# Matrix pencils with the numerical range equal to the whole complex plane

Vadym Koval, Patryk Pagacz  Faculty of Mathematics and Computer Science
Jagiellonian University
Łojasiewicza 6
30-348 Kraków
Poland<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
###### Abstract.

The main purpose of this article is to show that the numerical range of a
linear pencil $\lambda A+B$ is equal to $\mathbb{C}$ if and only if $0$
belongs to the convex hull of the joint numerical range of $A$ and $B$. We
also prove that if the numerical range of a linear pencil $\lambda A+B$ is
equal to $\mathbb{C}$ and $A+A^{*},B+B^{*}\geqslant 0$, then $A$ and $B$ have
a common isotropic vector.

Moreover, we improve the classical result which describes Hermitian linear
pencils.

###### Key words and phrases:

Keywords: linear pencil, numerical range, isotropic vector, matrix polynomial,
joint numerical range

###### 1991 Mathematics Subject Classification:

MSC2020: 15A22,15A60

##  1\. Introduction

Let $\mathbb{C}^{N\times N}$ and $\mathcal{H}(N)$ denote the algebra of
$N\times N$ complex matrices and the algebra of $N\times N$ Hermitian
matrices, respectively. By _a matrix polynomial_ $P(\lambda)$ we mean, a
polynomial with coefficients from $\mathbb{C}^{N\times N}$, i.e.

$P(\lambda)=A_{k}\lambda^{k}+A_{k-1}\lambda^{k-1}+\dots+A_{0},\quad
A_{0},A_{1},\dots,A_{k}\in\mathbb{C}^{N\times N}.$

For $k=1$, the matrix polynomial is called _a linear pencil_. In this paper we
deal with _a numerical range of matrix polynomial_ , i.e.

$W(P(\lambda)):=\\{\lambda\in\mathbb{C}:x^{*}P(\lambda)x=0,\textnormal{ for
some nonzero }x\in\mathbb{C}^{N}\\}.$

It is easy to observe that $W(\lambda I-A)$ is equal to _a numerical range of
matrix_ $A$, i.e. $W(A)=\\{x^{*}Ax:x^{*}x=1\\}$. More basic properties of the
numerical ranges of matrix polynomials one can find in [LR94].

The notion of numerical range of a matrix can be extended for $m$-tuple of
matrices. We define
$W(A_{1},A_{2},\ldots,A_{m}):=\\{(x^{*}A_{1}x,x^{*}A_{2}x,\dots,x^{*}A_{m}x):x^{*}x=1\\}$
and call it _a joint numerical range_. If it is convenient we will identify
$\mathbb{C}$ with $\mathbb{R}^{2}$. In particular, we identify $W(A,B)$ with
$W(A_{1},A_{2},B_{1},B_{2})$, where $A=A_{1}+iA_{2},B=B_{1}+iB_{2}$ and
$A_{1},A_{2},B_{1},B_{2}$ are Hermitian (self-adjoint) matrices.

Many matrix polynomials find their role in applications (see e.g. [BHM+13,
GLR09]). The study of numerical ranges of matrix polynomials has a crucial
role in the stability theory. In particular, the system represented by the
matrix polynomial is stable, if its numerical range is contained in the left
half-plane. Thus the shape of a numerical range was the topic of many papers
(see e.g. [IW16, CN02, NP01, Psa00, BdPNdP17, RS11, BdPE17, CNP02, MP97, MP98,
MP96, PT05, SW22, MMW21b]). Moreover, the classical approach to investigating
matrix polynomials is linearization, where the polynomial is converted into a
larger matrix pencil with the similar properties (see [MMMM06b, MMMM06a, MW02,
KL98, Lan08]). Thus, it is most natural to consider linear pencils, so we
mostly do that in this paper.

For matrices $A_{0},A_{1},\dots,A_{m}\in\mathbb{C}^{N\times N}$ a nonzero
vector $v\in\mathbb{C}^{N}$ is called _a common isotropic vector_ if

$v^{*}A_{0}v=v^{*}A_{1}v=\dots=v^{*}A_{m}v=0.$

It is easy to see that if the matrices
$A_{0},A_{1},\dots,A_{k}\in\mathbb{C}^{N\times N}$ have a common isotropic
vector, then $W(P(\lambda))=\mathbb{C}$, where
$P(\lambda)=A_{k}\lambda^{k}+A_{k-1}\lambda^{k-1}+\dots+A_{0}$. The Example
4.1 shows that the converse implication for quadratic polynomials is not true,
even if all matrices are Hermitian. On the other side one can deduce from
Theorem 4.1 [LR94] that two Hermitian matrices $A,B\in\mathbb{C}^{N\times N}$
have a common isotropic vector if and only if $W(A\lambda+B)=\mathbb{C}$.

Unfortunately, there is an error in Theorem 4.1(e) [LR94]. It was observed by
the authors of [BdPNdP17], but they also made an error in their improvement.

The main purpose of this paper is to adjudicate when the equality
$W(A\lambda+B)=\mathbb{C}$ implies that $A$ and $B$ have a common isotropic
vector.

The paper is divided into tree parts. In Section 2 we describe linear pencils
$P(\lambda)$ such that $W(P(\lambda))=\mathbb{C}$. Moreover, we solve an open
question proposed in [MMW21b] and inspired by the study on dissipative
Hamiltonian matrix pencils (see also [MMW18, SW22, MMW21a]). Namely, we prove
that for linear pencils $P(\lambda)=A\lambda+B$ such that
$A+A^{*},B+B^{*}\geqslant 0$ and $W(P(\lambda))=\mathbb{C}$, the matrices
$A,B$ have a common isotropic vector. Our Theorem 2.2 extends a partial result
Theorem 5.7 [MMW21b].

In Section 3 we consider matrix polynomials with Hermitian coefficients. We
improve the erroneous Theorem 4.1 from [LR94] (and also Theorem 1 from
[BdPNdP17]). In particular, we show that for linear pencils with Hermitian
coefficients the equality $W(P(\lambda))=\mathbb{R}$ is never satisfied.

