Inequalities for selected eigenvalues of the product of matrices

TL;DR

This paper derives bounds for the sums of eigenvalues of the product of a Hermitian matrix and a positive semidefinite matrix, providing insights into their spectral inequalities.

Contribution

It introduces new bounds for eigenvalue sums of matrix products involving Hermitian and positive semidefinite matrices, expanding spectral inequality theory.

Findings

Established bounds for eigenvalue sums of matrix products

Extended spectral inequality results to specific matrix classes

Provided theoretical tools for analyzing matrix eigenvalues

Abstract

The product of a Hermitian matrix and a positive semidefinite matrix has only real eigenvalues. We present bounds for sums of eigenvalues of such a product.

Equations

$$\sum _{t=1}^k{\lambda }_{i_t}(A+B)\le \sum _{t=1}^k{\lambda }_{i_t}(A)+\sum _{t=1}^k{\lambda }_t(B).$$

$$\sum _{t=1}^k{\lambda }_{i_t}(A)+\sum _{t=1}^k{\lambda }_{n-t+1}(B)\le \sum _{t=1}^k{\lambda }_{i_t}(A+B).$$

$$\sum _{t=1}^k{\lambda }_{i_t}(A){\lambda }_{n-t+1}(B)\le \sum _{t=1}^k{\lambda }_{i_t}(AB)\le \sum _{t=1}^k{\lambda }_{i_t}(A){\lambda }_t(B).$$

$$\sum _{t=1}^k{\lambda }_{i_t}(A)$$

$$n_A+n_{-A}={\pi }_{+}+{\nu }_{-}+2{\delta }_0=n+{\delta }_0.$$

$$\sum _{t=1}^k{\lambda }_{i_t}(AB)\le \sum _{t=1}^{k_A}{\lambda }_{i_t}(A){\lambda }_t(B)+\sum _{t=k_A+1}^k{\lambda }_{i_t}(A){\lambda }_{n-k+t}(B)$$

$$\sum _{t=1}^k{\lambda }_{i_t}(AB)\ge \sum _{t=1}^{k_A}{\lambda }_{i_t}(A){\lambda }_{n-t+1}(B)+\sum _{t=k_A+1}^k{\lambda }_{i_t}(A){\lambda }_{k-t+1}(B).$$

$$1\le j_1=n-i_k+1<\cdots <j_t=n-i_{k-t+1}+1<\cdots <j_k=n-i_1+1\le n.$$

$$\sum _{t=1}^k{\lambda }_{j_t}(-A){\lambda }_{n-t+1}(B)\le \sum _{t=1}^k{\lambda }_{j_t}(-AB)\le \sum _{t=1}^k{\lambda }_{j_t}(-A){\lambda }_t(B).$$

$$\sum _{t=1}^k{\lambda }_{i_{k-t+1}}(A){\lambda }_t(B)\le \sum _{t=1}^k{\lambda }_{i_{k-t+1}}(AB)\le \sum _{t=1}^k{\lambda }_{i_{k-t+1}}(A){\lambda }_{n-t+1}(B).$$

$$\sum _{r=1}^k{\lambda }_{i_r}(A){\lambda }_{k-r+1}(B)\le \sum _{r=1}^k{\lambda }_{i_r}(AB)\le \sum _{r=1}^k{\lambda }_{i_r}(A){\lambda }_{n-k+r}(B).$$

$$\sum _{t=1}^k{\lambda }_{i_t}(A){\lambda }_{n-t+1}(B)\le \sum _{t=1}^k{\lambda }_{i_t}(AB)\le \sum _{t=1}^k{\lambda }_{i_t}(A){\lambda }_t(B).$$

$$\sum _{t=1}^k{\lambda }_{i_t}(AB)\le \sum _{t=1}^k{\lambda }_{i_t}(A_{+}B)\le \sum _{t=1}^k{\lambda }_{i_t}(A_{+}){\lambda }_t(B)=\sum _{t=1}^k{\lambda }_{i_t}(A){\lambda }_t(B).$$

