Open Access

A new localization set for generalized eigenvalues

Journal of Inequalities and Applications20172017:113

https://doi.org/10.1186/s13660-017-1388-x

Received: 13 March 2017

Accepted: 1 May 2017

Published: 15 May 2017

Abstract

A new localization set for generalized eigenvalues is obtained. It is shown that the new set is tighter than that in (Numer. Linear Algebra Appl. 16:883-898, 2009). Numerical examples are given to verify the corresponding results.

Keywords

generalized eigenvalue inclusion set matrix pencil

MSC

15A45 15A48

1 Introduction

Let \(\mathbb{C}^{n\times n}\) denote the set of all complex matrices of order n. For the matrices \(A, B \in\mathbb{C}^{n\times n}\), we call the family of matrices \(A-zB\) a matrix pencil, which is parameterized by the complex number z. Next, we regard a matrix pencil \(A-zB\) as a matrix pair \((A, B)\) [1]. A matrix pair \((A, B)\) is called regular if \(\operatorname{det}(A-zB) \neq0 \), and otherwise singular. A complex number λ is called a generalized eigenvalue of \((A, B)\), if
$$\operatorname{det}(A-\lambda B)=0. $$
Furthermore, we call a nonzero vector \(x\in\mathbb{C}^{n}\) a generalized eigenvector of \((A, B)\) associated with λ if
$$Ax=\lambda Bx. $$
Let \(\sigma(A,B)=\{\lambda\in\mathbb{C}: \operatorname{det}(A-\lambda B)=0\}\) denote the generalized spectrum of \((A, B)\). Clearly, if B is an identity matrix, then \(\sigma(A,B)\) reduces to the spectrum of A, i.e. \(\sigma(A,B)=\sigma(A)\). When B is nonsingular, \(\sigma(A,B)\) is equivalent to the spectrum of \(B^{-1}A\), that is,
$$\sigma(A,B)=\sigma\bigl(B^{-1}A\bigr). $$
So, in this case, \((A, B)\) has n generalized eigenvalues. Moreover, if B is singular, then the degree of the characteristic polynomial \(\operatorname{det}(A-\lambda B)\) is \(d < n\), so the number of generalized eigenvalues of the matrix pair \((A, B)\) is d, and, by convention, the remaining \(n-d\) eigenvalues are ∞ [1, 2].
We now list some notation which will be used in the following. Let \(N=\{1,2,\ldots,n\}\). Given two matrices \(A=(a_{ij})\), \(B=(b_{ij})\in\mathbb{C}^{n\times n}\), we denote
$$\begin{gathered} r_{i}(A)=\sum_{k\in N,\atop k\neq i} \vert a_{ik} \vert ,\qquad r_{i}^{j}(A)=\sum _{k\in N,\atop k\neq i,j} \vert a_{ik} \vert , \\ R_{i}(A,B,z)=\sum_{k\in N,\atop k\neq i} \vert a_{ik}-zb_{ik} \vert ,\qquad R_{i}^{j}(A,B,z)= \sum_{k\in N,\atop k\neq i,j} \vert a_{ik}-zb_{ik} \vert , \\ \Gamma_{i}(A,B)= \bigl\{ z\in \mathbb{C}: \vert a_{ii}-zb_{ii} \vert \leq R_{i}(A,B,z) \bigr\} , \end{gathered} $$
and
$$\begin{aligned} \Phi_{ij}(A,B) =& \bigl\{ z\in \mathbb{C}: \bigl\vert (a_{ii}-zb_{ii}) (a_{jj}-zb_{jj})-(a_{ij}-zb_{ij}) (a_{ji}-zb_{ji}) \bigr\vert \\ & {}\leq \vert a_{jj}-zb_{jj} \vert R_{i}^{j}(A,B,z)+ \vert a_{ij}-zb_{ij} \vert R_{j}^{i}(A,B,z) \bigr\} . \end{aligned}$$

The generalized eigenvalue problem arises in many scientific applications; see [35]. Many researchers are interested in the localization of all generalized eigenvalues of a matrix pair; see [1, 2, 6, 7]. In [1], Kostić et al. provided the following Geršgorin-type theorem of the generalized eigenvalue problem.