Additionally in Section 4, we show that second-degree matrix polynomial
$P(\lambda)$ with semidefinite coefficients satisfy $W(P(\lambda))=\mathbb{C}$
if and only if matrix coefficients have a common isotropic vector. Moreover,
we give an example (Example 4.1) of three $3\times 3$ Hermitian matrices
$A,B,C$ without common isotropic vector such that $W(P(\lambda))=\mathbb{C}$,
where $P(\lambda)=A\lambda^{2}+B\lambda+C$.

##  2\. Linear Pencils with $W(P(\lambda))=\mathbb{C}$

The well-known Toeplitz-Hausdorff theorem(see e.g. [Dav71]) says that the
numerical range of a matrix is always convex. This result was extended in
[AYT83], namely the joint numerical range of three Hermitian matrices of
dimension larger than $2$ is convex. But the joint numerical range of four
Hermitian matrices often is not convex (see [AYP79]). In other words, the
joint numerical range of two (arbitrary) matrices does not have to be convex.

In Remark 1 [Psa00] the authors observed that the numerical range of a linear
pencil with one Hermitian matrix is equal $\mathbb{C}$ if $0$ belongs to the
convex hull of the joint numerical range of the linear pencil’s coefficients.
In view of results from [AYT83] the only example with nonconvex joint
numerical range can be found for $2\times 2$ matrices (see Corollary 2.1).

######  Example 2.1.

Let $A=\begin{bmatrix}0&0\\\ 2&0\end{bmatrix}$ and $B=\begin{bmatrix}1&0\\\
0&-1\end{bmatrix}$. Then the pencil $P(\lambda)=A\lambda+B$ has no common
isotropic vector, but $W(P(\lambda))=\mathbb{C}$.

###### Proof.

Indeed, if we take a vector $v=(re^{-i\phi},1)$, then
$W(P(\lambda))\ni-\frac{v^{*}Bv}{v^{*}Av}=\frac{1-r^{2}}{2r}\cdot e^{i\phi}$.
So $W(P(\lambda))=\mathbb{C}$.

On the other hand, for $v=(x,y)$ equality $v^{*}Bv=0$ holds if and only if
$|x|=|y|$ and equality $|v^{*}Av|=0$ if and only if $|x||y|=0$. ∎

Fortunately, Remark 1 from [Psa00] can be extended without any assumptions
about matrices.

######  Theorem 2.1.

Let $A,B\in\mathbb{C}^{N\times N}$. Then $W(A\lambda+B)=\mathbb{C}$ if and
only if $0$ belongs to the convex hull of $W(A,B)$.

###### Proof.

Let us denote $P(\lambda)=A\lambda+B$. The matrices $A,B$ have a common
isotropic vector if and only if $0\in W(A,B)$. Thus without loss of generality
we can assume that $A,B$ do not have a common isotropic vector.

($\Leftarrow$) Let us assume that $0$ belongs to the convex hull of $W(A,B)$.

Let $A=A_{1}+iA_{2}$ and $B=B_{1}+iB_{2}$, where
$A_{1},A_{2},B_{1},B_{2}\in\mathcal{H}(N)$. There are points $x,y\in W(A,B)$
such that $tx+(1-t)y=0\not\in W(A,B)$, for some $t\in(0,1)$.

Case $N>2$:
Let us consider an arbitrary three dimensional projection
$Q:\mathbb{R}^{4}\to\mathbb{R}^{4}$ ($\dim\ker Q=1$).

By [AYT83] a joint numerical range of three Hermitian operators is convex.
Thus it is easy to see that the set $QW(A_{1},A_{2},B_{1},B_{2})$ is convex as
well. Hence $0=Q0\in Q[x,y]=[Qx,Qy]\subset QW(A_{1},A_{2},B_{1},B_{2})$.

Hence there exists a point $0\not=w\in W(A,B)$ such that $Qw=0$. In other
words, the line $\ker Q$ has a joint point with $W(A,B)$. Since $Q$ was
arbitrary, all one dimensional subspaces (lines) intersect the joint numerical
range $W(A,B)$. In particular, for any $\lambda\in\mathbb{C}$ the two
dimensional subspace $\\{(z,w)\in\mathbb{C}^{2}:z\lambda+w=0\\}$ intersect
$W(A,B)$. Thus $W(P(\lambda))=\mathbb{C}$.

Case $N=2$:
It is well-known that the joint numerical range of three $2\times 2$ Hermitian
matrices is an ellipsoid (may be a degenerated one). Thus the set
$QW(A_{1},A_{2},B_{1},B_{2})$ is an ellipsoid for any three dimensional
projection $Q:\mathbb{R}^{4}\to\mathbb{R}^{4}$. Moreover,
$0\in\text{conv}(QW(A_{1},A_{2},B_{1},B_{2}))$, since
$0\in\text{conv}(W(A_{1},A_{2},B_{1},B_{2}))$. Thus
$QW(A_{1},A_{2},B_{1},B_{2})\cap QL_{\lambda}\not=\emptyset$, where
$L_{\lambda}=\\{(z,w)\in\mathbb{C}^{2}:z\lambda=w\\}$, as any line passing
through the point in the convex hull of an ellipsoid intersects it. Hence
$W(A,B)\cap L_{\lambda}\not=\emptyset$, as one can chose $Q$ to be a
projection parallel to a line $l$ such that $l\in L_{\lambda}$. Therefore,
$W(P(\lambda))=\mathbb{C}$.

($\Rightarrow$) Now, let us assume that $0$ does not belong to the convex hull
of $W(A,B)$.

Since $0\not\in\text{conv}(W(A,B))$, the set $W(A,B)$ can be separated from
$0$ by some hyper-plane, in other words there is a linear functional
$f:\mathbb{R}^{4}\ni(x_{1},x_{2},x_{3},x_{4})\mapsto\sum_{i=1}^{4}\alpha_{i}x_{i}\in\mathbb{R}$
and $\varepsilon>0$ such that $f(W(A,B))>2\varepsilon$.

