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Electronic Journal of Linear Algebra

Volume 3 ELA Volume 3 (1998) Article 9

1998

Positive definiteness of tridiagonal matrices via the

numerical range

Mao-Ting Chien

mtchien@math.math.scu.edu.tw Michael Neumann

neumann@math.uconn.edu

Follow this and additional works at:http://repository.uwyo.edu/ela

This Article is brought to you for free and open access by Wyoming Scholars Repository. It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized administrator of Wyoming Scholars Repository. For more information, please contactscholcom@uwyo.edu.

Recommended Citation

Chien, Mao-Ting and Neumann, Michael. (1998), "Positive definiteness of tridiagonal matrices via the numerical range", Electronic

Journal of Linear Algebra, Volume 3.

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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

ANDMICHAEL NEUMANN z

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.

Keywords. Positive de nite, tridiagonal matrices, numerical range, spread AMSsubject classi cations. 15A60, 15A57

1. Introduction.

In a paper on the question of unicity of best spline approxi-mations 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 ij)

2 R n;n

be a tridiagonal matrix with positive diagonal entries. If

a i;i,1 a i,1;i  1 4a ii a i,1;i,1  1 +  2 1 + 4n 2  ; 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 ij)

2 Rbe a

tridiagonal matrix with positive diagonal entries. If

a i;i,1 a i,1;i < 1 4a ii a i,1;i,1 1  cos  n+ 1  2 ; i= 2;:::;n;

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

The main purpose of this paper is to derive additional criteria, using the

numer-ical range

, for a symmetric tridiagonal matrixAwith 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

A2C

n;n is the set W(A) = fx  Axjx2C n ; jxj= 1g:

Many of our results here will be derived with the aid of the following recent charac-terization 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.

yDepartment 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.

zDepartment of Mathematics, University of Connecticut, Storrs, Connecticut 06269{3009, USA

(neumann@math.uconn.edu).

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94 Mao-Ting Chien and Michael Neumann

Theorem 1.3. [2, Theorem 1] LetA2C n;n

and supposeSbe a nonzero subspace

of C n . Then W(A) = [ x;y W(A xy) ;

wherexand y vary over all unit vectors inS and S

?, respectively, and where A xy =  x  Ax x  Ay y  Ax y  Ay  :

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 di erent criteria we have obtain in Section 2.

Some notations that we shall employ in this paper are as follows. Firstly, we shall usee

(k ) i ,

i= 1;:::;k, to denote thek{dimensionali{th unit coordinate vector, viz.,

thei{th entry ofe (k )

i is 1 and its remaining entries are 0.

Secondly, for a set f1;2;:::;ngdenote by 0=

f1;2;:::;ngn its complement

inf1;2;:::;ng. Thirdly, if and are subsets of f1;2;:::;ngand A 2C n;n

, 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 byA[ ].

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 thatAis annnHermitian matrix and partitionAinto A =  R B B  T  ;

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 Rin A, given by (A=R) = T ,B 

R ,1

B, 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 algorithmdue 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 ofAare positive de nite

which can be executed in parallel.

Theorem 2.1. LetA2C

n;nbe a Hermitian tridiagonal matrix and partition A into A = 2 6 4 A 1 a k ;k +1 e (k ) k 0 a k +1;k e (k )T k a k +1;k +1 a k +1;k +2 e (n,k ,1)T 1 0 a k +2;k +1 e (n,k ,1) 1 A 2 3 7 5 ;

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Positive De niteness of Tridiagonal Matrices 95 where1<k<n,A 1 2C k ;k , and A 2 2C n,k ,1;n,k ,1

. ThenA is positive de nite if

and only if the symmetric tridiagonal matrix

 A 1 0 0 A 2  , (1=a k +1;k +1)  " ja k ;k +1 j 2 e (k ) k e (k )T k a k ;k +1 a k +1;k +2 e (k ) k e (n,k ,1)T 1 a k +1;k a k +2;k +1 e (n,k ,1) 1 e (k )T k ja k +1;k +2 j 2 e (n,k ,1) 1 e (n,k ,1)T 1 # (1)

is positive de nite. Putting

C 1 = A 1 , ja k ;k +1 j 2 a k +1;k +1 e (k ) k e (k )T k ; (2)

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

A 1 and

A

2 are positive de nite and that the inequalities

1, ja k ;k +1 j 2 a k +1;k +1 , A ,1 1  k ;k > 0 (3) and 1, ja k +1;k +2 j 2+  ja k ;k +1 a k +1;k +2 j a k +1;k +1  2 (C ,1 1 ) k ;k ! , A ,1 2  1;1 > 0 (4) hold. In addition, if A 1 and A

2 are positive de nite, then any one of the following

three conditions is also sucient forAto be positive de nite.

