# A new localization set for generalized eigenvalues

## 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.

## 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 Î» 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 Î» 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 âˆž [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 [3â€“5]. Many researchers are interested in the localization of all generalized eigenvalues of a matrix pair; see [1, 2, 6, 7]. In [1], KostiÄ‡ et al. provided the following GerÅ¡gorin-type theorem of the generalized eigenvalue problem.

### Theorem 1

[1], TheoremÂ 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 GerÅ¡gorin set of a matrix pair $$(A, B)$$ and $$\Gamma_{i}(A,B)$$ the ith generalized GerÅ¡gorin 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 TheoremÂ 2).

### Theorem 2

[1], TheoremÂ 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 GerÅ¡gorin-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 TheoremÂ 1, the set is not tighter than that in TheoremÂ 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 TheoremÂ 1 (TheoremÂ 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 GerÅ¡gorin set in TheoremÂ 1.

### 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.â€ƒâ–¡

Since the matrix pairs $$(A,B)$$ and $$(A^{T},B^{T})$$ have the same generalized eigenvalues, we can obtain a theorem by applying TheoremÂ 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 $$\Phi_{ji}(A,B)$$.â€ƒâ–¡

### 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 $$\Phi_{ji}(A,B)$$.â€ƒâ–¡

### 2.2 Comparison with the generalized GerÅ¡gorin set

We now compare the set in TheoremÂ 3 with the generalized GerÅ¡gorin set in TheoremÂ 1. First, we observe two examples in which the generalized GerÅ¡gorin 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 TheoremÂ 2, we see that $$\Gamma (A,B)$$ is unbounded. However, the set $$\Phi(A,B)$$ in TheoremÂ 3 is compact. These sets are given by FigureÂ 1, where the actual generalized eigenvalues are plotted with asterisks. Clearly, $$\Phi(A,B)\subset\Gamma(A,B)$$.

### 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 TheoremÂ 2, we see that $$\Gamma (A,B)$$ is the entire complex plane, but the set $$\Phi(A,B)$$ in TheoremÂ 3 is not. $$\Phi(A,B)$$ is given by FigureÂ 2, where the actual generalized eigenvalues are plotted with asterisks.

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).â€ƒâ–¡

Since the matrix pairs $$(A,B)$$ and $$(A^{T},B^{T})$$ have the same generalized eigenvalues, we can obtain a theorem by applying TheoremÂ 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], ExampleÂ 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 TheoremÂ 2, we see that $$\Gamma(A,B)$$ is compact. $$\Gamma(A,B)$$ and $$\Phi(A,B)$$ are given by FigureÂ 3, where the exact generalized eigenvalues are plotted with asterisks. Clearly, $$\Phi(A,B)\subset\Gamma(A,B)$$.

### Remark 3

From Examples 1, 2 and 3, we see that the set in TheoremÂ 3 is tighter than that in TheoremÂ 1 (TheoremÂ 7 of [1]). In addition, note that A and B in ExampleÂ 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 TheoremÂ 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.

## References

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## Acknowledgements

This work was supported by National Natural Science Foundations of China (Grant No.[11601473]) and CAS â€˜Light of West Chinaâ€™ Program.

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Correspondence to Chaoqian Li.

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Gao, J., Li, C. A new localization set for generalized eigenvalues. J Inequal Appl 2017, 113 (2017). https://doi.org/10.1186/s13660-017-1388-x