ranks and determinants of the sum of matrices from unitary orbits · 2009. 3. 20. · in section 3....

Linear and Multilinear Algebra, Vol. 56, Nos. 1–2, Jan.–Mar. 2008, 105–130

Ranks and determinants of the sum ofmatrices from unitary orbits

CHI-KWONG LI*y, YIU-TUNG POONz and NUNG-SING SZEx

yDepartment of Mathematics, College of William & Mary,Williamsburg, VA 23185, USA

zDepartment of Mathematics, Iowa State University,Ames, IA 50011, USA

xDepartment of Mathematics, University of Connecticut,Storrs, CT 06269, USA

Communicated by C. Tretter

(Received 30 January 2007; in final form 27 March 2007)

The unitary orbit UðAÞ of an n�n complex matrix A is the set consisting of matrices unitarilysimilar to A. Given two n�n complex matrices A and B, ranks and determinants of matricesof the form Xþ Y with ðX,Y Þ2UðAÞ � UðBÞ are studied. In particular, a lower bound andthe best upper bound of the set RðA,BÞ ¼ frankðXþ YÞ : X2UðAÞ,Y2UðBÞg are determined.It is shown that �ðA,BÞ ¼ fdetðXþ YÞ : X2UðAÞ,Y2UðBÞg has empty interior if and onlyif the set is a line segment or a point; the algebraic structure of matrix pairs (A,B) with suchproperties are described. Other properties of the sets R(A,B) and �ðA,BÞ are obtained.The results generalize those of other authors, and answer some open problems. Extensionsof the results to the sum of three or more matrices from given unitary orbits are also considered.

Keywords: Rank; Determinant; Matrices; Unitary orbit

2000 Mathematics Subject Classifications: 15A03; 15A15

1. Introduction

Let A2Mn. The unitary orbit of A is denoted by

UðAÞ ¼ UAU� : U�U ¼ Inf g:

Evidently, if A is regarded as a linear operator acting on Cn, then UðAÞ consistsof the matrix representations of the same linear operator under different

*Corresponding author. Email: [email protected] to Professor Yik-Hoi Au-Yeung on the occasion of his 70th birthday.

Linear and Multilinear AlgebraISSN 0308-1087 print/ISSN 1563-5139 online � 2008 Taylor & Francis

http://www.tandf.co.uk/journalsDOI: 10.1080/03081080701369306

Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009

orthonormal bases. Naturally, UðAÞ captures many important features of theoperator A. For instance, A is normal if and only if UðAÞ has a diagonal matrix; A isHermitian (positive semi-definite) if and only if UðAÞ contains a (nonnegative) real diag-onal matrix; A is unitary if and only if UðAÞ has a diagonal matrix with unimodulardiagonal entries. There are also results on the characterization of diagonal entriesand submatrices of matrices in UðAÞ; see [14,20,23,30] and their references. In addition,the unitary orbit of A has a lot of interesting geometrical and algebraic properties,see [11].

Motivated by theory as well as applied problems, there has been a great dealof interest in studying the sum of two matrices from specific unitary orbits.For example, eigenvalues of UAU� þ VBV� for Hermitian matrices A and B werecompletely determined in terms of those of A and B (see [16] and its references); theoptimal norm estimate of UAU� þ VBV� was obtained (see [10] and its references);the range of values of detðUAU� þ VBV�Þ for Hermitian matrices A,B2Mn wasdescribed, see [15]. Later, Marcus and Oliveira [26,29] conjectured that if A,B2Mnare normal matrices with eigenvalues a1, . . . , an and b1, . . . , bn, respectively, then forany unitary U,V2Mn, detðUAU� þ VBV�Þ always lies in the convex hull of

PðA,BÞ ¼Ynj¼1

aj þ b�ðjÞ� �

: � is a permutation of f1, . . . , ng( )

: ð1:1Þ

In connection to this conjecture, researchers [2–5,7,8,12,13] considered the deter-minantal range of A,B2Mn defined by

�ðA,BÞ ¼ fdetðAþUBU�Þ: U is unitaryg:

This can be viewed as an analog of the generalized numerical range ofA and B defined by

WðA,BÞ ¼ ftrðAUBU�Þ: U is unitaryg,

which is a useful concept in pure and applied areas; see [18,21,25] and their references.In this article, we study some basic properties of the matrices in

UðAÞ þ UðBÞ ¼ fXþ Y : ðX,Y Þ 2 UðAÞ � UðBÞg:

The focus will be on the rank and the determinant of these matrices.Our article is organized as follows. In section 2, we obtain a lower bound and the best

upper bound for the set

RðA,BÞ ¼ frankðXþ Y Þ : X 2 UðAÞ,Y 2 UðBÞg;

moreover, we characterize those matrix pairs (A,B) such that R(A,B) is a singleton, andshow that the set R(A,B) has the form fk, kþ 1, . . . , ng if A and �B have no commoneigenvalues. On the contrary if A and �B are orthogonal projections, then therank values of matrices in UðAÞ þ UðBÞ will either be all even or all odd. In section 3,we characterize matrix pairs ðA,BÞ such that �ðA,BÞ has empty interior, which is

106 C.-K. Li et al.


possible only when �ðA,BÞ is a singleton or a nondegenerate line segment. This extendsthe results of other researchers who treated the case when A and B are normal [2–4,9].In particular, our result shows that it is possible to have normal matrices A and B suchthat �ðA,BÞ is a subset of a line, which does not pass through the origin. This disprovesa conjecture in [9]. In [3], the authors showed that if A,B2Mn are normal matrices suchthat the union of the spectra of A and �B consists of 2n distinct elements, then everynonzero sharp point of �ðA,BÞ is an element in P(A,B). (See the definition of sharppoint in section 3.) We showed that every (zero or nonzero) sharp point of �ðA,BÞbelongs to P(A,B) for arbitrary matrices A,B2Mn. In section 4, we consider the sumof three or more matrices from given unitary orbits, and matrix orbits correspondingto other equivalence relations.

In the literature, some authors considered the set

DðA,BÞ ¼ fdetðX� YÞ : X 2 UðAÞ,Y 2 UðBÞg

instead of �ðA,BÞ. Evidently, we have DðA,BÞ ¼ �ðA, �BÞ. It is easy to translateresults on D(A,B) to those on �ðA,BÞ, and vice versa. Indeed, for certain resultsand proofs, it is more convenient to use the formulation of D(A,B). We will do thatin section 3. On the other hand, it is more natural to use the summation formulationto discuss the extension of the results to matrices from three or more unitary orbits.

2. Ranks

2.1. Maximum and minimum rank

In [10], the authors obtained optimal norm bounds for matrices in UðAÞ þ UðBÞ for twogiven matrices A,B2Mn. By the triangle inequality, we have

maxfkUAU� þ VBV�k: U,V unitaryg � minfkA� �Ink þ kBþ �Ink: � 2 Cg:

It was shown in [10] that the inequality is actually an equality. For the rank functions,we have

maxfrankðUAU� þVBV�Þ :U,V unitaryg �minfrankðA��InÞþ rankðBþ�InÞ:� 2Cg:

Of course, the right side may be strictly larger than n, and thus equality may not hold ingeneral. It turns out that this obstacle can be overcome easily as shown in the following.

THEOREM 2.1 Let A,B2Mn and

m ¼ minfrankðA� �InÞ þ rankðBþ �InÞ: � 2 Cg¼ minfrankðA� �InÞ þ rankðBþ �InÞ: � is an eigenvalue of A��Bg:

Then

maxfrankðUAU� þ VBV�Þ : U,V unitaryg ¼ minfm, ng:

Ranks and determinants of the sum of matrices 107


Proof If � is an eigenvalue of A��B, then rankðA� �InÞ þ rankðBþ �InÞ�2n� 1;if � is not an eigenvalue of A��B, then rankðA� �InÞ þ rankðBþ �InÞ ¼ 2n: As aresult, rankðA� �InÞ þ rankðBþ �InÞ will attain its minimum at an eigenvalue � ofthe matrix A��B.

