Linear and Multilinear Algebra, Vol. 56, Nos. 1–2, Jan.–Mar. 2008, 105–130
Ranks and determinants of the sum of matrices 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 unitarily similar to A. Given two n n complex matrices A and B, ranks and determinants of matrices of the form X þ Y with ðX, Y Þ2UðAÞ UðBÞ are studied. In particular, a lower bound and the best upper bound of the set RðA, BÞ¼frankðX þ YÞ : X2UðAÞ, Y2UðBÞg are determined. It is shown that ðA, BÞ¼fdetðX þ YÞ : X2UðAÞ, Y2UðBÞg has empty interior if and only if the set is a line segment or a point; the algebraic structure of matrix pairs (A, B) with such properties 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. Extensions of the results to the sum of three or more matrices from given unitary orbits are also considered. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 Keywords: Rank; Determinant; Matrices; Unitary orbit
2000 Mathematics Subject Classifications: 15A03; 15A15
1. Introduction
Let A2Mn. The unitary orbit of A is denoted by
UðAÞ¼fgUAU : U U ¼ In :
Evidently, if A is regarded as a linear operator acting on Cn, then UðAÞ consists of the matrix representations of the same linear operator under different
*Corresponding author. Email: [email protected] Dedicated to Professor Yik-Hoi Au-Yeung on the occasion of his 70th birthday.
Linear and Multilinear Algebra ISSN 0308-1087 print/ISSN 1563-5139 online ß 2008 Taylor & Francis http://www.tandf.co.uk/journals DOI: 10.1080/03081080701369306 106 C.-K. Li et al.
orthonormal bases. Naturally, UðAÞ captures many important features of the operator A. For instance, A is normal if and only if UðAÞ has a diagonal matrix; A is Hermitian (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 unimodular diagonal entries. There are also results on the characterization of diagonal entries and 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 deal of 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 were completely determined in terms of those of A and B (see [16] and its references); the optimal 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 was described, see [15]. Later, Marcus and Oliveira [26,29] conjectured that if A, B2Mn are normal matrices with eigenvalues a1, ..., an and b1, ..., bn, respectively, then for any unitary U, V2Mn, detðUAU þ VBV Þ always lies in the convex hull of () Yn PðA, BÞ¼ aj þ b ðjÞ : is a permutation of f1, ..., ng : ð1:1Þ j¼1
In connection to this conjecture, researchers [2–5,7,8,12,13] considered the deter- minantal range of A, B2Mn defined by
ðA, BÞ¼fdetðA þ UBU Þ: U is unitaryg:
This can be viewed as an analog of the generalized numerical range of A and B defined by
WðA, BÞ¼ftrðAUBU Þ: U is unitaryg, Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 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
UðAÞþUðBÞ¼fX þ Y: ðX, Y Þ2UðAÞ 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, BÞ¼frankðX þ Y Þ: X 2UðAÞ, Y 2UðBÞg;
moreover, we characterize those matrix pairs (A, B) such that R(A, B) is a singleton, and show that the set R(A, B) has the form fk, k þ 1, ..., ng if A and B have no common eigenvalues. On the contrary if A and B are orthogonal projections, then the rank values of matrices in UðAÞþUðBÞ 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 Ranks and determinants of the sum of matrices 107
possible only when ðA, BÞ is a singleton or a nondegenerate line segment. This extends the 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 such that ðA, BÞ is a subset of a line, which does not pass through the origin. This disproves a conjecture in [9]. In [3], the authors showed that if A, B2Mn are normal matrices such that the union of the spectra of A and B consists of 2n distinct elements, then every nonzero sharp point of ðA, BÞ is an element in P(A, B). (See the definition of sharp point in section 3.) We showed that every (zero or nonzero) sharp point of ðA, BÞ belongs to P(A, B) for arbitrary matrices A, B2Mn. In section 4, we consider the sum of three or more matrices from given unitary orbits, and matrix orbits corresponding to other equivalence relations. In the literature, some authors considered the set
DðA, BÞ¼fdetðX YÞ: X 2UðAÞ, Y 2UðBÞg
instead of ðA, BÞ. Evidently, we have DðA, BÞ¼ ðA, BÞ. It is easy to translate results on D(A, B) to those on ðA, BÞ, and vice versa. Indeed, for certain results and proofs, it is more convenient to use the formulation of D(A, B). We will do that in section 3. On the other hand, it is more natural to use the summation formulation to 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 UðAÞþUðBÞ for two given matrices A, B2Mn. By the triangle inequality, we have
maxfkUA U þ VBV k: U, V unitaryg minfkA InkþkB þ Ink: 2 Cg: Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 It was shown in [10] that the inequality is actually an equality. For the rank functions, we have
maxfrankðUAU þ VBV Þ: U,V unitaryg minfrankðA InÞþrankðB þ InÞ: 2 Cg:
Of course, the right side may be strictly larger than n, and thus equality may not hold in general. 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 InÞþrankðB þ InÞ: 2 Cg
¼ minfrankðA InÞþrankðB þ InÞ: is an eigenvalue of A Bg:
Then
maxfrankðUAU þ VBV Þ : U, V unitaryg¼minfm, ng: 108 C.-K. Li et al.