Let us take (we can assume that $\alpha_{1}\not=0$)
$\tilde{\lambda}=\lambda_{1}+i\lambda_{2}\in\mathbb{C}$ such that

$\alpha_{1}+\alpha_{3}\lambda_{1}+\alpha_{4}\lambda_{2}=\frac{1}{k}$

and

$\alpha_{2}-\alpha_{3}\lambda_{2}+\alpha_{4}\lambda_{1}=\frac{1}{k}.$

Let us consider an intersection of the sets
$\\{(z,w)\in\mathbb{C}^{2}:\tilde{\lambda}z=w\\}=\\{(z_{1},z_{2},w_{1},w_{2})\in\mathbb{R}^{4}:w_{1}=\lambda_{1}z_{1}-\lambda_{2}z_{2},w_{2}=\lambda_{1}z_{2}+\lambda_{2}z_{1}\\}$
and $f^{-1}(\varepsilon)$. By simple computations one can get that this
intersection is equal

$I:=\\{(z_{1},z_{2},\lambda_{1}z_{1}-\lambda_{2}z_{2},\lambda_{1}z_{2}+\lambda_{2}z_{1}):z_{2}=-z_{1}+k\varepsilon\\}.$

Thus for any ball there is large enough $k$ such that $I$ has empty
intersection with that ball.

Let $\Omega:=\\{x\in f^{-1}(\varepsilon):tx\in W(A,B)\textnormal{ for some
}t>0\\}$. Since $W(A,B)$ is compact, $\Omega$ is compact as well. So for some
$k$ one can achieve $\Omega\cap I=\emptyset$. And as a consequence,
$\\{(z,w)\in\mathbb{C}^{2}:\tilde{\lambda}z=w\\}\cap W(A,B)=\emptyset$. Hence
$-\tilde{\lambda}\not\in W(P(\lambda))$. ∎

######  Remark 2.1.

The joint numerical range of two matrices does not have to be convex (see
[AYP79]). In fact, for any three independent with $Id$ Hermitian matrices
$A_{1},A_{2},B_{1}$ there exists a Hermitian matrix $B_{2}$ such that
$W(A_{1},A_{2},B_{1},B_{2})=W(A_{1}+iA_{2},B_{1}+iB_{2})$ is not convex (see
[LP99]).

The following corollary shows that Example 2.1 was not possible for larger
matrices.

######  Corollary 2.1.

Let $A\in\mathbb{C}^{N\times N}$ and $B\in\mathbb{C}^{N\times N}$ be Hermitian
matrices, where $N>2$. If $\mathbb{C}=W(A\lambda+B)$, then $A$ and $B$ have a
common isotropic vector.

###### Proof.

Without loss of generality let us assume that $B\in\mathcal{H}(N)$ and
$A=A_{1}+iA_{2}$, where $A_{1},A_{2}\in\mathcal{H}(N)$. Thus
$W(A,B)=W(A_{1},A_{2},B)\subset\mathbb{R}^{3}\times\\{0\\}$ and by [AYT83] it
is a convex set. Thus the thesis follows from Theorem 2.1. ∎

If we consider a linear pencil with two Hermitian coefficients, we can
strengthen the above thesis.

######  Corollary 2.2.

Let $A\in\mathbb{C}^{N\times N}$ and $B\in\mathbb{C}^{N\times N}$ be Hermitian
matrices. If $\mathbb{R}\subset W(A\lambda+B)$, then $A$ and $B$ have a common
isotropic vector.

###### Proof.

Let $A$ and $B$ do not have a common isotropic vector. Since
$W(A,B)=W(A+iB)\subset\mathbb{R}^{2}$ is convex, then by Theorem 2.1 $0\notin
W(A,B)$. Thus there is $\tilde{\lambda}\in\mathbb{R}$ such that
$L_{\tilde{\lambda}}\cap W(A,B)=\emptyset$, where
$L_{\tilde{\lambda}}=\\{x+iy\in\mathbb{C}:\tilde{\lambda}x=y\\}$. Therefore,
$-\tilde{\lambda}\not\in W(A,B)$. ∎

In Theorem 5.7 from [MMW21b] the authors showed that if some three of the four
matrices $R_{1},R_{2},J_{1},J_{2}$, such that

$R_{1}\geqslant 0,\quad R_{2}\geqslant 0,\quad J_{1}^{*}=-J_{1},\quad
J_{2}^{*}=-J_{2}$

do not have a common isotropic vector, then
$W(R_{1}+J_{1},R_{2}+J_{2})\not=\mathbb{C}$.

Now, we prove that it is enough to assume the four matrices
$R_{1},R_{2},J_{1},J_{2}$ do not have a common isotropic vector. And therefore
we solve an open question proposed in [MMW21b].

######  Theorem 2.2.

Let $N>2$ and $P(\lambda)=A\lambda+B$ be a pencil such that
$A=R_{1}+J_{1}\in\mathbb{C}^{N\times N}$ and
$B=R_{2}+J_{2}\in\mathbb{C}^{n\times n}$, where $R_{i}\geqslant 0$,
$J^{*}_{i}=-J_{i}$, for $i=1,2$ and $W(P(\lambda))=\mathbb{C}$ then $A,B$ have
a common isotropic vector.

###### Proof.

The joint numerical range $W(A,B)$ is naturally identified with
$W(R_{1},-iJ_{1},R_{2},-iJ_{2})$.

Let $Q:\mathbb{R}^{4}\to\mathbb{R}^{4}$ be an orthogonal projection onto the
space $lin\\{(1,0,1,0),(0,1,0,0),(0,0,0,1)\\}$. In other words,
$Q(x_{1},x_{2},x_{3},x_{4})=(\frac{x_{1}+x_{3}}{\sqrt{2}},x_{2},x_{4})$, where
the space $Q\mathbb{R}^{4}$ comes equipped with orthonormal basis
$\\{(\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2},0),(0,1,0,0),(0,0,0,1)\\}$.

Due to [AYT83] the joint numerical range
$W(\frac{R_{1}+R_{2}}{\sqrt{2}},-iJ_{1},-iJ_{2})$ is convex. Thus the set
$QW(A,B)=QW(R_{1},-iJ_{1},R_{2},-iJ_{2})=W(\frac{R_{1}+R_{2}}{\sqrt{2}},-iJ_{1},\frac{R_{1}+R_{2}}{\sqrt{2}},-iJ_{2})$
is convex as well.