Theorem 1

[1], Theorem 7

Let \(A,B\in\mathbb{C}^{n\times n}\), \(n\geq2\) and \((A,B)\) be a regular matrix pair. Then
$$\sigma(A,B)\subseteq\Gamma(A,B)=\bigcup_{i\in N} \Gamma_{i}(A,B). $$

Here, \(\Gamma(A,B)\) is called the generalized Geršgorin set of a matrix pair \((A, B)\) and \(\Gamma_{i}(A,B)\) the ith generalized Geršgorin set. As showed in [1], \(\Gamma(A,B)\) is a compact set in the complex plane if and only if B is strictly diagonally dominant (SDD) [8]. When B is not SDD, \(\Gamma(A,B)\) may be an unbounded set or the entire complex plane (see Theorem 2).

Theorem 2

[1], Theorem 8

Let \(A=(a_{ij})\), \(B=(b_{ij})\in\mathbb{C}^{n\times n}\), \(n\geq2\). Then the following statements hold:
  1. (i)

    Let \(i\in N\) be such that, for at least one \(j\in N\), \(b_{ij} \neq0\). Then \(\Gamma_{i} (A, B)\) is an unbounded set in the complex plane if and only if \(\vert b_{ii} \vert \leq r_{i}(B)\).

     
  2. (ii)

    \(\Gamma(A,B)\) is a compact set in the complex plane if and only if B is SDD, that is, \(\vert b_{ii} \vert > r_{i}(B)\).

     
  3. (iii)
    If there is an index \(i\in N\) such that both \(b_{ii} =0\) and
    $$\vert a_{ii} \vert \leq\sum_{k\in\beta(i), \atop k\neq i} \vert a_{ik} \vert , $$
    where \(\beta(i)=\{j\in N: b_{ij}=0\}\), then \(\Gamma_{i}(A, B)\), and consequently \(\Gamma(A, B)\), is the entire complex plane.
     

Recently, in [2], Nakatsukasa presented a different Geršgorin-type theorem to estimate all generalized eigenvalues of a matrix pair \((A,B)\) for the case that the ith row of either A (or B) is SDD for any \(i\in N\). Although the set obtained by Nakatsukasa is simpler to compute than that in Theorem 1, the set is not tighter than that in Theorem 1 in general.

In this paper, we research the generalized eigenvalue localization for a regular matrix pair \((A,B)\) without the restrictive assumption that the ith row of either A (or B) is SDD for any \(i\in N\). By considering \(Ax=\lambda Bx\) and using the triangle inequality, we give a new inclusion set for generalized eigenvalues, and then prove that this set is tighter than that in Theorem 1 (Theorem 7 of [1]). Numerical examples are given to verify the corresponding results.

2 Main results

In this section, a set is provided to locate all the generalized eigenvalue of a matrix pair. Next we compare the set obtained with the generalized Geršgorin set in Theorem 1.

2.1 A new generalized eigenvalue localization set

Theorem 3

Let \(A=(a_{ij})\), \(B=(b_{ij})\in\mathbb{C}^{n\times n}\), with \(n\geq2\) and \((A,B)\) be a regular matrix pair. Then
$$\sigma(A,B)\subseteq\Phi(A,B)=\bigcup_{i,j=1,\atop i\neq j}^{n} \Bigl\{ \Phi_{ij}(A,B)\cap\Phi_{ji}(A,B) \Bigr\} . $$