$$S_{t}={\rm{Span}}\big{(}B^{-\frac{1}{2}}{e}_{1},B^{-\frac{1}{2}}{e}_{2},\dots,B^{-\frac{1}{2}}{e}_{i_{t}}\big{)},\quad t=1,2,\dots,k.$$

$$\sum _{t=1}^k{\lambda }_{i_t}(AB)\ge \underset{\begin{array}{c}{u}_t\in S_t\\ (u_r,u_s)={\delta }_{rs}\end{array}}{\min }\sum _{t=1}^k{u}_t^{\ast }{B}^{\frac{1}{2}}A{B}^{\frac{1}{2}}{u}_t=\underset{\begin{array}{c}{u}_t\in S_t\\ (u_r,u_s)={\delta }_{rs}\end{array}}{\min }\sum _{t=1}^k\frac{u_t^{\ast }{B}^{\frac{1}{2}}A{B}^{\frac{1}{2}}{u}_t}{(B^{\frac{1}{2}}{u}_t{)}^{\ast }(B^{\frac{1}{2}}{u}_t)}\cdot u_t^{\ast }B{u}_t.$$

$$\frac{u_t^{\ast }{B}^{\frac{1}{2}}A{B}^{\frac{1}{2}}{u}_t}{(B^{\frac{1}{2}}{u}_t{)}^{\ast }(B^{\frac{1}{2}}{u}_t)}=\frac{1}{\sum _{j=1}^{i_t}|a_j{|}^2}\sum _{j=1}^{i_t}|a_j{|}^2{\lambda }_j\ge {\lambda }_{i_t}(A),t=1,2,\dots ,k.$$

$$\sum _{t=1}^k{\lambda }_{i_t}(AB)$$

$$\sum _{t=1}^k{\lambda }_{i_t}(AB)=\sum _{t=1}^{k_A}{\lambda }_{i_t}(AB)+\sum _{t=k_A+1}^k{\lambda }_{i_t}(AB).$$

$$\sum _{t=1}^{k_A}{\lambda }_{i_t}(A){\lambda }_{n-t+1}(B)\le \sum _{t=1}^{k_A}{\lambda }_{i_t}(AB)\le \sum _{t=1}^{k_A}{\lambda }_{i_t}(A){\lambda }_t(B).$$

$$\sum _{t=k_A+1}^k{\lambda }_{i_t}(AB)$$

$$\displaystyle A=\left({\begin{array}[]{*{20}c}1&2&0\\ 2&1&0\\ 0&0&-4\end{array}}\right),\quad B=\left({\begin{array}[]{*{20}c}2&-1&0\\ -1&2&0\\ 0&0&2\end{array}}\right).$$

$$\begin{array}[]{lll}\lambda_{1}(A)=3,&\lambda_{2}(A)=-1,&\lambda_{3}(A)=-4;\\ \lambda_{1}(B)=3,&\lambda_{2}(B)=2,&\lambda_{3}(B)=1;\\ \lambda_{1}(AB)=3,&\lambda_{2}(AB)=-3,&\lambda_{3}(AB)=-8.\end{array}$$

$$\sum _{t=1}^{k_A}{\lambda }_{i_t}(A){\lambda }_t(B)+\sum _{t=k_A+1}^k{\lambda }_{i_t}(A){\lambda }_{n-k+t}(B)$$

$$\sum _{t=1}^k{\lambda }_{i_t}(AB)=\{\begin{array}{ll}{\lambda }_1(AB)+{\lambda }_2(AB)=0,& \text{mbox}(I)\\ {\lambda }_1(AB)+{\lambda }_3(AB)=-5,& \text{mbox}(II)\\ {\lambda }_2(AB)+{\lambda }_3(AB)=-11,& \text{mbox}(III)\end{array}$$

$$\sum _{t=1}^{k_A}{\lambda }_{i_t}(A){\lambda }_{n-t+1}(B)+\sum _{t=k_A+1}^k{\lambda }_{i_t}(A){\lambda }_{k-t+1}(B)$$

$$\sum _{t=1}^n{\lambda }_t(A){\lambda }_{n-t+1}(B)\le \mathrm{tr}(AB)=\sum _{t=1}^n{\lambda }_t(AB)\le \sum _{t=1}^n{\lambda }_t(A){\lambda }_t(B).$$

