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Some generalizations of inequalities for sector matrices

Journal of Inequalities and Applications20182018:183

https://doi.org/10.1186/s13660-018-1786-8

  • Received: 27 April 2018
  • Accepted: 18 July 2018
  • Published:

Abstract

In this paper, we generalize some Schatten p-norm inequalities for accretive-dissipative matrices obtained by Kittaneh and Sakkijha. Moreover, we present some inequalities for sector matrices.

Keywords

  • Sector matrix
  • Schatten p-norm
  • Accretive-dissipative matrix
  • Inequality

MSC

  • 15A45
  • 15A15

1 Introduction

Throughout this paper, let \(\mathbb{M}_{n}\) be the set of all \(n\times n\) complex matrices. We denote by \(I_{n}\) the identity matrix in \(\mathbb{M}_{n}\). For two Hermitian matrices \(A, B\in\mathbb{M}_{n}\), we use \(A\ge B\) (\(B\le A\)) to mean that \(A-B\) is a positive semidefinite matrix. A matrix \(A\in\mathbb{M}_{n}\) is called accretive-dissipative if in its Cartesian (or Toeptliz) decomposition, \(A=\Re (A)+i \Im (A)\), the matrices \(\Re (A)\) and \(\Im (A)\) are positive semidefinite, where \(\Re(A)=\frac{A+A^{*}}{2}\), \(\Im(A)=\frac{A-A^{*}}{2i}\).

Let \(|\!|\!|\cdot |\!|\!|\) denote any unitarily invariant norm on \(\mathbb{M}_{n}\). Note that tr is the usual trace functional. For \(p>0\) and \(A\in\mathbb{M}_{n}\), let \(\Vert A \Vert _{p}=(\sum_{j=1}^{n} s_{j}^{p}(A))^{\frac{1}{p}}\), where \(s_{1}(A)\ge s_{2}(A)\ge\cdots\ge s_{n}(A)\) are the singular values of A. Thus, \(\Vert A \Vert _{p}=(\operatorname{tr}|A|^{p})^{\frac{1}{p}}\). For \(p\ge1\), this is the Schatten p-norm of A. For more information about the Schatten p-norms, see [1, p. 92].

A real-valued continuous function f on an interval I is called matrix concave of order n if \(f(\alpha A+(1-\alpha)B)\ge \alpha f(A)+(1-\alpha)f(B)\) for any two Hermitian matrices \(A, B\in\mathbb{M}_{n}\) with spectrum in I and all \(\alpha\in[0,1]\). Furthermore, f is called operator concave if f is matrix concave for all n.

The numerical range of \(A\in\mathbb{M}_{n}\) is defined by
$$ W(A)= \bigl\{ x^{*}Ax : x\in\mathbb{C}^{n}, x^{*}x=1 \bigr\} . $$
For \(\alpha\in [0,\frac{\pi}{2})\), \(S_{\alpha}\) denotes the sector in the complex plane as follows:
$$ S_{\alpha}= \bigl\{ z\in\mathbb{C} : \Re z\ge0, \vert \Im z \vert \le (\Re z)\tan\alpha \bigr\} . $$

Clearly, A is positive semidefinite if and only if \(W(A)\subset S_{0}\), and if \(W(A), W(B)\subset S_{\alpha}\) for some \(\alpha\in[0,\frac{\pi}{2})\), then \(W(A+B)\subset S_{\alpha}\). As \(0\notin S_{\alpha}\), if \(W(A)\subset S_{\alpha}\), then A is nonsingular.

In [7], Kittaneh and Sakkijha gave the following Schatten-p norm inequalities involving sums of accretive-dissipative matrices.