Proof

For any \(\lambda\in\sigma(A,B)\), let \(0\neq x=(x_{1},x_{2},\ldots,x_{n})^{T}\in \mathbb{C}^{n}\) be an associated generalized eigenvector, i.e.,
$$ Ax=\lambda Bx. $$
(1)
Without loss of generality, let
$$\vert x_{p} \vert \geq \vert x_{q} \vert \geq\max \bigl\{ \vert x_{i} \vert :i\in N, i\neq p,q\bigr\} . $$
Then \(x_{p}\neq0\).
(i) If \(x_{q}\neq0\), then from Equality (1), we have
$$a_{pp}x_{p}+ a_{pq}x_{q}+\sum _{k\in N, \atop k\neq p,q } a_{pk}x_{k}=\lambda b_{pp}x_{p}+ \lambda b_{pq}x_{q}+ \lambda \sum_{k\in N, \atop k\neq p,q } b_{pk}x_{k} $$
and
$$a_{qq}x_{q}+ a_{qp}x_{p}+\sum _{k\in N, \atop k\neq q,p } a_{qk}x_{k}=\lambda b_{qq}x_{q}+ \lambda b_{qp}x_{p}+ \lambda \sum_{k\in N, \atop k\neq q,p } b_{qk}x_{k}, $$
equivalently,
$$ (a_{pp}-\lambda b_{pp})x_{p}+ (a_{pq}-\lambda b_{pq})x_{q} = - \sum _{k\in N, \atop k\neq p,q} (a_{pk}-\lambda b_{pk})x_{k} $$
(2)
and
$$ (a_{qq}-\lambda b_{qq})x_{q}+ (a_{qp}-\lambda b_{qp})x_{p} = - \sum _{k\in N, \atop k\neq q,p} (a_{qk}-\lambda b_{qk})x_{k}. $$
(3)
Solving for \(x_{p}\) and \(x_{q}\) in (2) and (3), we obtain
$$\begin{aligned}& \bigl((a_{pp}-\lambda b_{pp}) (a_{qq}-\lambda b_{qq}) -(a_{pq}-\lambda b_{pq}) (a_{qp}-\lambda b_{qp}) \bigr)x_{p} \\& \quad = -(a_{qq}-\lambda b_{qq}) \sum _{k\in N, \atop k\neq p,q} (a_{pk}-\lambda b_{pk})x_{k} + (a_{pq}-\lambda b_{pq})\sum_{k\in N, \atop k\neq q,p} (a_{qk}-\lambda b_{qk})x_{k} \end{aligned}$$
(4)
and
$$\begin{aligned}& \bigl((a_{pp}-\lambda b_{pp}) (a_{qq}-\lambda b_{qq}) -(a_{pq}-\lambda b_{pq}) (a_{qp}-\lambda b_{qp}) \bigr)x_{q} \\& \quad = -(a_{pp}-\lambda b_{pp}) \sum _{k\in N, \atop k\neq q,p} (a_{qk}-\lambda b_{qk})x_{k} + (a_{qp}-\lambda b_{qp})\sum_{k\in N, \atop k\neq p,q} (a_{pk}-\lambda b_{pk})x_{k}. \end{aligned}$$
(5)
Taking absolute values of (4) and (5) and using the triangle inequality yield
$$\begin{aligned}& \bigl\vert (a_{pp}-\lambda b_{pp}) (a_{qq}-\lambda b_{qq}) -(a_{pq}-\lambda b_{pq}) (a_{qp}- \lambda b_{qp})\bigr\vert \vert x_{p}\vert \\& \quad \leq \vert a_{qq}-\lambda b_{qq} \vert \sum _{k\in N, \atop k\neq p,q} \vert a_{pk}-\lambda b_{pk}\vert\vert x_{k} \vert + \vert a_{pq}-\lambda b_{pq} \vert \sum_{k\in N, \atop k\neq q,p} \vert a_{qk}-\lambda b_{qk}\vert\vert x_{k} \vert \end{aligned}$$
