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].
Anzeige
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.
Anzeige
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:
and
where \(A, B\in\mathbb{M}_{n}\) and \(f : [0,\infty)\rightarrow [0,\infty)\) is an operator concave function.
$$ \prod_{j=1}^{k} s_{j}\bigl( \vert A+B \vert ^{r}\bigr)\le\prod_{j=1}^{k} s_{j}\bigl(I_{n}+ \vert A \vert ^{r}\bigr) \prod_{j=1}^{k} s_{j} \bigl(I_{n}+ \vert B \vert ^{r}\bigr)\quad \text{for }1\le k\le n, 1\le r\le2; $$
(1)
$$ \prod_{j=1}^{k} s_{j} \bigl(I_{n}+f\bigl( \vert A+B \vert \bigr)\bigr)\le\prod _{j=1}^{k} s_{j}\bigl(I_{n}+f \bigl( \vert A \vert \bigr)\bigr)\prod_{j=1}^{k} s_{j}\bigl(I_{n}+f\bigl( \vert B \vert \bigr)\bigr)\quad \text{for }1\le k\le n, $$
(2)
By letting \(A, B\ge0 \), \(r=1\) and \(f(X)=X\) for any \(X\in\mathbb{M}_{n}\) in (1) and (2), we have
and
$$ \prod_{j=1}^{k} s_{j}(A+B)\le \prod_{j=1}^{k} s_{j}(I_{n}+A) \prod_{j=1}^{k} s_{j}(I_{n}+B) \quad \text{for }1\le k\le n; $$
(3)
$$ \prod_{j=1}^{k} s_{j}(I_{n}+A+B) \le\prod_{j=1}^{k} s_{j}(I_{n}+A) \prod_{j=1}^{k} s_{j}(I_{n}+B) \quad \text{for }1\le k\le n. $$
(4)
2 Main results
Before we give the main results, let us present the following lemmas that will be useful later.
Lemma 2.1
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
which proves the first inequality.
$$\begin{aligned} \Biggl\Vert \sum_{j=1}^{n} A_{j} \Biggr\Vert ^{p}_{p}&= \Biggl\Vert \sum _{j=1}^{n} (B_{j}+iC_{j}) \Biggr\Vert ^{p}_{p} \\ &= \Biggl\Vert \sum_{j=1}^{n} {B_{j}}+i\sum_{j=1}^{n}{C_{j}} \Biggr\Vert ^{p}_{p} \\ &\ge2^{\frac{-p}{2}} \Biggl\Vert \sum_{j=1}^{n} {B_{j}}+\sum_{j=1}^{n}{C_{j}} \Biggr\Vert ^{p}_{p}\quad\text{(by Lemma 2.2)} \\ &=2^{\frac{-p}{2}} \Biggl\Vert \sum_{j=1}^{n} (B_{j}+C_{j}) \Biggr\Vert ^{p}_{p} \\ &\ge2^{\frac{-p}{2}}\sum_{j=1}^{n} \Vert B_{j}+C_{j} \Vert ^{p}_{p} \quad \text{(by Lemma 2.1)} \\ &\ge2^{\frac{-p}{2}}\sum_{j=1}^{n} \Vert B_{j}+iC_{j} \Vert ^{p}_{p} \quad \text{(by Lemma 2.2)} \\ &=2^{\frac{-p}{2}}\sum_{j=1}^{n} \Vert A_{j} \Vert ^{p}_{p}, \end{aligned}$$
To prove the second inequality, compute
which completes the proof. □
$$\begin{aligned} \Biggl\Vert \sum_{j=1}^{n} A_{j} \Biggr\Vert ^{p}_{p}&= \Biggl\Vert \sum _{j=1}^{n} (B_{j}+iC_{j}) \Biggr\Vert ^{p}_{p} \\ &= \Biggl\Vert \sum_{j=1}^{n} {B_{j}}+i\sum_{j=1}^{n}{C_{j}} \Biggr\Vert ^{p}_{p} \\ &\le \Biggl\Vert \sum_{j=1}^{n} {B_{j}}+\sum_{j=1}^{n}{C_{j}} \Biggr\Vert ^{p}_{p}\quad\text{(by Lemma 2.2)} \\ &= \Biggl\Vert \sum_{j=1}^{n} (B_{j}+C_{j}) \Biggr\Vert ^{p}_{p} \\ &\le n^{p-1}\sum_{j=1}^{n} \Vert B_{j}+C_{j} \Vert ^{p}_{p} \quad \text{(by Lemma 2.1)} \\ &\le n^{p-1}2^{\frac{p}{2}}\sum_{j=1}^{n} \Vert B_{j}+iC_{j} \Vert ^{p}_{p} \quad\text{(by Lemma 2.2)} \\ &=\frac{(2n^{2})^{\frac{p}{2}}}{n}\sum_{j=1}^{n} \Vert A_{j} \Vert ^{p}_{p}, \end{aligned}$$
The following lemma is the well-known Fan–Hoffman inequality.
