c
Applied Mathematics E-Notes, 9(2009), 289-296 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
Perron Complements Of Diagonally Dominant
Matrices And H-Matrices∗
Li Zeng†, Ming Xiao‡, Ting-Zhu Huang§
Received 27 November 2008
Abstract
In this paper, we consider properties of the Perron complements of diagonally
dominant matrices and H-matrices.
1
Introduction
Let A = (aij ) be an n × n matrix, and recall that A is (row) diagonally dominant if
X
|aii | ≥
|aij |, i = 1, 2, ..., n.
(1)
j6=i
A is further said to be strictly diagonally dominant if all the strict inequalities in (1)
hold. Obviously the principal submatrices of strictly diagonally matrices are strictly
diagonally dominant and thus A is nonsingular.
For A ∈ Z = {(aij ) ∈ Rn,n : aij ≤ 0, i 6= j}, if A = aI − B, B ≥ 0, a > ρ (B) , then
A is called an M-matrix. The comparison matrix µ (A) = (µij ) is defined by
− |aij | i 6= j,
µij =
|aij | i = j.
A ∈ C n,n is called an H-matrix if µ (A) is an M-matrix. If there exists a positive
diagonal matrix D = diag (d1 , ..., dn) such that D−1 AD is strictly diagonally dominant,
we call A a generalized diagonally dominant matrix. It is well-known that A is an Hmatrix is equivalent to A is generalized diagonally dominant.
The empty set is denoted by φ. Let α, β be nonempty ordered subsets of hni :=
{1, 2, ..., n}, both consisting of strictly increasing integers. By A (α, β) we shall denote
the submatrix of A lying in rows indexed by α and columns indexed by β. If, in
addition, α = β, then the principal submatrix A (α, α) is abbreviated to A (α).
∗ Mathematics
Subject Classifications: 15A48
of Computer Science and Technology, Southwest University for Nationalities, Chengdu,
Sichuan, 610041, P. R. China
‡ College of Computer Science and Technology, Southwest University for Nationalities, Chengdu,
Sichuan, 610041, P. R. China
§ School of Applied Mathematics, University of Electronic Science and Technology of China,
Chengdu, 610054, P. R. China
† College
289
290
Perron Complements of Matrices
Suppose that α ⊂ hni. If A (α) is nonsingular, then the Schur complement of A (α)
in A is given by
S (A/A (α)) = A (β) − A (β, α) [A (α)]
−1
A (α, β) ,
(2)
where β = hni \α. A well-known result due to Carlson and Markham [1] states that the
Schur complements of strictly diagonally dominant matrices are diagonally dominant.
For an n × n nonnegative and irreducible matrix A, Meyer [2,3] introduced the
notion of the Perron complement. Again, let α ⊂ hni and β = hni \α. Then the Perron
complement of A (α) in A is given by
−1
P (A/A (α)) = A (β) + A (β, α) [ρ (A) I − A (α)]
A (α, β) ,
(3)
where ρ (·) denotes the spectral radius of a matrix. Recall that as A is irreducible,
ρ (A) > ρ (A (α)), so that the expression on the right-hand side of (3) is well defined,
and we observe that ρ (A) I − A (α) is an M-matrix and thus (ρ (A) I − A (α))−1 ≥ 0.
Meyer [2,3] has derived several interesting and useful properties of P (A/A (α)), such
as P (A/A (α)) is also nonnegative and irreducible, and ρ(P (A/A (α))) = ρ (A). In
addition, the Perron complements of inverse M-matrices [4] have also been studied.
For any α ⊂ hni and for any t ≥ ρ (A), let the extended Perron complement at t be
the matrix
Pt (A/A (α)) = A (β) + A (β, α) [tI − A (α)]−1 A (α, β) ,
(4)
which is also well defined since t ≥ ρ (A) > ρ (A (α)).
In this paper, we shall show, in Section 2, that the Perron complement of a diagonally dominant and nonnegative irreducible matrix ,
−1
P (A/A (α)) = A (β) + A (β, α) [ρ (A) I − A (α)]
is diagonally dominant only if ρ (A) ≥ maxi∈α
n
P
A (α, β) ,
|aij |. In Section 3, we shall show a
j=1
similar result for H-matrices.
2
Perron Complements of Diagonally Dominant Matrices
First recall the following result proved in [2].
LEMMA 2.1 ([2]). If A is a nonnegative irreducible matrix with spectral radius
ρ (A), and let α ⊂ hni, α =
6 φ and β = hni \α. Then the Perron complement
P (A/A (α)) = A (β) + A (β, α) [ρ (A) I − A (α)]−1 A (α, β)
is also a nonnegative irreducible matrix with spectral radius ρ (A).
We are now in a position to state the main result of the paper on the Perron
complements of diagonally dominant matrices.
291
Zeng et al.
THEOREM 2.2. Let A be an n×n diagonally dominant and nonnegative irreducible
matrix with spectral radius ρ (A), and let α ⊂ hni, α 6= φ and β = hni \α. Then, for
n
P
ρ (A) ≥ maxi∈α
|aij |,
j=1
P (A/A (α)) = A (β) + A (β, α) [ρ (A) I − A (α)]−1 A (α, β)
is a diagonally dominant and nonnegative irreducible matrix.
PROOF. Let α = {i1 , i2 , ..., ik} and β = {j1 , j2 , ..., jl}, where k + l = n. Denote
|A| = (|aij |). Since A is a diagonally dominant matrix, we have, for any i ∈ hni,
X
|aii | ≥
|aij |
j6=i
or
l
X
|ajt jt | ≥
|ajt js | +
k
X
|ajt is |,
(5)
s=1
s=1,6=t
where jt , js ∈ β, is ∈ α. Note that A is an irreducible and nonnegative matrix, then
ρ (A) > ρ (A (α)), so that ρ (A) I − A (α) is an M-matrix. Then we have
(ρ (A) I − A (α))−1 ≥ 0 and aij ≥ 0.
By ρ (A) ≥ maxi∈α
n
P
(6)
|aij |, we have
j=1