(i)The matrices

A i , ja k ;k +1 j 2+ ja k +1;k +2 j 2 a k +1;k +1 I i ; i= 1;2; (5)

are positive de nite,

(ii) 1 > 1 a k +1;k +1  ja k ;k +1 j 2 , A ,1 1  k ;k+ ja k +1;k +2 j 2 , A ,1 2  1;1  ; (6) (iii) 1 > 1 a k +1;k +1  ja k ;k +1 j 2 d 1 +ja k +1;k +2 j 2 d k +2  ; (7) whered 1 and d

k +2 are the smallest eigenvalues of of A

1 and A

2, respectively.

Proof. LetSbe the subspace ofC

nspanned by e

(n)

k +1. Then, by Theorem 1.3, Ais

positive de nite if and only ifA

xyis positive de nite for all

x2Sandy2S ?. Assume now thatx= [0 0::: 0x k +10 ::: 0] t 2S and y= [y 1 ::: y k 0 y k +2 ::: y n] t 2S ?. Then A xy =  a k +1;k +1 a k +1;k x k +1 y k+ a k +1;k +2 x k +1 y k +2 a k ;k +1 y k x k +1+ a k +2;k +1 y k +2 x k +1 y  Ay 

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96 Mao-Ting Chien and Michael Neumann

is positive de nite if and only if

a k +1;k +1 y  Ay , (a k +1;k x k +1 y k+ a k +1;k +2 x k +1 y k +2)( a k ;k +1 y k x k +1+ a k +2;k +1) > 0: (8) Puty 0:= [ y 1 ::: y k y k +2 ::: y n] t 2C n,1. Then (8) becomes a k +1;k +1 y 0  A 1 0 0 A 2  y 0 , , ja k ;k +1 j 2 y k y k+ a k +1;k a k +1;k +2 y k y k +2 +a k +1;k a k +1;k +2 y k +2 y k+ ja k +1;k +2 j 2  y k +2 y k +2  = a k +1;k +1 y 0  A 1 0 0 A 2  y 0 ,y 0 " ja k ;k +1 j 2 e (k ) k e (k )T k a k ;k +1 a k +1;k +2 e (k ) k e (n,k ,1)T 1 a k +1;k a k +2;k +1 e (n,k ,1) 1 e (k ) T k ja k +1;k +2 j 2 e (n,k ,1) 1 e (n,k ,1)T 1 # y 0 >0

This proves the rst part of theorem. Namely, thatAis 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 matrixC

1 given

in (2) and by utilizing the fact that ifu;v2R `, then det, I+uv T  = 1 +v T u: (9)

Suppose now thatA 1and

A

2are positive de nite. We next establish (i). For this

purpose observe that the subtracted matrix in (1), which has rank 1, has eigenvalues 0 and, ja k ;k +1 j 2+ ja k +1;k +2 j 2  =a

k +1;k +1so that in the positive semide nite ordering,  A 1 0 0 A 2  , 1 a k +1;k +1  ja k ;k +1 j 2 e k e T k a k ;k +1 a k +1;k +2 e k e T 1 a k +1;k a k +2;k +1 e 1 e T k ja k +1;k +2 j 2 e 1 e T 1    A 1 0 0 A 2  , ja k ;k +1 j 2+ ja k +1;k +2 j 2 a k +1;k +1 I n,1 :

Thus if (5) holds, then Ais clearly positive de nite because then a submatrix ofA,

namelya

k ;k, as well as its Schur complement in

Agiven in (1) are positive de nite.