It is clear that

maxfrankðUAU� þ VBV�Þ: U,V unitaryg � minfm, ng:

It remains to show that there are U,V such that UAU� þ VBV� has rank equal tominfm, ng.

Suppose m�n and there is � such that rankðA� �InÞ ¼ k and rankðBþ�InÞ ¼ m� k. We may replace (A,B) by ðA� �In,Bþ �InÞ and assume that � ¼ 0.Furthermore, we may assume that k�m� k; otherwise, interchange A and B.

Let A ¼ XDY be such that X,Y2Mn are unitary, and D ¼ D1 � 0n�k with invertiblediagonal D1. Replace A by YAY*, we may assume that

A ¼ UD ¼U11 U12

U21 U22

� �D1 0

0 0n�k

� �

with U ¼ YX. Similarly, we may assume that

B ¼ VE ¼V11 V12

V21 V22

� �0k 0

0 E2

� �,

where V is a unitary matrix and E2 is a diagonal matrix with rank m� k. Let W be aunitary matrix such that the first k columns of WU together with the last n� kcolumns of V are linearly independent. That is, if

W ¼W11 W12

W21 W22

� �,

the matrix

W11U11 þW12U21 V12W21U11 þW22U21 V22

� �

is invertible. If W11 is invertible, then D1W�11 has rank k and so

WAW� þ B ¼W11U11 þW12U21 V12W21U11 þW22U21 V22

� �D1W

�11 D1W

�21

0 E2

� �

has rank m. If W11 is not invertible, we will replace W by ~W obtained as follows. By theCS decomposition, there are unitary matrices P1,Q12Mk and P2,Q22Mn�k such that

ðP1 � P2ÞWðQ1 �Q2Þ ¼C S

�S C

� �� In�2k,

108 C.-K. Li et al.


where C ¼ diagðc1, . . . , ckÞ with 1 � c1�� ck�0 and S ¼ diagðffiffiffiffiffiffiffiffiffiffiffiffiffi1� c21

q, . . . ,

ffiffiffiffiffiffiffiffiffiffiffiffiffi1� c2k

qÞ.

Then perturb the zero diagonal entries of C slightly to ~C ¼ diagð ~c1, . . . , ~ckÞ so that ~Cis invertible, and set ~S ¼ diagð

ffiffiffiffiffiffiffiffiffiffiffiffiffi1� ~c21

q, . . . ,

ffiffiffiffiffiffiffiffiffiffiffiffiffi1� ~c2k

qÞ. Then

~W ¼ ðP1 � P2Þ�~C ~S� ~S ~C

� �� In�2k

� �ðQ1 �Q2Þ�

will be a slight perturbation of W with invertible ~W11 ¼ P�1 ~CQ�1, which can be chosensuch that the matrix

~W11U11 þ ~W12U21 V12~W21U11 þ ~W22U21 V22

!

is still invertible. Then ~WA ~W� þ B has rank m.Now, assume that rankðA� �InÞ þ rankðBþ �InÞ�nþ 1 for every �2C.

Let a1, . . . , an and b1, . . . , bn be the eigenvalues of A and B. We consider two cases.First, suppose ai þ bj 6¼ 0 for some i, j. We may assume that i ¼ j ¼ 1. Applyingsuitable unitary similarity transforms, we may assume that A and B are unitarily similarto matrices in upper triangular form

a1 �0 A1

� �and

b1 �0 B1

� �:

Since rankðA� �InÞ þ rankðBþ �InÞ�nþ 1 for every �2C, it follows thatrankðA1 � �In�1Þ þ rankðB1 þ �In�1Þ�n� 1 for every �2C. By induction assumption,there is a unitary V1 such that detðA1 þ V1B1V�1Þ 6¼ 0. Let V ¼ ½1� � V1. ThendetðAþ VBV�Þ ¼ ða1 þ b1Þ detðA1 þ V1B1V�1Þ 6¼ 0.

Suppose ai þ bj ¼ 0 for all i, j2f1, . . . , ng. Replacing (A,B) by ðA� a1In,B� b1InÞ,we may assume that A and B are nilpotents. If A or B is normal, then it will be thezero matrix. Then rankðAÞ þ rankðBÞ

Suppose n�4. Applying suitable unitary similarity transforms, we may assumethat both A and B are in upper triangular form with nonzero (1, 2) entries;see [27, Lemma 1]. Modify B by interchanging its first two rows and columns. Then,A and B have the form

A11 A12

0 A22

� �and

B11 B12

0 B22

� �

respectively, so that

A11 ¼0 �

0 0

� �and B11 ¼

0 0

� 0

� �

with �� 6¼ 0, and A22,B22 are upper triangular nilpotent matrices. If rankAþrankB�nþ 2, then

rankðA22 � �In�2Þ þ rankðB22 þ �In�2Þ � rankA22 þ rankB22� rankAþ rankB� 4 � n� 2:

If rankAþ rankB ¼ nþ 1, we claim that by choosing a suitable unitary similaritytransform, we can further assume that rankA22 ¼ rankA� 1. Then

rankðA22 � �In�2Þ þ rankðB22 þ �In�2Þ � rankA22 þ rankB22� rankAþ rankB� 3 ¼ n� 2:

In both cases, by induction assumption, there is V2 such that detðA22 þ V2B22V�2Þ 6¼ 0.Let V ¼ I2 � V2. Then

detðAþ VBV�Þ ¼ �� detðA22 þ V2B22V�2Þ 6¼ 0:

Now it remains to verify our claim. Suppose A has rank k andrankAþ rankB ¼ nþ 1. Then k�ðnþ 1Þ=2. Let S be an invertible matrix such thatS�1AS ¼ J is the Jordan form of A. If J has a 2�2 Jordan block, then we canalways permute J so that

J ¼J11 0

0 J22

� �with J11 ¼

0 1

0 0

� �

and rankðJ22Þ ¼ p� 1. By QR factorization, write S ¼ U�T for some unitary matrix Uand invertible upper triangular matrix

T ¼T11 T12

0 T22

� �:

Then A is unitarily similar to

TJT�1 ¼ T11J11T

�111 �

0 T22J22T�122

!,

which has the described property.

110 C.-K. Li et al.


Suppose J does not contain any 2�2 block, then J must have an 1�1 Jordan block.Otherwise, k ¼ rankA�2n=3 and hence

rankAþ rankB � 2k � 4n=3 ¼ nþ n=3 > nþ 1:

Now we may assume that

J ¼02 J12

0 J22

� �

is strictly upper triangular matrix such that J12 has only a nonzero entry in the (1, 1)-thposition and rank J22 ¼ k� 1. Let Ŝ be obtained from In by replacing the (3, 2)-th entrywith one, then

Ŝ�1JŜ ¼ Ĵ ¼ Ĵ11 J120 J22

!

with

Ĵ11 ¼0 1

0 0

� �:

Applying QR factorization on SŜ ¼ U�T with unitary U and invertible upper triangularT, then A is unitarily similar to TĴT�1, which has the described form. g

The value m in Theorem 2.1 is easy to determine as we need only to focuson rankðA� �InÞ þ rankðBþ �InÞ for each eigenvalue � of A��B. In particular,if � is an eigenvalue of A, then rankðA� �InÞ ¼ n� k, where k is the geometric multi-plicity of �; otherwise, rankðA� �InÞ ¼ n. Similarly, one can determine rankðBþ �InÞ.The situation for normal matrices is even better as shown in the following.

COROLLARY 2.2 Suppose A and B are normal matrices such that ‘ is themaximum multiplicity of an eigenvalue of A��B. Then minfrankðA� �InÞþrankðBþ �InÞ : �2Cg equals 2n� ‘. Consequently,

maxfrankðUAU� þ VBV�Þ : U,V unitaryg ¼ minf2n� ‘, ng:

Here are some other consequences of Theorem 2.1.