$$\sum _{t=1}^k{\lambda }_{i_t}(A){\lambda }_{k-t+1}(B)\le \sum _{t=1}^k{\lambda }_{i_t}(AB)\le \sum _{t=1}^k{\lambda }_{i_t}(A){\lambda }_{n-k+t}(B).$$

$${\lambda }_s(A){\lambda }_n(B)+{\lambda }_t(A){\lambda }_1(B)\le {\lambda }_s(AB)+{\lambda }_t(AB)\le {\lambda }_s(A){\lambda }_1(B)+{\lambda }_t(A){\lambda }_n(B).$$

$${\lambda }_p(AB)-{\lambda }_q(AB)\le \left[{\lambda }_p(A)-{\lambda }_q(A)\right]{\lambda }_1(B).$$

Full text

The first author was supported in part by the National Natural Science Foundation of China Grant No. 11361038. The second author was supported in part by NSU Research Scholar grant.

1. Introduction

Let $A$ be an $n\times n$ Hermitian matrix with eigenvalues $\lambda_{1}(A)\geq\lambda_{2}(A)\geq\cdots\geq\lambda_{n}(A)$. If some, say $k$, of the eigenvalues of $A$ are selected, they may be indexed by a sequence $1\leq i_{1}<i_{2}<\dots<i_{k}\leq n$. Hence, $\lambda_{i_{1}}(A)\geq\lambda_{i_{2}}(A)\geq\cdots\geq\lambda_{i_{k}}(A)$.

A classical result of Wielandt [15] states that if $A$ and $B$ are $n\times n$ Hermitian matrices and $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n$, then

[TABLE]

A reversed inequality follows from (1.1) by replacing $A$ and $B$ with $-A$ and $-B$, respectively:

[TABLE]

There is a great amount of research on the partial sums of selected eigenvalues (see, e.g., [10, 12, 13] or [11, Chap. 9] and [2, Chap. III]) as well as on the characterization of the eigenvalues of the sum of Hermitian matrices (see, e.g., [3, 5, 7, 8]).

Inequalities analogous to (1.1) and (1.2) for the product of two matrices are presented in [14]: If $A$ and $B$ are $n\times n$ positive semidefinite matrices, then

[TABLE]

The eigenvalues of the product of two Hermitian matrices need not be real. For example, for $A=\left({0\atop 1}{1\atop 0}\right)$ and $B=\left({1\atop 1}{1\atop-1}\right),$ the eigenvalues of $AB$ are $1\pm i$. Thus, inequalities (1.3) do not extend to partial sums of eigenvalues of the product of two Hermitian matrices. However, requiring one matrix to be positive semidefinite (PSD) ensures that the eigenvalues of the product are all real; that is, if $A$ is Hermitian and $B$ is PSD, then $AB$ and $B^{1/2}AB^{1/2}$ have the same eigenvalues, so $AB$ has only real eigenvalues.

Eigenvalue problem is of central importance in matrix analysis and related areas. Usually, inequalities for selected eigenvalues involve two Hermitian matrices for sum and two positive semidefinite matrices for product. Nevertheless, the results on the partial sums of selected eigenvalues of the product of one PSD matrix and one Hermitian matrix are fragmentary. A result of this kind is, for example, a celebrated theorem of Ostrowski (see, e.g., [6, p. 283]) which states that, for Hermitian $A$ and positive definite $B$, $\lambda_{t}(AB)=\theta_{t}\lambda_{t}(A)$, where $\theta_{t}\in[\lambda_{n}(B),\lambda_{1}(B)]$.

The purpose of this paper is to present inequalities on the partial sums of selected eigenvalues of the product of a PSD matrix and a Hermitian matrix. Our results generalize some existing ones such as inequalities (1.3).

2. Eigenvalue inequalities for Hermitian and PSD matrices

In [17, Theorem 3], inequalities concerning $\sum_{t=1}^{k}\lambda_{t}(AB)$ are shown, where $A$ is Hermitian and $B$ is positive semidefinite. These inequalities are about the sum of the $k$ largest eigenvalues of $AB$. In this section, we show inequalities for selected eigenvalues; that is, we present inequalities concerning $\sum_{t=1}^{k}\lambda_{i_{t}}(AB)$.

