vol3_p93-102. 145KB Jun 04 2011 12:06:14 AM

The Electronic Journal of Linear Algebra.
A publication of the International Linear Algebra Society.
Volume 3, pp. 93-102, May 1998.
ISSN 1081-3810. http://math.technion.ac.il/iic/ela

ELA

POSITIVE DEFINITENESS OF TRIDIAGONAL MATRICES
VIA THE NUMERICAL RANGE
MAO-TING CHIENy AND MICHAEL NEUMANNz

Dedicated to Hans Schneider on the occasion of his seventieth birthday.

Abstract. The authors use a recent characterization of the numerical range to obtain several
conditions for a symmetric tridiagonal matrix with positive diagonal entries to be positive denite.
Key words. Positive denite, tridiagonal matrices, numerical range, spread
AMS subject classications. 15A60, 15A57

1. Introduction. In a paper on the question of unicity of best spline approximations for functions having a positive second derivative, Barrow, Chui, Smith, and
Ward proved the following result.
Proposition 1.1. [1, Proposition 1] Let A = (a ) 2 R

be a tridiagonal
n;n

ij

matrix with positive diagonal entries. If


2 
1

a ,1a ,1  4 a a ,1 ,1 1 + 1 + 4n2 ; i = 2; : : :; n;
then det(A) > 0.
Johnson, Neumann, and Tsatsomeros improved Proposition 1.1 as follows.
Proposition 1.2. [6, Proposition 2.5, Theorem 2.4] Let A = (a ) 2 R be a
i;i

i

;i

ii

i

;i

ij

tridiagonal matrix with positive diagonal entries. If

a ,1a ,1 < 41 a a ,1 ,1  1 2 ; i = 2; : : :; n;

cos n +
1
i;i

i

;i

ii

i

;i

then det(A) > 0. In addition, if A is (also) symmetric, then A is positive denite.

The main purpose of this paper is to derive additional criteria, using the numerical range, for a symmetric tridiagonal matrix A with positive diagonal entries to be
positive denite. An example of one of our results is Theorem 2.2.
Recall that the numerical range of a matrix A 2 C is the set
W (A) = fx Ax j x 2 C ; jxj = 1g:
n;n

n

Many of our results here will be derived with the aid of the following recent characterization of the numerical range due to Chien.
 Received by the editors on 23 December 1997. Final manuscript accepted on 1 May 1998.
Handling editor: Daniel Hershkowitz.

y Department
of Mathematics, Soochow University, Taipei, Taiwan 11102,
Taiwan (mtchien@math.math.scu.edu.tw). Supported in part by the National Science Council of
the Republic of China.
z Department of Mathematics, University of Connecticut, Storrs, Connecticut 06269{3009, USA
(neumann@math.uconn.edu).
93

ELA
94

Mao-Ting Chien and Michael Neumann

Theorem 1.3. [2, Theorem 1] Let A 2 C

n;n

of C . Then
n

and suppose S be a nonzero subspace

W (A) = [ W (A );
where x and y vary over all unit vectors in S and S ? , respectively, and where
 
Ay 
x
Ax
x
A = y Ax y Ay :
x;y

xy

xy

Our main results are developed in Section 2 which contains several conditions for
a symmetric tridiagonal matrix with positive diagonal entries to be positive denite.
In Section 3 we give some examples to illustrate dierent criteria we have obtain in
Section 2.

Some notations that we shall employ in this paper are as follows. Firstly, we shall
use e( ) , i = 1; : : :; k, to denote the k{dimensional i{th unit coordinate vector, viz.,
the i{th entry of e( ) is 1 and its remaining entries are 0.
Secondly, for a set   f1; 2; : : :; ng denote by 0 = f1; 2; : : :; ng n  its complement
in f1; 2; : : :; ng. Thirdly, if  and  are subsets of f1; 2; : : :; ng and A 2 C , then
we shall denote by A[;  ] the submatrix of A that is determined by the rows of
A indexed by  and by the columns indexed by  . If  =  , then A[;  ] will be
abbreviated by A[].
2. Main Results. We come now to the main thrust of this paper which is to
develop several criteria for a symmetric tridiagonal matrix with positive diagonal
entries to be positive denite.
Suppose rst that A is an n  n Hermitian matrix and partition A into


