On some Fischer-type determinantal inequalities for accretive-dissipative matrices

Abstract

In this note, we give some refinements of Fischer-type determinantal inequalities for accretive-dissipative matrices which are due to Lin (Linear Algebra Appl. 438:2808-2812, 2013) and Ikramov (J. Math. Sci. (N.Y.) 121:2458-2464, 2004).

1 Introduction

Let ${\mathbb{M}}_n(\mathbf{C})$ be the space of complex $n\times n$ matrices. For any $A\in {\mathbb{M}}_n(\mathbf{C})$, the conjugate transpose of A is denoted by $A^{\ast }$. $A\in {\mathbb{M}}_n(\mathbf{C})$ is said to be accretive-dissipative if it has the Hermitian decomposition

$$A=B+iC,B=B^{\ast },C=C^{\ast },$$

(1.1)

where both matrices B and C are positive definite. For simplicity, let A, B, C be partitioned as

$$\left(\begin{array}{cc}{A}_{11}& A_{12}\\ A_{21}& A_{22}\end{array}\right)=\left(\begin{array}{cc}{B}_{11}& B_{12}\\ B_{12}^{\ast }& B_{22}\end{array}\right)+i\left(\begin{array}{cc}{C}_{11}& C_{12}\\ C_{12}^{\ast }& C_{22}\end{array}\right)$$

(1.2)

such that the diagonal blocks $A_{11}$ and $A_{22}$ are of order k and l ($k>0$, $l>0$ and $k+l=n$), respectively, and let $m=min\{k,l\}$.

If $B=I_n$ in (1.1), then an accretive-dissipative matrix $A\in {\mathbb{M}}_n(\mathbf{C})$ is called a Buckley matrix.

If $A\in {\mathbb{M}}_n(\mathbf{C})$ is partitioned as

$$\left(\begin{array}{cc}{A}_{11}& A_{12}\\ A_{21}& A_{22}\end{array}\right),$$

where $A_{11}$ is a nonsingular submatrix, then the matrix $A/A_{11}:=A_{22}-A_{21}{A}_{11}^{-1}{A}_{12}$ is called the Schur complement of the submatrix $A_{11}$ in A. For a nonsingular matrix A, its condition number is denoted by $\kappa (A):=\sqrt{\frac{{\lambda }_{max}(A^{\ast }A)}{{\lambda }_{min}(A^{\ast }A)}}$ which is the ratio of largest and smallest singular values of A. For Hermitian matrices $B,C\in {\mathbb{M}}_n(\mathbf{C})$, we write $B\ge C$ if $B-C$ is positive-semidefinite.

If $A\in {\mathbb{M}}_n(\mathbf{C})$ is positive definite and partitioned as in (1.2), then the famous Fischer-type determinantal inequality is proved [[1], p.478]:

$$detA\le det{A}_{11}det{A}_{22}.$$

(1.3)

If $A\in {\mathbb{M}}_n(\mathbf{C})$ is an accretive-dissipative matrix and partitioned as in (1.2), Ikramov [2] first proved the determinantal inequality for A:

$$|detA|\le 3^m|det{A}_{11}||det{A}_{22}|.$$

(1.4)

Very recently, Lin [[3], Theorem 8] got a stronger result than (1.4) as follows.

If $A\in {\mathbb{M}}_n(\mathbf{C})$ is an accretive-dissipative matrix, then

$$|detA|\le 2^{\frac{3}{2}m}|det{A}_{11}||det{A}_{22}|.$$

(1.5)

For Buckley matrices, the stronger bound was obtained by Ikramov [2]:

$$|detA|\le {\left(\frac{1+\sqrt{17}}{4}\right)}^m|det{A}_{11}||det{A}_{22}|.$$

(1.6)

The purpose of this paper is to give refinements of (1.5) and (1.6). Our main results can be stated as follows.

Theorem 1 Let $A\in {\mathbb{M}}_n(\mathbf{C})$ be accretive-dissipative and partitioned as in (1.2). Then

$$|detA|\le 2^{\frac{1}{2}m}{(1+{\left(\frac{1-\kappa }{1+\kappa }\right)}^2)}^m|det{A}_{11}||det{A}_{22}|,$$

(1.7)

where κ is the maximum of the condition numbers of B and C.

