1 Introduction

Let \(A,B \in\mathbb{C}^{n\times n}\) be Hermitian matrices with B being positive definite. We now consider a perturbation problem for \(A\boldsymbol{x}= \lambda B\boldsymbol{x}\). It is known that the n generalized eigenvalues of the matrix pencil \(\langle A,B\rangle\) are real numbers and that the generalized eigenvalues of \(\langle A,B\rangle\) and the eigenvalues of \(AB^{-1}\) are the same. Without loss of generality, we can line up the eigenvalues of a Hermitian matrix A as

$$\lambda_{1}(A)\ge\lambda_{2}(A)\ge\dotsm\ge \lambda_{n}(A) $$

and order the generalized eigenvalues of \(\langle A,B\rangle\) by

$$\lambda_{1} \bigl(AB^{-1} \bigr)\ge\lambda_{2} \bigl(AB^{-1} \bigr)\ge \dotsm\ge\lambda_{n} \bigl(AB^{-1} \bigr). $$

For a standard Hermitian eigenvalue problem \(A\boldsymbol{x}= \lambda \boldsymbol{x}\), Weyl’s theorem [2] is perhaps the best-known perturbation result. We denote the spectral norm of a matrix by \(\|\cdot\|_{2}\) which is also called the largest singular value or the matrix 2-norm.

We now recall several known conclusions in the literature.

Theorem 1.1

([2, Weyl’s theorem])

Let \(A,E\in\mathbb{C}^{n\times n}\) be Hermitian matrices, and let \(\widetilde{A}=A+E\) be a perturbation of A, then

$$\max_{1\le i\le n} \bigl\vert \lambda_{i}(A)- \lambda_{i} (\widetilde {A} ) \bigr\vert \le \Vert E \Vert _{2}. $$

Theorem 1.2

([3])

Let \(A,E\in\mathbb{C}^{n\times n}\) be Hermitian matrices, and let \(\widetilde{A}=A+E\) be a perturbation of A, then

$$ \bigl\vert \lambda (\widetilde{A} )-\lambda(A) \bigr\vert \le\bigl( \Vert A \Vert _{2}+ \Vert E \Vert _{2}\bigr)^{1-1/n} \Vert E \Vert _{2}^{1/n}. $$

Theorem 1.3

([1, 4])

Let \(A,B \in\mathbb{C}^{n\times n}\) be Hermitian matrices, and let B be a positive definite Hermitian matrix. Then the equalities

$$\lambda_{i} \bigl(A B^{-1} \bigr)=\max_{\substack{S\subseteq\mathbb {C}^{n}\\ \dim S=i}} \min_{0\ne\boldsymbol{x}\in S} \biggl\{ \frac{\boldsymbol {x}^{*}A\boldsymbol{x}}{ \boldsymbol{x}^{*} B\boldsymbol{x}} \biggr\} =\min _{\substack{T\subseteq\mathbb{C}^{n}\\ \dim T=n-i+1}} \max_{0\ne\boldsymbol{x}\in T} \biggl\{ \frac{\boldsymbol{x}^{*}A\boldsymbol{x}}{\boldsymbol {x}^{*}B\boldsymbol{x}} \biggr\} $$

hold for \(1\le i\le n\). In particular, if \(B=I_{n}\), we have

$$ \lambda_{i}(A) =\max_{\substack{S\subseteq\mathbb{C}^{n}\\ \dim S=i}} \min _{0\ne\boldsymbol{x}\in S} \boldsymbol{x}^{*}A\boldsymbol{x} =\min _{\substack{T\subseteq\mathbb{C}^{n}\\ \dim T=n-i+1}} \max_{0\ne\boldsymbol{x}\in T} \boldsymbol{x}^{*}A\boldsymbol {x},\quad1\le i\le n. $$

Theorem 1.4

([5, p. 336])

Let \(A,B \in\mathbb{C}^{n\times n}\) be Hermitian matrices and \(i,j,k,\ell,\hbar\in\mathbb{N}\) with \(j+k-1\le i\le\ell+\hbar -n-1\). Then

$$ \lambda_{\ell}(A)+\lambda_{\hbar}(A)\le\lambda_{i}(A+B) \le\lambda _{j}(A)+\lambda_{k}(B). $$

In particular, we have

$$ \lambda_{i}(A)+\lambda_{n}(B)\le\lambda_{i}(A+B) \le\lambda _{i}(A)+\lambda_{1}(B). $$

Let \(A,E\in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix,

$$\widetilde{B}=B+E,\qquad\beta_{n}=\min_{1\le i\le n} \lambda_{i}(B), \qquad \mu=\frac{ \Vert E \Vert _{2}}{\beta_{n}}=\frac{ \Vert E \Vert _{2}}{\lambda_{n}(B)}. $$

Then μ is a sufficient condition for B̃ to be a Hermitian positive definite matrix.