A = BR BT ;
k

i

k

i

n;n

where R and T are square. A well known result due to Haynsworth [4] (see also
[5, Theorem 7.7.6]) is that A is positive denite if and only if R and the Schur
complement of R in A, given by (A=R) = T , B  R,1B , are positive denite. For
tridiagonal matrices, we now give a further proof of this fact based on Theorem
1.3 and in the spirit of the divide and conquer algorithm due to Sorensen and Tang
[9], but see also Gu and Eisenstat [3], in the sense that it is possible to reduce the
problem of determining whether A is positive denite into the problem of, rst of
all, determining whether two smaller principal submatrices of A are positive denite
which can be executed in parallel.
Theorem 2.1. Let A 2 C be a Hermitian tridiagonal matrix and partition A
n;n

into

2

A =

6
4

A1
a +1 e(
0
k

k )T
;k k

a
a

k;k +1

a

e(

k)

k

k +1;k +1

k +2;k +1

e1( , ,1)
n

k

a

k +1;k

3

0
( , ,1)
+2 e1
A2
n

k

T

7
5;

ELA
95

Positive Deniteness of Tridiagonal Matrices

where 1 < k < n, A1 2 C , and A2 2 C , ,1 , ,1. Then A is positive denite if
and only if the symmetric tridiagonal matrix
k;k



n

(1)

;n

k



A1 0 , (1=a +1 +1) 
0 A2
a
ja +1j2 e( )e( )
a +1 a +2 +1e1( , ,1)e( ) ja
k

"

k

;k

k

k;k

k

;k

k

k

k T

k;k +1

k
n k

k T

;k

k +1;k

k

is positive denite. Putting

C1 = A1 , jaa

(2)

k;k +1

e e1( , ,1)
, , e( , ,1)
+2 j e1
1

a

(k )
k +1;k +2 k
2 (n k 1)

n

k

n

T

k

#

T

j2 e( ) e( ) ;
k

k +1;k +1

k T

k

k

a necessary and sucient condition for the matrix in (1) to be positive denite is that
A1 and A2 are positive denite and that the inequalities

1 , jaa

(3)

k;k +1

j2 ,A,1

1

k +1;k +1

and

> 0

k;k

!

2
, ,1

j
a
+1a +1 +2 j
A2 1 1 > 0
(4)
(C1,1)
1 , ja +1 +2j +
a +1 +1
hold. In addition, if A1 and A2 are positive denite, then any one of the following
three conditions is also sucient for A to be positive denite.


k

2

;k

k;k

k

k;k

k

(i) The matrices

A , ja

(5)

;k

j2 + ja +1
a +1 +1

k;k +1

i

;

;k

k

k

;k +2

j2 I ; i = 1; 2;
i

;k

are positive denite,

(ii)

1 > a 1
+1

(6)



ja

;k +1

k

k;k +1

,


j2 A,1 1 + ja +1

(iii)

k

k;k

;

j2 + ja +1 +2j2  ;
d1
d +2
+1
where d1 and d +2 are the smallest eigenvalues of of A1 and A2 , respectively.
Proof. Let S be the subspace of C spanned by e( +1) . Then, by Theorem 1.3, A is
positive denite if and only if A is positive denite for all x 2 S and y 2 S ? . Assume
now that x = [0 0 : : : 0 x +1 0 : : : 0] 2 S and y = [y1 : : : y 0 y +2 : : : y ] 2 S ? .
1 > a 1
+1

(7)

k



,


j2 A2,1 1 1 ;

;k +2

ja

k;k +1

k

;k

;k

k

k

n

n

k

xy

Then

A

xy

t

k

=



a +1
a +1y x +1 + a
k

k;k

k

k

;k +1
k +2;k +1

k

y +2 x
k

k +1

a

k +1;k

x

k +1

k

y + a +1
y Ay
k

k

n

;k +2

t



x +1y
k

k +2

ELA
96

Mao-Ting Chien and Michael Neumann

is positive denite if and only if
a +1 +1 y Ay
(8)
, (a +1 x +1y + a
> 0:
k

;k

k

;k

k

k

k +1;k +2

x

k +1

y

k +2

) (a

k;k +1

y x
k

k +1

+a

k +2;k +1

)

