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On improvements of Fischer’s inequality and Hadamard’s inequality for K 0 -matrices

Abstract

In this paper, the class of K 0 -matrices, which includes positive definite matrices, totally positive matrices, M-matrices and inverse M-matrices, is first introduced and the refinements of Fischer’s inequality and Hadamard’s inequality for K 0 -matrices are obtained. Some previous well-known results for totally nonnegative matrices can be regarded as the special case of this paper.

MSC:15A15.

1 Introduction

All matrices considered in this paper are real. For an n×n matrix A,α,βn={1,2,,n}, the submatrix of A lying in rows indexed by α and the columns indexed by β will be denoted by A[α,β]. If α=β, then the principal submatrix A[α,α] is abbreviated to A[α]. For any αn, let α c denote the complement of α relative to n, and let |α| denote the cardinality of α. If α=, we define detA[]=1. We use S n for the symmetric group on n.

An n×n matrix A is called a P 0 -matrix (P-matrix) if all the principal minors of A are nonnegative (positive). A P-matrix A is called 1-minor symmetric if detA[α,β]detA[β,α]0, whenever |α|=|β|=1+|αβ|. Of course, each of the P-matrices, such as positive definite matrices (PD), totally positive matrices (TP), M-matrices (M) and inverse M-matrices ( M 1 ), is 1-minor symmetric, see [1]. We have known that the following multiplicative principal minor inequalities are classical for PD, M and M 1 matrices:

  • Hadamard: detA i = 1 n a i i ;

  • Fischer: detAdetA[S]detA[ S c ], for Sn;

  • Koteljanskii: detA[ST]detA[ST]detA[S]detA[T], for S,Tn.

These inequalities also hold for totally nonnegative matrices (TN), see [16].

The study of multiplicative principal minor inequalities has been actively going on for many years, many authors have done various wonderful works on this topic, see [711]. In [11], Zhang and Yang improved Hadamard’s inequality for totally nonnegative matrices as follows:

If A=[ a i j ] is an n×n totally nonnegative matrix with i = 1 n a i i >0, then

detAmin { i = 1 n a i i max 1 σ S n ( i = 1 n a i σ ( i ) a σ ( i ) i ) 1 / 2 , min k n a k k i = 1 i k n ( a i i a i k a k i a k k ) } .
(1.1)

Hadamard’s inequality for some subclasses of P 0 -matrices is an important inequality in matrix analysis, inequality (1.1) is the generalization of Hadamard’s inequality for totally nonnegative matrices. It is a noticeable problem to generalize inequality (1.1) for totally nonnegative matrices to other classes of matrices. In this paper we give some new upper bounds of Fischer’s inequality and Hadamard’s inequality for a subclass of P 0 -matrices and extend the corresponding results due to Zhang and Yang (see [11]).

2 Some lemmas

To avoid triviality, we always assume n>1. We will need important Sylvester’s identity for determinants (see [12]).

Lemma 2.1 [12]

Let A be an n×n matrix, αn, and suppose |α|=k (1kn1). Define the (nk)×(nk) matrix B=( b i j ), with i,j α c , by setting b i j =detA[α{i},α{j}] for every i,j α c . B is called the Sylvester matrix of A associated with A[α]. Then Sylvester’s identity states that for each δ,γ α c , with |δ|=|γ|=l,

detB[δ,γ]= ( det A [ α ] ) l 1 detA[αδ,αγ].
(2.1)

For convenience, we introduce the following definition.

Definition 2.1 A P 0 -matrix (P-matrix) A is called a K 0 -matrix (K-matrix) if every principal submatrix of A satisfies Koteljanskii’s inequality.

Obviously, each principal submatrix of a K 0 -matrix (K-matrix) is a K 0 -matrix (K-matrix). Of course, each of the matrices PD, TP, M and M 1 is a K-matrix, the totally nonnegative matrices are K 0 -matrices. In fact, an evident necessary and sufficient condition for a K-matrix was given in [1].