We borrow Wielandt’s min-max representation (see, e.g., [2, p. 67]) for the eigenvalues of Hermitian matrices, which is used in the proof of our main result.

Theorem 2.1** (Wielandt [15]).**

If $A\in\mathbb{C}^{n\times n}$ (the set of $n\times n$ complex matrices) is Hermitian and $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n$, then

[TABLE]

where $\delta_{rs}$ is the Kronecker delta and $x^{*}$ is the conjugate transpose of $x\in\mathbb{C}^{n}$.

Denote the inertia of an $n\times n$ Hermitian $A$ by $(\pi_{+},\nu_{-},\delta_{0})$, where $\pi_{+},\nu_{-},\delta_{0}$ are the numbers of positive, negative and zero eigenvalues of $A$, respectively (see, e.g., [16, p. 255]). Let $n_{A}$ be the number of nonnegative eigenvalues of $A$, namely, $n_{A}=\pi_{+}+\delta_{0}$. For any Hermitian matrix $A\in\mathbb{C}^{n\times n}$, we have

[TABLE]

Let $k_{A}$ be the number of nonnegative eigenvalues in $\{\lambda_{i_{1}}(A),\lambda_{i_{2}}(A),\dots,\lambda_{i_{k}}(A)\}$ for the given index sequence $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n$. So, if $k_{A}=0$, then all $\lambda_{i_{1}}(A),\lambda_{i_{2}}(A),\dots,\lambda_{i_{k}}(A)$ are negative; if $k_{A}=k$, then each of $\lambda_{i_{1}}(A),\lambda_{i_{2}}(A),$ $\dots,\lambda_{i_{k}}(A)$ is positive or zero. It is always true that $k_{A}\leq n_{A}$ and $\lambda_{k_{A}}\geq\lambda_{n_{A}}$.

Now we are ready to present our main theorem. In what follows, our convention is that a summation in the form $\sum_{t=p}^{q}$ over $t$ vanishes if $p>q$.

Theorem 2.2**.**

Let $A\in\mathbb{C}^{n\times n}$ be Hermitian, let $B\in\mathbb{C}^{n\times n}$ be positive semidefinite, and let $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n$. Then

[TABLE]

and

[TABLE]

Proof.

Note that if $k_{A}=0$, then the first terms (summations) on the right-hand sides of (2.2) and (2.3) are absent. If $k_{A}=k$, then the second terms disappear. For $1\leq k_{A}<k$, the first terms on the right-hand sides of (2.2) and (2.3) are nonnegative, while the second terms are nonpositive.

We divide the proof of the theorem into five cases.

Case (1). If $n_{A}=n$, then $A$ is PSD and $k_{A}=k$. (2.2) and (2.3) are (1.3).

Case (2). If $n_{A}=0$, then $k_{A}=0$ and $-A$ is positive definite. We set $j_{t}$ as follows:

[TABLE]

By inequalities (1.3), we have

[TABLE]

Because $\lambda_{j_{t}}(-A)=-\lambda_{n-j_{t}+1}(A)$ and $j_{t}=n-i_{k-t+1}+1$, we obtain

[TABLE]

Note that on the right hand side, $\lambda_{i_{k-t+1}}(A)$ is paired with (multiplied by) $\lambda_{n-t+1}(B)$, $t=1,2,\dots,k$, namely, $\lambda_{i_{r}}(A)$ is paired with $\lambda_{n-k+r}(B)$, $r=1,2,\dots,k$. Likewise, on the left, $\lambda_{i_{r}}(A)$ is paired with $\lambda_{k-r+1}(B)$, $r=1,2,\dots,k$. It follows that

[TABLE]

Inequalities (2.2) and (2.3) follow immediately.

If $A$ has no positive eigenvalues, then $-A$ is PSD which can be dealt with as above. We assume below that $A$ has both positive and negative eigenvalues.