Lemma 2.5
([12, p. 63])
Let
\(A\in\mathbb{M}_{n}\). Then
where
\(\lambda_{j}(\cdot)\)
denotes the
jth largest eigenvalue.
$$ \lambda_{j}(\Re A)\le s_{j}(A), $$
Lemma 2.6
Let
\(A\in\mathbb{M}_{n}\)
be such that
\(W(A)\subset S_{\alpha}\). Then
where
\(\lambda_{j}(\cdot)\)
denotes the
jth largest eigenvalue.
$$ s_{j}(A)\le \sec^{2}(\alpha)\lambda_{j}(\Re A), $$
Theorem 2.7
Let
\(A, B\in\mathbb{M}_{n}\)
be such that
\(W(A), W(B)\subset S_{\alpha}\). Then
and
$$ \prod_{j=1}^{k} s_{j}(A+B)\le \sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}(I_{n}+A)\prod_{j=1}^{k} s_{j}(I_{n}+B)\quad \textit{for }1\le k\le n; $$
(5)
$$ \prod_{j=1}^{k} s_{j}(I_{n}+A+B) \le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}(I_{n}+A)\prod_{j=1}^{k} s_{j}(I_{n}+B)\quad \textit{for }1\le k\le n. $$
(6)
Proof
We have
which proves (5).
$$\begin{aligned} \prod_{j=1}^{k} s_{j}(A+B)&\le \sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(A+B)\bigr)\quad\text{(by Lemma 2.6)} \\ &=\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(A)+\Re(B)\bigr) \\ &\le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(I_{n}+\Re(A)\bigr)\prod _{j=1}^{k} s_{j}\bigl(I_{n}+ \Re(B)\bigr)\quad\bigl(\text{by (3)}\bigr) \\ &=\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(I_{n}+A)\bigr)\prod _{j=1}^{k} s_{j}\bigl(\Re(I_{n}+B) \bigr) \\ &\le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}(I_{n}+A)\prod_{j=1}^{k} s_{j}(I_{n}+B)\quad\text{(by Lemma 2.5)} \end{aligned}$$
To prove (6), compute
which completes the proof. □
$$\begin{aligned} \prod_{j=1}^{k} s_{j}(I_{n}+A+B)& \le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(I_{n}+A+B)\bigr)\quad\text{(by Lemma 2.6)} \\ &=\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(I_{n}+\Re(A)+\Re(B)\bigr) \\ &\le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(I_{n}+\Re(A)\bigr)\prod _{j=1}^{k} s_{j}\bigl(I_{n}+ \Re(B)\bigr)\quad\bigl(\text{by (4)}\bigr) \\ &=\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}\bigl(\Re(I_{n}+A)\bigr)\prod _{j=1}^{k} s_{j}\bigl(\Re(I_{n}+B) \bigr) \\ &\le\sec^{2k}(\alpha)\prod_{j=1}^{k} s_{j}(I_{n}+A)\prod_{j=1}^{k} s_{j}(I_{n}+B),\quad\text{(by Lemma 2.5)} \end{aligned}$$
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
\(\mathbb{M}_{n}\),
and
$$ |\!|\!|A+B |\!|\!|\le\sec^{2}(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|; $$
$$ |\!|\!|I_{n}+A+B |\!|\!|\le\sec^{2}(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|. $$
Proof
From (5) and (6), we obtain
and
which is equivalent to the following inequalities:
and
$$ \prod_{j=1}^{k} s_{j}(A+B)\le \prod_{j=1}^{k} s_{j}\bigl(\sec( \alpha) (I_{n}+A)\bigr)s_{j}\bigl(\sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n; $$
$$ \prod_{j=1}^{k} s_{j}(I_{n}+A+B) \le\prod_{j=1}^{k} s_{j}\bigl( \sec(\alpha) (I_{n}+A)\bigr)s_{j}\bigl(\sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n, $$
$$ \prod_{j=1}^{k} s_{j}^{\frac{1}{2}}(A+B) \le\prod_{j=1}^{k} s_{j}^{\frac{1}{2}} \bigl(\sec(\alpha) (I_{n}+A)\bigr)s_{j}^{\frac{1}{2}}\bigl( \sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n; $$
$$ \prod_{j=1}^{k} s_{j}^{\frac{1}{2}}(I_{n}+A+B) \le\prod_{j=1}^{k} s_{j}^{\frac{1}{2}} \bigl(\sec(\alpha) (I_{n}+A)\bigr)s_{j}^{\frac{1}{2}}\bigl( \sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n. $$