l
k
X
X
X
X
ρ (A) ≥ maxi∈α 
|aij | +
|aij | = maxiv ∈α
aiv jt + maxiv ∈α
aiv it . (7)
If maxiv ∈α
l
P
t=1
j∈α
j∈β
t=1
aiv jt = 0, then P (A/A (α)) = A (β). Thus, the matrix P (A/A (α)) is
t=1
diagonally dominant. If maxiv ∈α
l
P
aiv jt > 0, then, by (7), we have
t=1
0 < maxiv ∈α
l
X
aiv jt ≤ ρ (A) − maxiv ∈α
t=1
k
X
ai v i t .
(8)
t=1
Thus
l
P
ai v j t
t=1
maxiv ∈α
ρ (A) −
k
P
≤ 1.
(9)
ai v i t
t=1
Denote
−1
x = (ρ (A) I − A (α))
l
X
s=1
ai1 js , ...,
l
X
s=1
ai k j s
!†
(10)
292
Perron Complements of Matrices
or
l
X
(ρ (A) I − A (α)) x =
ai1 js , ...,
s=1
l
X
ai k j s
s=1
!†
.
Letting xv = max{x1 , x2 , · · · , xk }, where xi is the i-th component of x, we obtain
l
P
ai v j s
=
(ρ (A) − aiv iv ) xv +
s=1
k
P
(−aiv it )xt
t=1,6=v
≥
(ρ (A) − aiv iv +
k
P
(−aiv it ))xv
t=1,6=v
=
(ρ (A) −
k
P
aiv it )xv .
t=1
By (8), we have
xv ≤
l
P
ai v j t
t=1
ρ (A) −
k
P
≤ maxiv ∈α
ai v i t
l
P
ai v j t
t=1
ρ (A) −
t=1
k
P
.
ai v i t
t=1
By (9), we have
xv ≤ 1.
(11)
Denote the (t, s)-entry of P (A/A (α)) by a0jt js . Then, for t = 1, 2, ..., l, we have
0
a
jt jt
=
l
X
0
−
a
jt js
s=1,6=t


ai 1 j t

−1  .
ajt jt + (ajt i1 , ..., ajtik ) (ρ (A) I − A (α))  ..

ai k j t 

ai 1 j s
l
X

−1  .
−
ajt js + (ajt i1 , ..., ajtik ) (ρ (A) I − A (α))  ..
 ,
s=1,6=t ai j
k s
293
Zeng et al.
so that
0
aj
≥
=
≥
t jt
l
X
0 −
aj j t s
s=1,6=t

|ai1 jt |

..