COROLLARY 2.3 Let A,B2Mn. Then UAU� þ VBV� is singular for all unitary U,V ifand only if there is �2C such that rankðA� �InÞ þ rankðBþ �InÞ

If k � � �>bn suchthat b1>n>b2>n� 1>b3> � � �>b1>1. Then (2.1) clearly holds andrankðAþ BÞ ¼ 1. In fact, by Theorem 2.7 subsequently, we know that for anym2f1, . . . , ng, there are unitary U,V2Mn such that rankðUAU� þ VBV�Þ ¼ m.

2.2. Additional results

Here we study other possible rank values of matrices in UðAÞ þ UðBÞ. The followingresult shows that if A and �B have disjoint spectra, then one can get every possiblerank values from the minimum to the maximum value, which is n.

THEOREM 2.7 Suppose A,B2Mn such that A and �B have disjoint spectra, and AþBhas rank k

argument again to get a unitary Û such that ÛUAU�Û� þ B has rank kþ 2. Repeatingthis procedure, we will get the desired conclusion.

If k ¼ n� 1. Then assume VAV* and WBW* are in upper triangular form. ForU ¼W�V, we have UAU� þ B ¼W�ðVAV� þWBW�ÞW is invertible.

For 1�k0 such thatfor U ¼ diagðei�, 1, . . . , 1Þ the matrix UAU� þ B has rank kþ 1 because removing itsfirst row has rank k, and adding the first row back will increase the rank by 1.

Now, suppose C21 ¼ 0. Then A21 6¼ 0. Otherwise, A and �B have common eigenva-lues since A22 ¼ B22. Now, C11 is invertible. Assume that the matrix obtained by remov-ing the first row and first column of C11 has rank k� 1. Otherwise, replace (A,B) byðVAV�,VBV�Þ by some unitary matrix V of the form V1 � In�k. Since A12 and A21are nonzero, we may further assume that vt ¼ ½a1, kþ1, . . . , a1n� 6¼ 0 andu ¼ ½akþ1, 1, . . . , an1�t 6¼ 0. Then there exists a small �>0 such that forU ¼ diagðei�, 1, . . . , 1Þ the matrix UAU� þ B has the form

Ĉ11 Ĉ12

Ĉ21 0n�k

!,

where Ĉ11 is invertible such that removing its first row and first column results in a rankk� 1 matrix, only the first row of Ĉ12 is nonzero and equal to ðei� � 1Þvt, only the firstcolumn of Ĉ21 is nonzero and equal to ðe�i� � 1Þu. Now, removing the first row andfirst column of UAU� þ B has rank k� 1; adding the column ðe�i� � 1Þu (to the left)will increase the rank by 1, and then adding the first row back will increase the rankby 1 more. So, UAU� þ B has rank kþ 1. g

Note that the assumption that A and �B have disjoint spectra is essential. Forexample, if A,B2M4 such that A and �B are rank 2 orthogonal projections, thenUAU� þ VBV� can only have ranks 0, 2, 4. More generally, we have the following.



PROPOSITION 2.8 Suppose A,B2Mn such that A and �B are orthogonal projectionsof rank p and q. Then k ¼ rank UAU� þ Bð Þ for a unitary matrix U2Mn, if and onlyif k ¼ jp� qj þ 2j with j�0 and k�minfpþ q, 2n� p� qg.

Proof SupposeUAU� ¼ Ip � 0n�p andVBV� ¼ 0j ��Iq � 0n�j�q. ThenUAU� þVBV�has rank k ¼ jp� qj þ 2j�minfpþ q, 2n� p� qg. Thus, V�UAU�Vþ B has rank kas well.

Conversely, consider UAU� þ B for a given unitary U. There is a unitary V such that

VUAU�V� ¼ Ip � 0n�p and VBV� ¼ � Ir � 0s �C2 CS

CS S2

� �� Iu � 0v

� ,

where C and S are invertible diagonal matrices with positive diagonal entries

such that C2 þ S2 ¼ It, rþ sþ t ¼ p and rþ tþ u ¼ q. (Evidently, the first rcolumns of V* span the intersection of the range spaces of UAU* and B, thenext s columns of V* span the intersection of the range space of UAU* and thenull space of B, the last v columns of V* span the intersection of the null spaceof UAU* and B, the u columns preceding those span the intersection of the rangespace of B and the null space of UAU*.) So, UAU� þ B has the asserted rankvalue. g

The following result was proved in [24].

THEOREM 2.9 Let A,B2Mn. Then UAU� þ VBV� is invertible for all unitary U,V2Mnif and only if there is �2C such that the singular values of A� �In and Bþ �In lie in twodisjoint closed intervals in ½0,1Þ.

Using this result, we can deduce the following.

THEOREM 2.10 Let A,B2Mn and k2f0, . . . , ng. Then rankðUAU� þ VBV�Þ ¼ k for allunitary U,V2Mn if and only if one of the following holds.

(a) One of the matrices A or B is scalar, and rankðAþ BÞ ¼ k.(b) k ¼ n and there is �2C such that the singular values of A� �In and Bþ �In lie

in two disjoint closed intervals in ½0,1Þ.

Proof If (a) holds, say, B ¼ �In, then rankðUAU� þ VBV�Þ ¼ rankðA� �InÞ ¼ k for allunitary U,V2Mn.

If (b) holds, then kðA� �InÞxk>kðBþ �InÞyk for all unit vectors x, y2Cn, orkðA� �InÞxk

3. Determinants

Let A,B2Mn with eigenvalues a1, . . . , an, and b1, . . . , bn, respectively. In this section westudy the properties of �ðA,BÞ and P(A,B). For notational convenience and easydescription of the results and proofs, we consider the sets

DðA,BÞ ¼ �ðA, �BÞ ¼ fdetðX� YÞ : X 2 UðAÞ,Y 2 UðBÞg

and

QðA,BÞ ¼ PðA, �BÞ ¼Ynj¼1

aj � b�ð j Þ� �

: � is a permutation of f1, . . . , ng( )

:

It is easy to translate the results on D(A,B) and Q(A,B) to those on �ðA,BÞ andP(A,B), and vice versa.

For any permutation ð�ð1Þ, . . . , �ðnÞÞ of ð1, . . . , nÞ, there are unitary matrices U and Vsuch that UAU* and VBV* are upper triangular matrices with diagonal entriesa1, . . . , an and b�ð1Þ, . . . , b�ðnÞ, respectively. It follows that

QðA,BÞ DðA,BÞ:

The elements in Q(A,B) are called �-points.Note also that if we replace (A,B) by ðUAU� � �In,VBV� � �InÞ for any �2C and

unitary U,V2Mn, the sets Q(A,B) and D(A,B) will be the same. Moreover,DðB,AÞ ¼ ð�1ÞnDðA,BÞ and QðB,AÞ ¼ ð�1ÞnQðA,BÞ.

The following result can be found in [9].

THEOREM 3.1 Suppose A,B2M2 have eigenvalues �1,�2 and �1,�2, respectively, andsuppose A� ðtrA=2ÞI2 and B� ðtrB=2ÞI2 have singular values a�b�0 and c�d�0.Then DðA,BÞ is an elliptical disk with foci ð�1 � �1Þð�2 � �2Þ and ð�1 � �2Þð�2 � �1Þwith length of minor axis equal to 2ðac� bd Þ. Consequently, D(A,B) is a singletonif and only if A or B is a scalar matrix; D(A,B) is a nondegenerate line segment if andonly if A and B are nonscalar normal matrices.

In the subsequent discussion, let

WðAÞ ¼ x�Ax : x 2 Cn, x�x ¼ 1

�

be the numerical range of A2Mn.

3.1. Matrices whose determinantal ranges have empty interior

THEOREM 3.2 Let A,B2Mn with n�3. Then DðA,BÞ ¼ f�g if and only if one ofthe following holds.

(a) � ¼ 0, and there is � 2 C such that rank ðA� �InÞ þ rank ðB� �InÞ < n.(b) � 6¼ 0, one of the matrices A or B is a scalar matrix, and detðA� BÞ ¼ �.



Proof If (a) or (b) holds, then clearly D(A,B) is a singleton. If DðA,BÞ ¼ f0g, thencondition (a) holds by Corollary 2.3.