Let us assume for a while that $0\notin QW(A,B)$. Thus there is a linear
functional
$f:Q\mathbb{R}^{4}\ni(x_{1},x_{2},x_{3})\mapsto\alpha_{1}x_{1}+\alpha_{2}x_{2}+\alpha_{3}x_{3}\in\mathbb{R}$
such that $f(QW(A,B))>0$. Since $QW(A,B)$ is compact we can assume that
$\alpha_{1}\alpha_{2}\alpha_{3}\not=0$.

Now, let us show that there is
$\tilde{\lambda}=\lambda_{1}+i\lambda_{2}\in\mathbb{C}$ such that
$QL_{\tilde{\lambda}}\cap QW(A,B)=\emptyset$, where
$L_{\tilde{\lambda}}:=\\{(z,w)\in\mathbb{C}^{2}:\tilde{\lambda}z=w\\}=\\{(z_{1},z_{2},w_{1},w_{2})\in\mathbb{R}^{4}:w_{1}=\lambda_{1}z_{1}-\lambda_{2}z_{2},w_{2}=\lambda_{1}z_{2}+\lambda_{2}z_{1}\\}$.
Indeed,

$QL_{\tilde{\lambda}}=\\{\Big{(}\frac{z_{1}+\lambda_{1}z_{1}-\lambda_{2}z_{2}}{\sqrt{2}},z_{2},\lambda_{1}z_{2}+\lambda_{2}z_{1}\Big{)}:z_{1},z_{2}\in\mathbb{R}\\}=$
$=\\{\Big{(}\big{(}\frac{1+\lambda_{1}}{\sqrt{2}\lambda_{2}}t-\frac{\lambda_{1}+\lambda_{1}^{2}+\lambda_{2}^{2}}{\sqrt{2}\lambda_{2}}z\big{)},z,t\Big{)}:z,t\in\mathbb{R}\\}=\\{\big{(}pt-
qz,z,t\big{)}:z,t\in\mathbb{R}\\},$

where $p=\frac{1+\lambda_{1}}{\sqrt{2}\lambda_{2}}$ and
$q=\big{(}p+\frac{1}{2p}\big{)}\lambda_{1}+\frac{1}{2p}.$

Thus an appropriate choice of $\tilde{\lambda}$ gives us an arbitrary nonzero
numbers $p,q\in\mathbb{R}$. Hence we can choose $\tilde{\lambda}$ such that
$f(QL_{\tilde{\lambda}})=0$, so $QL_{\tilde{\lambda}}\cap QW(A,B)=\emptyset$.
And finally, $L_{\tilde{\lambda}}\cap W(A,B)=\emptyset$. Hence
$-\tilde{\lambda}\notin W(P(\lambda))$. Thus $W(P(\lambda))\not=\mathbb{C}$,
which is a contradiction.

Therefore, $0\in QW(A,B)$, so there is a vector $v$ such that
$v^{*}\big{(}\frac{R_{1}+R_{2}}{\sqrt{2}}\big{)}v=0,v^{*}J_{1}v=0,v^{*}J_{2}v=0.$
However, $v^{*}R_{1}v\geqslant 0,v^{*}R_{2}v\geqslant 0$. So $v$ is a common
isotropic vector.

∎

######  Remark 2.2.

In Theorem 2.2 is not enough to assume that only one of the matrices
$R_{1},R_{2}$ is nonnegative.

In Theorem 4.2 [LP99], the authors show that for any matrices
$A_{1},A_{2},A_{3}$ such that $Id,A_{1},A_{2},A_{3}$ are linearly independent
there exists a matrix $X\in\mathbb{C}^{n\times 2}$ such that $X^{*}X=Id_{2}$
and $W(A_{1},A_{2},A_{3},XX^{*})$ is not convex. Moreover, by the proof of
Theorem 4.1 [LP99], it could be assumed that

$X^{*}A_{1}X=\begin{bmatrix}0&1\\\ 1&0\end{bmatrix},\quad
X^{*}A_{2}X=\begin{bmatrix}0&i\\\ -i&0\end{bmatrix}.$

Thus $W(X^{*}A_{1}X,X^{*}A_{2}X,X^{*}A_{3}X)$ is an ellipsoid with $0$ in its
interior. By Lemma 4.1 [Li88] $(a,b,c,0)\in W(A_{1},A_{2},A_{3},Id-XX^{*})$ if
and only if $(a,b,c)\in W(X^{*}A_{1}X,X^{*}A_{2}X,X^{*}A_{3}X)$ (see the proof
of Theorem 4.2 [LP99]).

Hence, for $R_{1}=Id-XX^{*}$, $R_{2}=A_{1}$, $J_{1}=iA_{2}$ and $J_{2}=iA_{3}$
there is

$0\not\in W(R_{1}+J_{1},R_{2}+J_{2}),\quad
0\in\text{conv}W(R_{1}+J_{1},R_{2}+J_{2}),\quad R_{1}\geqslant 0.$

Thus by Theorem 2.1 $W(R_{1}+J_{1},R_{2}+J_{2})=\mathbb{C}$ and
$R_{1}+J_{1},R_{2}+J_{2}$ have no common isotropic vector.

##  3\. Linear pencils with Hermitian coefficients

Theorem 4.1 [LR94] characterizes the numerical range of linear pencil with
Hermitian coefficients, i.e. matrix polynomial $P(\lambda)=A\lambda-B$, where
$A$ and $B$ are Hermitian matrices. Unfortunately, as it was observed in
[BdPNdP17], there is an error in Theorem 4.1(e) [LR94].

If $A\in\mathbb{C}^{N\times N}$ and $B\in\mathbb{C}^{N\times N}$ have common
isotropic vector, then $W(P(\lambda))=\mathbb{C}$. Otherwise, by [Tho76],
there is a non-singular matrix $X\in\mathbb{C}^{N\times N}$ such that

$X^{*}AX=I_{n}\oplus-I_{m}\oplus 0_{k},\quad
X^{*}BX=\text{diag}(a_{1},\ldots,a_{n},b_{1},\ldots,b_{m},c_{1}\ldots,c_{k}),$

for some
$a_{1},\ldots,a_{n},b_{1},\ldots,b_{m},c_{1}\ldots,c_{k}\in\mathbb{R}$.
Therefore, to characterize $W(A\lambda-B)$ it is enough to consider such
diagonal matrices. Moreover, since $W(A\lambda-B)=W(-(A\lambda-B))$ we can
assume that $n\geqslant m$.