Theorem 1.1

Let \(S, T\in\mathbb{M}_{n}\) be accretive-dissipative. Then
$$ 2^{\frac{-p}{2}}\bigl( \Vert S \Vert ^{p}_{p}+ \Vert T \Vert ^{p}_{p}\bigr)\le \Vert S+T \Vert ^{p}_{p}\le2^{\frac{3}{2}p-1}\bigl( \Vert S \Vert ^{p}_{p}+ \Vert T \Vert ^{p}_{p} \bigr) \quad \textit{for }p\ge1 . $$
In [5], Garg and Aujla showed the following inequalities:
$$ \prod_{j=1}^{k} s_{j}\bigl( \vert A+B \vert ^{r}\bigr)\le\prod_{j=1}^{k} s_{j}\bigl(I_{n}+ \vert A \vert ^{r}\bigr) \prod_{j=1}^{k} s_{j} \bigl(I_{n}+ \vert B \vert ^{r}\bigr)\quad \text{for }1\le k\le n, 1\le r\le2; $$
(1)
and
$$ \prod_{j=1}^{k} s_{j} \bigl(I_{n}+f\bigl( \vert A+B \vert \bigr)\bigr)\le\prod _{j=1}^{k} s_{j}\bigl(I_{n}+f \bigl( \vert A \vert \bigr)\bigr)\prod_{j=1}^{k} s_{j}\bigl(I_{n}+f\bigl( \vert B \vert \bigr)\bigr)\quad \text{for }1\le k\le n, $$
(2)
where \(A, B\in\mathbb{M}_{n}\) and \(f : [0,\infty)\rightarrow [0,\infty)\) is an operator concave function.
By letting \(A, B\ge0 \), \(r=1\) and \(f(X)=X\) for any \(X\in\mathbb{M}_{n}\) in (1) and (2), we have
$$ \prod_{j=1}^{k} s_{j}(A+B)\le \prod_{j=1}^{k} s_{j}(I_{n}+A) \prod_{j=1}^{k} s_{j}(I_{n}+B) \quad \text{for }1\le k\le n; $$
(3)
and
$$ \prod_{j=1}^{k} s_{j}(I_{n}+A+B) \le\prod_{j=1}^{k} s_{j}(I_{n}+A) \prod_{j=1}^{k} s_{j}(I_{n}+B) \quad \text{for }1\le k\le n. $$
(4)

In this paper, we give a generalization of Theorem 1.1. Moreover, we present some inequalities for sector matrices based on (3) and (4) which remove the absolute values in (1) and (2) from the right-hand side.

2 Main results

Before we give the main results, let us present the following lemmas that will be useful later.

Lemma 2.1

([2, 11])

Let \(A_{1}, \ldots, A_{n}\in\mathbb{M}_{n}\) be positive semidefinite. Then
$$ \sum_{j=1}^{n} \Vert A_{j} \Vert ^{p}_{p}\le \Biggl\Vert \sum _{j=1}^{n} A_{j} \Biggr\Vert ^{p}_{p}\le n^{p-1}\sum_{j=1}^{n} \Vert A_{j} \Vert ^{p}_{p} \quad \textit{for }p \ge1. $$

Lemma 2.2

([3])

Let \(A, B\in\mathbb{M}_{n}\) be positive semidefinite. Then
$$ \Vert A+iB \Vert _{p}\le \Vert A+B \Vert _{p}\le \sqrt{2} \Vert A+iB \Vert _{p} \quad \textit{for }p\ge1. $$

Our first main result is a generalization of Theorem 1.1.

Theorem 2.3

Let \(A_{1}, \ldots, A_{n}\in\mathbb{M}_{n}\) be accretive-dissipative. Then
$$ 2^{\frac{-p}{2}}\sum_{j=1}^{n} \Vert A_{j} \Vert ^{p}_{p}\le \Biggl\Vert \sum _{j=1}^{n} A_{j} \Biggr\Vert ^{p}_{p}\le\frac{(2n^{2})^{\frac{p}{2}}}{n}\sum _{j=1}^{n} \Vert A_{j} \Vert ^{p}_{p} \quad \textit{for }p\ge1. $$

