Kronecker canonical form, namely
explicitly displaying their complete eigenstructure. Hence, these are the
generic complete eigenstrustures of Hermitian $n\times n$ matrix pencils with
rank at most $r$, and the corresponding bundles are the generic bundles. The
case when $r=n$ is addressed separately, because it provides the generic
eigenstructures of general $n\times n$ Hermitian pencils (without any rank
constraint). In this case, all except one of the generic eigenstructures
contain non-real eigenvalues. However, this is not the case when $r<n$, where
the generic eigenstructures can only contain real eigenvalues (if any).

We have provided the number of generic bundles, which is larger than $1$.
Finally, we have obtained the (co)dimension of the generic bundles.

Acknowledgements. The work of F. De Terán and F. M. Dopico has been partially
supported by the Agencia Estatal de Investigación of Spain through grant
PID2019-106362GB-I00 MCIN/AEI/ 10.13039/501100011033/ and by the Madrid
Government (Comunidad de Madrid-Spain) under the Multiannual Agreement with
UC3M in the line of Excellence of University Professors (EPUC3M23), and in the
context of the V PRICIT (Regional Programme of Research and Technological
Innovation). The work of A. Dmytryshyn has been supported by the Swedish
Research Council (VR) under grant 2021-05393.

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