Because of $2^{\frac{m}{2}}{(1+{(\frac{\kappa -1}{\kappa +1})}^2)}^m\le 2^{\frac{3}{2}m}$, inequality (1.7) is a refinement of inequality (1.5).

Theorem 2 Let $A\in {\mathbb{M}}_n(\mathbf{C})$ be a Buckley matrix and partitioned as in (1.2). Then

$$|detA|\le {\left(\frac{d^2+d\sqrt{16+d^2}+8}{8}\right)}^{\frac{m}{2}}|det{A}_{11}||det{A}_{22}|,$$

(1.8)

where κ is the condition number of C and $d={(\frac{\kappa -1}{\kappa +1})}^2\in [0,1]$.

It is clear that inequality (1.8) improves (1.6). In fact, since the function

$$f(x)={\left(\frac{x^2+x\sqrt{16+x^2}+8}{8}\right)}^{\frac{m}{2}}$$

is increasing for $x\in [0,1]$, thus we have

$$f(d)={\left(\frac{d^2+d\sqrt{16+d^2}+8}{8}\right)}^{\frac{m}{2}}\le f(1)={\left(\frac{\sqrt{17}+1}{4}\right)}^m,$$

which implies that (1.8) is a refinement of (1.6).

2 Proofs of main results

To achieve the proofs of Theorem 1 and Theorem 2, we need the following lemmas.

Lemma 3 [[4], Property 6]

Let $A\in {\mathbb{M}}_n(\mathbf{C})$ be accretive-dissipative and partitioned as in (1.2). Then $A/A_{11}:=A_{22}-A_{21}{A}_{11}^{-1}{A}_{12}$, the Schur complement of $A_{11}$ in A is also accretive-dissipative.

Lemma 4 [[2], Lemma 1]

Let $A\in {\mathbb{M}}_n(\mathbf{C})$ be accretive-dissipative and partitioned as in (1.2). Then $A^{-1}=E-iF$ with $E={(B+C{B}^{-1}C)}^{-1}$ and $F={(C+B{C}^{-1}B)}^{-1}$.

Lemma 5 [[2], Lemma 5]

Let $A_1,A_2\in {\mathbb{M}}_n(\mathbf{C})$ be accretive-dissipative matrices and let

$$A_1=B_1+i{C}_1,A_2=B_2+i{C}_2$$

be the Hermitian decompositions of these matrices. If

$$B_1\le B_2,C_1\le C_2,$$

then

$$|det{A}_1|\le |det{A}_2|.$$

(2.1)

Lemma 6 [[3], Lemma 6]

Let $B,C\in {\mathbb{M}}_n(\mathbf{C})$ be positive definite. Then

$$|det(B+iC)|\le det(B+C)\le 2^{\frac{n}{2}}|det(B+iC)|.$$

(2.2)

Lemma 7 [[5], (6)]

Let $A\in {\mathbb{M}}_n(\mathbf{C})$ be positive definite. Then

$$A_{12}{A}_{22}^{-1}{A}_{21}\le {\left(\frac{{\lambda }_1-{\lambda }_n}{{\lambda }_1+{\lambda }_n}\right)}^2{A}_{11},$$

(2.3)

where ${\lambda }_1$ and ${\lambda }_n$ are the largest and the smallest eigenvalues of A.

Lemma 8 [[6], Lemma 3.2]

Let $B,C\in {\mathbb{M}}_n(\mathbf{C})$ be Hermitian and assume that $B>0$. Then

$$B+C{B}^{-1}C\ge 2C.$$

(2.4)

Remark 1 A stronger inequality than (2.4) was given in Lin [[7], Lemma 2.2]: Let $A>0$ and any Hermitian B. Then $A\mathrm{♯}(B{A}^{-1}B)\ge B$.

In what follows, we give the proofs of Theorem 1 and Theorem 2.

