Some results involving multiple matrices

Sheng Dong; Qingwen Wang; Ming Zhao; Zhiqiang Li; Sheng Dong; Qingwen Wang; Ming Zhao; Zhiqiang Li

doi:10.3934/math.2021702

AIMS Mathematics

2021, Volume 6, Issue 11: 12104-12113. doi: 10.3934/math.2021702

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Research article Special Issues

Some results involving multiple matrices

1.
School of Mathematics, Zhengzhou University of Aeronautics, Zhengzhou 450046, China
2.
Department of Mathematics, Shanghai University, Shanghai 200444, China
3.
Department of public basic education, Henan Police College, Zhengzhou 450046, China
4.
School of Mathematics and Information Science, Henan University of Economics and Law, Zhengzhou 450046, China

Received: 31 March 2021 Accepted: 17 August 2021 Published: 20 August 2021
MSC : 15A45, 15A60

In this paper, we first give an alternative proof for a result of Mao in [Linear Algebra Appl., 589 (2020) 96–102], then we present some results involving multiple positive definite matrices and multiple sector matrices.

Keywords:

Citation: Sheng Dong, Qingwen Wang, Ming Zhao, Zhiqiang Li. Some results involving multiple matrices[J]. AIMS Mathematics, 2021, 6(11): 12104-12113. doi: 10.3934/math.2021702

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Abstract

1. Introduction

Let $\mathbb{M}_{n}$ be the set of $n\times n$ complex matrices. We denote by $A^{(k)}, k = 1, \ldots, n-1,$ the $k$ -th leading principal submatrices of $A\in\mathbb{M}_{n}$ .

Let $A, B\in\mathbb{M}_{n}$ be positive definite. It is well known that

$\begin{equation} \det(A + B)\geq \det A+ \det B. \end{equation}$

(1.1)

In ^[4], Haynsworth proved the following refinement of (1.1),

$\begin{equation} \det(A + B)\geq \Big(1+\sum\limits^{n-1}_{k = 1}\frac{\det B^{(k)}}{\det A^{(k)}}\Big)\det A+ \Big(1+\sum\limits^{n-1}_{k = 1}\frac{\det A^{(k)}}{\det B^{(k)}}\Big)\det B. \end{equation}$

(1.2)

Later, Hartfiel ^[5] obtained an improvement of (1.2) as follows,

$\begin{equation} \begin{aligned} \det(A + B)\geq &\Big(1+\sum^{n-1}_{k = 1}\frac{\det B^{(k)}}{\det A^{(k)}}\Big)\det A+ \Big(1+\sum^{n-1}_{k = 1}\frac{\det A^{(k)}}{\det B^{(k)}}\Big)\det B \\&+ (2^{n}-2n)\sqrt{\det A B}.\; \end{aligned} \end{equation}$

(1.3)

And the author also gave an interesting result,

$\begin{equation} \begin{aligned} \det(A + B)\geq \det A+\det B &+ (2^{n}-2)\sqrt{\det A B}.\; \end{aligned} \end{equation}$

(1.4)

Extending results from two matrices case to multiple matrices case is a natural thought. Recently, by making use a result of Lin ^[11], Hou and Dong ^[7] gave the following extensions of Hartfiel's results (1.3) and (1.4) to the case of three matrices.

Theorem 1. Let $A, B, C\in\mathbb{M}_{n}$ be positive definite matrices. Then

$\begin{equation} \begin{aligned} \det(A+B+C)\geq &\Big(1+\sum^{n-1}_{k = 1}\frac{\det B^{(k)}+\det C^{(k)}}{\det A^{(k)}}\Big)\det A \\&+ \Big(1+\sum^{n-1}_{k = 1}\frac{\det A^{(k)}+\det C^{(k)}}{\det B^{(k)}}\Big)\det B \\&+ \Big(1+\sum^{n-1}_{k = 1}\frac{\det A^{(k)}+\det B^{(k)}}{\det C^{(k)}}\Big)\det C \\&+ (2^{n}-2n)\big(\sqrt{\det AB}+\sqrt{\det AC}+\sqrt{\det BC}\big).\; \end{aligned} \end{equation}$

(1.5)

Theorem 2. Let $A, B, C\in\mathbb{M}_{n}$ be positive definite matrices. Then

$\begin{equation} \begin{aligned} \det(A+B+C) \geq &\det A+\det B+\det C \\&+(2^{n}-2)\big(\sqrt{\det AB}+\sqrt{\det AC}+\sqrt{\det BC}\big). \end{aligned} \end{equation}$

(1.6)

Very recently, Li et al. ^[9] extended Haynsworth's result (1.2) to multiple matrices by induction. At the same time, by making use of [1, Corollary 4.4], Mao ^[16] gave the following extension of Hartfiel's inequality (1.3) for multiple matrices.