######  Example 3.1.

Let

$A=\begin{bmatrix}1&0\\\ 0&-1\end{bmatrix},\quad B=\begin{bmatrix}2&0\\\
0&-1\end{bmatrix}.$

Then $W(A\lambda-B)=\mathbb{R}\setminus(1,2)$. Indeed, the equation $\lambda
x^{*}Ax-x^{*}Bx=0$ give us

$\lambda|x_{1}|^{2}-\lambda|x_{2}|^{2}-2|x_{1}|^{2}+|x_{2}|^{2}=0,$

where $x=[x_{1},x_{2}]^{T}$. So

$\lambda=\frac{2|x_{1}|^{2}-|x_{2}|^{2}}{|x_{1}|^{2}-|x_{2}|^{2}}=1+\frac{|x_{1}|^{2}}{|x_{1}|^{2}-|x_{2}|^{2}}=2+\frac{|x_{2}|^{2}}{|x_{1}|^{2}-|x_{2}|^{2}}.$

Thus $\lambda\geqslant 2$ for $|x_{1}|>|x_{2}|$ and $\lambda\leqslant 1$ for
$|x_{1}|<|x_{2}|$.

Theorem 4.1(e) [LR94] claims that in the above example $W(A\lambda-B)$ should
be $\mathbb{R}$. The authors of [BdPNdP17] improve Theorem 4.1(e) and say that
$W(A\lambda-B)=\mathbb{R}$ or $W(A\lambda-B)=\mathbb{R}\setminus(a,b)$, for
some $a<b$. However, their reasoning is still erroneous. They claim that the
case $W(\lambda A-B)=\mathbb{R}$ may occur for some indefinite $A,B$, which is
not true (see Corollary 2.2). In fact, they consider the cases which they
earlier exclude as matrices with common isotropic vector.

In the following proposition, we improve a full characterisation of the
numerical range of Hermitian pencils given in [LR94].

######  Theorem 3.1.

Let $P(\lambda)=A\lambda-B$ be a matrix pencil with Hermitian coefficients
$A=I_{n}\oplus-I_{m}\oplus 0_{k}$, $B=B_{1}\oplus B_{2}\oplus
B_{3}=\text{diag}(a_{1},\ldots,a_{n},b_{1},\ldots,b_{m},c_{1}\ldots,c_{k})$,
where $n\geqslant m$. Suppose that $A$ and $B$ do not have a common non-zero
isotropic vector. Then exactly one of the following cases holds:

  1. a)

$A$ is positive definite (that is $A=I$) and $W(P(\lambda))=W(B)$. In
particular, $W(P(\lambda))$ is a singleton if and only if $B$ is a scalar
matrix; $W(P(\lambda))$ is a positive (nonnegative) line segment if and only
if $B$ is positive (semi)definite.

  2. b)

$A$ is a singular positive semidefinite matrix;
$W(P(\lambda))=[\alpha,\infty)$ if $B_{3}$ is positive definite, and
$W(P(\lambda))=(-\infty,\beta]$ if $B_{3}$ is negative definite, where
$W(B_{1})=[\alpha,\beta]$. In this case, $B_{3}$ must either be positive
definite or negative definite because $A$ and $B$ have no common isotropic
vector.

  3. c)

$A$ is indefinite and $B$ is positive (negative) definite and
$W(P(\lambda))=\mathbb{R}\setminus(\alpha,\beta)$, where
$W(B^{-\frac{1}{2}}AB^{-\frac{1}{2}})=[\frac{1}{\alpha},\frac{1}{\beta}]$
($W((-B^{-\frac{1}{2}})(-A)(-B^{-\frac{1}{2}}))=[\frac{1}{\alpha},\frac{1}{\beta}]$).
In this case, $W(P(\lambda))$ is the union of two disjoint unbounded intervals
and $0\not\in W(P(\lambda))$.

  4. d)

$A$ is indefinite and $B$ is a singular positive (negative) semidefinite
matrix, and $W(P(\lambda))=\\{\mu^{-1}:\mu\in W(B\lambda-A),\mu\neq
0\\}\cup\\{0\\}$ ($W(P(\lambda))=\\{\mu^{-1}:\mu\in
W((-B)\lambda-(-A)),\mu\neq 0\\}\cup\\{0\\}$). In this case, $W(P(\lambda))$
is the union of two disjoint unbounded intervals and $0\in W(P(\lambda))$.

  5. e)

Both $A$ and $B$ are indefinite and
$W(P(\lambda))=\mathbb{R}\setminus(\alpha,\beta)$ for some $\alpha<\beta$.
More precisely, let $W(B_{1})=[\alpha_{1},\beta_{1}]$,
$W(B_{2})=[\alpha_{2},\beta_{2}]$. If $k>0$ or $\alpha_{1}+\alpha_{2}>0$, then
$\alpha=-\alpha_{2}$, $\beta=\alpha_{1}$. Otherwise, $\alpha=\beta_{1}$,
$\beta=-\beta_{2}$.

###### Proof.

Proofs of (a)–(d) can be found in [LR94], so we will concentrate on case (e).

Without loss of generality we may assume that $a_{1}\leqslant
a_{2}\ldots\leqslant a_{n}$, $b_{1}\leqslant b_{2}\ldots\leqslant b_{m}$,
$c_{1}\leqslant c_{2}\ldots\leqslant c_{k}$. Obviously, if for some $i,j$
equality $a_{i}+b_{j}=0$ holds, then thr vector

$(0,\ldots,0,1,0,\ldots,0,1,0,\ldots,0),\quad\text{units are on the $i$-th and
$n+j$-th places}$

would be an isotropic vector for $A$ and $B$, thus such equalities cannot
hold.