Put y0 := [y1 : : : y y +2 : : : y ] 2 C ,1. Then (8) becomes


A
0
1
0
0 ,
2
a +1 +1 y
0 A2 y , ja +1j y y + a +1 a +1 +2y y +2


+ a +1 a +1 +2y +2 y + ja +1 +2j2y +2y +2


A
0
1
0
0
= a +1 +1 y
0 A2 y
#
"
( ) ( , ,1)
2 ( ) ( )
j
a
j
e
e
a
a
e
e
+1
+1
+1
+2
1
y0 > 0
, y0
T
a +1 a +2 +1e1( , ,1)e( ) ja +1 +2j2 e1( , ,1)e1( , ,1)
k

k

k

n

t

n

;k

k

k

k;k

;k

k

;k

k

k

k

k

;k

k

k

k

;k

k

;k

k

k

k

;k

k

k;k

k

n

k

;k

k

k T

k

k;k

k
k

k

;k

;k

k

k

n

k

T

k

n

k

k

n

k

T

;k

This proves the rst part of theorem. Namely, that A is positive denite if and only
if the (n , 1)  (n , 1) matrix in (1) is positive denite.
The equivalence of the positive deniteness of the matrix in (1) to the conditions
(3) and (4) is obtained by considering the Schur complement of the matrix C1 given
in (2) and by utilizing the fact that if u; v 2 R , then


,
(9)
det I + uv = 1 + v u:
Suppose now that A1 and A2 are positive denite. We next establish (i). For this
purpose
matrix in (1), which has rank 1, has eigenvalues


, observe that the subtracted
0 and ja +1j2 + ja +1 +2j2 =a +1 +1 so that in the positive semidenite ordering,
`

T

T

k;k



k

k

;k



;k



1

k

;k

ja



j2e e
a +1a +1 +2e e1
, a
a
a
e
e
ja +1 +2j2e1 e1
+1
+2 +1 1
+1 +1


2
2
 A01 A0 2 , ja +1aj + ja +1 +2j I ,1 :
+1 +1
Thus if (5) holds, then A is clearly positive denite because then a submatrix of A,
namely a , as well as its Schur complement in A given in (1) are positive denite.
A1 0
0 A2

k

k;k +1

;k

k

k

;k

T
k

k;k

T
k

k;k

k

k

k

;k

;k

k
T

T

;k

n

k

;k

k;k

To establish the validity of (ii), factor the rst matrix in (1) and compute the
determinant of the sum of the identity matrix and the rank 1 matrix using the formula
in (9). A simple inspection shows that condition (6) implies the positivity of that
determinant.
Finally, condition (7) implies ,condition
(6) because, by interlacing
properties of


,


symmetric matrices we have that A,1 1  (1=d1) > 0 and A,2 1 1 1  (1=d +1) >
0. This establishes (iii) and our proof is complete.
Next from Gershgorin theorem we know that a sucient condition for an Hermitian matrix A = (a ) with positive diagonal entries to be positive denite is that
k;k

i;j

;

k

ELA
97

Positive Deniteness of Tridiagonal Matrices

a > P =1 =6 ja j, i = 1; 2; : : :; n. However, what can be said if, for a tridiagonal
matrix, the condition fails to be satised for some rows? In this case, we obtain the
following criteria.
Theorem 2.2. Let A = (a ) 2 C
be an Hermitian tridiagonal matrix with
positive diagonal entries. Then A is positive denite if one of the following conditions
n

i;i

j

;

i;j

i

n;n

i;j

is satised.

(i)


i

max a
=1 3
; ;:::

 

, =1min3 a

i;i

i

; ;:::

 

jap+1 j + max jap,1 j 2 < min a :
+ =1
max
=3 5
3
=2 4
a
a
i

i;i

i

; ;:::

i

;i

i

i;i

(ii) =1min
a = =1
max
a , and
3
3
i

ii

; ;:::

i

i

i;i





2 max ja ,1 j2 + ja +1 j2 : i = 1; 3; : : : <
;i

i

;i

i

 

min a
=1 3
; ;:::

ii

i

min a
=2 4
; ;:::