Suppose DðA,BÞ ¼ f�g with � 6¼ 0. We claim that A or B is a scalar matrix. Suppose Aand B have eigenvalues a1, . . . , an and b1, . . . , bn, respectively. Assume that A has atleast two distinct eigenvalues a1, a2 and B also has two distinct eigenvalues b1, b2.Then

Qnj¼1ðaj � bjÞ and ða1 � b2Þða2 � b1Þ

Qnj¼3ðaj � bjÞ will be two distinct �-points,

which is a contradiction because QðA,BÞDðA,BÞ is also a singleton.So, we have a1 ¼ � � � ¼ an or b1 ¼ � � � ¼ bn. We may assume that the latter case holds;

otherwise, interchange the roles of A and B. Suppose neither A nor B is a scalar matrix.Applying a suitable unitary similarity transform to B, we may assume that B is inupper triangular form with nonzero (1, 2) entry. Also, we may assume that A is inupper triangular form so that the leading two-by-two matrix is not a scalar matrix. Let

A ¼ A11 A120 A22

� �and B ¼ B11 B12

0 B22

� �

with A11,B112M2, then DðA11,B11Þ is a nondegenerate circular disk by Theorem 3.1.Since

fdetðA22 � B22Þ� : � 2 DðA11,B11Þg DðA,BÞ,

we see that D(A,B) cannot be a nonzero singleton. g

THEOREM 3.3 Suppose A,B2Mn are such that D(A,B) is not a singleton. The followingconditions are equivalent.

(a) D(A,B) has empty interior.(b) D(A,B) is a nondegenerate line segment.(c) Q(A,B) is not a singleton, i.e., there are at least two distinct �-points, and one

of the following conditions holds.

(c.1) A and B are normal matrices with eigenvalues lying on the same straight lineor the same circle.

(c.2) There is �2C such that one of the matrices A� �In or B� �In is rank 1normal, and the other one is invertible normal so that the inverse matrixhas collinear eigenvalues.

(c.3) There is �2C such that A� �In is unitarily similar to ~A� 0n�k and B� �Inis unitarily similar to 0k � ~B so that ~A2Mk and ~B2Mn�k are invertible.

In [3], the authors conjectured that for normal matrices A,B2Mn, if D(A,B) iscontained in a line L, then L must pass through the origin. Using the above result,we see that the conjecture is not true. For example, if A ¼ diagð1, 1þ i, 1� iÞ�1 andB ¼ diagð�1, 0, 0Þ, then D(A,B) is a straight line segment joining the points 1� i=2and 1þ i=2; see Corollary 3.11. This shows that D(A,B) can be a subset of a straightline not passing through the origin. Of course, Theorem 3.3 covers more generalsituations. In (c.1), the line segment D(A,B) and the origin are collinear; in (c.3) theline segment D(A,B) has endpoints 0 and ð�1Þn�k detð ~AÞ detð ~BÞ; in (c.2) the line segmentand the origin may or may not be collinear.

116 C.-K. Li et al.


Since DðA,BÞ ¼ ð�1ÞnDðB,AÞ, D(A,B) has empty interior (is a line segment) if andonly if D(B,A) has empty interior (is a line segment). This symmetry will be used inthe following discussion. We establish several lemmas to prove the theorem.

Given a, b, c, d2C, with ad� bc 6¼ 0, let fðzÞ ¼ ðazþ bÞ=ðczþ dÞ be the fractionallinear transform on Cnf�d=cg (C, if c¼ 0). If A2Mn such that cAþ dIn is invertible,one can define fðAÞ ¼ ðaAþ bInÞðcAþ dInÞ�1. The following is easy to verify.

LEMMA 3.4 Suppose A,B2Mn, and fðzÞ ¼ ðazþ bÞ=ðczþ d Þ is a fractional lineartransform such that f(A) and f(B) are well defined. Then

Dð fðAÞ, fðBÞÞ ¼ detððcAþ dInÞðcBþ dInÞÞ�1ðad� bcÞnDðA,BÞ:

LEMMA 3.5 Let A,B2Mn with eigenvalues a1, . . . , an and b1, . . . , bn, respectively.If D(A,B) has empty interior, then

DðA,BÞ ¼ DðA, diagðb1, . . . , bnÞÞ ¼ Dðdiagða1, . . . , anÞ,BÞ¼ Dðdiagða1, . . . , anÞ, diagðb1, . . . , bnÞÞ:

Proof Assume that D(A,B) has empty interior. Applying a unitary similaritytransform we can assume that A ¼ ðarsÞ and B ¼ ðbrsÞ are upper triangular matrices.For any unitary matrix V2Mn, let D ¼ ðdrsÞ ¼ VBV�. For any �2½0, 2�Þ,let U� ¼ ½ei�� In�1. Denote by Xrs the ðn� 1Þ�ðn� 1Þ matrices obtained fromX2Mn by deleting its r-th row and the s-th column. Expanding the determinantdetðU�AU�� DÞ along the first row yields

detðU�AU�� DÞ ¼ ða11 � d11Þ detðA11 �D11Þ þXnj¼2ð�1Þ jþ1ðei�a1j � d1jÞ detðA1j �D1jÞ

¼ ða11 � d11Þ detðA11 �D11Þ �Xnj¼2ð�1Þ jþ1d1j detðA1j �D1jÞ

" #þ ei��,

where � ¼Pn

j¼2ð�1Þjþ1a1j detðA1j �D1jÞ: Thus,

CðA,VBV�Þ ¼ fdetðU�AU�� VBV�Þ: � 2 ½0, 2�Þg

is a circle with radius j�j. Suppose j�j 6¼ 0, i.e., CðA,VBV�Þ is a nondegenerate circle.Repeating the construction in the previous paragraph on V ¼ In, we get a degeneratecircle

CðA,BÞ ¼ fdetðA� BÞg:

Since the unitary group is path connected, there is a continuous function t�Vt fort2½0, 1� so that V0 ¼ V and V1 ¼ In. Thus, we have a family of circles CðA,VtBV�t Þ



in D(A,B) transforming CðA,VBV�Þ to C(A,B). Hence, all the points inside CðA,VBV�Þbelong to D(A,B). Thus, D(A,B) has nonempty interior. As a result,

� ¼Xnj¼2ð�1Þ jþ1a1j detðA1j �D1jÞ ¼ 0,

and

detðA�VBV�Þ ¼ detðA�DÞ ¼ ða11 � d11ÞdetðA11 �D11Þ �Xnj¼2ð�1Þ jþ1d1j detðA1j �D1jÞ

¼ detðA1 �DÞ ¼ detðA1 �VBV�Þ,

where A1 is the matrix obtained from A by changing all the nondiagonal entries inthe first row to zero. It follows that DðA,BÞ ¼ DðA1,BÞ. Inductively, by expandingthe determinant detðUjAjU�j �DÞ along the ( jþ 1)-th row with Uj ¼ Ij � ½ei�� In�j�1,we conclude that DðAj,BÞ ¼ DðAjþ1,BÞ where Ajþ1 is the matrix obtained from Aj bychanging all the nondiagonal entries in the ð jþ 1Þ-th row to zero. Therefore,

DðA,BÞ ¼ DðA1,BÞ ¼ DðA2,BÞ ¼ � � � ¼ DðAn�1,BÞ ¼ Dðdiagða11, . . . , annÞ,BÞ:

Note that a11, . . . , ann are the eigenvalues of A as A is in the upper triangular form.Similarly, we can argue that DðA,BÞ ¼ DðA, diagðb1, . . . , bnÞÞ. Now, we can apply theargument to Dðdiagða1, . . . , anÞ,BÞ to get the last set equality. g

LEMMA 3.6 Let A ¼ Â� 0n�k, where Â2Mk with k2f1, . . . , n� 1g is upper triangularinvertible. If B2Mn has rank n� k, then

DðA,BÞ ¼ fð�1Þn�k detðÂÞ detðX�BXÞ : X is n� ðn� kÞ, X�X ¼ In�kg:

If B ¼ 0k � B̂ so that B̂2Mn�k is invertible, then D(A,B) is the line segment joining 0 andð�1Þn�k detðÂÞ detðB̂Þ.