and
$$\begin{aligned}& \bigl\vert (a_{pp}-\lambda b_{pp}) (a_{qq}-\lambda b_{qq}) -(a_{pq}-\lambda b_{pq}) (a_{qp}- \lambda b_{qp})\bigr\vert \vert x_{q}\vert \\& \quad \leq \vert a_{pp}-\lambda b_{pp} \vert \sum _{k\in N, \atop k\neq q,p} \vert a_{qk}-\lambda b_{qk}\vert\vert x_{k} \vert + \vert a_{qp}-\lambda b_{qp} \vert \sum_{k\in N, \atop k\neq p,q} \vert a_{pk}-\lambda b_{pk}\vert\vert x_{k} \vert . \end{aligned}$$
Since \(x_{p}\neq0\) and \(x_{q}\neq0\) are, in absolute value, the largest and second largest components of x, respectively, we divide through by their absolute values to obtain
$$\begin{aligned}& \bigl\vert (a_{pp}-\lambda b_{pp}) (a_{qq}- \lambda b_{qq}) -(a_{pq}-\lambda b_{pq}) (a_{qp}-\lambda b_{qp}) \bigr\vert \\& \quad \leq \vert a_{qq}-\lambda b_{qq} \vert R_{p}^{q}(A,B,\lambda) + \vert a_{pq}-\lambda b_{pq} \vert R_{p}^{q}(A,B,\lambda) \end{aligned}$$
and
$$\begin{aligned}& \bigl\vert (a_{pp}-\lambda b_{pp}) (a_{qq}- \lambda b_{qq}) -(a_{pq}-\lambda b_{pq}) (a_{qp}-\lambda b_{qp}) \bigr\vert \\& \quad \leq \vert a_{pp}-\lambda b_{pp} \vert R_{q}^{p}(A,B,\lambda) + \vert a_{qp}-\lambda b_{qp} \vert R_{p}^{q}(A,B,\lambda). \end{aligned}$$
Hence,
$$\lambda\in \Bigl(\Phi_{pq}(A,B)\cap\Phi_{qp}(A,B) \Bigr) \subseteq \Phi(A,B). $$
(ii) If \(x_{q}= 0\), then \(x_{p}\) is the only nonzero entry of x. From equality (1), we have
$$A\left ( \begin{matrix} 0 \\ \vdots \\ 0 \\ x_{p}\\ 0\\ \vdots\\ 0 \end{matrix} \right )=\left ( \begin{matrix} a_{1p}x_{p} \\ \vdots \\ a_{p-1,p}x_{p} \\ a_{pp}x_{p} \\ a_{p+1,p}x_{p} \\ \vdots\\ a_{np}x_{p} \end{matrix} \right )=\lambda \left ( \begin{matrix} b_{1p}x_{p} \\ \vdots \\ b_{p-1,p}x_{p} \\ b_{pp}x_{p} \\ b_{p+1,p}x_{p} \\ \vdots\\ b_{np}x_{p} \end{matrix} \right ), $$
which implies that, for any \(i\in N\), \(a_{ip}=\lambda b_{ip}\), i.e., \(a_{ip}-\lambda b_{ip}=0\). Hence for any \(i\in N\), \(i\neq p\),
$$\lambda\in \Bigl(\Phi_{pi}(A,B)\cap\Phi_{ip}(A,B) \Bigr) \subseteq \Phi(A,B). $$
From (i) and (ii), \(\sigma(A,B)\subseteq\Phi(A,B)\). The proof is completed. □