.
|aik jt |


|ai1 js |
l
X 
−1  .
−
|ajt js | + (|ajt i1 | , ..., |ajt ik |) (ρ (A) I − A (α))  ..
s=1,6=t
|aik js |
 l
P
 s=1 ai1 js
l

X
−1 
aj t j t −
ajt js − (ajt i1 , ..., ajtik ) (ρ (A) I − A (α))  ...

s=1,6=t
l
 P
ai k j s
s=1


1
l
X


aj t j t −
ajt js − (ajt i1 , ..., ajtik )  ... 



−1 
|ajt jt | − (|ajt i1 | , ..., |ajt ik |) (ρ (A) I − A (α)) 
s=1,6=t
=
|ajt jt | −
l
X
s=1,6=t
≥
0.










1
|ajt js | −
k
X
|ajt is |
s=1
It follows that P (A/A (α)) is a diagonally dominant matrix. By Lemma 2.1, the matrix
P (A/A (α)) is nonnegative irreducible. This completes the proof.
By Theorem 2.2, we have several immediate results about the extended Perron
complements and the Perron complements of strictly diagonally dominant matrices.
COROLLARY 2.3. Let A be an n × n diagonally dominant and nonnegative irreducible matrix with spectral radius ρ (A), and let α ⊂ hni , α 6= φ and β = hni \α.
n
P
Then, for any t ∈ [ρ (A) , ∞) and ρ (A) ≥ max
|aij |,
i∈α j=1
−1
Pt (A/A (α)) = A (β) + A (β, α) [tI − A (α)]
A (α, β)
is a diagonally dominant and nonnegative irreducible matrix.
COROLLARY 2.4. Let A be an n × n strictly diagonally dominant and nonnegative irreducible matrix with spectral radius ρ (A), and let α ⊂ hni , α 6= φ and
n
P
β = hni \α. Then, for any t ∈ [ρ (A) , ∞) and ρ (A) ≥ maxi∈α
|aij |, P (A/A (α))
j=1
and Pt (A/A (α)) are strictly diagonally dominant and nonnegative irreducible matrices.
294
3
Perron Complements of Matrices
Perron Complements of H-matrices
In this section, we obtain a theorem of the Perron complements of H-matrices.
THEOREM 3.1. Let A be an n × n nonnegative irreducible H-matrix with spectral radius ρ (A), and let α ⊂ hni , α 6= φ and β = hni \α. Then, for ρ (A) ≥
n
P
maxi∈α
|aij | ≥ 2 |aii | , i ∈ α,
j=1
P (A/A (α)) = A (β) + A (β, α) [ρ (A) I − A (α)]−1 A (α, β)
is a nonnegative irreducible H-matrix.
PROOF. Let α = {i1 , i2 , ..., ik} and β = {j1 , j2 , ..., jl}, where k + l = n. Since A is
an H-matrix, then there exists a positive diagonal matrix
X = diag (x1 , x2 , ..., xn) > 0
such that X −1 AX is a strictly diagonally dominant matrix, i.e.,
|aii | >
X xj
xi
j6=i
|aij | , i ∈ hni .
Suppose that B = (bij ) = X −1 AX, we have ρ (B) = ρ (A) and B is a strictly diagonally
dominant matrix. Since
n
X
ρ (A) ≥ maxi∈α
|aij | ≥ 2 |aii | , i ∈ α
j=1
and
|aii | >
X xj
xi
j6=i
we have ρ (B) = ρ (A) ≥ maxi∈α
n
P
|aij | , i ∈ hni ,
|bij |. Then,by Corollary 2.4, P (B/B (α)) is a
j=1
strictly diagonally dominant matrix and
a11
 x1 a21
x2
B=
 ···
x1
xn an1

x2
x1 a12
a22
···
x2
xn an2

· · · xxn1 a1n
· · · xxn2 a2n 
.