Proof If is an eigenvalue of A B, then rankðA InÞþrankðB þ InÞ 2n 1; if is not an eigenvalue of A B, then rankðA InÞþrankðB þ InÞ¼2n: As a result, rankðA InÞþrankðB þ InÞ will attain its minimum at an eigenvalue of the matrix A B. It is clear that
maxfrankðUA U 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 to minfm, ng. Suppose m n and there is such that rankðA InÞ¼k and rankðB þ InÞ¼m k. We may replace (A, B)byðA In, B þ 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 invertible diagonal D1. Replace A by YAY*, we may assume that U11 U12 D1 0 A ¼ UD ¼ U21 U22 00n k
with U ¼ YX. Similarly, we may assume that V11 V12 0k 0 B ¼ VE ¼ , V21 V22 0 E2
where V is a unitary matrix and E2 is a diagonal matrix with rank m k. Let W be a unitary matrix such that the first k columns of WU together with the last n k columns of V are linearly independent. That is, if W11 W12 W ¼ ,
Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 W21 W22
the matrix W11U11 þ W12U21 V12
W21U11 þ W22U21 V22
is invertible. If W11 is invertible, then D1W11 has rank k and so W11U11 þ W12U21 V12 D1W D1W WAW þ B ¼ 11 21 W21U11 þ W22U21 V22 0 E2
has rank m.IfW11 is not invertible, we will replace W by W~ obtained as follows. By the CS decomposition, there are unitary matrices P1, Q1 2Mk and P2, Q2 2Mn k such that CS ðP1 P2ÞWðQ1 Q2Þ¼ In 2k, SC Ranks and determinants of the sum of matrices 109 qffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 where C ¼ diagðc1, ..., ckÞ with 1 c1 ck 0 and S ¼ diagð 1 c1, ..., 1 ck Þ. ~ ~ Then perturb the zero diagonalqffiffiffiffiffiffiffiffiffiffiffiffiffi entries ofqffiffiffiffiffiffiffiffiffiffiffiffiffiC slightly to C ¼ diagðc~1, ..., c~kÞ so that C ~ 2 2 is invertible, and set S ¼ diagð 1 c~1, ..., 1 c~k Þ. Then C~ S~ W~ ¼ðP P Þ I ðQ Q Þ 1 2 S~ C~ n 2k 1 2
~ ~ will be a slight perturbation of W with invertible W11 ¼ P1CQ1, which can be chosen such that the matrix ! W~ 11U11 þ W~ 12U21 V12
W~ 21U11 þ W~ 22U21 V22
is still invertible. Then WA~ W~ þ B has rank m. Now, assume that rankðA InÞþrankðB þ InÞ n þ 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. Applying suitable unitary similarity transforms, we may assume that A and B are unitarily similar to matrices in upper triangular form a1 b1 and : 0 A1 0 B1
Since rankðA InÞþrankðB þ InÞ n þ 1 for every 2C, it follows that rankðA1 In 1ÞþrankðB1 þ In 1Þ n 1 for every 2C. By induction assumption, there is a unitary V1 such that detðA1 þ V1B1V1Þ 6¼ 0. Let V ¼½1 V1. Then detðA þ VBV Þ¼ða1 þ b1Þ detðA1 þ V1B1V1Þ 6¼ 0. Suppose ai þ bj ¼ 0 for all i, j2f1, ..., ng. Replacing (A, B)byðA a1In, B b1InÞ, we may assume that A and B are nilpotents. If A or B is normal, then it will be the
Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 zero matrix. Then rankðAÞþrankðBÞ 00 0 2 3 0 such that 1, 3, 1, 3 are nonzero. Interchange the last two rows and the last two i columns of A to obtain A^. For U ¼ diagð1, 1, e Þ, we have 0 1 0 e i B 2 1 C ^ U AU ¼ @ 00 0A: i 0 3e 0 ^ Evidently, there is 2½0, 2 Þ such that detðU AU þ BÞ 6¼ 0. 110 C.