Case 1: $k\neq 0$.

We may assume that $c_{1}>0$ as all $c_{i}$ must have the same sign (otherwise
there will be an isotropic vector). Then equality $x^{*}(\lambda A-B)x=0$
becomes

(1)
$\lambda=\dfrac{\sum_{i=1}^{n}{|x_{i}|^{2}a_{i}}+\sum_{j=1}^{m}{|y_{j}|^{2}b_{j}}+\sum_{l=1}^{k}{|z_{l}|^{2}c_{l}}}{\sum_{i=1}^{n}|x_{i}|^{2}-\sum_{j=1}^{m}|y_{j}|^{2}},$

where $x=(x_{1},\ldots,x_{n},y_{1},\ldots,y_{m},z_{1}\ldots,z_{k})^{T}$. It is
easy to see that $a_{1}+b_{1}>0$ as otherwise vector
$(1,0,\ldots,0,1,0,\ldots,0,\sqrt{\frac{-a_{1}-b_{1}}{c_{1}}},0,\ldots,0)$
would be an isotropic one. Then we can rewrite (1) as

$\lambda=a_{1}-\dfrac{\sum_{i=1}^{n}{|x_{i}|^{2}(a_{i}-a_{1})}+\sum_{j=1}^{m}{|y_{j}|^{2}(b_{j}+a_{1})}+\sum_{l=1}^{k}{|z_{l}|^{2}c_{l}}}{\sum_{j=1}^{m}|y_{j}|^{2}-\sum_{i=1}^{n}|x_{i}|^{2}}.$

However, that means that $\lambda\not\in(-b_{1},a_{1})$ as

$\sum_{i=1}^{n}{|x_{i}|^{2}(a_{i}-a_{1})}+\sum_{j=1}^{m}{|y_{j}|^{2}(b_{j}+a_{1})}+\sum_{l=1}^{k}{|z_{l}|^{2}c_{l}}\geqslant(a_{1}+b_{1})(\sum_{j=1}^{m}|y_{j}|^{2}-\sum_{i=1}^{n}|x_{i}|^{2}).$

Furthermore, $\lambda$ can achieve any value from
$\mathbb{R}\setminus(-b_{1},a_{1})$ as $a_{1}+t$, for $t>0$, can be achieved
by the vector

$\Big{(}1,0,\ldots,0,\dfrac{t}{a_{1}+b_{1}},0,\ldots,0\Big{)},\quad\text{where
$\dfrac{t}{a_{1}+b_{1}}$ is on $(n+1)^{\text{st}}$ position}$

and $-b_{1}-t$, for $t>0$, can be achieved by the vector

$\Big{(}\dfrac{t}{t+a_{1}+b_{1}},0,\ldots,0,1,0,\ldots,0\Big{)},\quad\text{where
1 is on $(n+1)^{\text{st}}$ position}$

Case 2: $k=0$.

First of all, we note that (1) becomes

(2)
$\lambda=\dfrac{\sum_{i=1}^{n}{|x_{i}|^{2}a_{i}}+\sum_{j=1}^{m}{|y_{j}|^{2}b_{j}}}{\sum_{i=1}^{n}|x_{i}|^{2}-\sum^{m}_{j=1}|y_{j}|^{2}},$

in this case. Furthermore, we claim that if $A$ and $B$ have no common
isotropic vector, then either $a_{1}+b_{1}>0$ or $a_{n}+b_{m}<0$. Note that if
$n=m=1$, then the claim trivially holds, so we proceed with the assumption
$n\geqslant 2$, $m\geqslant 1$. If $a_{1}+b_{1}<0$ and $a_{n}+b_{m}>0$, then

$x=\bigg{(}\sqrt{\frac{-a_{n}-b_{m}}{a_{1}+b_{m}}},0,\ldots,0,1,0,\ldots,0,\sqrt{\frac{a_{1}-a_{n}}{a_{1}+b_{m}}}\bigg{)},\quad\text{$1$
is on the $n^{\text{th}}$ position}$

would be an isotropic vector if $a_{1}+b_{m}<0$ or

$x=\bigg{(}\sqrt{1-\frac{a_{1}+b_{1}}{a_{1}+b_{m}}},0,\ldots,0,1,0,\ldots,0,\sqrt{-\frac{a_{1}+b_{1}}{a_{1}+b_{m}}}\bigg{)},\quad\text{$1$
is on the $(n+1)^{\text{st}}$ position}$

would be an isotropic vector if $a_{1}+b_{m}>0$. Note that in this case
$m\geqslant 2$, as $a_{1}+b_{1}$ is already assumed to be negative.

If $a_{1}+b_{1}>0$, then the same argument as in the first case shows that
$W(P(\lambda))=\mathbb{R}\setminus(-b_{1},a_{1})$.

If $a_{n}+b_{m}<0$, rewrite (2) as

(3)
$\lambda=a_{n}+\dfrac{\sum^{n}_{i=1}{|x_{i}|^{2}(a_{n}-a_{i})}+\sum^{m}_{j=1}{|y_{j}|^{2}(-a_{n}-b_{j})}}{\sum^{m}_{j=1}|y_{j}|^{2}-\sum_{i=1}^{n}|x_{i}|^{2}},$

and note that

$\sum_{i=1}^{n}{|x_{i}|^{2}(a_{n}-a_{i})}+\sum_{j=1}^{m}{|y_{j}|^{2}(-a_{n}-b_{j})}\geqslant(-a_{n}-b_{m})(\sum^{m}_{j=1}|y_{j}|^{2}-\sum_{i=1}^{n}|x_{i}|^{2}),$

so $\lambda\not\in(a_{n},-b_{m})$. All other values can be reached as in the
first case and that ends the proof. ∎

As we mentioned at the beginning of this section, all cases considered in the
above theorem characterize the numerical range of any linear pencil
$A\lambda-B$, where $A,B$ are Hermitian and do not have a common isotropic
vector. Since in all cases $\mathbb{R}\not\subset W(A\lambda-B)$, one can
conclude Corollary 2.2 from Theorem 3.1

##  4\. Matrix polynomials with Hermitian coefficients

The following example explains that for matrix polynomials with degree larger
than $1$, the property $W(P(\lambda))=\mathbb{C}$ do not implies existence of
common isotropic vector even if the matrices are Hermitian and diagonal.