Proof

Let \(A_{j}=B_{j}+iC_{j}\) be the Cartesian decompositions of \(A_{j}\), \(j=1, \ldots, n\). Then we have
$$\begin{aligned} \Biggl\Vert \sum_{j=1}^{n} A_{j} \Biggr\Vert ^{p}_{p}&= \Biggl\Vert \sum _{j=1}^{n} (B_{j}+iC_{j}) \Biggr\Vert ^{p}_{p} \\ &= \Biggl\Vert \sum_{j=1}^{n} {B_{j}}+i\sum_{j=1}^{n}{C_{j}} \Biggr\Vert ^{p}_{p} \\ &\ge2^{\frac{-p}{2}} \Biggl\Vert \sum_{j=1}^{n} {B_{j}}+\sum_{j=1}^{n}{C_{j}} \Biggr\Vert ^{p}_{p}\quad\text{(by Lemma 2.2)} \\ &=2^{\frac{-p}{2}} \Biggl\Vert \sum_{j=1}^{n} (B_{j}+C_{j}) \Biggr\Vert ^{p}_{p} \\ &\ge2^{\frac{-p}{2}}\sum_{j=1}^{n} \Vert B_{j}+C_{j} \Vert ^{p}_{p} \quad \text{(by Lemma 2.1)} \\ &\ge2^{\frac{-p}{2}}\sum_{j=1}^{n} \Vert B_{j}+iC_{j} \Vert ^{p}_{p} \quad \text{(by Lemma 2.2)} \\ &=2^{\frac{-p}{2}}\sum_{j=1}^{n} \Vert A_{j} \Vert ^{p}_{p}, \end{aligned}$$
which proves the first inequality.
To prove the second inequality, compute
$$\begin{aligned} \Biggl\Vert \sum_{j=1}^{n} A_{j} \Biggr\Vert ^{p}_{p}&= \Biggl\Vert \sum _{j=1}^{n} (B_{j}+iC_{j}) \Biggr\Vert ^{p}_{p} \\ &= \Biggl\Vert \sum_{j=1}^{n} {B_{j}}+i\sum_{j=1}^{n}{C_{j}} \Biggr\Vert ^{p}_{p} \\ &\le \Biggl\Vert \sum_{j=1}^{n} {B_{j}}+\sum_{j=1}^{n}{C_{j}} \Biggr\Vert ^{p}_{p}\quad\text{(by Lemma 2.2)} \\ &= \Biggl\Vert \sum_{j=1}^{n} (B_{j}+C_{j}) \Biggr\Vert ^{p}_{p} \\ &\le n^{p-1}\sum_{j=1}^{n} \Vert B_{j}+C_{j} \Vert ^{p}_{p} \quad \text{(by Lemma 2.1)} \\ &\le n^{p-1}2^{\frac{p}{2}}\sum_{j=1}^{n} \Vert B_{j}+iC_{j} \Vert ^{p}_{p} \quad\text{(by Lemma 2.2)} \\ &=\frac{(2n^{2})^{\frac{p}{2}}}{n}\sum_{j=1}^{n} \Vert A_{j} \Vert ^{p}_{p}, \end{aligned}$$
which completes the proof. □

Remark 2.4

By letting \(n=2\) in Theorem 2.3, we thus get Theorem 1.1.

The following lemma is the well-known Fan–Hoffman inequality.

Lemma 2.5

([12, p. 63])

Let \(A\in\mathbb{M}_{n}\). Then
$$ \lambda_{j}(\Re A)\le s_{j}(A), $$
where \(\lambda_{j}(\cdot)\) denotes the jth largest eigenvalue.

In [4], Drury and Lin presented a reverse version of Lemma 2.5 as follows.

Lemma 2.6

Let \(A\in\mathbb{M}_{n}\) be such that \(W(A)\subset S_{\alpha}\). Then
$$ s_{j}(A)\le \sec^{2}(\alpha)\lambda_{j}(\Re A), $$
where \(\lambda_{j}(\cdot)\) denotes the jth largest eigenvalue.