··· ···
· · · ann
295
Zeng et al.
Let D = diag (xj1 , xj2 , ..., xjl) > 0. Then,
P (B/B (α))
=
−1
=
B (β) + B (β, α) [ρ (B) I − B (α)] B (α, β)


 
bj 1 i 1 · · · bj 1 i k
bj 1 j 1 · · · bj 1 j l
 ···
··· ··· + ···
··· ··· 
bj l j 1 · · · bj l j l
bj l i 1 · · · bj l i k



ρ (B) − bi1 i1 · · · −bi1 ik
bi 1 j 1 · · · bi 1 j l
 ···
··· ···
··· ··· 
× ···
−bik i1
· · · ρ (B) − bik ik
bi k j 1 · · · bi k j l

  xi

xi
xjl
1
aj 1 j 1
· · · x j aj 1 j l
aj1 i1 · · · xjk aj1 ik
x
j
1
1
  1


··· ···
··· ···
 ···
+ ···

xj1
xik
xi1
a
·
·
·
a
a
·
·
·
a
j
j
j
j
j
i
j
i
l
1
l
l
l
1
l
k
xj
xjl
xjl
 l
  xj

xjl
xik
1
ρ (A) − ai1 i1 · · · − xi ai1 ik
a
·
·
·
a
i
j
i
j
1
1
1
l
x
x
i1
1

  i1

··· ···
× ···
··· ···
 ···

xi1
xjl
xj1
· · · ρ (A) − aik ik
− x i ai k i 1
a
·
·
·
a
xik ik j1
xik ik jl
k


aj 1 j 1 · · · aj 1 j l
1
1 
···
· · · · · ·  diag(xj1 , ..., xjl)
diag
, ...,
xj1
xjl
aj l j 1 · · · aj l j l


aj 1 i 1 · · · aj 1 i k
1
1 
···
· · · · · ·  diag(xi1 , ..., xik )
+diag
, ...,
xj1
xjl
aj l i k · · · aj l i k


ρ (A) − ai1 i1 · · · −ai1 ik
1
1 
 diag(xi1 , ..., xik )
···
··· ···
×diag
, ...,
xi1
xik
−aik i1
· · · ρ (A) − aik ik


ai 1 j 1 · · · ai 1 j l
1 
1
···
· · · · · ·  diag(xj1 , ..., xjl)
, ...,
×diag
xi1
xik
ai k j 1 · · · ai k j l
=
D−1 P (A/A (α)) D.
=
=
=
D−1 A (β) D + D−1 A (β, α) [ρ (A) I − A (α)]−1 A (α, β) D
Note that the matrix
P (B/B (α)) = D−1 P (A/A (α)) D
is strictly diagonally dominant, then P (A/A (α)) is an H-matrix. By Lemma 2.1, we
have the matrix P (A/A (α)) is nonnegative irreducible. This completes the proof.
296
4
Perron Complements of Matrices
Example
Let

3
 1
A=
 2
1
1
3
1
2
1
0
4
1

0
1 
.
1 
4
Obviously, A is a diagonally dominant and nonnegative irreducible H-matrix. And,
ρ (A) = 6.3028 ≥ maxi∈α
n
X
|aij | ≥ 2 |aii | , i ∈ α, α = {1} or {1, 2}.
j=1
Then,


3.3028 0.3028 1
P (A/A (α)) =  1.6055 4.6055 1  ,
2.3028 1.3028 4
where α = {1}, is a diagonally dominant H-matrix. And,
4.7675 1.5351
P (A/A (α)) =
,
1.5351 4.7675
where α = {1, 2}, is a diagonally dominant H-matrix.
References
[1] D. Carlson and T. Markham, Schur complements of diagonally dominant matrices,
Czech. Math. J., 29(104)(1979), 246–251.
[2] C. D. Meyer, Uncoupling the Perron eigenvector problem, Linear Algebra Appl.,
114/115(1989), 69–94.
[3] C. D. Meyer, Stochastic complementation uncoupling Markov chains and the theory
of nearly reducible systems, SIAM Rev., 31(1989), 240–272.
[4] M. Neumann, Inverses of Perron complements of inverse M-matrices, Linear Algebra
Appl., 313(2000), 163–171.