Since the matrix pairs \((A,B)\) and \((A^{T},B^{T})\) have the same generalized eigenvalues, we can obtain a theorem by applying Theorem 3 to \((A^{T},B^{T})\).

Theorem 4

Let \(A=(a_{ij})\in\mathbb{C}^{n\times n}\), \(B=(b_{ij})\in \mathbb{C}^{n\times n}\), with \(n\geq2\), and \((A^{T},B^{T})\) be a regular matrix pair. Then
$$\sigma(A,B)\subseteq\Phi\bigl(A^{T},B^{T}\bigr). $$

Remark 1

If B is an identity matrix, then Theorems 3 and 4 reduce to the corresponding results of [9].

Remark 2

When all entries of the ith and jth rows of the matrix B are zero, then
$$\Phi_{ij}(A,B)= \bigl\{ z\in \mathbb{C}: \vert a_{ii}a_{jj}-a_{ij}a_{ji} \vert \leq \vert a_{jj} \vert r_{i}^{j}(A)+ \vert a_{ij} \vert r_{j}^{i}(A) \bigr\} $$
and
$$\Phi_{ji}(A,B)= \bigl\{ z\in \mathbb{C}: \vert a_{ii}a_{jj}-a_{ij}a_{ji} \vert \leq \vert a_{ii} \vert r_{j}^{i}(A)+ \vert a_{ji} \vert r_{i}^{j}(A) \bigr\} . $$
Hence, if
$$ \vert a_{ii}a_{jj}-a_{ij}a_{ji} \vert \leq \vert a_{jj} \vert r_{i}^{j}(A)+ \vert a_{ij} \vert r_{j}^{i}(A) $$
(6)
and
$$ \vert a_{ii}a_{jj}-a_{ij}a_{ji} \vert \leq \vert a_{ii} \vert r_{j}^{i}(A)+ \vert a_{ji} \vert r_{i}^{j}(A), $$
(7)
then
$$\Phi_{ij}(A,B)\cap\Phi_{ji}(A,B)=\mathbb{C}, $$
otherwise,
$$\Phi_{ij}(A,B)\cap\Phi_{ji}(A,B)=\emptyset. $$
Moreover, when inequalities (6) and (7) hold, the matrix B is singular, and \(\operatorname{det}(A-zB)\) has degree less than n. As we are considering regular matrix pairs, the degree of the polynomial \(\operatorname{det}(A-zB)\) has to be at least one; thus, at least one of the sets \(\Phi_{ij}(A,B)\cap\Phi_{ji}(A,B)\) has to be nonempty, implying that the set \(\Phi(A,B)\) of a regular matrix pair is always nonempty.

We now establish the following properties of the set \(\Phi(A,B)\).

Theorem 5

Let \(A=(a_{ij})\), \(B=(b_{ij})\in\mathbb{C}^{n\times n}\), with \(n\geq2\) and \((A,B)\) be a regular matrix pair. Then the set \(\Phi_{ij}(A,B)\cap\Phi_{ji}(A,B)\) contains zero if and only if inequalities (6) and (7) hold.

Proof

The conclusion follows directly from putting \(z=0\) in the inequalities of \(\Phi_{ij}(A,B)\) and \(\Phi_{ji}(A,B)\). □

Theorem 6

Let \(A=(a_{ij})\), \(B=(b_{ij})\in\mathbb{C}^{n\times n}\), with \(n\geq2\) and \((A,B)\) be a regular matrix pair. If there exist \(i,j \in N\), \(i\neq j\), such that
$$\begin{gathered} b_{ii}= b_{jj}=b_{ij}=b_{ji}=0, \\ \vert a_{ii}a_{jj}-a_{ij}a_{ji} \vert \leq \vert a_{jj} \vert \sum_{k\in\beta(i), \atop k\neq i,j} \vert a_{ik} \vert + \vert a_{ij} \vert \sum _{k\in\beta(j), \atop k\neq j,i} \vert a_{jk} \vert , \end{gathered} $$
and
$$\vert a_{ii}a_{jj}-a_{ij}a_{ji} \vert \leq \vert a_{ii} \vert \sum_{k\in\beta(j), \atop k\neq j,i} \vert a_{jk} \vert + \vert a_{ji} \vert \sum _{k\in\beta(i), \atop k\neq i,j} \vert a_{ik} \vert , $$
where \(\beta(i)=\{k\in N: b_{ik}=0\}\), then \(\Phi_{ij}(A,B)\cap\Phi_{ji}(A,B)\), and consequently \(\Phi(A, B)\) is the entire complex plane.

Proof

The conclusion follows directly from the definitions of \(\Phi _{ij}(A,B)\) and \(\Phi_{ji}(A,B)\). □

2.2 Comparison with the generalized Geršgorin set

We now compare the set in Theorem 3 with the generalized Geršgorin set in Theorem 1. First, we observe two examples in which the generalized Geršgorin set is an unbounded set or the entire complex plane.