Proof Suppose A ¼ ðarsÞ has columns A1, . . . ,An, and U�BU has columns B1, . . . ,Bn.Let C be obtained from A�U�BU by removing the first column, and let B22 beobtained from C by removing the first row. By linearity of the determinant functionon the first column,

detðA�U�BUÞ ¼ detð½A1jC �Þ � detð½B1jC �Þ ¼ �a11 detðB22Þ þ 0,

because ½B1jC � has rank at most n� 1. Inductively, we see that

detðA�U�BUÞ ¼ ð�1Þn�k detðÂÞ detðYÞ

where Y is obtained from U�BU by removing its first k rows and first k columns.

118 C.-K. Li et al.


Now if B ¼ 0k � B̂ so that B̂2Mn�k is invertible, then the setfdet X�BXð Þ : X�X ¼ In�kg is a line segment joining 0 and detðB̂Þ; e.g., see [6]. Thus,the last assertion follows. g

LEMMA 3.7 Suppose A and B are not both normal such that A� B has exactly n nonzeroeigenvalues. If D(A,B) has no interior point, then there exist �2C and 0�k�n such thatA� �In and B� �In are unitarily similar to matrices of the form ~A� 0n�k and 0k � ~B forsome invertible matrices ~A2Mk and ~B2Mn�k.

Proof Suppose A or B is a scalar matrix, say B ¼ �In. Under the given hypothesis,A� �In is invertible and B� �In ¼ 0. Thus, the result holds for k ¼ n. In the rest ofthe proof, assume that neither A nor B is a scalar matrix.

We may assume that

A ¼ ðarsÞ ¼A11 A12

0 A22

� �and B ¼ ðbrsÞ ¼

B11 B12

0 B22

� �ð3:1Þ

such that A11,B112Mm and A22,B222Mn�m are upper triangular matrices so thatA11,B22 are invertible, and A22,B11 are nilpotent.

If m¼ 0, then A ¼ A11 is nilpotent and B ¼ B22 is invertible. We are going to showthat A¼ 0. Hence, the lemma is satisfied with k¼ 0.

Suppose A 6¼ 0. We may assume that a12 6¼ 0. Let

X ¼0 a12

0 0

� �and Y ¼

b11 b12

0 b22

� �:

Since B is not a scalar matrix, we may assume that Y is not a scalar matrix either.Then D(X,Y) is a nondegenerate elliptical disk and

ð�1Þn�detðBÞb11b22

: � 2 DðX,Y Þ�

DðA,BÞ:

Therefore, D(A,B) has nonempty interior, a contradiction. Similarly, if m¼ n,then B¼ 0. Hence, we may assume that 1�m

submatrix are the same as A. Moreover, since we have shown that am,mþ1 ¼ 0 ¼ bm,mþ1,the principal submatrix of ÛAÛ� lying in rows m� 1,m,mþ 1 has the form

� � �0 amm 0

� � �

0B@

1CA:

A similar result is true for V̂BV̂�. Hence,

detðÛAÛ� � V̂BV̂�Þ ¼ � detðA� BÞ�=ðam�1,m�1 bmþ1,mþ1Þ:

As a result, D(A,B) contains the set

�ðam�1,m�1 bmþ1,mþ1Þ�1 detðA� BÞDðX,YÞ,

which is an elliptical disk. This is a contradiction.Next, we can show that am�2,mþ1 ¼ 0 ¼ bm�2,mþ1 and so forth, until we show that

a1,mþ1 ¼ 0 ¼ b1,mþ1. Note that it is important to show that aj,mþ1 ¼ 0 ¼ bj,mþ1 in theorder of j ¼ m,m� 1, . . . , 1. Remove the ðmþ 1Þ-th row and column from B and Ato get B̂ and Â. Then amþ1,mþ1DðÂ, B̂Þ is a subset of D(A,B) and has no interiorpoint. An inductive argument shows that the (1, 2) blocks of A and B are zero.Thus, A12 ¼ 0 ¼ B12.

If B11 and A22 are both equal to zero, then the desired conclusion holds. Suppose B11or A22 is nonzero. By symmetry, we may assume that B11 6¼ 0.

CLAIM (1) A11 ¼ �Im and (2) B22 ¼ �In�m for some � 6¼ 0.

If this claim is proved, then A� �In ¼ 0m � A22 � �In�mð Þ and B� �In ¼B11 � �Imð Þ � 0n�m. Thus, the result holds with k ¼ n�m.To prove our claim, suppose B11 6¼ 0. Then m�2 and we may assume that its leading

2�2 submatrix B0 is a nonzero strictly upper triangular matrix. If A11 is non-scalar,then we may assume that its leading 2�2 submatrix A0 is non-scalar. But thenDðA0,B0Þ will generate an elliptical disk in D(A,B), which is impossible. So,A11 ¼ �Im for some � 6¼ 0. This proves (1).

Now we prove (2). Suppose B22 6¼ �Im�n. Thus, n�m�2 and we may assume that4�4 submatrices of A and B lying at rows and columns labeled bym� 1,m,mþ 1,mþ 2 have the forms

A0 ¼ �I2 �0 �

0 0

� �and B0 ¼ B01 � B02

with �2C, a nonzero 2�2 nilpotent matrix B01 and a 2�2 matrix B02 such that B02 6¼ �I2.Let

P ¼ ½1� �0 1

1 0

� �� ½1�, V ¼ V1 � V2

with unitary V1,V22M2. Then detðPA0Pt � VB0V�Þ ¼ �1�2 with

�1 ¼ detðdiagð�, 0Þ � V1B01V�1Þ and �2 ¼ detðdiagð�, 0Þ � V2B02V�2Þ:

120 C.-K. Li et al.


Since B02 6¼ �I2, by Theorem 3.1, we can choose some unitary V2 such that �2 6¼ 0. Alsoas B01 is nonzero nilpotent, by Theorem 3.1, one can vary the unitary matrices V1 to getall values �1 in the nondegenerate circular disks Dðdiagð�, 0Þ,B01Þ. Hence,

ð�2bmþ1,mþ1 bmþ2,mþ2Þ�1�2 detðA� BÞDðdiagð�, 0Þ,B01Þ DðA,BÞ

so that D(A,B) also has nonempty interior, which is the desired contradiction. g

LEMMA 3.8 Let A;B 2Mn be normal matrices. Then QðA;BÞ ¼ f�g if and only if one ofthe following holds.

(a) � ¼ 0, and A� B has an eigenvalue with multiplicity at least nþ 1.(b) � 6¼ 0, and one of the matrix A or B is a scalar matrix, and detðA� BÞ ¼ �.

Proof Clearly if (a) or (b) holds, then QðA;BÞ is a singleton. Let A and B have eigen-values a1, . . . ,an and b1, . . . ,bn, respectively.

Suppose QðA;BÞ ¼ f�g. If both A and B are not scalar matrices, then A has at leasttwo distinct eigenvalues, say a1, a2 and B also has two distinct eigenvalues, say b1; b2.Then � ¼

Qni¼1ðai � biÞ ¼ ða1 � b2Þða2 � b1Þ

Qni¼3ðai � biÞ implies that � ¼ 0.

Now we claim that condition (a) holds. Suppose not, then every eigenvalue of A� Bhas multiplicity at most n. For k ¼ 1; 2; . . . ; n, let Sk ¼ fi : bi 6¼ akg. Suppose1 � k1 < k2 < � � � < km � n. Then i 62 [mj¼1Skj if and only if bi ¼ ak1 ¼ ak2 ¼ � � � ¼ akm .Therefore, there are at most n�m i not in [mj¼1Skj . Hence, [mj¼1Skj contains at leastm elements. By the theorem of P. Hall [19], there exist ik 2 Sk, k ¼ 1; . . . ; n, withik 6¼ ik0 for k 6¼ k0. Thus,

Qnk¼1ðak � bikÞ 6¼ 0, which contradicts the fact that � ¼ 0. g

LEMMA 3.9 Suppose A,B2Mn are normal matrices such that A has at least threedistinct eigenvalues, each eigenvalue of B has multiplicity at most n� 2, and each eigen-value of A� B has multiplicity at most n� 1. Then there are three distinct eigenvaluesa1, a2, a3 of A satisfying the following condition.

