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
and order the generalized eigenvalues of \(\langle A,B\rangle\) by
$$\lambda_{1}(A)\ge\lambda_{2}(A)\ge\dotsm\ge \lambda_{n}(A) $$
$$\lambda_{1} \bigl(AB^{-1} \bigr)\ge\lambda_{2} \bigl(AB^{-1} \bigr)\ge \dotsm\ge\lambda_{n} \bigl(AB^{-1} \bigr). $$
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.
Anzeige
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])
Anzeige
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
Let
\(A,B \in\mathbb{C}^{n\times n}\)
be Hermitian matrices, and let
B
be a positive definite Hermitian matrix. Then the equalities
hold for
\(1\le i\le n\). In particular, if
\(B=I_{n}\), we have
$$\lambda_{i} \bigl(A B^{-1} \bigr)=\max_{\substack{S\subseteq\mathbb {C}^{n}\\ \dim S=i}} \min_{0\ne\boldsymbol{x}\in S} \biggl\{ \frac{\boldsymbol {x}^{*}A\boldsymbol{x}}{ \boldsymbol{x}^{*} B\boldsymbol{x}} \biggr\} =\min _{\substack{T\subseteq\mathbb{C}^{n}\\ \dim T=n-i+1}} \max_{0\ne\boldsymbol{x}\in T} \biggl\{ \frac{\boldsymbol{x}^{*}A\boldsymbol{x}}{\boldsymbol {x}^{*}B\boldsymbol{x}} \biggr\} $$
$$ \lambda_{i}(A) =\max_{\substack{S\subseteq\mathbb{C}^{n}\\ \dim S=i}} \min _{0\ne\boldsymbol{x}\in S} \boldsymbol{x}^{*}A\boldsymbol{x} =\min _{\substack{T\subseteq\mathbb{C}^{n}\\ \dim T=n-i+1}} \max_{0\ne\boldsymbol{x}\in T} \boldsymbol{x}^{*}A\boldsymbol {x},\quad1\le i\le n. $$
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
In particular, we have
$$ \lambda_{\ell}(A)+\lambda_{\hbar}(A)\le\lambda_{i}(A+B) \le\lambda _{j}(A)+\lambda_{k}(B). $$
$$ \lambda_{i}(A)+\lambda_{n}(B)\le\lambda_{i}(A+B) \le\lambda _{i}(A)+\lambda_{1}(B). $$
Let \(A,E\in\mathbb{C}^{n\times n}\) be Hermitian matrices, B be a positive definite Hermitian matrix,
Then μ is a sufficient condition for B̃ to be a Hermitian positive definite matrix.
$$\widetilde{B}=B+E,\qquad\beta_{n}=\min_{1\le i\le n} \lambda_{i}(B), \qquad \mu=\frac{ \Vert E \Vert _{2}}{\beta_{n}}=\frac{ \Vert E \Vert _{2}}{\lambda_{n}(B)}. $$
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
is valid for all
\(1\le i\le n\).
$$ (1-\mu)\lambda_{i} \bigl(AB^{-1} \bigr)+ \lambda_{n} \bigl(HB^{-1} \bigr) \le\lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) \le\frac{\lambda_{i} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\mu} $$
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
is valid for all
\(1\le i\le n\).
$$ (1-\varepsilon)\lambda_{i} \bigl(AB^{-1} \bigr)+ \lambda_{n} \bigl(HB^{-1} \bigr) \le\lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) \le\frac{\lambda_{i} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\varepsilon} $$
Remark 1.1
Let
Then
Let
Then
These two examples demonstrate that Theorems 1.5 and 1.6 are not necessarily true.
$$A =\operatorname {diag}(-3, -2),\qquad B=\operatorname {diag}(3, 4),\qquad H =I_{2}, \qquad E=\operatorname {diag}(2, 1). $$
$$ \lambda_{2} \bigl(HB^{-1} \bigr)+(1-\mu)\lambda_{2} \bigl(AB^{-1} \bigr)=\frac{1}{3} >0 = \lambda_{2} \bigl((A+H)\widetilde{B}^{-1} \bigr). $$
$$A=\operatorname {diag}(-3,-2),\qquad B=\operatorname {diag}(3,4),\qquad H =-2I_{n},\qquad E=\operatorname {diag}(2,1). $$
$$ \lambda_{1} \bigl((A+H)\widetilde{B}^{-1} \bigr) =- \frac{4}{5} >-3=\frac{\lambda_{1} (AB^{-1} )+\lambda_{1} (HB^{-1} )}{1-\varepsilon}. $$
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.
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.