Theorem 1.5

([4])

Let \(A,B,H,E \in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix, and \(\widetilde{B}=B+E\). If \(\mu =\frac{\|E\|_{2}}{\lambda_{n}(B)}<1\), then the double inequality

$$ (1-\mu)\lambda_{i} \bigl(AB^{-1} \bigr)+ \lambda_{n} \bigl(HB^{-1} \bigr) \le\lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) \le\frac{\lambda_{i} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\mu} $$

is valid for all \(1\le i\le n\).

Theorem 1.6

([4])

Let \(A,B,H,E \in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix, and \(\widetilde{B}=B+E\). If \(\varepsilon\triangleq\max_{1\le i\le n} |\lambda_{i} (EB^{-1} ) |<1\), then the double inequality

$$ (1-\varepsilon)\lambda_{i} \bigl(AB^{-1} \bigr)+ \lambda_{n} \bigl(HB^{-1} \bigr) \le\lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) \le\frac{\lambda_{i} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\varepsilon} $$

is valid for all \(1\le i\le n\).

Remark 1.1

Let

$$A =\operatorname {diag}(-3, -2),\qquad B=\operatorname {diag}(3, 4),\qquad H =I_{2}, \qquad E=\operatorname {diag}(2, 1). $$

Then

$$ \lambda_{2} \bigl(HB^{-1} \bigr)+(1-\mu)\lambda_{2} \bigl(AB^{-1} \bigr)=\frac{1}{3} >0 = \lambda_{2} \bigl((A+H)\widetilde{B}^{-1} \bigr). $$

Let

$$A=\operatorname {diag}(-3,-2),\qquad B=\operatorname {diag}(3,4),\qquad H =-2I_{n},\qquad E=\operatorname {diag}(2,1). $$

Then

$$ \lambda_{1} \bigl((A+H)\widetilde{B}^{-1} \bigr) =- \frac{4}{5} >-3=\frac{\lambda_{1} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\varepsilon}. $$

These two examples demonstrate that Theorems 1.5 and 1.6 are not necessarily true.

In this paper, we will establish some inequalities of perturbation problems for generalized eigenvalues.

2 Main results

We are now in a position to state and prove our main results in this paper.

Theorem 2.1

Let \(A,B,H,E \in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix, and \(\widetilde{B}=B+E\). If \(\mu =\frac{\|E\|_{2}}{\lambda_{n}(B)}<1\) and \(i,j,k,\ell,\hbar\in\mathbb {N}\) with \(j+k-1\le i\le\ell+\hbar-n-1\), then

  1. 1.

    when \(\lambda_{i}(A+H)\ge0\), we have

    $$ \frac{\lambda_{\ell}(AB^{-1} )+\lambda_{\hbar} (HB^{-1} )}{1+\mu} \le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \le\frac{\lambda_{j} (AB^{-1} )+\lambda_{k} (HB^{-1} )}{1-\mu}; $$
  2. 2.

    when \(\lambda_{i}(A+H)\le0\), we have

    $$ \frac{\lambda_{j} (AB^{-1} )+\lambda_{k} (HB^{-1} )}{1-\mu} \le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \le\frac{\lambda_{\ell}(AB^{-1} )+\lambda_{\hbar} (HB^{-1} )}{1+\mu}. $$

Proof

Since \(B^{-1/2}(A+H)B^{-1/2}\) is a Hermitian matrix, then there exists an orthogonal matrix \(U=(\boldsymbol{u}_{1},\boldsymbol{u}_{2},\dotsc,\boldsymbol{u}_{n})\in \mathbb{C}^{n\times n}\) such that

$$ B^{-1/2}(A+H)B^{-1/2}=U^{*}\operatorname {diag}\bigl(\lambda_{1} \bigl((A+H)B^{-1} \bigr),\dotsc,\lambda_{n} \bigl((A+H)B^{-1} \bigr) \bigr)U. $$