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  k ;k (1=d 1) >0 and , A ,1 2  1;1 (1=d k +1) >

0. This establishes (iii) and our proof is complete.

Next from Gershgorin theorem we know that a sucient condition for an Her-mitian matrixA= (a

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Positive De niteness of Tridiagonal Matrices 97

a i;i > P n j=1;6=i ja i;j

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 i;j)

2C n;n

be an Hermitian tridiagonal matrix with positive diagonal entries. ThenAis positive de nite if one of the following conditions

is satis ed. (i)  max i=1;3;::: a i;i  ,  min i=1;3;::: a i;i  +  max i=1;3;::: ja i+1;i j p a i;i + max i=3;5;::: ja i,1;i j p a i;i  2 < min i=2;4;::: a i;i : (ii) min i=1;3;::: a ii= max i=1;3;::: a ii, and 2 max ja i,1;i j 2+ ja i+1;i j 2: i= 1;3;::: <  min i=1;3;::: a ii   min i=2;4;::: a ii  : (iii) min i=1;3;::: a ii 6 = max i=1;3;::: a ii, 2  max i=1;3;::: ja i,1;i j 2+ ja i+1;i j 2 , min i=1;3;::: ja i,1;i j 2+ ja i+1;i j 2  >  max i=1;3;::: fa i;i g, min i=1;3;::: fa i;i g  2 ; and 2 max ja i,1;i j 2+ ja i+1;i j 2: i= 1;3;::: <  min i=1;3;::: a i;i   min i=2;4;::: a i;i  : (iv) min i=1;3;::: a i;i 6 = max i=1;3;::: a i;i, 2  max i=1;3;::: ja i,1;i j 2+ ja i+1;i j 2 , min i=1;3;::: ja i,1;i j 2+ ja i+1;i j 2    max i=1;3;::: fa i;i g, min i=1;3;::: fa i;i g  2 ; and 1 4  min i=1;3;::: a i;i , max i=1;3;::: a i;i  2 + max i=1;3;::: ja i,1;i j 2+ ja i+;i j 2+ min i=1;3;::: ja i,1;i j 2+ ja i+1;i j 2 +  max1;3;::: ja i,1;i j 2+ ja i+1;i j 2 ,min i=1;3;::: ja i,1;i j 2+ ja i+1;i j 2 maxi=1;3;::: a i;i ,min i=1;3;::: a i;i  2 <  min i=1;3;::: fa i;i g) ( min i=2;4;::: fa i;i g  :

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98 Mao-Ting Chien and Michael Neumann

Proof. We rst note that Bessel's inequality (see, e.g., [7, p.488]) implies that if

xandy aren{vectors which are orthogonal to each other, then jx  Ay j 2  kAxk 2 2 ,jx  Axj 2 : (10)

Next letSbe the subspace generated bye 1

;e 3

;e 5

;:::and observe that vectors x2S

andy2S

?have the form x= [x 10 x 30 x 5 ] T and y= [0y 20 y 4 ] T, respectively.

Then by Theorem 1.3,Ais positive de nite if and only ifA

xy is positive de nite for x2S andy2S ?, which is equivalent to x  Axy  Ay > jx  Ay j 2 : (11)

Assume now that condition (i) is satis ed. We rst recall the following inequality. Letp

1 ;;p

nbe positive numbers. Then for any real numbers q 1 ;q 2 ;:::;q n, (q 1+ q 2+ +q n) =(p 1+ p 2+ +p n)  maxfq i =p i: i= 1;2;:::;ng (12)

Then by (12) we have that

P i=1;3;::: ja i;i x i j 2 P i=1;3;::: a i;i jx i j 2  max i=1;3;::: a i;i ; (13) and P i=1;3;::: ja i+1;i x i+ a i+1;i+2 x i+2 j 2 P i=1;3;::: a i;i jx i j 2  s P i=1;3;::: ja i+1;i x i j 2 P i=1;3;::: a i;i jx i j 2 + s P i=1;3;::: ja i+1;i+2 x i+2 j 2 P i=1;3;::: a i;i jx i j 2 ! 2  s max i=1;3;::: ja i+1;i j 2 a i;i + s max i=3;5;::: ja i,1;i j 2 a i;i ! 2 =  max i=1;3;::: ja i+1;i j p a i;i + max i=3;5;::: ja i,1;i j p a i;i  2 : (14)