For any eigenvalue b of B with b =2fa1, a2, a3g, there exist eigenvalues b1, b2, b3 of B withb1 =2 fb2, b3g, b3 ¼ b, and the remaining eigenvalues can be labeled so that

Qnj¼4ðaj � bjÞ 6¼ 0.

Moreover, if A has more than three distinct eigenvalues, and B has exactly two distincteigenvalues, then we can replace a3 by any eigenvalue of A different from a1, a2, a3, andget the same conclusion.

Proof Let A and B have k distinct common eigenvalues �1, �2, . . . , �k so that � jhas multiplicity mj in the matrix A� B for j ¼ 1, . . . , k, with m1�� mk. By ourassumption, n� 1�m1.

The choices for ai and bi depend on k. We illustrate the different cases in thefollowing table.

k¼ 0 k¼ 1 k¼ 2 k� 3

a1 * �1 �1 �1a2 * * �2 �2a3 * * * �3b1 6¼b �1 �1 �1b2 6¼b1 6¼b1 �2 �2b3 b b b b



where � denotes any choice subject to the condition that a1, a2, a3 are distinct eigenva-lues of A. For any eigenvalue b of B with b =2fa1, a2, a3g, set b3 ¼ b and choose b1 ¼ �1if k�1 and b1 to be any eigenvalue of B not equal to b. Since the multiplicity of b1is �n� 2, there is always a third eigenvalue b2 of B, with b2 6¼ b1. Furthermore,we can choose b2 ¼ �2 if k�2.

Use the remaining eigenvalues of A and B to construct the matricesÂ ¼ diagða4, . . . , anÞ and B̂ ¼ diagðb4, . . . , bnÞ. By Lemma 3.8, the proof will be com-pleted if we can prove the following:

CLAIM If � is a common eigenvalue of Â and B̂ then the multiplicity of � in the matrixÂ� B̂ is at most n� 3.

To verify our claim, let � be a common eigenvalue of Â and B̂. Then � ¼ �r withr2f1, . . . , kg.

If r2f1, 2g, then two of the entries ða1, a2, a3, b1, b2, b3Þ equals �r by our construction.Since n� 1�mr, the multiplicity of �r in Â� B̂ equals mr � 2�n� 3.

If r¼ 3, then b3 6¼�i for i ¼ 1, 2, 3. Thus, m3�ðm1 þm2 þm3Þ=3� ð2n� 1Þ=3½ ��n� 2,where ½t� is the integral part of the real number t. Since one of the entriesin ða1, a2, a3, b1, b2, b3Þ equals �3, we see that the multiplicity of �3 in Â� B̂ equalsm3 � 1�n� 3.

Suppose r¼ 4. If n¼ 4 then ða1, a2Þ ¼ ðb1, b2Þ ¼ ð�1, �2Þ, a3 ¼ �3 and b3 ¼ �4 by ourconstruction. Thus, a4 � b4 ¼ �4 � �3 6¼ 0. If n�5, then the multiplicity of �r inÂ� B̂ is at most m4� 2n=4½ ��n� 3.

If r�5, then n�r�5 and the multiplicity of �r in Â� B̂ is at most mr�ð2n=5Þ�n� 3.By the above arguments, the claim holds.Note that if B has exactly two distinct eigenvalues then k�2 and a3 can be chosen to

be any eigenvalue different from a1, a2 in our construction. Thus, the last assertion ofthe lemma follows. g

LEMMA 3.9 Let A ¼ diagða1, a2, a3Þ and B ¼ diagðb1, b2, b3Þ with aj 6¼ ak for 1� j

It follows that the set

R ¼ tr C jursj2� ��

: ðursÞ is unitary

�

has empty interior. Let S0 be the 3�3 matrix with all entries equal to 1=3. For�,�2½0, 1=5�, let

S ¼ Sð�,�Þ ¼

1

3� � 1

3þ � 1

3þ ð�� Þ

1

3þ � 1

3� � 1

3� ð�� Þ

1

3

1

3

1

3

0BBBBBB@

1CCCCCCA¼ S0 þ

1

�10

0B@

1CAð�� Þ:

Since

4

15�

ffiffiffiffiffiffiffiffiffiffiffiffiffi1

9� �2

r,

ffiffiffiffiffiffiffiffiffiffiffiffiffi1

9� �2

r,

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

9� ð�� Þ2

r� 1

3,

by the result in [1], there is a unitary ðursÞ such that ðjursj2Þ ¼ S. Direct calculation showsthat

trðCSÞ ¼ trðCS0Þ þ ðb1 � b2Þ½�ðab3 þ aÞ � �ðb3 þ aÞ�:

The set R having empty interior implies that ðab3 þ aÞ and ðb3 þ aÞ are linearlyindependent over reals, which is possible only when b3 is real. Thus,fa1, a2, a3, b3gR and the result follows. g

Proof of Theorem 3.3 The implication (b)) (a) is clear.Suppose (c) holds. If (c.1) holds, then D(A,B) is a line segment on a line passing

through origin as shown in [3].If (c.2) holds, then we can assume that A� �In ¼ diagða, 0, . . . , 0Þ, and B� �In has

full rank and the eigenvalues of ðB� �InÞ�1 are collinear. We may replace (A,B) byðA� �In,B� �InÞ and assume that � ¼ 0. Since B�1 is normal with collinear eigenva-lues, the numerical range WðB�1Þ of B�1 is a line segment.

Let U2Mn be unitary, and

U�BU ¼b11 �� B22

� �

with B222Mn�1. Then detðB22Þ is the (1, 1) entry of detðBÞU�B�1U. Hence,detðB22Þ=detðBÞ2WðB�1Þ. Thus,

detðUAU� � BÞ ¼ detðA�U�BUÞ ¼ að�1Þn�1 detðB22Þ þ ð�1Þn detðU�BUÞ2 að�1Þn�1 detðBÞWðB�1Þ þ ð�1Þn detðBÞ:



If (c.3) holds, then D(A,B) is the line segment joining 0 and ð�1Þn�k detð ~AÞ detð ~BÞby Lemma 3.6. Thus, we have (c)) (b).

Finally, suppose (a) holds, i.e., D(A,B) has empty interior. Since D(A,B) is not asingleton, neither A nor B is a scalar matrix. Suppose A and B have eigenvaluesa1, . . . , an and b1, . . . , bn. Let A

0 ¼ diagða1, . . . , anÞ and B0 ¼ diagðb1, . . . , bnÞ.By Lemma 3.5, DðA0,B0Þ ¼ DðA,BÞ and hence DðA0,B0Þ is not singleton. It then followsfrom Corollary 2.2 and Lemma 3.8 that QðA0,B0Þ is not a singleton, so as QðA,BÞ.Now we show that one of (c.1)–(c.3) holds.

Suppose A and B have k distinct common eigenvalues �1, �2, . . . , �k such that � jhas multiplicity mj in the matrix A� B for j ¼ 1, . . . , k, with m1�� mk. SinceQðA,BÞ ¼ QðA0,B0Þ 6¼ f0g, we have m1�n.

If m1 ¼ n, then ðA� �1IÞ � ðB� �1IÞ has exactly n nonzero eigenvalues, and hence(c.3) follows from Lemma 3.7.

Suppose k¼ 0 or mj�n� 1 for all 1� j�k. We claim that both A and B are normal.If it is not true, we may assume that A is not normal. Otherwise, interchange the rolesof A and B. Then we may assume that A is in upper triangular form with nonzero (1, 2)entry by the result in [27]. Suppose A has diagonal entries a1, . . . , an. Let A1 be theleading 2�2 principal submatrix of A. We can also assume that B is uppertriangular with diagonal b1, . . . , bn, where b1 6¼ b2 satisfies the following additionalassumptions:

(1) If fa1, a2g \ f�1, . . . , �kg ¼ 6 0, then b1 and b2 are chosen so that �j2fb1, b2g for1� j�minfk, 2g.