Theorem 3. Suppose $A_{i} \in\mathbb{M}_{n}, i = 1, \ldots, m$ , are positive definite. Then

$\begin{equation} \det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)\geq \sum\limits^{m}_{i = 1}\Bigg(1+\sum\limits^{n-1}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\det A_{i}+(2^{n}-2n)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}}. \end{equation}$

(1.7)

For any $A\in \mathbb{M}_{n}$ , $W(A) = \{x^{*}Ax|x \in\mathbb{C}^{n}, x^{*}x = 1 \}$ is defined as the numerical range of $A$ . For $\theta \in [0, \frac{\pi}{2})$ , the sector in the complex plane is given by $S_{\theta} = \{z\in \mathbb{C}|\mathfrak{R}z > 0, |\mathfrak{S}z|\leq (\mathfrak{R}z)\tan(\theta) \}.$ A matrix $A$ is called a sector matrix if $W(A)\subset S_{\theta}$ . This class of matrices, as a natural generalization of positive definite matrix, has been extensively investigated; see ^{[3,12,13,14,17]} and reference therein.

In 2015, Lin ^[14] first extended (1.3) to sector matrix, his result was improved by Zheng et al. ^[17], Dong and Wang ^[2] at the same time. And their improvements are the special case of the following inequality ^[7], which is the generalization of Theorem 1 for sector matrices.

Theorem 4. Let $A, B, C\in \mathbb{M}_{n}$ , $W(A), W(B), W(C)\subset S_{\theta}.$ Then

$\begin{equation*} \begin{aligned} \sec^{n}(\theta)|\det(A+B+C)|\geq &\Big(1+\sum^{n-1}_{k = 1}\cos^{k}(\theta)\frac{|\det B^{(k)}|+|\det C^{(k)}|}{|\det A^{(k)}|}\Big)|\det A| \\&+ \Big(1+\sum^{n-1}_{k = 1}\cos^{k}(\theta)\frac{|\det A^{(k)}|+|\det C^{(k)}|}{|\det B^{(k)}|}\Big)|\det B| \\&+ \Big(1+\sum^{n-1}_{k = 1}\cos^{k}(\theta)\frac{|\det A^{(k)}|+|\det B^{(k)}|}{|\det C^{(k)}|}\Big)|\det C| \\&+ (2^{n}-2n)\Big(\sqrt{|\det AB|}+\sqrt{|\det AC|}+\sqrt{|\det BC|}\Big).\; \end{aligned} \end{equation*}$

In ^[17], Zheng et al. gave the following complement of Theorem 4.

Theorem 5. Let $A, B, C\in \mathbb{M}_{n}$ , $W(A), W(B), W(C)\subset S_{\theta}$ . Then

$\begin{equation*} \begin{aligned} |\det(A+B+C)|\geq &\prod^{m}_{k = 1}\Big( \frac{|\det A^{(k)}|}{|\det A^{(k-1)}|}+\frac{|\det B^{(k)}|}{|\det B^{(k-1)}|}\Big)\cos^{k}(\theta) \\&+ \prod^{m}_{k = 1}\Big( \frac{|\det A^{(k)}|}{|\det A^{(k-1)}|}+\frac{|\det C^{(k)}|}{|\det C^{(k-1)}|}\Big)\cos^{k}(\theta) \\&+ \prod^{m}_{k = 1}\Big( \frac{|\det B^{(k)}|}{|\det B^{(k-1)}|}+\frac{|\det C^{(k)}|}{|\det C^{(k-1)}|}\Big)\cos^{k}(\theta) \\&-(|\det A|+|\det B|+|\det C|).\; \end{aligned} \end{equation*}$

The extension of Theorem 4 for multiple matrices is proved by Mao ^[16] as follows, which is the generation of Theorem 3 for sector matrices.