By the property of majorization [1, p. 42], we have
and
$$ \sum_{j=1}^{k} s_{j}^{\frac{1}{2}}(A+B) \le\sum_{j=1}^{k} s_{j}^{\frac{1}{2}} \bigl(\sec(\alpha) (I_{n}+A)\bigr)s_{j}^{\frac{1}{2}}\bigl( \sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n; $$
$$ \sum_{j=1}^{k} s_{j}^{\frac{1}{2}}(I_{n}+A+B) \le\sum_{j=1}^{k} s_{j}^{\frac{1}{2}} \bigl(\sec(\alpha) (I_{n}+A)\bigr)s_{j}^{\frac{1}{2}}\bigl( \sec(\alpha) (I_{n}+B)\bigr)\quad \text{for }1\le k\le n. $$
Now, by the Cauchy–Schwarz inequality, we obtain
and
which is equivalent to the following inequalities:
and
$$ \sum_{j=1}^{k} s_{j}^{\frac{1}{2}}(A+B) \le\Biggl(\sum_{j=1}^{k} s_{j} \bigl(\sec(\alpha) (I_{n}+A)\bigr)\Biggr)^{\frac{1}{2}}\Biggl(\sum _{j=1}^{k}s_{j}\bigl(\sec(\alpha) (I_{n}+B)\bigr)\Biggr)^{\frac{1}{2}}\quad \text{for }1\le k\le n; $$
$$ \begin{aligned} &\sum_{j=1}^{k} s_{j}^{\frac{1}{2}}(I_{n}+A+B) \le\Biggl(\sum_{j=1}^{k} s_{j} \bigl(\sec(\alpha) (I_{n}+A)\bigr)\Biggr)^{\frac{1}{2}}\Biggl(\sum _{j=1}^{k}s_{j}\bigl(\sec(\alpha) (I_{n}+B)\bigr)\Biggr)^{\frac{1}{2}}\\ &\quad \text{for }1\le k\le n, \end{aligned} $$
$$ \bigl\Vert \vert A+B \vert ^{\frac{1}{2}} \bigr\Vert ^{2}_{k}\le \bigl\Vert \sec(\alpha) (I_{n}+A) \bigr\Vert _{k} \bigl\Vert \sec(\alpha) (I_{n}+B) \bigr\Vert _{k}; $$
$$ \bigl\Vert \vert I_{n}+A+B \vert ^{\frac{1}{2}} \bigr\Vert ^{2}_{k}\le \bigl\Vert \sec(\alpha) (I_{n}+A) \bigr\Vert _{k} \bigl\Vert \sec(\alpha) (I_{n}+B) \bigr\Vert _{k}. $$
By the generalization of Fan dominance theorem [8], we have
and
$$ \bigl|\!\bigl|\!\bigl|\vert A+B \vert ^{\frac{1}{2}} \bigr|\!\bigr|\!\bigr|^{2} \le \bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+B) \bigr|\!\bigr|\!\bigr|; $$
(7)
$$ \bigl|\!\bigl|\!\bigl|\vert I_{n}+A+B \vert ^{\frac{1}{2}} \bigr|\!\bigr|\!\bigr|^{2}\le \bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\sec(\alpha) (I_{n}+B) \bigr|\!\bigr|\!\bigr|. $$
(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
which completes the proof. □
$$ |\!|\!|I_{n}+A+B |\!|\!|\le\sec^{2}(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|, $$
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
and
$$ \bigl\vert \det(A+B) \bigr\vert \le\sec^{2n}(\alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert ; $$
$$ \bigl\vert \det(I_{n}+A+B) \bigr\vert \le\sec^{2n}( \alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert . $$
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
\(\mathbb{M}_{n}\),
and
$$ |\!|\!|A+B |\!|\!|\le\sec(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|; $$
(9)
$$ |\!|\!|I_{n}+A+B |\!|\!|\le\sec(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|. $$
(10)
Proof
By (3), (4), and the proof of Corollary 2.8, we obtain
and
$$ \bigl|\!\bigl|\!\bigl|\Re(A)+\Re(B) \bigr|\!\bigr|\!\bigr|\le \bigl|\!\bigl|\!\bigl|I_{n}+ \Re(A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|I_{n}+\Re(B) \bigr|\!\bigr|\!\bigr|; $$
(11)
$$ \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A)+\Re(B) \bigr|\!\bigr|\!\bigr|\le \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|I_{n}+\Re(B) \bigr|\!\bigr|\!\bigr|. $$
(12)
Hence
which proves (9).