(2) If fa1, a2g \ f�1, . . . , �kg 6¼ 6 0, then b1 and b2 are chosen so that �j2fa1, a2, b1, b2gfor 1� j�minfk, 3g.

Then b3, . . . , bn can be arranged so that p ¼Qn

j¼3 aj � bj� �

6¼ 0. It follows fromTheorem 3.1 that fp� : �2DðA1, diagðb1, b2ÞÞg is a nondegenerate elliptical disk inD(A,B), which is a contradiction.

Now, suppose both A and B are normal, and assume that k¼ 0 or mj�n� 1 for all1� j�k.

Case 1 Suppose A or B has an eigenvalue with multiplicity n� 1.

Interchanging the role of A and B, if necessary, we may assume thatA ¼ diagða, 0, . . . , 0Þ þ a2In. We can further set a2 ¼ 0; otherwise, replace (A,B)by ðA� a2In,B� a2InÞ. Since mj�n� 1, we see that B is invertible. Moreover, forany unitary matrix U, let u be the first column of U, and let ~U be obtained from Uby removing u. Then

detðA�U�BU Þ ¼ ð�1Þn detðBÞ � a detð ~U�B ~U Þ� �

:

Note that detð ~U�B ~U Þ= detðBÞ is the (1, 1) entry of ðU�BUÞ�1, and equals u�B�1u. So,

DðA,BÞ ¼ ð�1Þn detðBÞ � a detðBÞu�B�1u� �

: u 2 Cn, u�u ¼ 1

�

:

Since D(A,B) is a set with empty interior and so is the numerical range WðB�1Þ of B�1.Thus, B�1 has collinear eigenvalues; see [21]. Hence condition (c.2) holds.

124 C.-K. Li et al.


Case 2 Suppose both A and B have two distinct eigenvalues and each eigenvalue of Aand B has multiplicity at most n� 2.

Let A and B have two distinct eigenvalues, say, a1, a2 and b1, b2, respectively.We claim that a1, a2, b1, b2 are on the same straight line or circle, i.e., condition (c.1)holds. Suppose it is not true. Assume that a1, a2 and b1 are not collinear and b2is not on the circle passing through a1, a2 and b1. Then there is a factional lineartransform f(z) such that f(A) and f(B) has eigenvalues 1, 0 and a , b, respectively,where a2R nf1, 0g and b =2R. By Lemma 3.4, D(A,B) has empty interior if and onlyif Dð fðAÞ, fðBÞÞ has empty interior. We may replace (A,B) by ð fðAÞ, fðBÞÞ and assumethat A ¼ diagð1, 0, 1, 0Þ � A2, B ¼ diagða, b, a, bÞ � B2 with detðA2 � B2Þ 6¼ 0.By Theorem 3.1,

Dðdiagð1, 0Þ, diagða, bÞÞ ¼ ð1� sÞaðb� 1Þ þ sbða� 1Þ : s 2 ½0, 1�

�¼ aðb� 1Þ þ sða� bÞ : s 2 ½0, 1�

�

:

Hence, D(A,B) contains the set

R ¼ detðA2 � B2Þðaðb� 1Þ þ sða� bÞÞðaðb� 1Þ þ tða� bÞÞ : s, t 2 ½0, 1�

�¼ detðA2 þ B2Þðaðb� 1ÞÞ2 1þ ðsþ tÞ

a� baðb� 1Þ þ st

a� baðb� 1Þ

� �2 !: s, t 2 ½0, 1�

( ):

Note that fðst, sþ tÞ : s, t2½0, 1�g has nonempty interior. Let r ¼ ða� bÞ=aðb� 1Þ.

Then ðarþ 1Þb ¼ að1þ rÞ and so r cannot be real. Therefore, the complex numbers rand r2 are linearly independent over reals. Hence the mapping ðu, vÞ� 1þ urþ vr2 isan invertible map from R2 to C. Thus, the set R DðA,BÞ has nonempty interior,which is a contradiction.

Case 3 Suppose each eigenvalue of A and B has multiplicity at most n� 2 and one ofthe matrices has at least three distinct eigenvalues.

Assume that A has at least three distinct eigenvalues. Otherwise, interchange the rolesof A and B. By Lemma 3.8, there are three distinct eigenvalues of A, say, a1, a2, a3, suchthat the conclusion of the lemma holds. Applying a fractional linear transformation,if necessary, we may assume that a1, a2, a3 are collinear. For any eigenvalue b of Bwith b =2fa1, a2, a3g we can get b1; b2 and b3 ¼ b satisfying the conclusion ofLemma 3.8. Therefore, D(A,B) contains the set f�

Qnj¼4ðaj � bjÞ : �2Dðdiagða1, a2, a3Þ,

diagðb1, b2, b3ÞÞg. Since D(A,B) has empty interior, Lemma 3.9 ensures that a1, a2, a3and b are collinear. Therefore, all eigenvalues of B lie on the line L passing througha1, a2, a3.

Suppose B has three distinct eigenvalues. We can interchange the roles of A and Band conclude that the eigenvalues of A lie on the same straight line L. Suppose B hasexactly two eigenvalues, and a is an eigenvalue of A with a =2fa1, a2, a3g such that ais not an eigenvalue of B. By Lemma 3.8, we may replace a3 by a and showthat a1, a2, a and the two eigenvalues of B belong to the same straight line. Hence,all eigenvalues of A and B are collinear and (c.1) holds in this case. g



3.2. Sharp points

A boundary point � of a compact set S in C is a sharp point if there exists d>0 and0� t1< t2< t1þ � such that

S \ fz 2 C : jz� �j � dg f�þ ei� : 2 ½0, d�, � 2 ½t1, t2�g:

It was shown [3, Theorem 2] that for two normal matrices A,B2Mn such that the unionof the spectra of A and B has 2n distinct elements, a nonzero sharp point of D(A,B) isa �-point, that is, an element in Q(A,B). More generally, we have the following.

THEOREM 3.10 Let A,B2Mn. Every sharp point of D(A,B) is a �-point.

Proof Using the idea in [2], we can show that a nonzero sharp point detðUAU� � BÞis a �-point as follows. For simplicity, assume U ¼ In so that detðA� BÞ is a sharppoint of D(A,B). For each Hermitian H2Mn, consider the following one parametercurve in D(A,B):

�� det A� e�i�HBei�H� �

¼ detðA� BÞ 1þ i�tr ððA� BÞ�1½H,B�Þ þO �2� � �

,

where ½X,Y� ¼ XY� YX. Since detðA� BÞ is a sharp point,

0 ¼ trðA� BÞ�1½H,B� ¼ trH½B, ðA� BÞ�1� for all Hermitian H,

and hence

0 ¼ ½B, ðA� BÞ�1� ¼ BðA� BÞ�1 � ðA� BÞ�1B:

Consequently, 0 ¼ ðA� BÞB� BðA� BÞ, equivalently, AB ¼ BA. Thus, there existsa unitary V such that both VAV* and VBV* are in triangular form. As a result,detðA� BÞ ¼ detðVðA� BÞV�Þ is a �-point.