Theorem 6. Suppose $A_{i} \in\mathbb{M}_{n}$ , $W(A_{i})\subset S_{\theta}, i = 1, \ldots, m$ . Then

$\begin{equation} \begin{aligned} \sec^{n}(\theta)|\det \Big(\sum^{m}_{i = 1}A_{i}\Big)|\geq & \sum^{m}_{i = 1}\Bigg(1+\sum^{n-1}_{k = 1}\cos^{k}(\theta)\frac{\sum^{m}_{j = 1,j\neq i}|\det A^{(k)}_{j}|}{|\det A^{(k)}_{i}|}\Bigg)|\det A_{i}|\\&+(2^{n}-2n)\sum\limits_{1\leq i < j\leq m}\sqrt{|\det A_{i} A_{j}|}. \end{aligned} \end{equation}$

(1.8)

In this paper, we first give an alternative proof of Mao' result (1.7), then we present some interesting results for multiple positive definite matrices, this is done in section 2. In section 3, we show some generations of the results previously proved for sector matrices, and give an complement of Theorem 6 by extending Theorem 5 to multiple sector matrices.

2. Results involving multiple positive definite matrices

In ^[16], the way that Mao proved Theorem 3 is original. Considering that alternative proof may provide new perspectives to the elegant result, we take the chance to do it here.

The following lemma is useful for our new proof.

Lemma 1. [9, Lemma 6] Suppose $A_{i} \in\mathbb{M}_{n}, i = 1, \ldots, m$ , are positive definite. Let $A^{(k)}_{i}, k = 1, \ldots, n-1,$ denote the $k$ -th leading principal submatrices of $A_{i}$ . Then

$\begin{equation*} \det\Bigg( \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)/\Big(\sum\limits^{m}_{i = 1}A^{(k)}_{i}\Big)\Bigg)\geq \sum\limits^{m}_{i = 1} \frac{\det A_{i}}{\det A^{(k)}_{i}}. \end{equation*}$

Proof of Theorem 3. We prove the theorem by induction on $n$ .

For $n = 2$ , we get

$\begin{equation*} \begin{aligned} \det \Big(\sum^{m}_{i = 1}A_{i}\Big) = &\det \Big(\sum^{m}_{i = 1}A^{(1)}_{i}\Big)\cdot \det\Bigg( \Big(\sum^{m}_{i = 1}A_{i}\Big)/\Big(\sum^{m}_{i = 1}A^{(1)}_{i}\Big)\Bigg) \\ \geq &\det\Big(\sum^{m}_{i = 1}A^{(1)}_{i}\Big)\cdot \sum^{m}_{i = 1} \frac{\det A_{i}}{\det A^{(1)}_{i}} \\ \geq &\Big(\sum^{m}_{i = 1}\det A^{(1)}_{i}\Big) \cdot \sum^{m}_{i = 1} \frac{\det A_{i}}{\det A^{(1)}_{i}} \\ = &\sum^{m}_{i = 1}\Bigg(1+\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(1)}_{j}}{\det A^{(1)}_{i}}\Bigg)\det A_{i}, \end{aligned} \end{equation*}$

where the first inequality is by Lemma 1. This proves (1.7) for $n = 2$ .

Suppose the inequality (1.7) holds for all $A_{i}$ of order less than or equal to $n-1$ . If $A_{i}, i = 1, \ldots, m$ , are of order $n$ , by Lemma 1, we have

$\begin{equation*} \begin{aligned} \det \Big(\sum^{m}_{i = 1}A_{i}\Big) = &\det \Big(\sum^{m}_{i = 1}A^{(n-1)}_{i}\Big)\cdot \det\Bigg( \Big(\sum^{m}_{i = 1}A_{i}\Big)/\Big(\sum^{m}_{i = 1}A^{(n-1)}_{i}\Big)\Bigg) \\ \geq &\det\Big(\sum^{m}_{i = 1}A^{(n-1)}_{i}\Big)\cdot \sum^{m}_{i = 1} \frac{\det A_{i}}{\det A^{(n-1)}_{i}}. \end{aligned} \end{equation*}$