(iii) =1min
a =
6 =1
max
a ,
3
3
i

ii

; ;:::



i

2 =1
max
ja ,1 j2 + ja +1 j2 , =1min3 ja ,1 j2 + ja +1 j2
3

>

i

; ;:::



i

;i

i

;i

i

i;i

; ;:::

i

i

; ;:::

;i

i

i

;i

;i

i

;i



< =1min
a
3
i

; ;:::

 
i;i

i

min a

=2 4

; ;:::

(iv) =1min
a =
6 =1
max
a ,
3
3


i


and



i

; ;:::

i

;i

i

;i

i

i

; ;:::

;i

i

;i

2

max fa g , =1min
fa g ;
3

=1 3

i;i

; ;:::

i

i;i

; ;:::





i



i;i



1 min a , max a 2
=1 3
4 =1 3
2
+ =1
max
ja ,1 j + ja + j2 + =1min3 ja ,1 j2 + ja +1 j2
3
i



i;i

; ;:::

2 =1
max
ja ,1 j2 + ja +1 j2 , =1min3 ja ,1 j2 + ja +1 j2
3
i



i;i

; ;:::


2 max ja ,1 j2 + ja +1 j2 : i = 1; 3; : : :
i;i

; ;:::

; ;:::

i;i

i

i

; ;:::

;i

i

i;i

;i

i

; ;:::

i

;i

i

;i

2 + ja +1 j2 , min =1 3 ja ,1 j2 + ja +1 j2 2
max
j
a
j
1
3
,
1
+
max =1 3 a , min =1 3 a


< =1min
fa g) ( =2min
fa g :
3
4
; ;:::

i

;i

i

i

i

:

max fa g , =1min
fa g ;
3

=1 3



; ;:::

ii

2

and

i



ii

; ;:::

i

; ;:::

ii

; ;:::

i

;i

; ;:::

; ;:::

i;i

i

;i

; ;:::

; ;:::

i;i

i

i;i

; ;:::

i

; ;:::

i

i;i

;i

i

;i

:

i;i

ELA
98

Mao-Ting Chien and Michael Neumann

Proof. We rst note that Bessel's inequality (see, e.g., [7, p.488]) implies that if
x and y are n{vectors which are orthogonal to each other, then
(10)
jx Ayj2  kAxk22 , jx Axj2:
Next let S be the subspace generated by e1 ; e3; e5; : : : and observe that vectors x 2 S
and y 2 S ? have the form x = [x1 0 x3 0 x5


] and y = [0 y2 0 y4


] , respectively.
Then by Theorem 1.3, A is positive denite if and only if A is positive denite for
x 2 S and y 2 S ? , which is equivalent to
(11)
x Ax y Ay > jx Ayj2 :
Assume now that condition (i) is satised. We rst recall the following inequality.
Let p1 ;


; p be positive numbers. Then for any real numbers q1; q2; : : :; q ,
(12) (q1 + q2 +


+ q )=(p1 + p2 +


+ p )  maxfq =p : i = 1; 2; : : :; ng
Then by (12) we have that
P
ja x j2
P =1 3 a jx j2  =1
(13)
max
a ;
3
=1 3
and
P
ja +1 x + a +1 +2x +2 j2
=1 3 P
=1 3 a jx j2
sP
sP
!2
j
a +1 x j2
j
a +1 +2x +2 j2
=1
3
=1
3
P
P
+

=1 3 a jx j2
=1 3 a jx j2
s
s
!2
2
2
j
a
j
j
a
j
+1
,
1
+ =3

max a
max
=1 3
5
a


ja +1 j + max ja ,1 j 2 :
= =1
max
(14)
3 pa
=3 5 pa
From (13), (14), and the hypothesis we now obtain,
kAxk22 , jxAxj2
xAx
P
2 P
ja +1 x + a +1 +2 x +2j2 X
j
a
x
j
=1 3 P
=1
3
a jx j2
+
,
= P
=1 3 a jx j2
=1 3 a jx j2
=1 3


2
ja +1 j + max ja ,1 j , min fa g
 =1
max
fa g + =1
max
3
3 pa
=3 5 pa
=1 3
X
< =2min
fa g 
a jy j2 = y Ay:
4
T

T

xy

n

n

n

i

n

i

; ;:::

i

; ;:::

i

; ;:::

i;i

i

;i

;i

i

i

i;i

; ;:::

i

; ;:::

i

; ;:::

i

; ;:::

i

; ;:::