Example 1

Let
$$A=(a_{ij})=\left ( \begin{matrix} -1 &1 &0 &0.2 \\ 0 &1 &0.4 &0 \\ 0 &0 &i &1 \\ 0.2 &0 &0 &-i \end{matrix} \right ),\qquad B=(b_{ij})=\left ( \begin{matrix} 0.3 &0.1 &0.1 &0.1 \\ 0 &-1 &0.1 &0.1 \\ 0 &0 &i &0.1 \\ 0.1 &0 &0 &-0.2i \end{matrix} \right ). $$
It is easy to see that \(b_{12}=0.1>0\) and
$$\vert b_{11} \vert =\sum_{k=2,3,4} \vert b_{1k} \vert =0.3. $$
Hence, from the part (i) of Theorem 2, we see that \(\Gamma (A,B)\) is unbounded. However, the set \(\Phi(A,B)\) in Theorem 3 is compact. These sets are given by Figure 1, where the actual generalized eigenvalues are plotted with asterisks. Clearly, \(\Phi(A,B)\subset\Gamma(A,B)\).
Figure 1

\(\pmb{\Gamma(A,B)}\) of Example 1 on the left, and \(\pmb{\Phi(A,B)}\) on the right.

Example 2

Let
$$A=(a_{ij})=\left ( \begin{matrix} -1 &1 &0 &0.2 \\ 0 &1 &0.4 &0 \\ 0 &0 &i &1 \\ 0.2 &0 &0 &-i \end{matrix} \right ),\qquad B=(b_{ij})=\left ( \begin{matrix} 0 &0 &0.1 &0.1 \\ 0 &-1 &0.1 &0.1 \\ 0 &0 &i &0.1 \\ 0.1 &0 &0 &-0.2i \end{matrix} \right ). $$
It is easy to see that \(b_{11}=0\), \(\beta(1)=\{2\}\) and
$$\vert a_{11} \vert =\sum_{k\in\beta(1), \atop k\neq 1} \vert a_{1k} \vert = \vert a_{12} \vert =1. $$
Hence, from the part (iii) of Theorem 2, we see that \(\Gamma (A,B)\) is the entire complex plane, but the set \(\Phi(A,B)\) in Theorem 3 is not. \(\Phi(A,B)\) is given by Figure 2, where the actual generalized eigenvalues are plotted with asterisks.
Figure 2

\(\pmb{\Phi(A,B)}\) of Example 2 .

We establish their comparison in the following.

Theorem 7

Let \(A=(a_{ij})\in\mathbb{C}^{n\times n}\), \(B=(b_{ij})\in\mathbb{C}^{n\times n}\), with \(n\geq2\) and \((A,B)\) be a regular matrix pair. Then
$$\Phi(A,B)\subseteq\Gamma(A,B). $$