$$\begin{aligned} |\!|\!|A+B |\!|\!|&\le\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(A+B) \bigr|\!\bigr|\!\bigr|\quad\text{(by Lemma 2.10)} \\ &=\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(A)+\Re(B) \bigr|\!\bigr|\!\bigr|\\ &\le\sec(\alpha) \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|I_{n}+\Re(B) \bigr|\!\bigr|\!\bigr|\quad\bigl(\text{by (11) }\bigr) \\ &=\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\Re(I_{n}+B) \bigr|\!\bigr|\!\bigr|\\ &\le\sec(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|, \end{aligned}$$
To prove (10), compute
which completes the proof. □
$$\begin{aligned} |\!|\!|I_{n}+A+B |\!|\!|&\le\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(I_{n}+A+B) \bigr|\!\bigr|\!\bigr|\quad\text{(by Lemma 2.10)} \\ &=\sec(\alpha) \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A)+\Re(B) \bigr|\!\bigr|\!\bigr|\\ &\le\sec(\alpha) \bigl|\!\bigl|\!\bigl|I_{n}+\Re(A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|I_{n}+\Re(B) \bigr|\!\bigr|\!\bigr|\quad\bigl(\text{by (12) }\bigr) \\ &=\sec(\alpha) \bigl|\!\bigl|\!\bigl|\Re(I_{n}+A) \bigr|\!\bigr|\!\bigr|\bigl|\!\bigl|\!\bigl|\Re(I_{n}+B) \bigr|\!\bigr|\!\bigr|\\ &\le\sec(\alpha) |\!|\!|I_{n}+A |\!|\!||\!|\!|I_{n}+B |\!|\!|, \end{aligned}$$
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
and
$$ \bigl\vert \det(A+B) \bigr\vert \le\sec^{n}(\alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert ; $$
(13)
$$ \bigl\vert \det(I_{n}+A+B) \bigr\vert \le\sec^{n}( \alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert . $$
(14)
Proof
Letting \(k=n\) in (3) and (4), we have
and
$$ \det\bigl(\Re(A)+\Re(B)\bigr)\le\det\bigl(I_{n}+\Re(A)\bigr) \det \bigl(I_{n}+\Re(B)\bigr); $$
(15)
$$ \det\bigl(I_{n}+\Re(A)+\Re(B)\bigr)\le\det\bigl(I_{n}+\Re(A) \bigr) \det\bigl(I_{n}+\Re(B)\bigr). $$
(16)
Thus
which proves (13).
$$\begin{aligned} \bigl\vert \det(A+B) \bigr\vert &\le\sec^{n}(\alpha)\det\bigl( \Re(A+B)\bigr)\quad\text{(by Lemma 2.13)} \\ &=\sec^{n}(\alpha)\det\bigl(\Re(A)+\Re(B)\bigr) \\ &\le\sec^{n}(\alpha)\det\bigl(I_{n}+\Re(A)\bigr) \det \bigl(I_{n}+\Re(B)\bigr)\quad\bigl(\text{by (15)}\bigr) \\ &=\sec^{n}(\alpha)\det\bigl(\Re(I_{n}+A)\bigr) \det\bigl( \Re(I_{n}+B)\bigr) \\ &\le\sec^{n}(\alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert \quad\text{(by Lemma 2.12)} \end{aligned}$$
To prove (14), compute
which completes the proof. □