By the induction hypothesis

$\begin{equation*} \begin{aligned} \det\Big(\sum^{m}_{i = 1}A^{(n-1)}_{i}\Big)\geq &\sum^{m}_{i = 1}\Bigg(1+\sum^{n-2}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\det A^{(n-1)}_{i}\\&+\big(2^{n-1}-2(n-1)\big)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A^{(n-1)}_{i}\det A^{(n-1)}_{j}}, \end{aligned} \end{equation*}$

we get

$\begin{equation*} \begin{aligned} \det \Big(\sum^{m}_{i = 1}A_{i}\Big) \geq &\sum^{m}_{i = 1}\Bigg(1+\sum^{n-2}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\det A^{(n-1)}_{i}\cdot \sum^{m}_{h = 1} \frac{\det A_{h}}{\det A^{(n-1)}_{h}}\\ &+\big(2^{n-1}-2(n-1)\big)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A^{(n-1)}_{i}\det A^{(n-1)}_{j}}\cdot \sum^{m}_{h = 1} \frac{\det A_{h}}{\det A^{(n-1)}_{h}}\\ = &\sum^{m}_{i = 1,h = 1,i = h}\Bigg(1+\sum^{n-2}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\det A^{(n-1)}_{i}\cdot \frac{\det A_{h}}{\det A^{(n-1)}_{h}}\\ &+\sum^{m}_{i = 1,h = 1,i\neq h}\Bigg(1+\sum^{n-2}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\det A^{(n-1)}_{i}\cdot \frac{\det A_{h}}{\det A^{(n-1)}_{h}} \\ &+\big(2^{n-1}-2n+2\big)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A^{(n-1)}_{i}\det A^{(n-1)}_{j}}\cdot \sum^{m}_{h = 1} \frac{\det A_{h}}{\det A^{(n-1)}_{h}}\\ \geq &\sum^{m}_{i = 1}\Bigg(1+\sum^{n-2}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\cdot \det A_{i}\\ &+\sum^{m}_{i = 1,h = 1,i\neq h}\det A^{(n-1)}_{i}\cdot \frac{\det A_{h}}{\det A^{(n-1)}_{h}} \\ &+\sum^{m}_{i = 1,h = 1,i\neq h}\sum^{n-2}_{k = 1}\frac{\det A^{(k)}_{h}}{\det A^{(k)}_{i}}\det A^{(n-1)}_{i}\cdot \frac{\det A_{h}}{\det A^{(n-1)}_{h}}\\ &+\big(2^{n-1}-2n+2\big)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A^{(n-1)}_{i}\det A^{(n-1)}_{j}}\cdot \Bigg(\frac{\det A_{i}}{\det A^{(n-1)}_{i}}+\frac{\det A_{j}}{\det A^{(n-1)}_{j}}\Bigg)\\ \geq &\sum^{m}_{i = 1}\Bigg(1+\sum^{n-1}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\cdot \det A_{i}+2(n-2)\sum\limits_{1\leq i < h\leq m}\sqrt{\det A_{i}\det A_{h}}\\ &+\big(2^{n}-4n+4\big)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i}\det A_{j}}\\ = &\sum^{m}_{i = 1}\Bigg(1+\sum^{n-1}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\det A_{i}+(2^{n}-2n)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i}\det A_{j}} . \end{aligned} \end{equation*}$

This completes the proof.

The next theorem is an interesting result for multiple positive definite matrices. Clearly, when m = 2, (2.1) becomes (1.4); when m = 3, (2.1) reduces to (1.6).

Theorem 7. Let $A_{i} \in\mathbb{M}_{n}, i = 1, \ldots, m$ , be positive definite. Then

$\begin{equation} \det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)\geq \sum\limits^{m}_{i = 1}\det A_{i}+(2^{n}-2)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}}. \end{equation}$

(2.1)

