The Estimates of Diagonally Dominant Degree and Eigenvalue Inclusion Regions for the Schur Complement of Matrices ()
1. Introduction
Let denote the set of all complex matrices, and. We write
We know that A is called a strictly diagonally dominant matrix if
A is called an Ostrowski matrix (see [1] ) if
and will be used to denote the sets of all strictly diagonally dominant matrices and the sets all Ostrowski matrices, respectively.
As shown in [2] , for and, we call, and the i-th diagonally, α-diagonally and product α-diagonally dominant degree of A, respectively.
For, denote by the cardinality of β and. If, then is the submatrix of A with row indices in β and column indices in. In particular, is abbreviated to. If is nonsingular,
is called the Schur complement of A with respect to.
The comparison matrix of A, , is defined by
A matrix is called an M-matrix, if there exist a nonnegative matrix B and a real number, where is the spectral radius of B, such that. It is known that A is an h-matrix if and only if is an m-matrix, and if A is an m-matrix, then the Schur complement of A is also an m-matrix and (see [3] ). We denote by Hn and Mn the sets of h-matrices and m-matrices, respectively.
The Schur complement of matrix is an important part of matrix theory, which has been proved to be useful tools in many fields such as control theory, statistics and computational mathematics. A lot of work has been done on it (see [4] -[8] ). We know that the Schur complements of strictly diagonally dominant matrices are strictly diagonally dominant matrices, and the Schur complements of Ostrowski matrices are Ostrowski matrices. These properties have been used for deriving matrix inequalities in matrix analysis and for the convergence of iterations in numerical analysis (see [9] -[12] ). More importantly, studying the locations for the eigenvalues of the Schur complement is of great significance, as shown in [2] [6] [13] -[18] .
The paper is organized as follows. In Section 2, we give some new estimates of diagonally dominant degree on the Schur complement of matrices. In Section 3, we present several new eigenvalue inclusion regions for the Schur complement of matrices. In Section 4, we give a numerical example to illustrate the advantages of our derived results.
2. The Diagonally Dominant Degree for the Schur Complement
In this section, we present several new estimates of diagonally, α-diagonally and product α-diagonally dominant degree on the Schur complement of matrices.
Lemma 1. [3] If, then.
Lemma 2. [3] If or, then, i.e.,.
Lemma 3. [6] If or and, then the Schur complement of A is in or, where is the complement of β in N and is the cardinality of.
Lemma 4. [16] Let, , and. Then
Theorem 1. Let, , , and. Then for all,
(1)
and
(2)
where
Proof. Since, then and. From Lemma 1 and Lemma 2, we have
Thus, for any and, we obtain
For any, denote
If
then there exists sufficiently small positive number such that
(3)
Construct a positive diagonal matrix, where
Let. For, by (3), we have
And for, by, , we obtain
Thus, , and so. Note that, then
(4)
Let x be in. Then
Since, by (4), we have
Let. Then we obtain (1). Similarly, we can prove (2). □
Remark 1. Note that
This shows that Theorem 1 improves Theorem 2 of [17] and [2] , respectively.
Next, we present some new estimates of α-diagonally and product α-diagonally dominant degree of the Schur complement.
Theorem 2. Let, , , and. Then for all, ,
(5)
and
(6)
where for any,
Proof. By Lemma 1 and Lemma 2, we have. Thus, for all, , we have
Let
Similar as the proof of Theorem 1, we can prove
Similarly, we have
By Lemma 4, we have
Hence, (5) holds. Similarly, we can prove (6).
Remark 2. Note that
This shows that Theorem 3 improves Theorem 4 of [2] .
Similar as the proof of Theorem 2, we can prove the following theorem immediately, which improves Theorem 2 of [2] .
Theorem 3. Let, , , and. Then for all, ,
and
3. Eigenvalue Inclusion Regions of the Schur Complement
In this section, based on these derived results in Section 2, we present new eigenvalue inclusion regions for the Schur complement of matrices.
Theorem 4. Let, , , and and be eigenvalue of. Then there exists such that
(7)
Proof. By Gerschgorin Circle Theorem, we know that there exists such that. Thus, by Lemma 1 and Lemma 2, we have
i.e.,
Thus, (7) holds.
Lemma 5. [2] Let and. Then for any eigenvalue of A, there exists such that
Theorem 5. Let, , , , and be eigenvalue of. Then for any, there exists such that
(8)
Proof. By Lemma 5, we know that there exists such that
Therefore,
Similar as the proof of Theorem 2, we can prove
Thus, we have
Further, we obtain (8).
4. A Numerical Example
In this section, we present a numerical example to illustrate the advantages of our derived results.
Example 1. Let
By calculation with Matlab 7.1, we have that
Since, by Theorem 4, the eigenvalue inclusion set of is
From Theorem 4 of [2] , the eigenvalue inclusion set of is
We use Figure 1 to illustrate. And the eigenvalues of are denoted by “+” in Figure 1. The blue dotted line and green dashed line denote the corresponding discs and respectively.
Meanwhile, since, by taking in Theorem 5, the eigenvalue inclusion set of is
Figure 1. The blue dotted line and green dashed line denote the corresponding discs and, respectively.
Figure 2. The blue dotted line and green dashed line denote the corresponding discs and, respectively.
From Theorem 5 of [2] , the eigenvalue inclusion set of is
We use Figure 2 to illustrate. And the eigenvalues of are denoted by “+” in Figure 2. The blue dotted line and green dashed line denote the corresponding discs and respectively. It is clear that and.