Next, we refine the previous argument to treat the case when detðA� BÞ ¼ 0 isa sharp point. If the spectra of A and B overlap, then 0 is a �-point. So, we assumethat A and B has disjoint spectra. By Theorem 2.7, there is unitary U such thatUAU� � B has rank n� 1. Assume U ¼ In so that A� B has rank n� 1 anddetðA� BÞ ¼ 0 is a sharp point. Then for any Hermitian H and 1�k�n,

det A� e�i�HBei�H� �

¼ det A� Bþ i�ðHB� BHÞ þ �2M� �

¼ detðA� BÞ þ i�Xnj¼1

rkjsjk

!þO �2

� �,

where adjðA� BÞ ¼ R ¼ ðrpqÞ andHB� BH ¼ S ¼ ðspqÞ. Thus, for k ¼ 1, . . . , n we havePnj¼1 rkjsjk ¼ 0, and hence

0 ¼ trRS ¼ trðadjðA� BÞðHB� BHÞÞ ¼ trH ½BadjðA� BÞ � adjðA� BÞB�

for every Hermitian H. So, B and adjðA� BÞ commute. Since A� B has rank n� 1, thematrix adjðA� BÞ has rank 1, and equals uvt for some nonzero vectors u , v. Comparingthe columns of the matrices on left and right sides of the equality Buvt ¼ uvtB, we seethat Bu ¼ bu for some b2C. Similarly, we have Auvt ¼ uvtA and hence Au ¼ au for

126 C.-K. Li et al.


some a2C. Consequently, 0 ¼ ðA� BÞadjðA� BÞ ¼ ðA� BÞuvt ¼ ða� bÞuvt. Thus,a� b ¼ 0, i.e., the spectra of A and B overlap, which is a contradiction. g

Clearly, if D(A,B) is a line segment, then the end points are sharp points.By Theorem 3.3 and the above theorem, we have the following corollary showing thatMarcus–Oliveira conjecture holds if D(A,B) has empty interior.

COROLLARY 3.11 Let A,B2Mn. If D(A,B) has empty interior, then D(A,B) equals theconvex hull of Q(A,B).

By Theorem 2.9, 02DðA,BÞ if for every �2C, the singular values of A� �In andB� �In do not lie in two disjoint closed intervals in ½0,1Þ. Following is a sufficientcondition forA,B2Mn to have 0 as a sharp point ofD(A,B) in terms ofW(A) andW(B).

PROPOSITION 3.12 Let A,B2Mn be such that 02DðA,BÞ and

WðAÞ [Wð�BÞ rei� : r � 0, � 2 ð��=ð2nÞ,�=ð2nÞÞ

�

:

Then

DðA,BÞ rei� : r � 0, � 2 ð��=2,�=2Þ

�

:

Proof Note that for any unitary U and V, there is a unitary matrix R such that

RðUAU� � VBV�ÞR� ¼ apq� �

� bpq� �

is in upper triangular form. Hence, app � bqq ¼ rpei�p with rp�0 and �p2ð��=ð2nÞ,�=ð2nÞÞ for p ¼ 1, . . . , n. So, detðUAU� � VBV�Þ ¼

Qnp¼1ðapp � bppÞ ¼ rei� with r�0

and �2ð��=2,�=2Þ. g

4. Further extensions

There are many related topics and problems which deserve further investigation.One may ask whether the results can be extended to the sum of k matrices from k

different unitary orbits for k>2.For the norm problem, in an unpublished manuscript Li and Choi have extended

the norm bound result to k matrices A1, . . . ,Ak2Mn for k�2 to

max kX1 þ � � � þ Xkk : Xj 2 UðAjÞ, j ¼ 1, . . . , k

�¼ min

Xkj¼1kAj � �jInk : �j 2 C, j ¼ 1, . . . , k,

Xkj¼1

�j ¼ 0( )

:

However, we are not able to extend the maximum rank result in section 2 to k matriceswith k>2 at this point. In any event, it is easy to show that for any �1, . . . ,�k2Csatisfying

Pkj¼1 �j ¼ 0,

min rankðX1 þ � � � þXkÞ : Xj 2 UðAjÞ, j ¼ 1, . . . ,k

�

�max rankðAj ��jInÞ : j ¼ 1, . . . ,k

�

:



It is challenging to determine all the possible rank values of matrices inUðA1Þ þ � � � þ UðAkÞ.

For Hermitian matrices A1, . . . ,Ak, there is a complete description of the eigenvaluesof the matrices in UðA1Þ þ � � � þ UðAkÞ; see [16]. Evidently, the set

�ðA1, . . . ,AkÞ ¼ detXkj¼1

Xj

!: Xj 2 UðAjÞ, j ¼ 1, . . . , k

( )

is a real line segment. When k¼ 2, the end points of the line segment have the formdetðX1 þ X2Þ for some diagonal matrices X12UðA1Þ and X22UðA2Þ; see [15].However, this is not true if k>2 as shown in the following example.

Example 4.1 Let

A ¼3 0

0 1

� �, B ¼

3 1

1 1

� �, C ¼

1 �1�1 5

� �:

Then for any unitary U,V,W2M2, the matrix UAU� þ VBV� þWCW� is positivedefinite with eigenvalues 7þ d and 7� d with d2½0, 7Þ. Hence

detðUAU� þ VBV� þWCW�Þ � 72 ¼ detðAþ Bþ CÞ:

Thus, the right end point of the set �ðA,B,CÞ is not of the formða1 þ b�ð1Þ þ cð1ÞÞða2 þ b�ð2Þ þ cð2ÞÞ for permutations � and of (1, 2).

It is interesting to determine the set �ðA1, . . . ,AkÞ for Hermitian, normal, or generalmatrices A1, . . . ,Ak2Mn. Inspired by the Example 4.1, we have the followingobservations.

(1) Suppose A1, . . . ,Ak are positive semi-definite matrices. If there are unitaryU1, . . . ,Uk such that

Xkj¼1

UjAU�j ¼ �In ð4:1Þ

for some scalar �, then max�ðA1, . . . ,AkÞ ¼ �n. The necessary and sufficientconditions for (4.1) to hold can be found in [15].

(2) Let A1,A2,A3 be Hermitian matrices such that detðA1 þ A2 þ A3Þ ¼max�ðA1,A2,A3Þ. Then there exist unitary U and V such that UA1U� andVðA2 þ A3ÞV� are diagonal and

detðUA1U� þ VðA2 þ A3ÞV�Þ ¼ detðA1 þ A2 þ A3Þ:

Proof Let U be unitary matrix such that UA1U� ¼ D1 is diagonal. Suppose A2 þ A3

has eigenvalues �1, . . . , �n. By the result of [15], there exists a permutation matrixP such that

detðD1 þ P diagð�1, . . . , �nÞP�Þ � detðA1 þ ðA2 þ A3ÞÞ:

128 C.-K. Li et al.


So if V is unitary such that VðA2 þ A3ÞV� ¼ P diagð�1, . . . , �nÞP�, then

detðA1 þ A2 þ A3Þ ¼ max�ðA1,A2,A3Þ � detðUA1U� þ VðA2 þ A3ÞV�Þ� detðA1 þ A2 þ A3Þ

and hence the above inequalities become equalities. g

Besides the unitary orbits, one may consider orbits of matrices under other groupactions. For example, we can consider the usual similarity orbit of A2Mn

SðAÞ ¼ SAS�1 : S 2Mn is invertible

�

;

the unitary equivalence orbit of A2Mn

VðAÞ ¼ fUAV : U,V 2Mn are unitaryg;

the unitary congruence orbit of A2Mn

UtðAÞ ¼ UAUt : U 2Mn is unitary

�

:

It is interesting to note that for any A,B2Mn,

maxfrankðUAU� þ VBV�Þ : U,V 2Mn are unitaryg

� max rank SAS�1 þ TBT�1� �

: S,T 2Mn are invertible

�

� minfrankðAþ �InÞ þ rankðB� �InÞ : � 2 Cg:

By our result in section 2, the inequalities are equalities.One may consider the ranks, determinants, eigenvalues, and norms of the sum, the

product, the Lie product, the Jordan product of matrices from different orbits;[17,22,28]. One may also consider similar problems for matrices over arbitrary fieldsor rings. Some problems are relatively easy. For example, the setfdetðSAS�1 þ TBT�1Þ : S,T are invertibleg is either a singleton or C. But some ofthem seem very hard. For example, it is difficult to determine when

0 2 det S1A1S�11 þ S2A2S�12 þ S3A3S�13� �

: S1,S2,S3 are invertible

�

:

Acknowledgement

The research of first author was supported by a USA NSF grant and a HK RCG grant.Research of Nung-Sing Sze was supported by a HK RCG grant. The authors would liketo thank the referee for some helpful comments. In particular, in an earlier versionof the article, the implication (a)) (b) in Theorem 3.3 was only a conjecture. Ourfinal proof of the result was stimulated by a different proof of the referee sketched inthe report.



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