Proof. By (1.7), we get

$\begin{equation*} \begin{aligned} &\det \Big(\sum^{m}_{i = 1}A_{i}\Big)\\ \geq& \sum^{m}_{i = 1}\Bigg(1+\sum^{n-1}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\Bigg)\det A_{i}+(2^{n}-2n)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}}\\ = &\sum^{m}_{i = 1}\det A_{i}+\sum^{m}_{i = 1}\Bigg(\sum^{n-1}_{k = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\cdot \det A_{i}\Bigg)+(2^{n}-2n)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}}\\ = &\sum^{m}_{i = 1}\det A_{i}+\sum^{n-1}_{k = 1}\Bigg(\sum^{m}_{i = 1}\frac{\sum^{m}_{j = 1,j\neq i}\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\cdot \det A_{i}\Bigg)+(2^{n}-2n)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}} \\ = &\sum^{m}_{i = 1}\det A_{i}+\sum^{n-1}_{k = 1}\Bigg(\sum^{m}_{i,j = 1,i\neq j}\Big(\frac{\det A^{(k)}_{j}}{\det A^{(k)}_{i}}\cdot \det A_{i}+\frac{\det A^{(k)}_{i}}{\det A^{(k)}_{j}}\cdot \det A_{j}\Big)\Bigg)\\&+(2^{n}-2n)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}} \\ \geq &\sum^{m}_{i = 1}\det A_{i}+\sum^{n-1}_{k = 1}\sum^{m}_{i,j = 1,i\neq j}2\sqrt{\det A_{i} A_{j}}+(2^{n}-2n)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}} \\ = &\sum^{m}_{i = 1}\det A_{i}+(2^{n}-2)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}}, \end{aligned} \end{equation*}$

the proof is completed.

Remark 1. We mention that Theorem 7 can be obtained by inequality (1.4) and the following inequality, which is a special case of [1, Corollary 4.4].

$\begin{equation} \det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)+(m-2)\sum\limits^{m}_{i = 1}\det A_{i}\geq\sum\limits_{1\leq i < j\leq m}\det (A_{i}+A_{j}). \end{equation}$

(2.2)

The following is equivalent to Theorem 7.

Corollary 1. Let $A_{i} \in\mathbb{M}_{n}, i = 1, \ldots, m$ , be positive definite. Then

$\begin{equation*} \det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)\geq \Big(\sum\limits^{m}_{i = 1}\sqrt{\det A_{i}}\Big)^{2} + (2^{n}-4)\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}}. \end{equation*}$

Proof. Note that

$\begin{equation*} \Big(\sum^{m}_{i = 1}\sqrt{\det A_{i}}\Big)^{2} = \sum^{m}_{i = 1}\det A_{i}+ 2\sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}}. \end{equation*}$

By Theorem 7, the desired result follows.

The next result can be derived from Theorem 7.

Corollary 2. Let $A_{i} \in\mathbb{M}_{n}, i = 1, \ldots, m$ , be positive definite. Then

$\begin{equation*} \det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)\geq \sum\limits^{m}_{i = 1}\det A_{i}+ m(m-1)(2^{n-1}-1)(\prod\limits^{m}_{i = 1}\det A_{i})^{\frac{1}{m}}. \end{equation*}$

Proof. By the arithmetic-geometric inequality, we get

$\begin{equation*} \begin{aligned} \sum\limits_{1\leq i < j\leq m}\sqrt{\det A_{i} A_{j}}\geq & \frac{m(m-1)}{2}\Big(\big(\prod^{m}_{i = 1}\sqrt{\det A_{i}}\big)^{m-1}\Big)^{\frac{1}{m(m-1)/2}}\\ = &\frac{m(m-1)}{2}(\prod^{m}_{i = 1}\det A_{i})^{\frac{1}{m}}. \end{aligned} \end{equation*}$

The desired result follows by Theorem 7 and the inequality above.

3. Results involving multiple sector matrices

We present two lemmas which are going to be needed in our proofs. The first lemma is known as the Ostrowski-Taussky inequality.

Lemma 2. [6, Theorem 7.8.19] Let $A\in \mathbb{M}_{n}$ . If $\mathfrak{R}(A)$ is positive definite, then

$\begin{equation*} \det(\mathfrak{R}A)\leq |\det A|. \end{equation*}$

The next lemma is a reverse of the Ostrowski-Taussky inequality proved by Lin ^[14].

Lemma 3. [14, Lemma 2.6] Let $A\in \mathbb{M}_{n}$ with $W(A)\subset S_{\theta}$ . Then

$\begin{equation*} \sec^{n}(\theta)\det(\mathfrak{R}A)\geq |\det A|. \end{equation*}$

Now, we give the following generation of Theorem 7 for sector matrices.