Theorem 2.7

Let \(A, B\in\mathbb{M}_{n}\) be such that \(W(A), W(B)\subset S_{\alpha}\). Then
$$ \prod_{j=1}^{k} s_{j}(A+B)\le \sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}(I_{n}+A)\prod_{j=1}^{k} s_{j}(I_{n}+B)\quad \textit{for }1\le k\le n; $$
(5)
and
$$ \prod_{j=1}^{k} s_{j}(I_{n}+A+B) \le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}(I_{n}+A)\prod_{j=1}^{k} s_{j}(I_{n}+B)\quad \textit{for }1\le k\le n. $$
(6)

Proof

We have
$$\begin{aligned} \prod_{j=1}^{k} s_{j}(A+B)&\le \sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(A+B)\bigr)\quad\text{(by Lemma 2.6)} \\ &=\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(A)+\Re(B)\bigr) \\ &\le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(I_{n}+\Re(A)\bigr)\prod _{j=1}^{k} s_{j}\bigl(I_{n}+ \Re(B)\bigr)\quad\bigl(\text{by (3)}\bigr) \\ &=\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(I_{n}+A)\bigr)\prod _{j=1}^{k} s_{j}\bigl(\Re(I_{n}+B) \bigr) \\ &\le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}(I_{n}+A)\prod_{j=1}^{k} s_{j}(I_{n}+B)\quad\text{(by Lemma 2.5)} \end{aligned}$$
which proves (5).
To prove (6), compute
$$\begin{aligned} \prod_{j=1}^{k} s_{j}(I_{n}+A+B)& \le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(I_{n}+A+B)\bigr)\quad\text{(by Lemma 2.6)} \\ &=\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(I_{n}+\Re(A)+\Re(B)\bigr) \\ &\le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(I_{n}+\Re(A)\bigr)\prod _{j=1}^{k} s_{j}\bigl(I_{n}+ \Re(B)\bigr)\quad\bigl(\text{by (4)}\bigr) \\ &=\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(I_{n}+A)\bigr)\prod _{j=1}^{k} s_{j}\bigl(\Re(I_{n}+B) \bigr) \\ &\le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}(I_{n}+A)\prod_{j=1}^{k} s_{j}(I_{n}+B),\quad\text{(by Lemma 2.5)} \end{aligned}$$
which completes the proof. □

Corollary 2.8

Let \(A, B\in\mathbb{M}_{n}\) be such that \(W(A), W(B)\subset S_{\alpha}\). Then, for all unitarily invariant norms \(|\!|\!|\cdot |\!|\!|\) on \(\mathbb{M}_{n}\),
$$ |\!|\!|A+B |\!|\!|\le\sec^{2}(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|; $$
and
$$ |\!|\!|I_{n}+A+B |\!|\!|\le\sec^{2}(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|. $$