-K. Li et al. Suppose n 4. Applying suitable unitary similarity transforms, we may assume that 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 B11 B12 and 0 A22 0 B22 respectively, so that 0 00 A11 ¼ and B11 ¼ 00 0 with 6¼ 0, and A22, B22 are upper triangular nilpotent matrices. If rank A þ rank B n þ 2, then rankðA22 In 2ÞþrankðB22 þ In 2Þ rank A22 þ rank B22 rank A þ rank B 4 n 2: If rank A þ rank B ¼ n þ 1, we claim that by choosing a suitable unitary similarity transform, we can further assume that rank A22 ¼ rank A 1. Then rankðA22 In 2ÞþrankðB22 þ In 2Þ rank A22 þ rank B22 rank A þ rank B 3 ¼ n 2: In both cases, by induction assumption, there is V2 such that detðA22 þ V2B22V2Þ 6¼ 0. Let V ¼ I2 V2. Then detðA þ VBV Þ¼ detðA22 þ V2B22V2Þ 6¼ 0: Now it remains to verify our claim. Suppose A has rank k and Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 rank A þ rank B ¼ n þ 1. Then k ðn þ 1Þ=2. Let S be an invertible matrix such that S 1AS ¼ J is the Jordan form of A.IfJ has a 2 2 Jordan block, then we can always permute J so that J11 0 01 J ¼ with J11 ¼ 0 J22 00 and rankðJ22Þ¼p 1. By QR factorization, write S ¼ U T for some unitary matrix U and invertible upper triangular matrix T11 T12 T ¼ : 0 T22 Then A is unitarily similar to ! T J T 1 TJT 1 ¼ 11 11 11 , 1 0 T22J22T22 which has the described property. Ranks and determinants of the sum of matrices 111 Suppose J does not contain any 2 2 block, then J must have an 1 1 Jordan block. Otherwise, k ¼ rank A 2n=3 and hence rank A þ rank B 2k 4n=3 ¼ n þ n=3 > n þ 1: Now we may assume that 02 J12 J ¼ 0 J22 is strictly upper triangular matrix such that J12 has only a nonzero entry in the (1, 1)-th ^ position and rank J22 ¼ k 1. Let S be obtained from In by replacing the (3, 2)-th entry with one, then ! J^ J S^ 1JS^ ¼ J^ ¼ 11 12 0 J22 with 01 J^11 ¼ : 00 Applying QR factorization on SS^ ¼ U T with unitary U and invertible upper triangular T, then A is unitarily similar to TJT^ 1, which has the described form. g The value m in Theorem 2.1 is easy to determine as we need only to focus on rankðA InÞþrankðB þ InÞ for each eigenvalue of A B. In particular, if is an eigenvalue of A, then rankðA InÞ¼n k, where k is the geometric multi- plicity of ; otherwise, rankðA InÞ¼n. Similarly, one can determine rankðB þ InÞ. The situation for normal matrices is even better as shown in the following. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 COROLLARY 2.2 Suppose A and B are normal matrices such that ‘ is the maximum multiplicity of an eigenvalue of A B. Then minfrankðA InÞþ rankðB þ InÞ: 2Cg equals 2n ‘. Consequently, maxfrankðUAU þ VBV Þ: U, V unitaryg¼minf2n ‘, 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, Vif and only if there is 2C such that rankðA InÞþrankðB þ InÞ COROLLARY 2.4 Let A2Mn, and k ¼ minfrankðA InÞ: is an eigenvalue of Ag: Then maxfrankðUAU VAV Þ: U, V unitary g¼minfn,2kg: 112 C.-K. Li et al. If k Partition Mn as the disjoint union of unitary orbits. We can define a metric on the set of unitary orbits by dðUðAÞ, UðBÞÞ ¼ minfrankðX Y Þ: X 2UðAÞ, Y 2UðBÞg: For example, if A and B are two orthogonal projections of rank p and q, respectively, then dðUðAÞ, UðBÞÞ ¼ jp qj; see Proposition 2.8. So, the minimum rank of the sum or difference of matrices from two different unitary orbits has a geometrical meaning. However, it is not so easy to determine the minimum rank for matrices in UðAÞþUðBÞ in general. We have the following observation. PROPOSITION 2.5 Let A, B2Mn and 2C be such that rankðA InÞ¼p and rankðB þ InÞ¼q. Then minfrankðUAU þ VBV Þ : U, V unitaryg maxfp, qg: The inequality becomes equality if A In and B þ In are positive semi-definite. Proof There exist unitary U, V such that the last n p columns of UðA InÞU are zero, and the last n q columns of VðB þ InÞV are zero. Then rankðUAU þ VBV Þ maxfp, qg. g The upper bound in the above proposition is rather weak. For example, we may have A and B such that maxfrankðA InÞ, rankðB þ InÞ: 2 Cg¼n 1 ð2:1Þ and rankðUAU þ VBV Þ¼1. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 Example 2.6 Let A ¼ diagð1, 2, ..., nÞ and B ¼ J A, where J2Mn is the matrix having all entries equal to 1/n. Then B has distinct eigenvalues b1 > >bn such that b1 >n>b2 >n 1>b3 > >b1 >1. Then (2.1) clearly holds and rankðA þ BÞ¼1. In fact, by Theorem 2.7 subsequently, we know that for any m2f1, ..., ng, there are unitary U, V2Mn such that rankðUAU þ VBV Þ¼m. 2.2. Additional results Here we study other possible rank values of matrices in UðAÞþUðBÞ. The following result shows that if A and B have disjoint spectra, then one can get every possible rank 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 A þ B has rank k Proof Let A, B2Mn satisfy the hypotheses of the theorem. We need only to show that there is a unitary U such that UAU þ B has rank k þ 1. Then we can apply the Ranks and determinants of the sum of matrices 113 argument again to get a unitary U^ such that UUAU^ U^ þ B has rank k þ 2. Repeating this procedure, we will get the desired conclusion. If k ¼ n 1. Then assume VAV* and WBW* are in upper triangular form. For U ¼ W V, we have UAU þ B ¼ W ðVAV þ WBW ÞW is invertible. For 1 k Let A11 A12 B11 B12 A ¼ and B ¼ A21 A22 B21 B22 with A11, B11 2Mk. Note that A12 6¼ 0. Otherwise, A and B have common eigenvalues since A22 ¼ B22. Assume C21 6¼ 0. We may replace A þ B by VðA þ BÞV for some permutation matrix V2Mn of the form V1 In k so that the matrix obtained by removing the first row of VðA þ BÞV still has rank k. For notational simplicity, we may assume that V ¼ In. Since A12 6¼ 0, we may assume that the first row of A12 6¼ 0. Otherwise, replace (A, B) by ðVAV , VBV Þ for some unitary V ¼ V1 In k. Here we still assume that removing the first row of A þ B results in a rank k matrix. Then there exists a small >0 such that for U ¼ diagðei ,1,...,1Þ the matrix UAU þ B has rank k þ 1 because removing its first 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)by ðVAV , VBV Þ by some unitary matrix V of the form V1 In k. Since A12 and A21 are nonzero, we may further assume that vt ¼½a , ..., a 6¼ 0 and Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 1, kþ1 1n t u ¼½akþ1, 1, ..., an1 6¼ 0. Then there exists a small >0 such that for U ¼ diagðei ,1,...,1Þ the matrix UAU þ B has the form ! C^ C^ 11 12 , C^21 0n k where C^11 is invertible such that removing its first row and first column results in a rank i t k 1 matrix, only the first row of C^12 is nonzero and equal to ðe 1Þv , only the first i column of C^21 is nonzero and equal to ðe 1Þu. Now, removing the first row and first 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 rank by 1 more. So, UAU þ B has rank k þ 1. g Note that the assumption that A and B have disjoint spectra is essential. For example, if A, B2M4 such that A and B are rank 2 orthogonal projections, then UAU þ VBV can only have ranks 0, 2, 4. More generally, we have the following. 