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
Let
By virtue of Theorems 1.3 and 1.4, if \(j+k-1\le i\le\ell+\hbar-n-1\), we have
Similarly, we have
The proof of Theorem 2.1 is complete. □
$$ B^{-1/2}(A+H)B^{-1/2}=U^{*}\operatorname {diag}\bigl(\lambda_{1} \bigl((A+H)B^{-1} \bigr),\dotsc,\lambda_{n} \bigl((A+H)B^{-1} \bigr) \bigr)U. $$
$$ T_{i}=\operatorname{Span} (\boldsymbol{u}_{i}, \boldsymbol{u}_{i+1},\dotsc,\boldsymbol{u}_{n} ),\quad1\le i\le n. $$
$$ \begin{aligned}[b] \lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) &\le\max_{0\ne\boldsymbol{x}\in T} \biggl\{ \frac{\boldsymbol{x}^{*}B^{-1/2}(A+H)B^{-1/2}\boldsymbol{x}}{ \boldsymbol{x}^{*} (I_{n}+B^{-1/2}EB^{-1/2} )\boldsymbol {x}} \biggr\} \\ &\le \textstyle\begin{cases} \frac{1}{1-\mu}\max_{0\ne\boldsymbol{x}\in T} \{\frac{\boldsymbol{x}^{*}B^{-1/2}(A+H)B^{-1/2}\boldsymbol{x}}{\boldsymbol{x}^{*}\boldsymbol{x}} \},&\lambda_{i}(A+H)\ge0; \\ \frac{1}{1+\mu}\max_{0\ne\boldsymbol{x}\in T} \{\frac{\boldsymbol{x}^{*}B^{-1/2}(A+H)B^{-1/2}\boldsymbol{x}}{\boldsymbol{x}^{*}\boldsymbol{x}} \},&\lambda_{i}(A+H)< 0 \end{cases}\displaystyle \\ &= \textstyle\begin{cases} \frac{1}{1-\mu}\lambda_{i} ((A+H)B^{-1} ),&\lambda _{i}(A+H)\ge0; \\ \frac{1}{1+\mu}\lambda_{i} ((A+H)B^{-1} ),&\lambda _{i}(A+H)< 0 \end{cases}\displaystyle \\ &\le \textstyle\begin{cases} \frac{\lambda_{j} (AB^{-1} )+\lambda_{k} (HB^{-1} )}{1-\mu},&\lambda_{i}(A+H)\ge0; \\ \frac{\lambda_{\ell}(AB^{-1} )+\lambda_{\hbar} (HB^{-1} )}{1+\mu},&\lambda_{i}(A+H)< 0. \end{cases}\displaystyle \end{aligned} $$
(2.1)
$$ \lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) \ge \textstyle\begin{cases} \frac{\lambda_{\ell}(AB^{-1} )+\lambda_{\hbar} (HB^{-1} )}{1+\mu}, &\lambda_{i}(A+H)\ge0;\\ \frac{\lambda_{j} (AB^{-1} )+\lambda_{k} (HB^{-1} )}{1-\mu}, &\lambda_{i}(A+H)< 0. \end{cases} $$
(2.2)
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.
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.
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.
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.
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.
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.
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
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
for
\(1\le i\le n\), where
$$\begin{aligned} \beta_{i}(A)\lambda_{i} \bigl(AB^{-1} \bigr)+\beta_{n}(H)\lambda_{n} \bigl(HB^{-1} \bigr) &\le\lambda_{i} \bigl((A+H)\widetilde{B}^{-1} \bigr) \\ &\le\alpha_{i} (A)\lambda_{i} \bigl(AB^{-1} \bigr)+\alpha_{1}(H)\lambda _{1} \bigl(HB^{-1} \bigr)\end{aligned} $$
$$ \alpha_{i} (A)= \textstyle\begin{cases} \frac{1}{1-\mu}, &\lambda_{i} (A)\ge0;\\ \frac{1}{1+\mu}, &\lambda_{i} (A)< 0 \end{cases}\displaystyle \quad \textit{and}\quad \beta_{i} (A)= \textstyle\begin{cases} \frac{1}{1-\mu}, &\lambda_{i}(A)< 0;\\ \frac{1}{1+\mu}, &\lambda_{i}(A)\ge0. \end{cases} $$
Proof
Since
for \(1\le i\le n\). From inequalities in (2.1) and (2.2), it follows that
for \(1\le i\le n\). The proof of Theorem 2.3 is complete. □
$$\begin{aligned} \lambda_{i} \bigl(\widetilde{B}^{-1/2} A\widetilde{B}^{-1/2} \bigr) +\lambda_{n} \bigl(\widetilde{B}^{-1/2} H \widetilde{B}^{-1/2} \bigr) &\le\lambda_{i} \bigl((A+H) \widetilde{B}^{-1} \bigr) \\ &\le\lambda_{i} \bigl(\widetilde{B}^{-1/2} A\widetilde {B}^{-1/2} \bigr) +\lambda_{1} \bigl(\widetilde{B}^{-1/2} H\widetilde{B}^{-1/2} \bigr)\end{aligned} $$
$$\begin{gathered} \beta_{i} (A)\lambda_{i} \bigl(AB^{-1} \bigr) \le \lambda_{i} \bigl(\widetilde{B}^{-1/2}A\widetilde{B}^{-1/2} \bigr) =\lambda_{i} \bigl(A\widetilde{B}^{-1} \bigr)\le \alpha_{i} (A)\lambda _{i} \bigl(AB^{-1} \bigr), \\ \beta_{n}(H)\lambda_{n} \bigl(HB^{-1} \bigr)\le \lambda_{n} \bigl(H\widetilde{B}^{-1} \bigr),\qquad \lambda_{1} \bigl(H\widetilde{B}^{-1} \bigr)\le \alpha_{1}(H)\lambda _{1} \bigl(AB^{-1} \bigr) \end{gathered}$$
Acknowledgements
The authors appreciate the anonymous referees for their careful corrections to and valuable comments on the original version of this paper.
Competing interests
The authors declare that they have no competing interests.
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