######  Example 4.1.

The matrix polynomial

$P(\lambda)=\lambda^{2}\left(\begin{array}[]{ccc}1&0&0\\\ 0&-1&0\\\
0&0&0\end{array}\right)+\lambda\left(\begin{array}[]{ccc}4&0&0\\\ 0&-3&0\\\
0&0&-4\end{array}\right)+\left(\begin{array}[]{ccc}-1&0&0\\\ 0&2&0\\\
0&0&-9\end{array}\right)$

satisfies $W(P(\lambda))=\mathbb{C}$ and its coefficients do not have a common
isotropic vector.

###### Proof.

In order to show the first part of the claim, it would be sufficient to show
that for any $a,b\in\mathbb{R}$ there exists a non-zero vector
$v\in\mathbb{C}^{3}$ such that $\lambda=a+bi$ is a solution to the equation
$v^{*}P(\lambda)v=0$. Let $v=(\sqrt{x},\sqrt{y},\sqrt{z})$, where
$x,y,z\in\mathbb{R}_{\geqslant 0}$. Then the equation becomes

$\lambda^{2}(x-y)+\lambda(4x-3y-4z)+(-x+2y-9z)=0.$

The quadratic formula gives us solutions

$\lambda_{1,2}=\dfrac{-4x+3y+4z\pm\sqrt{(4x-3y-4z)^{2}-4(x-y)(-x+2y-9z)}}{2(x-y)},$

so it will be enough to find $x,y,z$ such that

$a=\dfrac{-4x+3y+4z}{2(x-y)},\quad-b^{2}=\dfrac{(4x-3y-4z)^{2}-4(x-y)(-x+2y-9z)}{4(x-y)^{2}},$

which is equivalent to

$\displaystyle a=\dfrac{-4x+3y+4z}{2(x-y)}$
$\displaystyle,\quad-b^{2}=a^{2}-\dfrac{(-x+2y-9z)}{(x-y)}$ $\displaystyle
a+2-\dfrac{2z}{x-y}=-\dfrac{y}{x-y}$
$\displaystyle,\quad-b^{2}-a^{2}-\dfrac{9z}{x-y}-1=-\dfrac{y}{(x-y)}$
$\displaystyle\dfrac{z}{x-y}=\dfrac{-b^{2}-a^{2}-a-3}{7}$
$\displaystyle,\quad\dfrac{y}{x-y}=\dfrac{-2b^{2}-2a^{2}-9a-20}{7}.$

It is easy to see that $\dfrac{-b^{2}-a^{2}-a-3}{7}<0$ and
$\dfrac{-2b^{2}-2a^{2}-9a-20}{7}<-1$, so for every $y>0$ one can find $x>0$,
which satisfies the second equation. However, for every such pair $x-y<0$,
thus there exists $z>0$ which satisfies the first equation, showing that
$W(P(\lambda))=\mathbb{C}$.

In order to see that the coefficients do not have a common isotropic vector,
suppose that $v=(z_{1},z_{2},z_{3})$ would be a such one. Then we get

$|z_{1}|^{2}-|z_{2}|^{2}=0,\quad
4|z_{1}|^{2}-3|z_{2}|^{2}-4|z_{3}|^{2}=0,\quad-|z_{1}|^{2}+2|z_{2}|^{2}-9|z_{3}|^{2}=0,$

but this system’s only solution is $z_{1}=z_{2}=z_{3}=0$, thus proving the
rest of the claim. ∎

If we assume that matrices $A$, $B$, $C$ are semidefinite, then no such
example exist.

######  Proposition 4.1.

If $A,B,C$ are semidefinite matrices such that $W(P(\lambda))=\mathbb{C}$,
where $P(\lambda)=A\lambda^{2}+B\lambda+C$, then $A,B,C$ have a common
isotropic vector.

###### Proof.

Let us assume that $A,B,C$ do not have a common isotropic vector.

Since $W(P(\lambda))=W(-P(\lambda))$, it is enough to consider four cases:
(1) $A\geqslant 0,B\geqslant 0,C\geqslant 0$
Then $\mathbb{R}_{+}\not\subset W(P(\lambda))$, because for $\lambda>0$ we
have

$\lambda^{2}x^{*}Ax+x^{*}Cx\geqslant 0\geqslant-\lambda x^{*}Bx,$

and since $A,B,C$ has no common isotropic vector at least one inequality is
strict.
(2) $A\geqslant 0,B\leqslant 0,C\geqslant 0$

Then $\mathbb{R}_{-}\not\subset W(P(\lambda))$, because for $\lambda<0$ we
have

$\lambda^{2}x^{*}Ax+x^{*}Cx\geqslant 0\geqslant-\lambda x^{*}Bx,$

and since $A,B,C$ has no common isotropic vector at least one inequality is
strict.
(3) $A\geqslant 0,B\geqslant 0,C\leqslant 0$ and (4) $A\geqslant 0,B\leqslant
0,C\leqslant 0$

Since

$(x^{*}Bx)^{2}\geqslant 0\geqslant x^{*}Ax\cdot x^{*}Cx,$

all solutions of the equation

$\lambda^{2}x^{*}Ax+\lambda x^{*}Bx+x^{*}Cx=0,$

are real. Thus $W(P(\lambda))\subset\mathbb{R}$. ∎

## Acknowledgment

The authors are sincerely grateful to Michał Wojtylak for bringing their
attention to this subject and for his all inspiring comments.