Proof of Theorem 1 By Lemma 4, we obtain

$$\begin{array}{rcl}A/A_{11}& =& A_{22}-A_{21}{A}_{11}^{-1}{A}_{12}\\ =& B_{22}+i{C}_{22}-(B_{12}^{\ast }+i{C}_{12}^{\ast }){(B_{11}+i{C}_{11})}^{-1}(B_{12}+i{C}_{12})\\ =& B_{22}+i{C}_{22}-(B_{12}^{\ast }+i{C}_{12}^{\ast })(E_k-i{F}_k)(B_{12}+i{C}_{12}).\end{array}$$

Furthermore, we have

$$E_k={(B_{11}+C_{11}{B}_{11}^{-1}{C}_{11})}^{-1},F_k={(C_{11}+B_{11}{C}_{11}^{-1}{B}_{11})}^{-1},$$

where $E_k$ and $F_k$ are positive definite.

By Lemma 8 and the operator reverse monotonicity of the inverse, we get

$$E_k\le \frac{1}{2}{C}_{11}^{-1},F_k\le \frac{1}{2}{B}_{11}^{-1}.$$

(2.5)

Set $A/A_{11}=R+iS$ with $R=R^{\ast }$, $S=S^{\ast }$. By Lemma 3, it is easy to know that R, S are positive definite. A simple calculation shows

$$\begin{array}{c}R=B_{22}-B_{12}^{\ast }{E}_k{B}_{12}+C_{12}^{\ast }{E}_k{C}_{12}-B_{12}^{\ast }{F}_k{C}_{12}-C_{12}^{\ast }{F}_k{B}_{12},\hfill \\ S=C_{22}+B_{12}^{\ast }{F}_k{B}_{12}-C_{12}^{\ast }{F}_k{C}_{12}-C_{12}^{\ast }{E}_k{B}_{12}-B_{12}^{\ast }{E}_k{C}_{12}.\hfill \end{array}$$

By the inequalities

$$(B_{12}^{\ast }\pm C_{12}^{\ast })F_k(B_{12}\pm C_{12})\ge 0,(B_{12}^{\ast }\pm C_{12}^{\ast })E_k(B_{12}\pm C_{12})\ge 0,$$

it can be proved that

$$\begin{array}{c}\pm (B_{12}^{\ast }{F}_k{C}_{12}+C_{12}^{\ast }{F}_k{B}_{12})\le B_{12}^{\ast }{F}_k{B}_{12}+C_{12}^{\ast }{F}_k{C}_{12},\hfill \\ \pm (C_{12}^{\ast }{E}_k{B}_{12}+B_{12}^{\ast }{E}_k{C}_{12})\le B_{12}^{\ast }{E}_k{B}_{12}+C_{12}^{\ast }{E}_k{C}_{12}.\hfill \end{array}$$

Thus

$$R+S\le B_{22}+2B_{12}^{\ast }{F}_k{B}_{12}+C_{22}^{\ast }+2C_{12}^{\ast }{E}_k{C}_{12}.$$

(2.6)

Since B, C are positive definite, we have by Lemma 7

$$B_{12}{B}_{22}^{-1}{B}_{12}^{\ast }\le {\left(\frac{{\lambda }_1-{\lambda }_n}{{\lambda }_1+{\lambda }_n}\right)}^2{B}_{11},C_{12}{C}_{22}^{-1}{C}_{12}^{\ast }\le {\left(\frac{{\lambda }_1^{\mathrm{\prime }}-{\lambda }_n^{\mathrm{\prime }}}{{\lambda }_1^{\mathrm{\prime }}+{\lambda }_n^{\mathrm{\prime }}}\right)}^2{C}_{11}.$$

(2.7)

By (2.7), it is easy to know that

$$B_{12}^{\ast }{B}_{11}^{-1}{B}_{12}\le {\left(\frac{{\lambda }_1-{\lambda }_n}{{\lambda }_1+{\lambda }_n}\right)}^2{B}_{22},C_{12}^{\ast }{C}_{11}^{-1}{C}_{12}\le {\left(\frac{{\lambda }_1^{\mathrm{\prime }}-{\lambda }_n^{\mathrm{\prime }}}{{\lambda }_1^{\mathrm{\prime }}+{\lambda }_n^{\mathrm{\prime }}}\right)}^2{C}_{22}.$$

(2.8)

In (2.7) and (2.8), ${\lambda }_1$ and ${\lambda }_n$ (${\lambda }_1^{\mathrm{\prime }}$ and ${\lambda }_n^{\mathrm{\prime }}$) are the largest and the smallest eigenvalues of B (C), respectively.