; ;:::

i;i

i

i

i

;i

i

i

i

i

i

; ;:::

;i

i

i

i

i

; ;:::

; ;:::

i

; ;:::

i

;i

i;i

i;i

; ;:::

i

i;i

i

i

i

i

i;i

; ;:::

i

i

i;i

i

i

;i

; ;:::

i;i

;i

i

i

i

;i

i;i

;i

i;i

i

;i

i

i

; ;:::

i

ii

i;i

i

i

i

; ;:::

i

; ;:::

i;i

i

i;i

i

; ;:::

=2 4

; ;:::

;i

i;i

i;i

ii

i

i

i

i

; ;:::

i

; ;:::

;i

i;i

i

; ;:::

i;i

i

ELA
Positive Deniteness of Tridiagonal Matrices

99

The result now follows from (11) and (10).
Let us now set 1 := mini=1;3;::: fai;ig; 1 := maxi=1;3;::: fai;ig;
2 := mini=1;3;::: jai,1;ij2 + jai+1;ij2, and 2 := maxi=1;3;::: jai,1;ij2 + jai+1;ij2 . Then
2 , jxAxj2
kAxkX
X
X
=
jai;ixij2 +
jai+1;ixi + ai+1;i+2xi+2 j2 ,
aii jxij2



i=1;3;:::

X

i=1;3;:::

jai;ixij2 ,

i=1;3;:::

X

i=1;3;:::

i=1;3;:::

ai;ijxij2

0s
12
s X
X
+@
jai+1;ij2jxi j2 +
jai+1;i+2j2 jxi+2j2A
i=1;3;:::
i=1;3;:::
2
3
X
X
X
2
2
2
2
2
ai;ijxij + 2 4
(jai,1;ij + jai+1;ij )jxij 5
jai;ixij ,


i=1;3;:::
i=1;3;:::
2
2
2
= t1 + (1 , t)1 , [t1 + (1 , t)1 ] + 2[t2 + (1 , t)2 ]
= t(1 , t)(1 , 1 )2+ 2(t2 + (1 , t)2 ) 
(15) = , (1 , 1 )2 t2 + (1 , 1 )2 + 2(2 , 2 ) t + 22 ;
i=1;3;:::

for some t 2 [0; 1]. If 1 = 1 then the value of (15) is 2(2 , 2 )t + 22 : If
2 )2 at
1 6= 1 then (15) assumes its maximum 41 (1 , 1 )2 + (2 + 2 ) + ( 2 ,
1 , 1

,

1
2
2
2
t = 2 + 2( ,  )2 when 2(2 , 2)  (1 , 1) and assumes it maximum 22
1
1
when 2(2 , 2) > (1 , 1)2 : Therefore

81
 
>
2 + (2 + 2 ) +  , 2 ;
(

,

)
1
1
>
4
 ,
<
2

2
if

=
6

and
2(

kAxk2 ,jx Axj  >
1
1
2 , 2)  (1 , 1 )2;
>
:
2
1

2
1

22 ; if 1 = 1 or 1 6= 1 and 2(2 , 2 ) > (1 , 1)2 :
(16)
If one of the conditions (ii), (iii), or (iv) is satised, then by (16),

kAxk22 , jxAxj2 < (i=1min
a ) ( min a )  x Ax y Ay:
;3;::: ii i=2;4;::: ii
Thus, by (11) and (10), A is positive denite.
We remark that a similar result can be obtained by switching the odd and even
indices in statement of theorem and in the proof.
Recall now that for a Hermitian matrix A 2 C n;n with eigenvalues 1  : : :  n ,
the spread of A is 1 , n and is denoted by Sp(A). For references on eigenvalue
spreads, see, e.g., Nylen and Tam [8], and Thompson [10].
Theorem 2.3. Let A = (ai;j ) 2 C n;n be an Hermitian tridiagonal matrix with

ELA
100

Mao-Ting Chien and Michael Neumann

positive diagonal entries. If

i

min fa g
=1 3
; ;:::

 

i;i



i

n(n , 1) X ,ja , a j2 + 4ja j2
;
min
f
a
g
>
=2 4
8
i;i

; ;:::

i;i

j;j

i;j

i