Lemma 2.2 [1]

A P-matrix satisfies Koteljanskii’s inequality if and only if it is 1-minor symmetric.

There are K-matrices that lie in none of the classes PD, TP, M and M 1 , e.g.,

( 4 2 3 1 3 1 1 1 2 ).

Lemma 2.3 If an n×n matrix A=[ a i j ] is a K 0 -matrix, then a i j a j i 0 for i,jn.

Proof For i,jn, i<j, we consider the submatrix A[{i,j}] of the K 0 -matrix A. Since A[{i,j}] is a K 0 -matrix, by the Hadamard’s inequality, we have

detA [ { i , j } ] = a i i a j j a i j a j i a i i a j j ,

therefore

a i j a j i 0for i,jn.

This completes the proof. □

Lemma 2.4 Let A be a K-matrix, B be the Sylvester matrix of A associated with A[α], then B is a K-matrix.

Proof For δ α c , with |δ|=l, by Sylvester’s identity (2.1), we have

detB[δ,δ]= ( det A [ α ] ) l 1 detA[αδ,αδ]>0,

this means that B is a P-matrix.

For δ , γ α c , with | δ |=|γ|=1+| δ γ|= k 1 , by Sylvester’s identity (2.1), we have

det B [ δ , γ ] det B [ γ , δ ] = ( det A [ α ] ) 2 k 1 2 det A [ α δ , α γ ] det A [ α γ , α δ ] > 0 .
(2.2)

Obviously,

| α δ | = | α | + | δ | = 1 + | α | + | δ γ | = 1 + | ( α δ ) ( α γ ) | , | α γ | = | α | + | γ | = 1 + | α | + | δ γ | = 1 + | ( α δ ) ( α γ ) | .

That is,

|α δ |=|αγ|=1+| ( α δ ) (αγ)|.

By Lemma 2.2 and (2.2), we conclude that B is 1-minor symmetric, therefore B is a K-matrix. □

Lemma 2.5 If A=[ a i j ] is an n×n K 0 -matrix, then

i = 1 n a i i [ i = 1 n a i σ ( i ) a σ ( i ) i ] 1 2 for each σ S n .
(2.3)

Proof For each σ S n , by Lemma 2.3, we obtain

a i σ ( i ) a σ ( i ) i 0for i=1,2,,n.

Since each 2×2 principal minor of A is nonnegative, we have

a i i a σ ( i ) i σ ( i ) a i σ ( i ) a σ ( i ) i 0for i=1,2,,n,

thus

( i = 1 n a i i ) 2 i = 1 n a i σ ( i ) a σ ( i ) i ,

hence inequality (2.3) follows. □

3 Main results

In this section, we give some new upper bounds for Fischer’s inequality and Hadamard’s inequality, and extend the corresponding results due to Zhang and Yang (see [11]).

Theorem 3.1 If A=[ a i j ] is an n×n K 0 -matrix, with i = 1 n a i i >0, then

detAdetA[α]detA [ α c ] ( 1 max i α j α c a i j a j i a i i a j j ) .
(3.1)

Proof If A is singular, (3.1) is valid obviously. If A is a nonsingular K 0 -matrix, by Fisher’s inequality, we know that A is a P-matrix, therefore A is a K-matrix. Now we prove (3.1) for the K-matrix by induction on n. When n=2, it is very easy to see that the result is valid. We suppose that the result is valid for all m×m (m<n and n3) K-matrices, let k be the cardinality of α. Note that α and α c are symmetric in inequality (3.1). Without loss of generality, we assume k2. Let

max i α j α c a i j a j i a i i a j j = a i 1 j 1 a j 1 i 1 a i 1 i 1 a j 1 j 1 ,

where i 1 α and j 1 α c .