Proof

Let \(z\in\Phi(A,B)\). Then there are \(i,j \in N\), \(i \neq j\) such that
$$z\in \Bigl(\Phi_{ij}(A,B) \cap\Phi_{ji}(A,B) \Bigr). $$
Next, we prove that
$$ \Phi_{ij}(A,B)\subseteq \Bigl(\Gamma_{i}(A,B) \cup \Gamma_{j}(A,B) \Bigr) $$
(8)
and
$$ \Phi_{ji}(A,B)\subseteq \Bigl(\Gamma _{i}(A,B) \cup \Gamma_{j}(A,B) \Bigr). $$
(9)
(i) For \(z\in\Phi_{ij}(A,B)\), then \(z\in\Gamma_{i}(A,B)\) or \(z\notin \Gamma_{i}(A,B)\). If \(z\in\Gamma_{i}(A,B)\), then (8) holds. If \(z\notin\Gamma_{i}(A,B)\), that is,
$$ \vert a_{ii}-zb_{ii} \vert >R_{i}(A,B,z), $$
(10)
then
$$\begin{aligned}& \vert a_{jj}-z b_{jj} \vert R_{i}^{j}(A,B,z) + \vert a_{ij}-z b_{ij} \vert R_{i}^{j}(A,B,z) \\& \quad \geq \bigl\vert (a_{ii}-z b_{ii}) (a_{jj}-z b_{jj}) -(a_{ij}-zb_{ij}) (a_{ji}-z b_{ji}) \bigr\vert \\& \quad \geq \vert a_{ii}-z b_{ii} \vert\vert a_{jj} -z b_{jj} \vert - \vert a_{ij}-zb_{ij} \vert\vert a_{ji}-z b_{ji} \vert . \end{aligned}$$
(11)
Note that \(R_{i}^{j}(A,B,z)=R_{i}(A,B,z)- \vert a_{ij}-z b_{ij} \vert \) and \(R_{j}^{i}(A,B,z)=R_{j}(A,B,z)- \vert a_{ji}-z b_{ji} \vert \). Then from inequalities (10) and (11), we have
$$\begin{aligned}& \vert a_{jj}-z b_{jj} \vert \bigl(R_{i}(A,B,z)- \vert a_{ij}-z b_{ij} \vert \bigr) + \vert a_{ij}-z b_{ij} \vert \bigl(R_{j}(A,B,z)- \vert a_{ji}-z b_{ji} \vert \bigr) \\& \quad \geq \vert a_{jj}-z b_{jj} \vert R_{i}(A,B,z) -\vert a_{ij}-zb_{ij}\vert\vert a_{ji}-z b_{ji}\vert, \end{aligned}$$
which implies that
$$ \vert a_{ij}-z b_{ij} \vert R_{j}(A,B,z)\geq \vert a_{ij}-zb_{ij}\vert\vert a_{jj}-z b_{jj} \vert . $$
(12)
If \(a_{ij}=z b_{ij}\), then from \(z\in\Phi_{ij}(A,B)\), we have
$$\vert a_{ii}-z b_{ii}\vert\vert a_{jj}-z b_{jj} \vert \leq \vert a_{jj}-z b_{jj} \vert R_{i}^{j}(A,B,z)\leq \vert a_{jj}-z b_{jj} \vert R_{i}(A,B,z). $$
Moreover, from inequality (10), we obtain \(\vert a_{jj}-z b_{jj} \vert =0\). It is obvious that
$$z\in\Gamma_{j}(A,B)\subseteq \Bigl(\Gamma_{i}(A,B)\cup \Gamma_{j}(A,B) \Bigr). $$
If \(a_{ij}\neq z b_{ij}\), then from inequality (12), we have
$$\vert a_{jj}-z b_{jj} \vert \leq R_{j}(A,B,z), $$
that is,
$$z\in\Gamma_{j}(A,B)\subseteq \Bigl(\Gamma_{i}(A,B)\cup \Gamma_{j}(A,B) \Bigr). $$
Hence, (8) holds.

(ii) Similar to the proof of (i), we also see that, for \(z\in \Phi_{ji}(A,B)\), (9) holds.

The conclusion follows from (i) and (ii). □

Since the matrix pairs \((A,B)\) and \((A^{T},B^{T})\) have the same generalized eigenvalues, we can obtain a theorem by applying Theorem 7 to \((A^{T},B^{T})\).

Theorem 8

Let \(A=(a_{ij})\in\mathbb{C}^{n\times n}\), \(B=(b_{ij})\in\mathbb{C}^{n\times n}\), with \(n\geq2\) and \((A^{T},B^{T})\) be a regular matrix pair. Then
$$\Phi\bigl(A^{T},B^{T}\bigr)\subseteq\Gamma \bigl(A^{T},B^{T}\bigr). $$