Proof

From (5) and (6), we obtain
$$ \prod_{j=1}^{k} s_{j}(A+B)\le \prod_{j=1}^{k} s_{j}\bigl(\sec( \alpha) (I_{n}+A)\bigr)s_{j}\bigl(\sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n; $$
and
$$ \prod_{j=1}^{k} s_{j}(I_{n}+A+B) \le\prod_{j=1}^{k} s_{j}\bigl( \sec(\alpha) (I_{n}+A)\bigr)s_{j}\bigl(\sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n, $$
which is equivalent to the following inequalities:
$$ \prod_{j=1}^{k} s_{j}^{\frac{1}{2}}(A+B) \le\prod_{j=1}^{k} s_{j}^{\frac{1}{2}} \bigl(\sec(\alpha) (I_{n}+A)\bigr)s_{j}^{\frac{1}{2}}\bigl( \sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n; $$
and
$$ \prod_{j=1}^{k} s_{j}^{\frac{1}{2}}(I_{n}+A+B) \le\prod_{j=1}^{k} s_{j}^{\frac{1}{2}} \bigl(\sec(\alpha) (I_{n}+A)\bigr)s_{j}^{\frac{1}{2}}\bigl( \sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n. $$
By the property of majorization [1, p. 42], we have
$$ \sum_{j=1}^{k} s_{j}^{\frac{1}{2}}(A+B) \le\sum_{j=1}^{k} s_{j}^{\frac{1}{2}} \bigl(\sec(\alpha) (I_{n}+A)\bigr)s_{j}^{\frac{1}{2}}\bigl( \sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n; $$
and
$$ \sum_{j=1}^{k} s_{j}^{\frac{1}{2}}(I_{n}+A+B) \le\sum_{j=1}^{k} s_{j}^{\frac{1}{2}} \bigl(\sec(\alpha) (I_{n}+A)\bigr)s_{j}^{\frac{1}{2}}\bigl( \sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n. $$
Now, by the Cauchy–Schwarz inequality, we obtain
$$ \sum_{j=1}^{k} s_{j}^{\frac{1}{2}}(A+B) \le\Biggl(\sum_{j=1}^{k} s_{j} \bigl(\sec(\alpha) (I_{n}+A)\bigr)\Biggr)^{\frac{1}{2}}\Biggl(\sum _{j=1}^{k}s_{j}\bigl(\sec(\alpha) (I_{n}+B)\bigr)\Biggr)^{\frac{1}{2}}\quad \text{for }1\le k\le n; $$
and
$$ \begin{aligned} &\sum_{j=1}^{k} s_{j}^{\frac{1}{2}}(I_{n}+A+B) \le\Biggl(\sum_{j=1}^{k} s_{j} \bigl(\sec(\alpha) (I_{n}+A)\bigr)\Biggr)^{\frac{1}{2}}\Biggl(\sum _{j=1}^{k}s_{j}\bigl(\sec(\alpha) (I_{n}+B)\bigr)\Biggr)^{\frac{1}{2}}\\ &\quad \text{for }1\le k\le n, \end{aligned} $$
which is equivalent to the following inequalities:
$$ \bigl\Vert \vert A+B \vert ^{\frac{1}{2}} \bigr\Vert ^{2}_{k}\le \bigl\Vert \sec(\alpha) (I_{n}+A) \bigr\Vert _{k} \bigl\Vert \sec(\alpha) (I_{n}+B) \bigr\Vert _{k}; $$
and
$$ \bigl\Vert \vert I_{n}+A+B \vert ^{\frac{1}{2}} \bigr\Vert ^{2}_{k}\le \bigl\Vert \sec(\alpha) (I_{n}+A) \bigr\Vert _{k} \bigl\Vert \sec(\alpha) (I_{n}+B) \bigr\Vert _{k}. $$
By the generalization of Fan dominance theorem [8], we have
$$ \bigl|\!\bigl|\!\bigl|\vert A+B \vert ^{\frac{1}{2}} \bigr|\!\bigr|\!\bigr|^{2} \le \bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+B) \bigr|\!\bigr|\!\bigr|; $$
(7)
and
$$ \bigl|\!\bigl|\!\bigl|\vert I_{n}+A+B \vert ^{\frac{1}{2}} \bigr|\!\bigr|\!\bigr|^{2}\le \bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+B) \bigr|\!\bigr|\!\bigr|. $$
(8)
Let \(A+B=U|A+B|\), \(I_{n}+A+B=V|I_{n}+A+B|\) be the polar decomposition of \(A+B\) and \(I_{n}+A+B\), respectively, where U and V are unitary matrices. Thus, by (7), we have
$$\begin{aligned} |\!|\!|A+B |\!|\!|&= \bigl|\!\bigl|\!\bigl|U \vert A+B \vert \bigr|\!\bigr|\!\bigr|\\ &= \bigl|\!\bigl|\!\bigl|\bigl( \vert A+B \vert ^{\frac{1}{2}}\bigr)^{2} \bigr|\!\bigr|\!\bigr|\\ &\le \bigl|\!\bigl|\!\bigl|\vert A+B \vert ^{\frac{1}{2}} \bigr|\!\bigr|\!\bigr|^{2} \\ &\le \bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+B) \bigr|\!\bigr|\!\bigr|\\ &=\sec(\alpha)^{2} \bigl|\!\bigl|\!\bigl|(I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|(I_{n}+B) \bigr|\!\bigr|\!\bigr|. \end{aligned}$$
Similarly, by (8) we have
$$ |\!|\!|I_{n}+A+B |\!|\!|\le\sec^{2}(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|, $$
which completes the proof. □

Taking \(k=n\) in Theorem 2.7, we obtain the following corollary.