$$\begin{aligned} \bigl\vert \det(I_{n}+A+B) \bigr\vert &\le\sec^{n}( \alpha)\det\bigl(\Re(I_{n}+A+B)\bigr)\quad\text{(by Lemma 2.13)} \\ &=\sec^{n}(\alpha)\det\bigl(I_{n}+\Re(A)+\Re(B)\bigr) \\ &\le\sec^{n}(\alpha)\det\bigl(I_{n}+\Re(A)\bigr) \det \bigl(I_{n}+\Re(B)\bigr)\quad\bigl(\text{by (16)}\bigr) \\ &=\sec^{n}(\alpha)\det\bigl(\Re(I_{n}+A)\bigr) \det\bigl( \Re(I_{n}+B)\bigr) \\ &\le\sec^{n}(\alpha) \bigl\vert \det(I_{n}+A) \bigr\vert \bigl\vert \det(I_{n}+B) \bigr\vert \quad\text{(by Lemma 2.12)} \end{aligned}$$
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
and
$$ \bigl\vert \det{(A+B)} \bigr\vert \le2^{n} \bigl\vert \det{(I_{n}+A)} \bigr\vert \bigl\vert \det{(I_{n}+B)} \bigr\vert ; $$
(18)
$$ \bigl\vert \det{\bigl(I_{n}+\mu(A+B)\bigr)} \bigr\vert \le2^{n} \bigl\vert \det{(I_{n}+\mu A)} \bigr\vert \bigl\vert \det{(I_{n}+\mu B)} \bigr\vert . $$
(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
and
$$ \det{(A_{1}+A_{2}+B_{1}+B_{2})}\le \det{(I_{n}+A_{1}+A_{2})} \det{(I_{n}+B_{1}+B_{2})}; $$
(20)
$$ \det\bigl({I_{n}+\mu(A_{1}+A_{2}+B_{1}+B_{2})} \bigr)\le\det{\bigl(I_{n}+\mu(A_{1}+A_{2})\bigr)} \det{\bigl(I_{n}+\mu(B_{1}+B_{2})\bigr)}. $$
(21)
Hence
which proves (18).
$$\begin{aligned} \bigl\vert \det{(A+B)} \bigr\vert &= \bigl\vert \det{(A_{1}+iA_{2}+B_{1}+iB_{2})} \bigr\vert \\ &= \bigl\vert \det{\bigl((A_{1}+B_{1})+i(A_{2}+B_{2}) \bigr)} \bigr\vert \\ &\le\det{(A_{1}+B_{1}+A_{2}+B_{2})} \quad\text{(by Lemma 2.15)} \\ &=\det{(A_{1}+A_{2}+B_{1}+B_{2})} \\ &\le\det{(I_{n}+A_{1}+A_{2})} \det{(I_{n}+B_{1}+B_{2})}\quad\bigl(\text{by (20)}\bigr) \\ &\le2^{n} \bigl\vert \det{(I_{n}+A_{1}+iA_{2})} \bigr\vert \bigl\vert \det{(I_{n}+B_{1}+iB_{2})} \bigr\vert \quad\text{(by Lemma 2.15)} \\ &=2^{n} \bigl\vert \det{(I_{n}+A)} \bigr\vert \bigl\vert \det{(I_{n}+B)} \bigr\vert , \end{aligned}$$
To prove (19), compute
which completes the proof. □
$$\begin{aligned} &\bigl\vert \det{\bigl(I_{n}+\mu(A+B)\bigr)} \bigr\vert \\ &\quad = \bigl\vert \det{\bigl(I_{n}+\mu(A_{1}+iA_{2}+B_{1}+iB_{2}) \bigr)} \bigr\vert \\ &\quad = \bigl\vert \det{\bigl(I_{n}+\mu(A_{1}+B_{1})+ \mu i(A_{2}+B_{2})\bigr)} \bigr\vert \\ &\quad \le\det{\bigl(I_{n}+\mu(A_{1}+B_{1}+A_{2}+B_{2}) \bigr)}\quad\text{(by Lemma 2.15)} \\ &\quad =\det{\bigl(I_{n}+\mu(A_{1}+A_{2}+B_{1}+B_{2}) \bigr)} \\ &\quad \le\det{\bigl(I_{n}+\mu(A_{1}+A_{2})\bigr)} \det{\bigl(I_{n}+\mu(B_{1}+B_{2})\bigr)}\quad\bigl(\text{ by (21)}\bigr) \\ &\quad \le2^{n} \bigl\vert \det{\bigl(I_{n}+ \mu(A_{1}+iA_{2})\bigr)} \bigr\vert \bigl\vert \det{ \bigl(I_{n}+\mu(B_{1}+iB_{2})\bigr)} \bigr\vert \quad\text{(by Lemma 2.15)} \\ &\quad =2^{n} \bigl\vert \det{(I_{n}+\mu A)} \bigr\vert \bigl\vert \det{(I_{n}+\mu B)} \bigr\vert , \end{aligned}$$
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