114 C.-K. Li et al. PROPOSITION 2.8 Suppose A, B2Mn such that A and B are orthogonal projections of rank p and q. Then k ¼ rankðÞUAU þ B for a unitary matrix U2Mn, if and only if k ¼jp qjþ2j with j 0 and k minfp þ q,2n p qg. Proof Suppose UAU ¼ Ip 0n p and VBV ¼ 0j Iq 0n j q. Then UAU þVBV has rank k ¼jp qjþ2j minfp þ q,2n p qg. Thus, V UAU V þ B has rank k as well. Conversely, consider UAU þ B for a given unitary U. There is a unitary V such that 2 C CS VUAU V ¼ Ip 0n p and VBV ¼ Ir 0s Iu 0v , CS S2 where C and S are invertible diagonal matrices with positive diagonal entries 2 2 such that C þ S ¼ It, r þ s þ t ¼ p and r þ t þ u ¼ q. (Evidently, the first r columns of V* span the intersection of the range spaces of UAU* and B, the next s columns of V* span the intersection of the range space of UAU* and the null space of B, the last v columns of V* span the intersection of the null space of UAU* and B, the u columns preceding those span the intersection of the range space of B and the null space of UAU*.) So, UAU þ B has the asserted rank value. g The following result was proved in [24]. THEOREM 2.9 Let A, B2Mn. Then UAU þ VBV is invertible for all unitary U, V2Mn if and only if there is 2C such that the singular values of A In and B þ In lie in two disjoint 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 Þ¼k for all unitary U, V2Mn if and only if one of the following holds. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 (a) One of the matrices A or B is scalar, and rankðA þ BÞ¼k. (b) k ¼ n and there is 2C such that the singular values of A In and B þ In lie in two disjoint closed intervals in ½0, 1Þ. Proof If (a) holds, say, B ¼ In, then rankðUAU þ VBV Þ¼rankðA InÞ¼k for all unitary U, V2Mn. n If (b) holds, then kðA InÞxk>kðB þ InÞyk for all unit vectors x, y2C ,or n kðA InÞxk 3. Determinants Let A, B2Mn with eigenvalues a1, ..., an, and b1, ..., bn, respectively. In this section we study the properties of ðA, BÞ and P(A, B). For notational convenience and easy description of the results and proofs, we consider the sets DðA, BÞ¼ ðA, BÞ¼fdetðX YÞ: X 2UðAÞ, Y 2UðBÞg and () Yn QðA, BÞ¼PðA, BÞ¼ aj b ð j Þ : is a permutation of f1, ..., ng : j¼1 It is easy to translate the results on D(A, B) and Q(A, B) to those on ðA, BÞ and P(A, B), and vice versa. For any permutation ð ð1Þ, ..., ðnÞÞ of ð1, ..., nÞ, there are unitary matrices U and V such that UAU* and VBV* are upper triangular matrices with diagonal entries a1, ..., 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)byðUAU 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, AÞ¼ð 1ÞnDðA, BÞ and QðB, AÞ¼ð 1ÞnQðA, BÞ. The following result can be found in [9]. THEOREM 3.1 Suppose A, B2M2 have eigenvalues 1, 2 and 1, 2, respectively, and suppose A ðtr A=2ÞI2 and B ðtr B=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 1Þ with length of minor axis equal to 2ðac bd Þ. Consequently, D(A, B) is a singleton Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 if and only if A or B is a scalar matrix; D(A, B) is a nondegenerate line segment if and only if A and B are nonscalar normal matrices. In the subsequent discussion, let ÈÉ WðAÞ¼ 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, BÞ¼f g if and only if one of the following holds. (a) ¼ 0, and there is 2 C such that rank ðA InÞþrank ðB InÞ < n. (b) 6¼ 0, one of the matrices A or B is a scalar matrix, and detðA BÞ¼ . 