From (13), (14), and the hypothesis we now obtain,

kAxk 2 2 ,jx  Axj 2 x  Ax = P i=1;3;::: ja i;i x i j 2 P i=1;3;::: a i;i jx i j 2+ P i=1;3;::: ja i+1;i x i+ a i+1;i+2 x i+2 j 2 P i=1;3;::: a ii jx i j 2 , X i=1;3;::: a ii jx i j 2  max i=1;3;::: fa i;i g+  max i=1;3;::: ja i+1;i j p a i;i + max i=3;5;::: ja i,1;i j p a i;i  2 , min i=1;3;::: fa i;i g < min i=2;4;::: fa i;i g  X i=2;4;::: a i;i jy i j 2 = y  Ay :

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Positive De niteness of Tridiagonal Matrices 99

The result now follows from (11) and (10). Let us now set

1:= mini=1;3;::: fa i;i g; 1:= maxi=1;3;::: fa i;i g; 2:= mini=1;3;::: ja i,1;i j 2+ ja i+1;i j 2, and 2:= maxi=1;3;::: ja i,1;i j 2+ ja i+1;i j 2. Then kAxk 2 ,jx  Axj 2 = X i=1;3;::: ja i;i x i j 2 + X i=1;3;::: ja i+1;i x i+ a i+1;i+2 x i+2 j 2 , X i=1;3;::: a ii jx i j 2  X i=1;3;::: ja i;i x i j 2 , X i=1;3;::: a i;i jx i j 2 + 0 @ s X i=1;3;::: ja i+1;i j 2 jx i j 2+ s X i=1;3;::: ja i+1;i+2 j 2 jx i+2 j 2 1 A 2  X i=1;3;::: ja i;i x i j 2 , X i=1;3;::: a i;i jx i j 2+ 2 2 4 X i=1;3;::: (ja i,1;i j 2+ ja i+1;i j 2) jx i j 2 3 5 = t 2 1+ (1 ,t) 2 1 ,[t 1+ (1 ,t) 1] 2+ 2[ t 2+ (1 ,t) 2] = t(1,t)( 1 , 1) 2+ 2( t 2+ (1 ,t) 2) = ,( 1 , 1) 2 t 2+  ( 1 , 1) 2+ 2( 2 , 2)  t+ 2 2 ; (15) for some t 2 [0;1]. If 1 =

1 then the value of (15) is 2( 2 , 2) t + 2 2 : If 1 6 =

1 then (15) assumes its maximum 14( 1 , 1) 2+ ( 2+ 2) + ( 2 , 2 1 , 1 )2 at t= 12 + 2 , 2 2( 1 , 1) 2 when 2( 2 , 2)  ( 1 , 1)

2 and assumes it maximum 2 2 when 2( 2 , 2) >( 1 , 1) 2 :Therefore kAxk 2 2 ,jx  Axj 2  8 > > > < > > > : 1 4( 1 , 1) 2+ ( 2+ 2) +  2 , 2 1, 1  2 ; if 1 6 = 1 and2( 2 , 2) ( 1 , 1) 2 ; 2 2 ; if 1= 1 or 1 6 = 1 and2( 2 , 2) >( 1 , 1) 2 : (16)

If one of the conditions (ii), (iii), or (iv) is satis ed, then by (16),

kAxk 2 2 ,jx  Axj 2 <( min i=1;3;::: a ii) ( min i=2;4;::: a ii) x  Axy  Ay :

Thus, by (11) and (10),Ais 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 matrixA2C 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 = (a i;j)

2C n;n

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100 Mao-Ting Chien and Michael Neumann

positive diagonal entries. If

 min i=1;3;::: fa i;i g   min i=2;4;::: fa i;i g  > n(n,1) 8 X i<j , ja i;i ,a j;j j 2+ 4 ja i;j j 2  ;

thenA is positive de nite.

Proof. Let S be the subspace generated bye 1 ;e 3 ;:::. Forx 2 S and y 2S ?, y  x= 0. By Remark 1 in Tsing [11] , W 0( A)fu  Av ju;v2C n ; juj=jv j= 1; u  v= 0g

is a circular disk centered at the origin of radius ( 1 , n) =2. Thus jy  Axj 2  1 4(  1 , n) 2= 1 4 Sp(A) 2. By [8, Corollary 4], Sp(A) 2  n (n,2) 2 n X i=1 Sp(A[fig 0])2 ; (17)

But then from (17),

Sp(A) 2  n (n,2) 2 n,1 (n,3) 2 n,2 (n,4) 2  5 32 4 22 3 12 X i;j ((n,2)!)Sp(A[fi;jg]) 2 ;

from which we now get that 1 4Sp(A) 2  n(n,1) 8 X i<j , ja i;i ,a j;j j 2+ 4 ja i;j j 2  :

It now follows immediately from (11) and the fact that

x  Axy  Ay   min i=1;3;::: fa i;i g   min i=2;4;::: fa i;i g  ;

thatA is positive de nite.