Note that $f(x)={(\frac{x-1}{x+1})}^m$ ($m\ge 1$) is increasing for $x\in [1,\mathrm{\infty })$. Without loss of generality, assume $m=l$. Then we have

$$\begin{array}{rcl}|detA/A_{11}|& =& |detR+iS|\\ \le & det(R+S)\text{(by Lemma 6)}\\ \le & det(B_{22}+2B_{12}^{\ast }{F}_k{B}_{12}+C_{22}+2C_{12}^{\ast }{E}_k{C}_{12})\text{(by (2.6))}\\ \le & det(B_{22}+B_{12}^{\ast }{B}_{11}^{-1}{B}_{12}+C_{22}+C_{12}^{\ast }{C}_{11}^{-1}{C}_{12})\text{(by (2.5))}\\ \le & det(B_{22}+C_{22}+{\left(\frac{{\lambda }_1-{\lambda }_n}{{\lambda }_1+{\lambda }_n}\right)}^2{B}_{22}+{\left(\frac{{\lambda }_1^{\mathrm{\prime }}-{\lambda }_n^{\mathrm{\prime }}}{{\lambda }_1^{\mathrm{\prime }}+{\lambda }_n^{\mathrm{\prime }}}\right)}^2{C}_{22})\text{(by (2.8))}\\ =& det(B_{22}+C_{22}+{\left(\frac{\frac{{\lambda }_1}{{\lambda }_n}-1}{\frac{{\lambda }_1}{{\lambda }_n}+1}\right)}^2{B}_{22}+{\left(\frac{\frac{{\lambda }_1^{\mathrm{\prime }}}{{\lambda }_n^{\mathrm{\prime }}}-1}{\frac{{\lambda }_1^{\mathrm{\prime }}}{{\lambda }_n^{\mathrm{\prime }}}+1}\right)}^2{C}_{22})\\ \le & {(1+{\left(\frac{\kappa -1}{\kappa +1}\right)}^2)}^{m}det(B_{22}+C_{22})\\ \le & 2^{\frac{m}{2}}{(1+{\left(\frac{\kappa -1}{\kappa +1}\right)}^2)}^m|det(B_{22}+i{C}_{22})|\text{(by Lemma 6)}\\ =& 2^{\frac{m}{2}}{(1+{\left(\frac{\kappa -1}{\kappa +1}\right)}^2)}^m|det{A}_{22}|,\end{array}$$

where $\kappa =max(\frac{{\lambda }_1}{{\lambda }_n},\frac{{\lambda }_1^{\mathrm{\prime }}}{{\lambda }_n^{\mathrm{\prime }}})\ge 1$, i.e., the maximum of the condition numbers of B and C.

By noting

$$|detA|=|det{A}_{11}||det(A/A_{11})|,$$

the proof is completed. □

Remark 2 In fact, a reverse direction to the inequality of Theorem 1 has been given in Lin [[8], Theorem 1.2].

Proof of Theorem 2 The proof is similar to Theorem 1. By Lemma 4, we obtain

$$\begin{array}{rcl}A/A_{11}& =& A_{22}-A_{21}{A}_{11}^{-1}{A}_{12}\\ =& I_l+i{C}_{22}-C_{12}^{\ast }{(I_k+i{C}_{11})}^{-1}{C}_{12}\\ =& I_l+i{C}_{22}+C_{12}^{\ast }(E_k-i{F}_k)C_{12}\end{array}$$

with

$$E_k={(I_k+C_{11}^2)}^{-1},F_k={(C_{11}+C_{11}^{-1})}^{-1}.$$

By Lemma 8 and the operator reverse monotonicity of the inverse, we get

$$E_k\le \frac{1}{2}{C}_{11}^{-1},F_k\le \frac{1}{2}{I}_k.$$

(2.9)