Case (3). Let $1\leq\pi_{+}$, $1\leq\nu_{-}$, and $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n_{A}$. Then all $\lambda_{i_{1}}(A),\lambda_{i_{2}}(A),$ $\dots,\lambda_{i_{k}}(A)$ are nonnegative, namely, $k_{A}=k$. We show that

[TABLE]

We may assume that $A$ is a diagonal matrix as the inequalities are invariant under unitary similarity. For the upper bound, using the approach of splitting (see [9, p. 381] or [6, p. 250]), we write $A=A_{+}+A_{-}$, where $A_{+}$ is the diagonal matrix with all the positive (if any) eigenvalues of $A$ on the main diagonal (plus some zeros), and $A_{-}$ is the diagonal matrix with all the negative (if any) eigenvalues of $A$ on the main diagonal (plus some zeros). Then $A_{+}$ is positive semidefinite and $\lambda_{i_{1}}(A),\lambda_{i_{2}}(A),$ $\dots,\lambda_{i_{k}}(A)$ are all contained on the main diagonal of $A_{+}$ as $i_{k}\leq n_{A}$.

Since $A\leq A_{+}$ (i.e., $A_{+}-A$ is PSD), we have $\lambda_{t}(AB)\leq\lambda_{t}(A_{+}B)$ for every $t$, and moreover, $\lambda_{i_{t}}(A_{+})=\lambda_{i_{t}}(A)$ for $t\leq k$. For the upper bound, by (1.3), we get

[TABLE]

For the lower bound, we need to use the Wielandt’s min-max representation.

Bear in mind that $\lambda_{t}(A)\geq 0$, $t=1,2,\dots,n_{A}$, and $\lambda_{i_{t}}(A)\geq 0$, $t=1,2,\dots,k$. Let $A={\rm{diag}}(\lambda_{1}(A),\lambda_{2}(A),\dots,\lambda_{n}(A))$. The standard column unit vectors $e_{i}=(0,\dots,0,1,0\cdots,0)^{T}$ are eigenvectors corresponding to $\lambda_{i}(A)$, $i=1,2,\dots,n$.

We assume $B$ is nonsingular (or use continuity with $B_{\varepsilon}=B+\varepsilon I$, $\varepsilon>0$). Let

[TABLE]

Then $S_{1}\subset S_{2}\subset\cdots\subset S_{k}$ and $\dim(S_{t})=i_{t}$. The min-max representation reveals

[TABLE]

For ${u}_{t}\in S_{t}$, let $u_{t}=\sum\limits_{j=1}^{i_{t}}a_{j}B^{-\frac{1}{2}}{e}_{j}$, $a_{1},a_{2},\dots,a_{i_{t}}\in\mathbb{C}$, $t=1,2,\dots,k$. Then we have

[TABLE]

Let $C=\sum_{t=1}^{k}\lambda_{i_{t}}(A)u_{t}u_{t}^{*}\in\mathbb{C}^{n\times n}$, where $\{u_{1},u_{2},\dots,u_{k}\}$ is an orthonormal set in $\mathbb{C}^{n}$. Then $C$ is PSD and $\lambda_{t}(C)=\lambda_{i_{t}}(A)$ for each $t$. We obtain

[TABLE]

We note here that it would be nice if we could derive the lower bound from the upper bound with $A$ replaced by $-A$, without using the Wielandt’s min-max representation. However, this approach doesn’t work as $\lambda_{i_{t}}(-A)\leq 0$. Moreover, like the lower bound, the upper bound can also be obtained by using the min-max representation with suitably selected subspaces.

Case (4). Let $1\leq\pi_{+}$, $1\leq\nu_{-}$, and $n_{A}<i_{1}<i_{2}<\cdots<i_{k}\leq n$. Then none of $\lambda_{i_{1}}(A),\lambda_{i_{2}}(A),\dots,\lambda_{i_{k}}(A)$ is positive. Consider $-A$ with $1\leq j_{1}=n-i_{k}+1<\cdots<j_{k}=n-i_{1}+1\leq n$. Since $j_{k}=n-i_{1}+1\leq n-n_{A}\leq n_{-A}$ (see (2.1)), we apply case (3) to $-A$ and $B$ to get the desired inequalities (2.5) as in case (2).