For any rα and r i 1 , let ω=α{r}, and S=( S i j ) be the (nk+1)×(nk+1) Sylvester matrix of A associated with A[ω]. By Lemma 2.4, S is a K-matrix. Clearly, S r r =detA[α], then, by Fisher’s inequality, we have

detS S r r detS [ α c ] =detA[α]detS [ α c ] ,
(3.2)

where α c =nα. It follows from Sylvester’s identity (2.1) and (3.2) that

detA= det S ( det A [ ω ] ) n k det A [ α ] det S [ α c ] ( det A [ ω ] ) n k .
(3.3)

Let β=n{r}, from Sylvester’s identity (2.1), we obtain

detS [ α c ] = ( det A [ ω ] ) n k 1 detA[β].
(3.4)

Thus, by the inductive hypothesis, we have

detA[β]detA[ω]detA [ α c ] ( 1 max i ω j α c a i j a j i a i i a j j ) .
(3.5)

From (3.3), (3.4) and (3.5), it follows that

detAdetA[α]detA [ α c ] ( 1 max i ω j α c a i j a j i a i i a j j ) .
(3.6)

Clearly,

max i ω j α c a i j a j i a i i a j j = max i α j α c a i j a j i a i i a j j .
(3.7)

Combining (3.6) and (3.7), we obtain inequality (3.1). □

Example 3.1 Now, we consider a 4×4 K-matrix that lies in none of the classes PD, TP, M and M 1 . Let

A=( 15 5 0 3 1 3 6 9 0 3 7 11 1 5 15 30 ).

Then A is 1-minor symmetric and A is a P-matrix. By Lemma 2.2, we know that A is a K-matrix. Let α={1,2}, then α c ={3,4}. By calculating, we have

detA=36<detA [ { 1 , 2 } ] detA [ { 3 , 4 } ] ( 1 max i { 1 , 2 } j { 3 , 4 } a i j a j i a i i a j j ) = 1 , 800 7 ,

therefore inequality (3.1) holds.

Let A[α]=A[{k}] (kn), by Theorem 3.1 and the induction, we can obtain the following conclusion.

Corollary 3.2 If A=[ a i j ] is an n×n K 0 -matrix, with i = 1 n a i i >0, then

detA min k n a k k i = 1 i k n ( a i i a i k a k i a k k ) .

If A=[ a i j ] is a totally nonnegative matrix with i = 1 n a i i >0, Corollary 3.2 is certainly valid, so Corollary 3.2 is the generalization of Theorem 3 in [11].

Theorem 3.3 If A=[ a i j ] is an n×n K 0 -matrix, then

detA i = 1 n a i i max 1 σ S n ( i = 1 n a i σ ( i ) a σ ( i ) i ) 1 / 2 .
(3.8)

Proof If A is singular, by Lemma 2.3 and Lemma 2.5, we know (3.8) is valid. If A is a nonsingular K 0 -matrix, then A is a K-matrix. We suppose that A is a K-matrix in the following.

If n=2, then equality in (3.8) holds. If n>2, then first we prove the following conclusion:

For any 1σ S n , there exists k 0 σ( k 0 ) such that

[ i = 1 n a i σ ( i ) a σ ( i ) i ] 1 2 a k 0 σ ( k 0 ) a σ ( k 0 ) k 0 i = 1 , i k 0 , σ ( k 0 ) n a i i .
(3.9)

Case 1. If i = 1 n a i σ ( i ) a σ ( i ) i =0. For any 1σ S n , by Lemma 2.3, we know that (3.9) certainly holds.

Case 2. If i = 1 n a i σ ( i ) a σ ( i ) i >0. Let T={k|σ(k)k}, then |T|2, it is easy to see that (3.9) is true even with the equality sign for |T|=2. Now we assume that |T|>2.