Set $A/A_{11}=R+iS$ with $R=R^{\ast }$, $S=S^{\ast }$. By Lemma 3, it is easy to know that R and S are positive definite. A simple calculation shows

$$\begin{array}{c}R=I_l+C_{12}^{\ast }{E}_k{C}_{12},\hfill \\ S=C_{22}-C_{12}^{\ast }{F}_k{C}_{12},\hfill \end{array}$$

where $F_k$ is positive definite. Therefore

$$S\le C_{22}.$$

By (2.9), we have

$$R\le I_l+\frac{1}{2}{C}_{12}^{\ast }{C}_{11}^{-1}{C}_{12}.$$

As C is positive definite, we get by (2.8)

$$C_{12}^{\ast }{C}_{11}^{-1}{C}_{12}\le {\left(\frac{{\lambda }_1^{\mathrm{\prime }}-{\lambda }_n^{\mathrm{\prime }}}{{\lambda }_1^{\mathrm{\prime }}+{\lambda }_n^{\mathrm{\prime }}}\right)}^2{C}_{22},$$

(2.10)

where ${\lambda }_1^{\mathrm{\prime }}$, ${\lambda }_n^{\mathrm{\prime }}$ are the largest and the smallest eigenvalues of C. So we have

$$R\le I_l+\frac{1}{2}{\left(\frac{{\lambda }_1^{\mathrm{\prime }}-{\lambda }_n^{\mathrm{\prime }}}{{\lambda }_1^{\mathrm{\prime }}+{\lambda }_n^{\mathrm{\prime }}}\right)}^2{C}_{22}.$$

Without loss of generality, assume $m=l$. Thus we get

$$\begin{array}{rcl}|detA/A_{11}|& =& |detR+iS|\\ \le & det(I_l+\frac{1}{2}{\left(\frac{{\lambda }_1^{\mathrm{\prime }}-{\lambda }_n^{\mathrm{\prime }}}{{\lambda }_1^{\mathrm{\prime }}+{\lambda }_n^{\mathrm{\prime }}}\right)}^2{C}_{22}+i{C}_{22})\text{(by Lemma 5)}\\ =& det(I_l+\frac{1}{2}{\left(\frac{\frac{{\lambda }_1^{\mathrm{\prime }}}{{\lambda }_n^{\mathrm{\prime }}}-1}{\frac{{\lambda }_1^{\mathrm{\prime }}}{{\lambda }_n^{\mathrm{\prime }}}+1}\right)}^2{C}_{22}+i{C}_{22})\\ =& det(I_l+\frac{1}{2}{\left(\frac{\kappa -1}{\kappa +1}\right)}^2{C}_{22}+i{C}_{22}),\end{array}$$

where $\kappa =\frac{{\lambda }_1^{\mathrm{\prime }}}{{\lambda }_n^{\mathrm{\prime }}}$.

Let ${\gamma }_1\ge {\gamma }_2\ge \cdots \ge {\gamma }_l$ be the eigenvalues of $C_{22}$ and we denote $d={(\frac{\kappa -1}{\kappa +1})}^2$. Then it is easy to know that

$$|detA/A_{11}|\le |det(I_l+\frac{1}{2}{\left(\frac{\kappa -1}{\kappa +1}\right)}^2{C}_{22}+i{C}_{22})|$$

(2.11)

$$=\prod _{j=1}^l|(1+\frac{1}{2}d{\gamma }_j)+i{\gamma }_j|$$

(2.12)

$$=\prod _{j=1}^l{[{(1+\frac{1}{2}d{\gamma }_j)}^2+{\gamma }_j^2]}^{\frac{1}{2}}.$$

(2.13)

On the other hand,

$$|det{A}_{22}|=|det(I+i{C}_{22})|=\prod _{j=1}^l{(1+{\gamma }_j^2)}^{\frac{1}{2}}.$$

(2.14)

By (2.13) and (2.14), we have

$$\begin{array}{rcl}|detA|& =& |det{A}_{11}||det(A/A_{11})|\\ =& \frac{|det(A/A_{11})|}{|det{A}_{22}|}|det{A}_{11}||det{A}_{22}|\\ \le & \frac{{\prod }_{j=1}^l{[{(1+\frac{1}{2}d{\gamma }_j)}^2+{\gamma }_j^2]}^{\frac{1}{2}}}{{\prod }_{j=1}^l{(1+{\gamma }_j^2)}^{\frac{1}{2}}}|det{A}_{11}||det{A}_{22}|.\end{array}$$

By noting

$$\underset{x\ge 0}{max}\frac{{(1+\frac{d}{2}x)}^2+x^2}{1+x^2}=\frac{d^2+d\sqrt{16+d^2}+8}{8},0\le d\le 1,$$

the proof is completed. □