Theorem 8. Let $A_{i} \in\mathbb{M}_{n}, W(A_{i})\subset S_{\theta}, i = 1, \ldots, m$ . Then

$\begin{equation*} |\det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)|\geq \cos^{n}(\theta) \sum\limits^{m}_{i = 1}|\det A_{i}|+(2^{n}-2)\cos^{n}(\theta)\sum\limits^{m}_{1\leq i < j\leq m}\sqrt{|\det A_{i} A_{j}|}. \end{equation*}$

Proof. Compute

where the first inequality above is by Lemma 2; the second is due to Theorem 7 and the last inequality holds by Lemma 3. $\square$

Similarly to Theorem 8, the generations of Corollary 1 and Corollary 2 for sector matrices are as follows.

Corollary 3. Let $A_{i} \in\mathbb{M}_{n}, W(A_{i})\subset S_{\theta}, i = 1, \ldots, m$ . Then

$\begin{equation*} |\det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)|\geq \cos^{n}(\theta) \Big(\sum\limits^{m}_{i = 1}\sqrt{|\det A_{i}|}\Big)^{2}+(2^{n}-4)\cos^{n}(\theta)\sum\limits_{1\leq i < j\leq m}\sqrt{|\det A_{i} A_{j}|}. \end{equation*}$

Corollary 4. Let $A_{i} \in\mathbb{M}_{n}, W(A_{i})\subset S_{\theta}, i = 1, \ldots, m$ . Then

$\begin{equation*} |\det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)|\geq \cos^{n}(\theta) \sum\limits^{m}_{i = 1}|\det A_{i}|+m(m-1)(2^{n-1}-1)\cos^{n}(\theta)|\prod\limits^{m}_{i = 1}\det A_{i}|^{\frac{1}{m}}. \end{equation*}$

For $A \in\mathbb{M}_{n}$ , recall the Cartesian decomposition (see, e.g., [, p.7]) $A = \mathfrak{R}A +i \mathfrak{S}A,$ where $\mathfrak{R}A = \frac{1}{2}(A+A^{*}), \; \; \mathfrak{S}A = \frac{1}{2i}(A-A^{*}).$ If $\mathfrak{R}A$ and $\mathfrak{S}A$ are positive definie, We say $A$ is accretive-dissipative. For more details about this class of matrices, we refer to ^[8,10,15]. Note that if $A$ is accretive-dissipative, then $W(e^{-i\pi/4}A)\subset S_{\pi/4}$ . Thus, we have the following corollaries.

Corollary 5. Let $A_{i} \in\mathbb{M}_{n}, i = 1, \ldots, m$ , be accretive-dissipative. Then

$\begin{equation*} |\det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)|\geq 2^{-\frac{n}{2}}\sum\limits^{m}_{i = 1}|\det A_{i}|+(2^{\frac{n}{2}}-2^{1-\frac{n}{2}})\sum\limits_{1\leq i < j\leq m}\sqrt{|\det A_{i}A_{j}|}. \end{equation*}$

Corollary 6. Let $A_{i} \in\mathbb{M}_{n}, i = 1, \ldots, m$ , be accretive-dissipative. Then

$\begin{equation*} |\det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)|\geq 2^{-\frac{n}{2}} \Big(\sum\limits^{m}_{i = 1}\sqrt{|\det A_{i}|}\Big)^{2}+(2^{\frac{n}{2}}-2^{2-\frac{n}{2}})\sum\limits_{1\leq i < j\leq m}\sqrt{|\det A_{i} A_{j}|}. \end{equation*}$

Corollary 7. Let $A_{i} \in\mathbb{M}_{n}, i = 1, \ldots, m$ , be accretive-dissipative. Then

$\begin{equation*} |\det \Big(\sum\limits^{m}_{i = 1}A_{i}\Big)|\geq 2^{-\frac{n}{2}} \sum\limits^{m}_{i = 1}|\det A_{i}|+m(m-1)(2^{\frac{n}{2}-1}-2^{-\frac{n}{2}} ) |\prod\limits^{m}_{i = 1}\det A_{i}|^{\frac{1}{m}}. \end{equation*}$

Next, We give the following complement of Theorem 6. Clearly, when m = 3, Theorem 9 becomes Theorem 5.

