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Some determinantal inequalities for accretive-dissipative matrices

Abstract

This article aims to obtain some determinantal inequalities for accretive-dissipative matrices which are generalizations of the determinantal inequalities presented by Lin (Linear Algebra Appl. 438:2808-2812, 2013). At the same time, we give some numerical examples which show the effectiveness of our results.

1 Introduction

Let M n (C) be the space of complex matrices of size n×n matrices. A M n (C) is said to be accretive-dissipative, if, in its Toeplitz decomposition

A=B+iC,B= B ,C= C ,
(1.1)

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

( A 11 A 12 A 21 A 22 )=( B 11 B 12 B 12 B 22 )+i( C 11 C 12 C 12 C 22 )
(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 A M n (C) is partitioned as

( A 11 A 12 A 21 A 22 ),

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.

If A M n (C) is positive definite and partitioned as in (1.2), then the inequalities [[1], Lemma 6] hold:

|detA|= | det ( B + i C ) | | det ( B + C ) | 2 n 2 | det ( B + i C ) | .
(1.3)

If A M n (C) is positive definite and partitioned as in (1.2), then the famous Fischer-type determinantal inequality is proved [[2], p.478]:

detAdet A 11 det A 22 .
(1.4)

If A M n (C) is an accretive-dissipative matrix and partitioned as in (1.2), Ikramov [3] first proved the determinantal inequality for A:

|detA| 3 m |det A 11 ||det A 22 |.
(1.5)

Lin [[1], Theorem 8] got a stronger result than (1.5) as follows:

If A M n (C) is an accretive-dissipative matrix, then

|detA| 2 3 2 m |det A 11 ||det A 22 |.
(1.6)

The purpose of this paper is to give some generalizations of (1.3) and (1.6). Our main results can be stated as follows.

Theorem 1 Let B,C M n (C) be positive definite and x, y be positive real numbers. Then

| det ( B + i C ) | det(B+C) ( x 2 + y 2 ) n 2 | det ( B x + i C y ) | .
(1.7)

When x=y, the inequality det(B+C) 2 n 2 |det(B+iC)| is a special case of Theorem 1. Thus (1.5) is a generalization of the inequality |det(B+iC)|det(B+C) 2 n 2 |det(B+iC)| [[1], Lemma 6].

Theorem 2 Let A M n (C) be accretive-dissipative and partitioned as in (1.2), and let x, y be positive real numbers. Then

|detA| ( x 2 + y 2 ) n 2 | det ( B 11 x + i C 11 y ) | | det ( B 22 x + i C 22 y ) | .
(1.8)

When x=y, we get the inequality |detA| 2 n 2 |det A 11 ||det A 22 | [[1], (3.1)], which is a special case of Theorem 2.

Theorem 3 Let A M n (C) be accretive-dissipative and partitioned as in (1.2), and let x, y be positive real numbers. Then

|detA| 2 m ( x 2 + y 2 ) m 2 |det A 11 | | det ( B 22 x + i C 22 y ) | .
(1.9)

When x=y, we get the inequality [[1], (3.2)]

|detA| 2 3 2 m |det A 22 ||det A 11 |,

which is a special case of (1.7).

2 Proofs of main results

To achieve the proofs of Theorem 1, Theorem 2 and Theorem 3, we need the following lemmas.

Lemma 4 [[4], Property 6]

Let A M n (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 5 [[3], Lemma 1]

Let A M n (C) be accretive-dissipative and partitioned as in (1.2). Then A 1 =EiF with E= ( B + C B 1 C ) 1 and F= ( C + B C 1 B ) 1 .

Lemma 6 [[3], Lemma 4]

Let B,C M n (C) be Hermitian and assume that B>0. Then

B+C B 1 C2C.
(2.1)

Remark 1 A stronger inequality than (2.4) was given in Lin [[5], Lemma 2.2]: Let A>0 and any Hermitian B. Then A(B A 1 B)B.

Proof of Theorem 1 Let λ j , j=1,,n, be the eigenvalues of B 1 2 C B 1 2 , where B 1 2 means the unique positive definite square root of B. Then we have

|1+i λ j ||1+ λ j | x 2 + y 2 | 1 x + i λ j y | .
(2.2)

The first inequality follows from [[6], Theorem 2.2], while the second one we prove is as follows:

| det ( B + i C ) | | det ( B + C ) | = | det B 1 2 ( I + B 1 2 C B 1 2 ) B 1 2 | = | det B 1 2 | | det B 1 2 | | det ( I + B 1 2 C B 1 2 ) | = | det B | | j = 1 n | 1 + λ j | | det B | | j = 1 n x 2 + y 2 | 1 x + i λ j y | (by (2.2)) = ( x 2 + y 2 ) n 2 | det B | | det ( I x + i y B 1 2 C B 1 2 ) | = ( x 2 + y 2 ) n 2 | det ( B x + i y C ) | .

The proof is completed. □

Proof of Theorem 2

| det A | det ( B + C ) (by Theorem 1) det ( B 11 + C 11 ) det ( B 22 + C 22 ) (by (1.3)) ( x 2 + y 2 ) k 2 | det B 11 x + i y C 11 | ( x 2 + y 2 ) l 2 | det B 22 x + i y C 22 | (by (1.6)) = ( x 2 + y 2 ) n 2 | det B 11 x + i y C 11 | | det B 22 x + i y C 22 | .