Case (5). Let $1\leq\pi_{+}$, $1\leq\nu_{-}$, and $1\leq i_{1}<\cdots<i_{k_{A}}\leq n_{A}<i_{k_{A}+1}<\cdots<i_{k}\leq n$. Then

[TABLE]

For $1\leq i_{1}<\cdots<i_{k_{A}}\leq n_{A}$, using (2.6), we have

[TABLE]

For $n_{A}<i_{k_{A}+1}<\cdots<i_{k}\leq n$, set $j_{t}=i_{k_{A}+t}$, $t=1,2,\dots,k^{\prime}$, where $k^{\prime}=k-k_{A}$. Then $n_{A}<j_{1}<\cdots<j_{k^{\prime}}\leq n$. By case (4), we have (for the upper bound)

[TABLE]

and (for the lower bound)

[TABLE]

Combing inequalities (2.7), (2.8) and (2.9) results in (2.2) and (2.3). ∎

Remark 2.3*.*

Li and Mathias [9, Theorem 2.3] showed some inequalities in partial products of the form $\prod_{t=1}^{k}\frac{\lambda_{i_{t}}(AB)}{\lambda_{i_{t}}(A)}$, where $A$ is Hermitian and $B=$ is PSD.

It is known that for positive numbers, inequalities of partial products imply those of partial sums. Simply put in the language of majorization, log-majorization implies weak-majorization (see, e.g., [4, p. 232] or [16, p. 345]). Thus, it is tempting to obtain the results on partial sums from Li and Mathias’ result on partial products. However, this is only possible for $|{\lambda_{i_{t}}(AB)}|$ and $|{\lambda_{i_{t}}(A)}|$ by observing that $\frac{\lambda_{i_{t}}(AB)}{\lambda_{i_{t}}(A)}>0$. Since ${\lambda_{i_{t}}(AB)}$ and ${\lambda_{i_{t}}(A)}$ are paired with the same sign (maybe both negative), we don’t see how the absolute values are dropped so that ${\lambda_{i_{t}}(AB)}$ and ${\lambda_{i_{t}}(A)}$ appear in a partial sum without being paired as a quotient.

Remark 2.4*.*

Hoffman’s min-max representation (see, e.g., [1, 2.17]) in the product form $\prod_{t=1}^{k}\lambda_{i_{t}}(\cdot)$ for positive semidefinite matrices does not generalize to Hermitian matrices as Li and Mathias showed by example [9, pp. 411-412]; that is, there is no multiplicative analog of Wielandt’s min-max representation (Theorem 2.1) for Hermitian matrices. Therefore, it is impossible to derive sum inequalities $\sum_{t=1}^{k}\lambda_{i_{t}}(\cdot)$ from product inequalities $\prod_{t=1}^{k}\lambda_{i_{t}}(\cdot)$ for Hermitian matrices through majorization. In view of this, our Theorem 2.2 appears to be important.

Example 2.5*.*

Let $n=3$, $k=2$, and let

[TABLE]

Then

[TABLE]

Thus, $n_{A}=1$. We consider the cases (I) $i_{1}=1,$ $i_{2}=2,$ $k_{A}=1$; (II) $i_{1}=1,$ $i_{2}=3,$ $k_{A}=1$; and (III) $i_{1}=2,$ $i_{2}=3,$ $k_{A}=0$, to get, respectively,

[TABLE]

[TABLE]

and

[TABLE]

Recall the trace inequality (see, e.g., [2, p. 78] or [6, p. 255]) for Hermitian $A,B$:

[TABLE]

As we noted in Section 1, the product of two Hermitian matrices may have nonreal eigenvalues. So, the trace in (2.10) cannot be replaced by partial sums in general. Theorem 2.2 presents partial sum inequalities for the product of one Hermitian matrix and one PSD matrix.

We present a few results that are immediate from Theorem 2.2.

Corollary 2.6*.*

Let $A\in\mathbb{C}^{n\times n}$ be a stable Hermitian matrix (i.e., the eigenvalues of $A$ are located on the left half-plane) and $B\in\mathbb{C}^{n\times n}$ be positive semidefinite. Then

[TABLE]

Proof.