Theorem 9. Let $A_{i} \in\mathbb{M}_{n}, W(A_{i})\subset S_{\theta}, i = 1, \ldots, m$ . Then

$\begin{equation*} \begin{aligned} |\det \Big(\sum^{m}_{i = 1}A_{i}\Big)|\geq & \sum\limits_{1\leq i < j\leq m}\prod^{m}_{k = 1}\bigg(\frac{\left|\det A^{(k)}_{i}\right|}{\left|\det A^{(k-1)}_{i}\right|}+\frac{\left|\det A^{(k)}_{j}\right|}{\left|\det A^{(k-1)}_{j}\right|}\bigg)\cos^{k}(\theta)\\ &-(m-2)\sum^{m}_{i = 1}\left|\det A_{i}\right|. \end{aligned} \end{equation*}$

Proof. Let $B, C\in\mathbb{M}_{n}$ be positive definite. Haynsworth ^[4] proved

$\begin{equation*} \begin{aligned} \frac{\det( B^{(k)}+ C^{(k)})}{\det (B^{(k-1)}+ C^{(k-1)})} \geq \frac{\det B^{(k)}}{\det B^{(k-1)}}+\frac{\det C^{(k)}}{\det C^{(k-1)}}. \end{aligned} \end{equation*}$

Taking products for $k$ from $1$ to $n$ yields

$\begin{equation} \det( B+ C)\geq \prod\limits^{m}_{k = 1}\Big( \frac{\det B^{(k)}}{\det B^{(k-1)}}+\frac{\det C^{(k)}}{\det C^{(k-1)}}\Big). \end{equation}$

(3.1)

Then, we have

$\begin{equation*} \begin{aligned} |\det \Big(\sum^{m}_{i = 1}A_{i}\Big)|\geq &\det \Big(\mathfrak{R}\big(\sum^{m}_{i = 1}A_{i}\big)\Big)\\ = &\det \Big(\sum^{m}_{i = 1}\mathfrak{R}A_{i}\Big) \\ \geq & \sum\limits_{1\leq i < j\leq m}\det (\mathfrak{R}A_{i}+\mathfrak{R}A_{j})-(m-2)\sum^{m}_{i = 1}\det \mathfrak{R}A_{i}\\ \geq &\sum\limits_{1\leq i < j\leq m}\prod^{m}_{k = 1}\bigg(\frac{\det \mathfrak{R}A^{(k)}_{i}}{\det \mathfrak{R}A^{(k-1)}_{i}}+\frac{\det \mathfrak{R}A^{(k)}_{j}}{\det \mathfrak{R}A^{(k-1)}_{j}}\bigg)-(m-2)\sum^{m}_{i = 1}\det \mathfrak{R}A_{i} \\ \geq &\sum\limits_{1\leq i < j\leq m}\prod^{m}_{k = 1}\bigg(\frac{\det \mathfrak{R}A^{(k)}_{i}}{\left|\det A^{(k-1)}_{i}\right|}+\frac{\det \mathfrak{R}A^{(k)}_{j}}{\left|\det A^{(k-1)}_{j}\right|}\bigg)-(m-2)\sum^{m}_{i = 1}\left|\det A_{i}\right| \\ \geq &\sum\limits_{1\leq i < j\leq m}\prod^{m}_{k = 1}\bigg(\frac{\left|\det A^{(k)}_{i}\right|}{\left|\det A^{(k-1)}_{i}\right|}+\frac{\left|\det A^{(k)}_{j}\right|}{\left|\det A^{(k-1)}_{j}\right|}\bigg)\cos^{k}(\theta)\\ &-(m-2)\sum^{m}_{i = 1}\left|\det A_{i}\right|, \end{aligned} \end{equation*}$

where the first inequality is by Lemma 2; the second is by (2.2); the third is by (3.1); the fourth is by Lemma 2 and the last inequality is due to Lemma 3.

Acknowledgments

The work was supported by National Natural Science Foundation of China (NNSFC) [grant number 11971294], the Key Research Project of Henan Higher Education Institutions under Grant 20A120003 and the Young Talents Fund of HUEL under Grant hncjzfdxqnbjrc201607.

Conflict of interest

We declare no conflict of interest.

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AIMS Mathematics

Some results involving multiple matrices

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1. Introduction

2. Results involving multiple positive definite matrices

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AIMS Mathematics

Some results involving multiple matrices

Related Papers:

Abstract

1. Introduction

2. Results involving multiple positive definite matrices

3. Results involving multiple sector matrices

Acknowledgments

Conflict of interest

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1. Introduction

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