# Some generalizations of inequalities for sector matrices

## 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.

## 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 , 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 , 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.

## 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

()

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 , 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 , 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

()

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

()

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

()

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. □

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## Funding

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

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Correspondence to Chaojun Yang.

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