116 C.-K. Li et al. Proof If (a) or (b) holds, then clearly D(A, B) is a singleton. If DðA, BÞ¼f0g, then condition (a) holds by Corollary 2.3. Suppose DðA, BÞ¼f g with 6¼ 0. We claim that A or B is a scalar matrix. Suppose A and B have eigenvalues a1, ..., an and b1, ..., bn, respectively. Assume that A has at least twoQ distinct eigenvalues a1, a2 and QB also has two distinct eigenvalues b1, b2. n n Then j¼1ðaj bjÞ and ða1 b2Þða2 b1Þ j¼3ðaj bjÞ will be two distinct -points, which is a contradiction because QðA, BÞ DðA, BÞ is also a singleton. So, we have a1 ¼ ¼an or b1 ¼ ¼bn. 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 in upper triangular form with nonzero (1, 2) entry. Also, we may assume that A is in upper triangular form so that the leading two-by-two matrix is not a scalar matrix. Let A A B B A ¼ 11 12 and B ¼ 11 12 0 A22 0 B22 with A11, B11 2M2, 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 following conditions 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. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 (c.1) A and B are normal matrices with eigenvalues lying on the same straight line or the same circle. (c.2) There is 2C such that one of the matrices A In or B In is rank 1 normal, and the other one is invertible normal so that the inverse matrix has collinear eigenvalues. (c.3) There is 2C such that A In is unitarily similar to A~ 0n k and B In is unitarily similar to 0k B~ so that A~ 2Mk and B~ 2Mn k are invertible. In [3], the authors conjectured that for normal matrices A, B2Mn,ifD(A, B)is contained 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, 1 þ i,1 iÞ 1 and B ¼ diagð 1, 0, 0Þ, then D(A, B) is a straight line segment joining the points 1 i=2 and 1 þ i=2; see Corollary 3.11. This shows that D(A, B) can be a subset of a straight line not passing through the origin. Of course, Theorem 3.3 covers more general situations. In (c.1), the line segment D(A, B) and the origin are collinear; in (c.3) the line segment D(A, B) has endpoints 0 and ð 1Þn k detðA~Þ detðB~Þ; in (c.2) the line segment and the origin may or may not be collinear. Ranks and determinants of the sum of matrices 117 Since DðA, BÞ¼ð 1ÞnDðB, AÞ, D(A, B) has empty interior (is a line segment) if and only if D(B, A) has empty interior (is a line segment). This symmetry will be used in the following discussion. We establish several lemmas to prove the theorem. Given a, b, c, d2C, with ad bc 6¼ 0, let fðzÞ¼ðaz þ bÞ=ðcz þ dÞ be the fractional linear transform on Cnf d=cg (C,ifc ¼ 0). If A2Mn such that cA þ dIn is invertible, 1 one can define fðAÞ¼ðaA þ bInÞðcA þ dInÞ . The following is easy to verify. LEMMA 3.4 Suppose A, B2Mn, and fðzÞ¼ðaz þ bÞ=ðcz þ d Þ is a fractional linear transform such that f(A) and f(B) are well defined. Then 1 n Dð fðAÞ, fðBÞÞ ¼ detððcA þ dInÞðcB þ dInÞÞ ðad bcÞ DðA, BÞ: LEMMA 3.5 Let A, B2Mn with eigenvalues a1, ..., an and b1, ..., bn, respectively. If D(A, B) has empty interior, then DðA, BÞ¼DðA, diagðb1, ..., bnÞÞ ¼ Dðdiagða1, ..., anÞ, BÞ ¼ Dðdiagða1, ..., anÞ, diagðb1, ..., bnÞÞ: Proof Assume that D(A, B) has empty interior. Applying a unitary similarity transform we can assume that A ¼ðarsÞ and B ¼ðbrsÞ are upper triangular