3. Examples.

In this section, we give some examples to illustrate di erent cri-teria obtained in the paper.

Example 3.1. Consider the matrix

A = 2 6 6 4 a 2 0 0 2 b 3 0 0 3 4 5 0 0 5 b 3 7 7 5 ;

where aand b are positive numbers. Table 1 lists the applicability for positive

de -niteness ofAwitha= 4 so that min i=1;3;:::

a

ii= max i=1;3;:::

a

ii, while Table 2 shows criteria

fora= 3:9 and min i=1;3;::: a ii 6 = max i=1;3;::: a ii.

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Positive De niteness of Tridiagonal Matrices 101

Table1

Test for positive de niteness ofA(Example 3.1)

Applicability a=4, b=18 a=4, b=17 a=4, b=16.1

Theorem 2.2 (i) yes yes yes

Theorem 2.2 (ii) yes no no

Proposition 1.2 yes yes no

Gershgorin Theorem no no no

Table2

Test for positive de niteness ofA(Example 3.1)

Applicability a=3.9, b=18 a=3.9, b=17 a=3.9, b=16.2

Theorem 2.2 (i) yes yes yes

Theorem 2.2 (iii) yes no no

Proposition 1.2 yes yes no

Gershgorin Theorem no no no

Example 3.2. Consider the matrix

A = 2 4 10 0:5 0 0:5 8 8:5 0 8:5 10 3 5 :

It is easy to verify the principal minors ofAare positive,Ais positive de nite. Table

3 compares three other criteria, and shows that the condition Theorem 2.3 works.

Table3

Test for positive de niteness ofA(Example 3.2)

Applicability

Theorem 2.3 yes

Proposition 1.2 no

Gershgorin Theorem no

REFERENCES

[1] David L. Barrow, Charles K. Chui, Philip W. Smith, and Joseph D. Ward. Unicity of best mean approximation by second order splines with variable notes.Mathematics of Computation, 32:1131{1143, 1978.

[2] Mao-Ting Chien. On the numerical range of tridiagonal operators. Linear Algebra and its Applications, 246:203{214, 1996.

[3] Ming Gu and Stanley C. Eisenstat. A divide-and-conquer algorithm for the symmetric tridi-agonal eigenproblem. SIAM Journal on Matrix Analysis and Applications, 16:172{191, 1995.

[4] Emilie V. Haynsworth. Determination of the inertia of a partitioned Hermitian matrix.Linear Algebra and its Application, 1:73{81, 1968.

[5] Roger A. Horn and Charles R. Johnson. Matrix Analysis. Cambridge University Press, New York, 1985.

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ELA

102 Mao-Ting Chien and Michael Neumann

[6] Charles R. Johnson, Michael Neumann, and Michael J. Tsatsomeros. Conditions for the posi-tivity of determinants.Linear and Multilinear Algebra, 40:241{248, 1996.

[7] Ben Noble and James W.Daniel. Applied Linear Algebra. Prentice{Hall, Englewood Cli s, New Jersey, 1988.

[8] Peter Nylen and Tin-Yau Tam. On the spread of a Hermitian matrix and a conjecture of Thompson.Linear and Multilinear Algebra, 37:3{11, 1994.

[9] Danny C. Sorensen and Pink-Tak-Peter Tang. On the orthogonality of eigenvectors computed by divide{and{conquer techniques.SIAM Journal on Numerical Analysis, 28:1752{1775, 1991.

[10] Robert C. Thompson. The eigenvalue spreads of a Hermitian matrix and its principal subma-trices.Linear and Multilinear Algebra, 32:327:333, 1992.

[11] Nam-Kiu, Tsing. The constrained bilinear form and theC-numerical range. Linear Algebra

References

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