Suppose that there exists 1σ S n , for any kσ(k), such that

[ i = 1 n a i σ ( i ) a σ ( i ) i ] 1 2 > a k σ ( k ) a σ ( k ) k i = 1 , i k , σ ( k ) n a i i .
(3.10)

Then multiplying all the possible inequalities in (3.10) yields

[ i = 1 n a i σ ( i ) a σ ( i ) i ] | T | 2 > k T [ a k σ ( k ) a σ ( k ) k i = 1 , i k , σ ( k ) n a i i ] .
(3.11)

Let S=nT. If S, by (3.11) we have

[ i T a i σ ( i ) a σ ( i ) i ] | T | 2 [ i S a i i ] | T | > [ i T a i σ ( i ) a σ ( i ) i ] [ i S a i i ] | T | i T a i i | T | 2 .
(3.12)

If S=, by (3.11) we have

[ i T a i σ ( i ) a σ ( i ) i ] | T | 2 > [ i T a i σ ( i ) a σ ( i ) i ] i T a i i | T | 2 .
(3.13)

By (3.12) and (3.13), we have

[ i T a i σ ( i ) a σ ( i ) i ] 1 2 > i T a i i .
(3.14)

On the other hand, since the principal submatrix A[T] of A is a K-matrix, applying Lemma 2.5 to A[T] yields

[ i T a i σ ( i ) a σ ( i ) i ] 1 2 i T a i i ,
(3.15)

which is a contradiction to (3.14), therefore (3.9) holds.

By Fisher’s inequality (3.1) and (3.9), we have

det A det A [ { k 0 , σ ( k 0 ) } ] det A [ { k 0 , σ ( k 0 ) } c ] [ a k 0 k 0 a σ ( k 0 ) σ ( k 0 ) a σ ( k 0 ) k 0 a k 0 σ ( k 0 ) ] i = 1 , i k 0 , σ ( k 0 ) a i i i = 1 n a i i [ i = 1 n a i σ ( i ) a σ ( i ) i ] 1 2 ,

therefore (3.8) holds. □

If A=[ a i j ] is a totally nonnegative matrix, Theorem 3.3 is certainly valid too, so Theorem 3.3 is the generalization of Theorem 4 in [11].

Example 3.2 Now we consider the previous K-matrix. Let

A=( 4 2 3 1 3 1 1 1 2 ).

By calculating we have

detA=2 i = 1 3 a i i max 1 σ S 3 [ i = 1 3 a i σ ( i ) a σ ( i ) i ] 1 2 =24max{4,9, 6 ,4, 6 }=15,

therefore inequality (3.8) holds.

By Corollary 3.2 and Theorem 3.3, we get the following result.

Corollary 3.4 If A=[ a i j ] is an n×n K 0 -matrix, with i = 1 n a i i >0, then

detAmin { i = 1 n a i i max 1 σ S n ( i = 1 n a i σ ( i ) a σ ( i ) i ) 1 / 2 , min k n a k k i = 1 i k n ( a i i a i k a k i a k k ) } .

If A is a totally nonnegative matrix, Corollary 3.4 is valid, so we obtain the result in [11].

Corollary 3.5 [11]

If A=[ a i j ] is an n×n totally nonnegative matrix, with i = 1 n a i i >0, then

detAmin { i = 1 n a i i max 1 σ S n ( i = 1 n a i σ ( i ) a σ ( i ) i ) 1 / 2 , min k n a k k i = 1 i k n ( a i i a i k a k i a k k ) } .

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Acknowledgements

The author expresses his deep gratitude to the referees for their many very valuable suggestions and comments. The research of this paper was supported by the Natural Science Foundation of Shandong Province of China (ZR2010AL017).

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Li, Y. On improvements of Fischer’s inequality and Hadamard’s inequality for K 0 -matrices. J Inequal Appl 2013, 460 (2013). https://doi.org/10.1186/1029-242X-2013-460

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Keywords

  • totally nonnegative matrix
  • K 0 -matrix
  • Fischer’s inequality
  • Hadamard’s inequality