Corollary 2.9

Let \(A, B\in\mathbb{M}_{n}\) be such that \(W(A), W(B)\subset S_{\alpha}\). Then
$$ \bigl\vert \det(A+B) \bigr\vert \le\sec^{2n}(\alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert ; $$
and
$$ \bigl\vert \det(I_{n}+A+B) \bigr\vert \le\sec^{2n}( \alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert . $$

Lemma 2.10

([13])

Let \(A\in\mathbb{M}_{n}\) be such that \(W(A)\subset S_{\alpha}\). Then, for all unitarily invariant norms \(|\!|\!|\cdot |\!|\!|\) on \(\mathbb{M}_{n}\),
$$ |\!|\!|A |\!|\!|\le \sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(A) \bigr|\!\bigr|\!\bigr|. $$

Next we give an improvement of Corollary 2.8.

Theorem 2.11

Let \(A, B\in\mathbb{M}_{n}\) be such that \(W(A), W(B)\subset S_{\alpha}\). Then, for all unitarily invariant norms \(|\!|\!|\cdot |\!|\!|\) on \(\mathbb{M}_{n}\),
$$ |\!|\!|A+B |\!|\!|\le\sec(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|; $$
(9)
and
$$ |\!|\!|I_{n}+A+B |\!|\!|\le\sec(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|. $$
(10)

Proof

By (3), (4), and the proof of Corollary 2.8, we obtain
$$ \bigl|\!\bigl|\!\bigl|\Re(A)+\Re(B) \bigr|\!\bigr|\!\bigr|\le \bigl|\!\bigl|\!\bigl|I_{n}+ \Re(A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|I_{n}+\Re(B) \bigr|\!\bigr|\!\bigr|; $$
(11)
and
$$ \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A)+\Re(B) \bigr|\!\bigr|\!\bigr|\le \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|I_{n}+\Re(B) \bigr|\!\bigr|\!\bigr|. $$
(12)
Hence
$$\begin{aligned} |\!|\!|A+B |\!|\!|&\le\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(A+B) \bigr|\!\bigr|\!\bigr|\quad\text{(by Lemma 2.10)} \\ &=\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(A)+\Re(B) \bigr|\!\bigr|\!\bigr|\\ &\le\sec(\alpha) \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|I_{n}+\Re(B) \bigr|\!\bigr|\!\bigr|\quad\bigl(\text{by (11) }\bigr) \\ &=\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\Re(I_{n}+B) \bigr|\!\bigr|\!\bigr|\\ &\le\sec(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|, \end{aligned}$$
which proves (9).
To prove (10), compute
$$\begin{aligned} |\!|\!|I_{n}+A+B |\!|\!|&\le\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(I_{n}+A+B) \bigr|\!\bigr|\!\bigr|\quad\text{(by Lemma 2.10)} \\ &=\sec(\alpha) \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A)+\Re(B) \bigr|\!\bigr|\!\bigr|\\ &\le\sec(\alpha) \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|I_{n}+\Re(B) \bigr|\!\bigr|\!\bigr|\quad\bigl(\text{by (12) }\bigr) \\ &=\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\Re(I_{n}+B) \bigr|\!\bigr|\!\bigr|\\ &\le\sec(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|, \end{aligned}$$
which completes the proof. □

The following lemma can be obtained by Lemma 2.5.

Lemma 2.12

([6, p. 510])

If \(A\in\mathbb{M}_{n}\) has positive definite real part, then
$$ \det{(\Re A)}\le \vert \det{A} \vert . $$

Lemma 2.13

([10])

Let \(A\in\mathbb{M}_{n}\) be such that \(W(A)\subset S_{\alpha}\). Then
$$ \sec^{n}(\alpha)\det(\Re A)\ge \vert \det A \vert . $$

Now we are ready to give an improvement of Corollary 2.9.