Second author acknowledges the financial support by the JU (POB DigiWorld)
grant, decision No. PSP: U1U/P06/NO/02.12

## References

  * [AYP79] Yik-Hoi Au-Yeung and Yiu-Tung Poon.  A remark on the convexity and positive definiteness concerning hermitian matrices.  Southeast Asian Bull. Math, 3(2):85–92, 1979.
  * [AYT83] Yik-Hoi Au-Yeung and Nam-Kiu Tsing.  An extension of the hausdorff-toeplitz theorem on the numerical range.  Proceedings of the American Mathematical Society, 89(2):215–218, 1983.
  * [BdPE17] Natália Bebiano, João da Providência, and Fatemeh Esmaeili.  The characteristic polynomial of linear pencils of small size and the numerical range.  In Natália Bebiano, editor, Applied and Computational Matrix Analysis, pages 181–197, Cham, 2017. Springer International Publishing.
  * [BdPNdP17] Natália Bebiano, João da Providência, Ana Nata, and João P. da Providência.  Fields of values of linear pencils and spectral inclusion regions.  In Natália Bebiano, editor, Applied and Computational Matrix Analysis, pages 165–179, Cham, 2017. Springer International Publishing.
  * [BHM+13] Timo Betcke, Nicholas J. Higham, Volker Mehrmann, Christian Schröder, and Françoise Tisseur.  Nlevp: A collection of nonlinear eigenvalue problems.  ACM Trans. Math. Softw., 39(2), feb 2013.
  * [CN02] Mao-Ting Chien and Hiroshi Nakazato.  The numerical range of linear pencils of 2-by-2 matrices.  Linear Algebra and its Applications, 341(1):69–100, 2002.  Special issue dedicated to Professor T. Ando.
  * [CNP02] Mao-Ting Chien, Hiroshi Nakazato, and Panayiotis J. Psarrakos.  Point equation of the boundary of the numerical range of a matrix polynomial.  Linear Algebra and its Applications, 347(1):205–217, 2002.
  * [Dav71] Chandler Davis.  The Toeplitz-Hausdorff theorem explained.  Canadian Mathematical Bulletin, 14(2):245–246, 1971.
  * [GLR09] Israel Gohberg, Peter Lancaster, and Leiba Rodman.  Matrix polynomials, volume 58 of Classics in Applied Mathematics.  Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 2009.  Reprint of the 1982 original [ MR0662418].
  * [IW16] Naoharu Ito and Harald K. Wimmer.  Self-inversive Hilbert space operator polynomials with spectrum on the unit circle.  Journal of Mathematical Analysis and Applications, 436(2):683–691, 2016.
  * [KL98] Ilya Krupnik and Peter Lancaster.  Linearizations, realization, and scalar products for regular matrix polynomials.  Linear Algebra and its Applications, 272(1):45–57, 1998.
  * [Lan08] Peter Lancaster.  Linearization of regular matrix polynomials.  Electron. J. Linear Algebra, 17:21–27, 2008.
  * [Li88] Chi-Kwong Li.  Matrices with some extremal properties.  Linear Algebra Appl., 101:255–267, 1988.
  * [LP99] Chi-Kwong Li and Yiu-Tung Poon.  Convexity of the joint numerical range.  SIAM J. Matrix Anal. Appl., 21(2):668–678, 1999.
  * [LR94] Chi-Kwong Li and Leiba Rodman.  Numerical range of matrix polynomials.  SIAM J. Matrix Anal. Appl., 15(4):1256–1265, 1994.
  * [MMMM06a] D. Steven Mackey, Niloufer Mackey, Christian Mehl, and Volker Mehrmann.  Structured polynomial eigenvalue problems: Good vibrations from good linearizations.  SIAM Journal on Matrix Analysis and Applications, 28(4):1029–1051, 2006.
  * [MMMM06b] D. Steven Mackey, Niloufer Mackey, Christian Mehl, and Volker Mehrmann.  Vector spaces of linearizations for matrix polynomials.  SIAM Journal on Matrix Analysis and Applications, 28(4):971–1004, 2006.
  * [MMW18] Christian Mehl, Volker Mehrmann, and Michal Wojtylak.  Linear algebra properties of dissipative hamiltonian descriptor systems.  SIAM Journal on Matrix Analysis and Applications, 39(3):1489–1519, 2018.
  * [MMW21a] Christian Mehl, Volker Mehrmann, and Michal Wojtylak.  Distance problems for dissipative Hamiltonian systems and related matrix polynomials.  Linear Algebra Appl., 623:335–366, 2021.
  * [MMW21b] Christian Mehl, Volker Mehrmann, and Michal Wojtylak.  Matrix pencils with coefficients that have positive semidefinite hermitian part, 2021.
  * [MP96] John Maroulas and Panayiotis J. Psarrakos.  Geometrical properties of numerical range of matrix polynomials.  Computers & Mathematics with Applications, 31(4):41–47, 1996.  Selected Topics in Numerical Methods.
  * [MP97] John Maroulas and Panayiotis J. Psarrakos.  The boundary of the numerical range of matrix polynomials.  Linear Algebra and its Applications, 267:101–111, 1997.
  * [MP98] John Maroulas and Panayiotis J. Psarrakos.  On the connectedness of numerical range of matrix polynomials.  Linear Algebra and its Applications, 280(2):97–108, 1998.
  * [MW02] Volker Mehrmann and David Watkins.  Polynomial eigenvalue problems with Hamiltonian structure.  Electron. Trans. Numer. Anal., 13:106–118, 2002.
  * [NP01] Hiroshi Nakazato and Panayiotis J. Psarrakos.  On the shape of numerical range of matrix polynomials.  Linear Algebra and its Applications, 338(1):105–123, 2001.
  * [Psa00] Panayiotis J. Psarrakos.  Numerical range of linear pencils.  Linear Algebra and its Applications, 317(1):127–141, 2000.
  * [PT05] Panayiotis J. Psarrakos and Charalampos Tsitouras.  Numerical approximation of the boundary of numerical range of matrix polynomials.  Applied Numerical Analysis & Computational Mathematics, 2(1):126–133, 2005.
  * [RS11] Leiba Rodman and Ilya M. Spitkovsky.  Ratio numerical ranges of operators.  Integr. Equ. Oper. Theory, 71(2):245–257, 2011.
  * [SW22] Oskar Jakub Szymański and Michał Wojtylak.  Stability of matrix polynomials in one and several variables, 2022.
  * [Tho76] Robert C. Thompson.  The characteristic polynomial of a principal subpencil of a hermitian matrix pencil.  Linear Algebra and its Applications, 14(2):135–177, 1976.