The proof is completed. □

Proof of Theorem 3 By Lemma 4, we obtain

A / A 11 = A 22 A 21 A 11 1 A 12 = B 22 + i C 22 ( B 12 + i C 12 ) ( B 11 + i C 11 ) 1 ( B 12 + i C 12 ) = B 22 + i C 22 ( B 12 + i C 12 ) ( E k i F k ) ( B 12 + i C 12 ) .

Furthermore, by Lemma 5, 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 a simple computation, we obtain

A/ A 11 =R+iS.

By Lemma 4, it is easy to know that R, S are positive definite and we have

R = B 22 B 12 E k B 12 + C 12 E k C 12 B 12 F k C 12 C 12 F k B 12 , S = C 22 + B 12 F k B 12 C 12 F k C 12 C 12 E k B 12 B 12 E k C 12 .

By the inequalities

( B 12 ± C 12 ) F k ( B 12 ± C 12 )0, ( B 12 ± C 12 ) E k ( B 12 ± C 12 )0,

it can be proved that

± ( B 12 F k C 12 + C 12 F k B 12 ) B 12 F k B 12 + C 12 F k C 12 , ± ( C 12 E k B 12 + B 12 E k C 12 ) B 12 E k B 12 + C 12 E k C 12 .

Thus

R+S B 22 +2 B 12 F k B 12 + C 22 +2 C 12 E k C 12 .
(2.3)

By Lemma 6 and the operator reverse monotonicity of the inverse, we get

E k 1 2 C 11 1 , F k 1 2 B 11 1 .
(2.4)

As B, C are positive definite, we also have

B 22 > B 12 B 11 1 B 12 , C 22 > C 12 C 11 1 C 12 .
(2.5)

Without loss of generality, assume m=l. Then we have

| det A / A 11 | = | det R + i S | det ( R + S ) (by (1.7)) det ( B 22 + 2 B 12 F k B 12 + C 22 + 2 C 12 E k C 12 ) (by (2.3)) det ( B 22 + B 12 B 11 1 B 12 + C 22 + C 12 C 11 1 C 12 ) (by (2.4)) < det ( 2 ( B 22 + C 22 ) ) (by (2.5)) 2 m ( x 2 + y 2 ) m 2 | det ( B 22 x + i C 22 y ) | (by (1.6)).

By noting

|detA|=|det A 11 | | det ( A / A 11 ) | ,

the proof is completed. □

3 Numerical examples

There are many upper bounds for the determinant of the accretive-dissipative matrices which are due to (1.6), (1.8) and (1.9). However, these bounds are incomparable.

In this section, we give some numerical examples to show that (1.8) and (1.9) are better than (1.6) in some cases.

Example 3.1 Let

A = B + i C = ( 1.01 1 1 1.01 ) + i ( 1.01 1 1 1.01 ) = ( 1.01 + 1.01 i 1 + i 1 + i 1.01 + 1.01 i ) .

We calculate that |detA|=4.0402.

By the upper bound of |detA| in (1.6), we have

2 3 2 m |det A 11 ||det A 22 |=5.7706,

where A 11 =1.01+1.01i, A 22 =1.01+1.01i.

Let x=4, y=5. From the upper bound of |detA| in (1.8), we have

( x 2 + y 2 ) n 2 | det ( B 11 x + i C 11 y ) | | det ( B 22 x + i C 22 y ) | =4.2870,

where B 11 =1.01, B 22 =1.01, C 11 =1.01, C 22 =1.01.

Meanwhile, by the upper bound of |detA| in (1.9), we get

2 m ( x 2 + y 2 ) m 2 |det A 11 | | det ( B 22 x + i C 22 y ) | =5.9148.

Example 3.2 Let

A = B + i C = ( 5 1 3 1 2 2 3 2 3 ) + i ( 5 3 1 3 2 1 1 1 4 ) = ( 5 + 5 i 1 + 3 i 3 + i 1 + 3 i 2 + 2 i 2 + i 3 + i 2 + i 3 + 4 i ) .

We calculate that |detA|=119.0378.

By the upper bound of |detA| in (1.6), we have

2 3 2 m |det A 11 ||det A 22 |=384.7077,

where A 11 = ( 5 + 5 i 1 + 3 i 1 + 3 i 2 + 2 i ) , A 22 =3+4i.

Let x=4, y=5. Then, by the upper bound of |detA| in (1.8), we have

( x 2 + y 2 ) n 2 | det ( B 11 x + i C 11 y ) | | det ( B 22 x + i C 22 y ) | =403.3473,

where B 11 = ( 5 1 1 2 ) , C 11 = ( 5 3 3 2 ) , B 22 =3, C 22 =4.

Meanwhile, by the upper bound of |detA| in (1.9), we get

2 m ( x 2 + y 2 ) m 2 |det A 11 | | det ( B 22 x + i C 22 y ) | =382.0149.

From the two examples above, we can obtain that (1.8) and (1.9) are better than (1.6) in some cases.

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Correspondence to Junjian Yang.

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Yang, J. Some determinantal inequalities for accretive-dissipative matrices. J Inequal Appl 2013, 512 (2013). https://doi.org/10.1186/1029-242X-2013-512

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