Theorem 2.14

Let \(A, B\in\mathbb{M}_{n}\) be such that \(W(A), W(B)\subset S_{\alpha}\). Then
$$ \bigl\vert \det(A+B) \bigr\vert \le\sec^{n}(\alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert ; $$
(13)
and
$$ \bigl\vert \det(I_{n}+A+B) \bigr\vert \le\sec^{n}( \alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert . $$
(14)

Proof

Letting \(k=n\) in (3) and (4), we have
$$ \det\bigl(\Re(A)+\Re(B)\bigr)\le\det\bigl(I_{n}+\Re(A)\bigr) \det \bigl(I_{n}+\Re(B)\bigr); $$
(15)
and
$$ \det\bigl(I_{n}+\Re(A)+\Re(B)\bigr)\le\det\bigl(I_{n}+\Re(A) \bigr) \det\bigl(I_{n}+\Re(B)\bigr). $$
(16)
Thus
$$\begin{aligned} \bigl\vert \det(A+B) \bigr\vert &\le\sec^{n}(\alpha)\det\bigl( \Re(A+B)\bigr)\quad\text{(by Lemma 2.13)} \\ &=\sec^{n}(\alpha)\det\bigl(\Re(A)+\Re(B)\bigr) \\ &\le\sec^{n}(\alpha)\det\bigl(I_{n}+\Re(A)\bigr) \det \bigl(I_{n}+\Re(B)\bigr)\quad\bigl(\text{by (15)}\bigr) \\ &=\sec^{n}(\alpha)\det\bigl(\Re(I_{n}+A)\bigr) \det\bigl( \Re(I_{n}+B)\bigr) \\ &\le\sec^{n}(\alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert \quad\text{(by Lemma 2.12)} \end{aligned}$$
which proves (13).
To prove (14), compute
$$\begin{aligned} \bigl\vert \det(I_{n}+A+B) \bigr\vert &\le\sec^{n}( \alpha)\det\bigl(\Re(I_{n}+A+B)\bigr)\quad\text{(by Lemma 2.13)} \\ &=\sec^{n}(\alpha)\det\bigl(I_{n}+\Re(A)+\Re(B)\bigr) \\ &\le\sec^{n}(\alpha)\det\bigl(I_{n}+\Re(A)\bigr) \det \bigl(I_{n}+\Re(B)\bigr)\quad\bigl(\text{by (16)}\bigr) \\ &=\sec^{n}(\alpha)\det\bigl(\Re(I_{n}+A)\bigr) \det\bigl( \Re(I_{n}+B)\bigr) \\ &\le\sec^{n}(\alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert \quad\text{(by Lemma 2.12)} \end{aligned}$$
which completes the proof. □

Lemma 2.15

([9])

Let \(A, B\in\mathbb{M}_{n}\) be positive semidefinite. Then
$$ \bigl\vert \det{(A+iB)} \bigr\vert \le\det{(A+B)}\le2^{\frac{n}{2}} \bigl\vert \det{(A+iB)} \bigr\vert . $$
We remark that (2) extends the well-known Rotfel’d inequality:
$$ \det{\bigl(I_{n}+\mu \vert A+B \vert ^{p}\bigr)}\le\det{ \bigl(I_{n}+\mu \vert A \vert ^{p}\bigr)} \det{ \bigl(I_{n}+\mu \vert B \vert ^{p}\bigr)}\quad\text{for } \mu>0, 0\le p\le1. $$
(17)

Finally, we present two inequalities for accretive-dissipative matrices.

Theorem 2.16

Let \(A, B\in\mathbb{M}_{n}\) be accretive-dissipative and \(\mu>0\). Then
$$ \bigl\vert \det{(A+B)} \bigr\vert \le2^{n} \bigl\vert \det{(I_{n}+A)} \bigr\vert \bigl\vert \det{(I_{n}+B)} \bigr\vert ; $$
(18)
and
$$ \bigl\vert \det{\bigl(I_{n}+\mu(A+B)\bigr)} \bigr\vert \le2^{n} \bigl\vert \det{(I_{n}+\mu A)} \bigr\vert \bigl\vert \det{(I_{n}+\mu B)} \bigr\vert . $$
(19)
In particular,
$$ \bigl\vert \det{(I_{n}+A+B)} \bigr\vert \le2^{n} \bigl\vert \det{(I_{n}+A)} \bigr\vert \bigl\vert \det{(I_{n}+B)} \bigr\vert . $$