Let

$$ T_{i}=\operatorname{Span} (\boldsymbol{u}_{i}, \boldsymbol{u}_{i+1},\dotsc,\boldsymbol{u}_{n} ),\quad1\le i\le n. $$

By virtue of Theorems 1.3 and 1.4, if \(j+k-1\le i\le\ell+\hbar-n-1\), we have

$$ \begin{aligned}[b] \lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) &\le\max_{0\ne\boldsymbol{x}\in T} \biggl\{ \frac{\boldsymbol{x}^{*}B^{-1/2}(A+H)B^{-1/2}\boldsymbol{x}}{ \boldsymbol{x}^{*} (I_{n}+B^{-1/2}EB^{-1/2} )\boldsymbol {x}} \biggr\} \\ &\le \textstyle\begin{cases} \frac{1}{1-\mu}\max_{0\ne\boldsymbol{x}\in T} \{\frac{\boldsymbol{x}^{*}B^{-1/2}(A+H)B^{-1/2}\boldsymbol{x}}{\boldsymbol{x}^{*}\boldsymbol{x}} \},&\lambda_{i}(A+H)\ge0; \\ \frac{1}{1+\mu}\max_{0\ne\boldsymbol{x}\in T} \{\frac{\boldsymbol{x}^{*}B^{-1/2}(A+H)B^{-1/2}\boldsymbol{x}}{\boldsymbol{x}^{*}\boldsymbol{x}} \},&\lambda_{i}(A+H)< 0 \end{cases}\displaystyle \\ &= \textstyle\begin{cases} \frac{1}{1-\mu}\lambda_{i} ((A+H)B^{-1} ),&\lambda _{i}(A+H)\ge0; \\ \frac{1}{1+\mu}\lambda_{i} ((A+H)B^{-1} ),&\lambda _{i}(A+H)< 0 \end{cases}\displaystyle \\ &\le \textstyle\begin{cases} \frac{\lambda_{j} (AB^{-1} )+\lambda_{k} (HB^{-1} )}{1-\mu},&\lambda_{i}(A+H)\ge0; \\ \frac{\lambda_{\ell}(AB^{-1} )+\lambda_{\hbar} (HB^{-1} )}{1+\mu},&\lambda_{i}(A+H)< 0. \end{cases}\displaystyle \end{aligned} $$
(2.1)

Similarly, we have

$$ \lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) \ge \textstyle\begin{cases} \frac{\lambda_{\ell}(AB^{-1} )+\lambda_{\hbar} (HB^{-1} )}{1+\mu}, &\lambda_{i}(A+H)\ge0;\\ \frac{\lambda_{j} (AB^{-1} )+\lambda_{k} (HB^{-1} )}{1-\mu}, &\lambda_{i}(A+H)< 0. \end{cases} $$
(2.2)

The proof of Theorem 2.1 is complete. □

Corollary 2.1

Let \(A,B,H,E \in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix, and \(\widetilde{B}=B+E\). If \(\mu =\frac{\|E\|_{2}}{\lambda_{n}(B)}<1\), then

  1. 1.

    when \(\lambda_{i}(A+H)\ge0\) for \(1\le i\le n\),

    $$ \frac{\lambda_{i} (AB^{-1} )+\lambda_{n} (HB^{-1} )}{1+\mu} \le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \le\frac{\lambda_{i} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\mu}; $$
  2. 2.

    when \(\lambda_{i}(A+H)\le0\) for \(1\le i\le n\),

    $$ \frac{\lambda_{i} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\mu} \le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \le\frac{\lambda_{i} (AB^{-1} )+\lambda_{n} (HB^{-1} )}{1+\mu}. $$

Corollary 2.2

Let \(A,B,H,E \in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix, and \(\widetilde{B}=B+E\). If \(\mu =\frac{\|E\|_{2}}{\lambda_{n}(B)}<1\), then

  1. 1.

    when \(\lambda_{i}(A+H)\ge0\) for \(1\le i\le n\), then

    $$ \frac{1}{1+\mu} \biggl[\lambda_{i} \bigl(AB^{-1} \bigr)-\frac{\|H\| }{\lambda_{n}(B)} \biggr] \le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \le\frac{1}{1-\mu} \biggl[\lambda_{i} \bigl(AB^{-1} \bigr)+\frac{\| H\|}{\lambda_{n}(B)} \biggr]; $$
  2. 2.