Note that $k_{A}=0$ or $\lambda_{i_{t}}(A)=0$ for $t=1,2,\dots,k_{A}$ in (2.2) and (2.3). ∎

Corollary 2.7*.*

Let $A\in\mathbb{C}^{n\times n}$ be Hermitian and $B\in\mathbb{C}^{n\times n}$ be positive semidefinite. Then for any positive eigenvalue $\lambda_{s}(AB)$ and negative eigenvalue $\lambda_{t}(AB)$ (if any)

[TABLE]

Proof.

Take $k=2$ and $1\leq i_{1}=s<i_{2}=t\leq n$. Then $\lambda_{s}(A)>0>\lambda_{t}(A)$. In (2.2), there is one term in each of the two summations; the same is true for (2.3). ∎

Corollary 2.8*.*

Let $A\in\mathbb{C}^{n\times n}$ be Hermitian and $B\in\mathbb{C}^{n\times n}$ be positive semidefinite. If $\lambda_{p}(AB)$ is the smallest positive eigenvalue of $AB$ and $\lambda_{q}(AB)$ is the largest negative eigenvalue of $AB$ (if any), then their distance (gap near 0) is bounded as

[TABLE]

Proof.

Since $\lambda_{p}(AB)>0$ and $\lambda_{q}(AB)<0$, we have $\lambda_{p}(A)>0>\lambda_{q}(A)$. Setting $k=1$ and $i_{1}=t$, for each $t=1,2,\dots,n$ in Theorem 2.2, we obtain

[TABLE]

and

[TABLE]

In particular,

[TABLE]

The desired inequality follows immediately by subtraction. ∎

Corollary 2.8 provides an estimate for the gap between two eigenvalues of $AB$ near zero (on both sides) in terms of the eigenvalues of $A$ and $B$. The ratio $[\lambda_{p}(AB)-\lambda_{q}(AB)]/[\lambda_{p}(A)-\lambda_{q}(A)]$ is bounded above by the spectral norm of $B$. Thus, if neither $A$ nor $-A$ is stable Hermitian, then an application (multiplication) of a strictly contractive PSD matrix $B$ to $A$ narrows the gap between the positive and negative eigenvalues of $A$. (A matrix is said to be strictly contractive if its spectral norm is less than one.)

3. Comparison of the bounds

Recall from the proof of Theorem 2.2 the splitting of the real diagonal matrix $A$: $A=A_{+}+A_{-}$, where $A_{+}$ is the diagonal matrix with all positive (if any) eigenvalues of $A$ on the main diagonal (plus some zeros), and $A_{-}$ is the diagonal matrix with all negative (if any) eigenvalues of $A$ on the main diagonal (plus some zeros). It is natural to ask what inequalities would be derived if one applies (1.1), (1.2), and (1.3) to $AB=A_{+}B+A_{-}B$. We present the inequalities obtained by the splitting approach in the following proposition; we then show that these inequalities are in general weaker than the ones in Theorem 2.2. We show the case for upper bound.

Proposition 3.1*.*

Let $A\in\mathbb{C}^{n\times n}$ be Hermitian and let $B\in\mathbb{C}^{n\times n}$ be positive semidefinite. For $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n$, we have

[TABLE]

Proof.

Note that $A_{+}$ is positive semidefinite and all $\lambda_{i_{1}}(A),\lambda_{i_{2}}(A),$ $\dots,\lambda_{i_{k}}(A)$ are contained on the main diagonal of $A_{+}$. So, $\lambda_{i_{t}}(A_{+})=\lambda_{i_{t}}(A)$ for $t=1,2,\dots,k_{A}$, and $\lambda_{i_{t}}(A_{+})=0$ for $t>k_{A}$. Observe $\lambda_{t}(A_{-})=0$ for $t\leq n_{A}$. It follows that

[TABLE]

In comparison, the first term on the right hand side of (3.1) is the same as that of (2.2), while the second terms are different. We denote the second term in (2.2) by $T_{1}$ and the second term in (3.1) by $T_{2}$. We show that $T_{1}\leq T_{2}$ as follows.

[TABLE]

Acknowledgement. The second author appreciates discussions with Roger Horn, Chi-Kwong Li, and Ren-cang Li during the preparation of the manuscript.