Proof

Let \(A=A_{1}+iA_{2}\) and \(B=B_{1}+iB_{2}\) be the Cartesian decompositions of A and B. By (3) and (17), we obtain
$$ \det{(A_{1}+A_{2}+B_{1}+B_{2})}\le \det{(I_{n}+A_{1}+A_{2})} \det{(I_{n}+B_{1}+B_{2})}; $$
(20)
and
$$ \det\bigl({I_{n}+\mu(A_{1}+A_{2}+B_{1}+B_{2})} \bigr)\le\det{\bigl(I_{n}+\mu(A_{1}+A_{2})\bigr)} \det{\bigl(I_{n}+\mu(B_{1}+B_{2})\bigr)}. $$
(21)
Hence
$$\begin{aligned} \bigl\vert \det{(A+B)} \bigr\vert &= \bigl\vert \det{(A_{1}+iA_{2}+B_{1}+iB_{2})} \bigr\vert \\ &= \bigl\vert \det{\bigl((A_{1}+B_{1})+i(A_{2}+B_{2}) \bigr)} \bigr\vert \\ &\le\det{(A_{1}+B_{1}+A_{2}+B_{2})} \quad\text{(by Lemma 2.15)} \\ &=\det{(A_{1}+A_{2}+B_{1}+B_{2})} \\ &\le\det{(I_{n}+A_{1}+A_{2})} \det{(I_{n}+B_{1}+B_{2})}\quad\bigl(\text{by (20)}\bigr) \\ &\le2^{n} \bigl\vert \det{(I_{n}+A_{1}+iA_{2})} \bigr\vert \bigl\vert \det{(I_{n}+B_{1}+iB_{2})} \bigr\vert \quad\text{(by Lemma 2.15)} \\ &=2^{n} \bigl\vert \det{(I_{n}+A)} \bigr\vert \bigl\vert \det{(I_{n}+B)} \bigr\vert , \end{aligned}$$
which proves (18).
To prove (19), compute
$$\begin{aligned} &\bigl\vert \det{\bigl(I_{n}+\mu(A+B)\bigr)} \bigr\vert \\ &\quad = \bigl\vert \det{\bigl(I_{n}+\mu(A_{1}+iA_{2}+B_{1}+iB_{2}) \bigr)} \bigr\vert \\ &\quad = \bigl\vert \det{\bigl(I_{n}+\mu(A_{1}+B_{1})+ \mu i(A_{2}+B_{2})\bigr)} \bigr\vert \\ &\quad \le\det{\bigl(I_{n}+\mu(A_{1}+B_{1}+A_{2}+B_{2}) \bigr)}\quad\text{(by Lemma 2.15)} \\ &\quad =\det{\bigl(I_{n}+\mu(A_{1}+A_{2}+B_{1}+B_{2}) \bigr)} \\ &\quad \le\det{\bigl(I_{n}+\mu(A_{1}+A_{2})\bigr)} \det{\bigl(I_{n}+\mu(B_{1}+B_{2})\bigr)}\quad\bigl(\text{ by (21)}\bigr) \\ &\quad \le2^{n} \bigl\vert \det{\bigl(I_{n}+ \mu(A_{1}+iA_{2})\bigr)} \bigr\vert \bigl\vert \det{ \bigl(I_{n}+\mu(B_{1}+iB_{2})\bigr)} \bigr\vert \quad\text{(by Lemma 2.15)} \\ &\quad =2^{n} \bigl\vert \det{(I_{n}+\mu A)} \bigr\vert \bigl\vert \det{(I_{n}+\mu B)} \bigr\vert , \end{aligned}$$
which completes the proof. □

Declarations

Funding

This research is supported by the National Natural Science Foundation of P.R. China (No. 11571247).

Authors’ contributions

All authors contributed almost the same amount of work to the manuscript. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

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, Soochow University, Suzhou, P.R. China

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