    when \(\lambda_{i}(A+H)\le0\) for \(1\le i\le n\), then

    $$ \frac{1}{1-\mu} \biggl[\lambda_{i} \bigl(AB^{-1} \bigr)-\frac{\|H\| }{\lambda_{n}(B)} \biggr] \le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \le\frac{1}{1+\mu} \biggl[\lambda_{i} \bigl(AB^{-1} \bigr)+\frac{\| H\|}{\lambda_{n}(B)} \biggr]. $$

Theorem 2.2

Let \(A,B,H,E \in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix, and \(\widetilde{B}=B+E\). If \(\varepsilon=\max_{1\le i\le n} |\lambda_{i} (EB^{-1} ) |<1\), then

  1. 1.

    when \(\lambda_{i}(A+H)\ge0\) for \(1\le i\le n\),

    $$ \frac{\lambda_{i} (AB^{-1} )+\lambda_{n} (HB^{-1} )}{1+\varepsilon} \le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \le\frac{\lambda_{i} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\varepsilon}; $$
  2. 2.

    when \(\lambda_{i}(A+H)\le0\) for \(1\le i\le n\),

    $$ \frac{\lambda_{i} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\varepsilon} \le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \le\frac{\lambda_{i} (AB^{-1} )+\lambda_{n} (HB^{-1} )}{1+\varepsilon}. $$

Proof

Using inequalities (2.1) and (2.2), we obtain the required results. The proof of Theorem 2.2 is thus complete. □

Theorem 2.3

Let \(A,B,H,E \in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix, and \(\widetilde{B}=B+E\). If \(\mu =\frac{\|E\|_{2}}{\lambda_{n}(B)}<1\), then

$$\begin{aligned} \beta_{i}(A)\lambda_{i} \bigl(AB^{-1} \bigr)+\beta_{n}(H)\lambda_{n} \bigl(HB^{-1} \bigr) &\le\lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) \\ &\le\alpha_{i} (A)\lambda_{i} \bigl(AB^{-1} \bigr)+\alpha_{1}(H)\lambda _{1} \bigl(HB^{-1} \bigr)\end{aligned} $$

for \(1\le i\le n\), where

$$ \alpha_{i} (A)= \textstyle\begin{cases} \frac{1}{1-\mu}, &\lambda_{i} (A)\ge0;\\ \frac{1}{1+\mu}, &\lambda_{i} (A)< 0 \end{cases}\displaystyle \quad \textit{and}\quad \beta_{i} (A)= \textstyle\begin{cases} \frac{1}{1-\mu}, &\lambda_{i}(A)< 0;\\ \frac{1}{1+\mu}, &\lambda_{i}(A)\ge0. \end{cases} $$

Proof

Since

$$\begin{aligned} \lambda_{i} \bigl(\widetilde{B}^{-1/2} A\widetilde{B}^{-1/2} \bigr) +\lambda_{n} \bigl(\widetilde{B}^{-1/2} H \widetilde{B}^{-1/2} \bigr) &\le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \\ &\le\lambda_{i} \bigl(\widetilde{B}^{-1/2} A\widetilde {B}^{-1/2} \bigr) +\lambda_{1} \bigl(\widetilde{B}^{-1/2} H\widetilde{B}^{-1/2} \bigr)\end{aligned} $$

for \(1\le i\le n\). From inequalities in (2.1) and (2.2), it follows that

$$\begin{gathered} \beta_{i} (A)\lambda_{i} \bigl(AB^{-1} \bigr) \le \lambda_{i} \bigl(\widetilde{B}^{-1/2}A\widetilde{B}^{-1/2} \bigr) =\lambda_{i} \bigl(A\widetilde{B}^{-1} \bigr)\le \alpha_{i} (A)\lambda _{i} \bigl(AB^{-1} \bigr), \\ \beta_{n}(H)\lambda_{n} \bigl(HB^{-1} \bigr)\le \lambda_{n} \bigl(H\widetilde{B}^{-1} \bigr),\qquad \lambda_{1} \bigl(H\widetilde{B}^{-1} \bigr)\le \alpha_{1}(H)\lambda _{1} \bigl(AB^{-1} \bigr) \end{gathered}$$

for \(1\le i\le n\). The proof of Theorem 2.3 is complete. □