Example 3

[1], Example 1

Let
$$A=(a_{ij})=\left ( \begin{matrix} 1 &1 &0 &0.2 \\ 0 &-1 &0.4 &0 \\ 0 &0 &i &1 \\ 0.2 &0 &0 &-i \end{matrix} \right ),\qquad B=(b_{ij})=\left ( \begin{matrix} 0.5 &0.1 &0.1 &0.1 \\ 0 &-1 &0.1 &0.1 \\ 0 &0 &i &0.1 \\ 0.1 &0 &0 &-0.5i \end{matrix} \right ). $$
It is easy to see that B is SDD. Hence, from the part (ii) of Theorem 2, we see that \(\Gamma(A,B)\) is compact. \(\Gamma(A,B)\) and \(\Phi(A,B)\) are given by Figure 3, where the exact generalized eigenvalues are plotted with asterisks. Clearly, \(\Phi(A,B)\subset\Gamma(A,B)\).
Figure 3

\(\pmb{\Gamma(A,B)}\) of Example 3 on the left, and \(\pmb{\Phi(A,B)}\) on the right.

Remark 3

From Examples 1, 2 and 3, we see that the set in Theorem 3 is tighter than that in Theorem 1 (Theorem 7 of [1]). In addition, note that A and B in Example 1 satisfy
$$\vert a_{11} \vert =1< \sum_{k=2,3,4} \vert a_{1k} \vert =1.2 $$
and
$$\vert b_{11} \vert =\sum_{k=2,3,4} \vert b_{1k} \vert =0.3, $$
respectively. Hence, we cannot use the method in [2] to estimate the generalized eigenvalues of the matrix pair (A,B). However, the set we obtain is very compact.

3 Conclusions

In this paper, we present a new generalized eigenvalue localization set \(\Phi(A,B)\), and we establish the comparison of the sets \(\Phi(A,B)\) and \(\Gamma(A,B)\) in Theorem 7 of [1], that is, \(\Phi(A,B)\) captures all generalized eigenvalues more precisely than \(\Gamma(A,B)\), which is shown by three numerical examples.

Declarations

Acknowledgements

This work was supported by National Natural Science Foundations of China (Grant No.[11601473]) and CAS ‘Light of West China’ Program.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Guangzhou Vocational College of Technology & Business
(2)
School of Mathematics and Statistics, Yunnan University

References

  1. Kostić, V, Cvetković, LJ, Varga, RS: Geršgorin-type localizations of generalized eigenvalues. Numer. Linear Algebra Appl. 16, 883-898 (2009) MathSciNetView ArticleMATHGoogle Scholar
  2. Nakatsukasa, Y: Gerschgorin’s theorem for generalized eigenvalue problem in the Euclidean metric. Math. Comput. 80(276), 2127-2142 (2011) MathSciNetView ArticleMATHGoogle Scholar
  3. Hua, Y, Sarkar, TK: On SVD for estimating generalized eigenvalues of singular matrix pencil in noise. IEEE Trans. Signal Process. 39(4), 892-900 (1991) View ArticleGoogle Scholar
  4. Mangasarian, OL, Wild, EW: Multisurface proximal support vector machine classification via generalized eigenvalues. IEEE Trans. Pattern Anal. Mach. Intell. 28(1), 69-74 (2006) View ArticleGoogle Scholar
  5. Qiu, L, Davison, EJ: The stability robustness of generalized eigenvalues. IEEE Trans. Autom. Control 37(6), 886-891 (1992) MathSciNetView ArticleMATHGoogle Scholar
  6. Hochstenbach, ME: Fields of values and inclusion region for matrix pencils. Electron. Trans. Numer. Anal. 38, 98-112 (2011) MathSciNetMATHGoogle Scholar
  7. Gershgorin, SGW: Theory for the generalized eigenvalue problem \(Ax=\lambda Bx\). Math. Comput. 29(130), 600-606 (1975) MathSciNetGoogle Scholar
  8. Varga, RS: Geršgorin and His Circles. Springer, Berlin (2004) View ArticleMATHGoogle Scholar
  9. Melmana, A: An alternative to the Brauer set. Linear Multilinear Algebra 58, 377-385 (2010) MathSciNetView ArticleGoogle Scholar

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