matrices. For any unitary matrix V2Mn, let D ¼ðdrsÞ¼VBV . For any 2½0, 2 Þ, i let U ¼½e In 1. Denote by Xrs the ðn 1Þ ðn 1Þ matrices obtained from X2Mn by deleting its r-th row and the s-th column. Expanding the determinant detðU AU DÞ along the first row yields Xn jþ1 i detðU AU DÞ¼ða11 d11Þ detðA11 D11Þþ ð 1Þ ðe a1j d1jÞ detðA1j D1jÞ "#j¼2 Xn Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 jþ1 i ¼ða11 d11Þ detðA11 D11Þ ð 1Þ d1j detðA1j D1jÞ þ e , j¼2 P n jþ1 where ¼ j¼2ð 1Þ a1j detðA1j D1jÞ: Thus, CðA, VBV Þ¼fdetðU AU VBV Þ: 2½0, 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 degenerate circle CðA, BÞ¼fdetðA BÞg: Since the unitary group is path connected, there is a continuous function t ° Vt for t2½0, 1 so that V0 ¼ V and V1 ¼ In. Thus, we have a family of circles CðA, VtBVt Þ 118 C.-K. Li et al. in D(A, B) transforming CðA, VBV Þ to C(A, B). Hence, all the points inside CðA, VBV Þ belong to D(A, B). Thus, D(A, B) has nonempty interior. As a result, Xn jþ1 ¼ ð 1Þ a1j detðA1j D1jÞ¼0, j¼2 and Xn jþ1 detðA VBV Þ¼detðA DÞ¼ða11 d11Þ detðA11 D11Þ ð 1Þ d1j detðA1j D1jÞ j¼2 ¼ detðA1 DÞ¼detðA1 VBV Þ, where A1 is the matrix obtained from A by changing all the nondiagonal entries in the first row to zero. It follows that DðA, BÞ¼DðA1, BÞ. Inductively, by expanding i the determinant detðUjAjUj DÞ along the ( j þ 1)-th row with Uj ¼ Ij ½e In j 1, we conclude that DðAj, BÞ¼DðAjþ1, BÞ where Ajþ1 is the matrix obtained from Aj by changing all the nondiagonal entries in the ð j þ 1Þ-th row to zero. Therefore, DðA, BÞ¼DðA1, BÞ¼DðA2, BÞ¼ ¼DðAn 1, BÞ¼Dðdiagða11, ..., 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, BÞ¼DðA, diagðb1, ..., bnÞÞ. Now, we can apply the argument to Dðdiagða1, ..., anÞ, BÞ to get the last set equality. g LEMMA 3.6 Let A ¼ A^ 0n k, where A^ 2Mk with k2f1, ..., n 1g is upper triangular invertible. If B2Mn has rank n k, then n k DðA, BÞ ¼ fð 1Þ detðA^Þ detðX BXÞ: X is n ðn kÞ, X X ¼ In kg: Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 If B ¼ 0k B^ so that B^ 2Mn k is invertible, then D(A, B) is the line segment joining 0 and ð 1Þn k detðA^Þ detðB^Þ. Proof Suppose A ¼ðarsÞ 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 be obtained from C by removing the first row. By linearity of the determinant function on the first column, detðA U BUÞ¼detð½A1jC Þ detð½B1jC Þ ¼ a11 detðB22Þþ0, because ½B1jC has rank at most n 1. Inductively, we see that detðA U BUÞ¼ð 1Þn k detðA^Þ detðYÞ where Y is obtained from U BU by removing its first k rows and first k columns. Ranks and determinants of the sum of matrices 119 Now if B ¼ 0k B^ so that B^ 2Mn k is invertible, then the set fdetðÞX BX : X X ¼ In kg is a line segment joining 0 and detðB^Þ; 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 nonzero eigenvalues. If D(A, B) has no interior point, then there exist 2C and 0 k n such that A In and B In are unitarily similar to matrices of the form A~ 0n k and 0k B~ for some invertible matrices A~ 2Mk and B~ 2Mn 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 of the proof, assume that neither A nor B is a scalar matrix. We may assume that A11 A12 B11 B12 A ¼ðarsÞ¼ and B ¼ðbrsÞ¼ ð3:1Þ 0 A22 0 B22 such that A11, B11 2Mm and A22, B22 2Mn m are upper triangular matrices so that A11, B22 are invertible, and A22, B11 are nilpotent. If m ¼ 0, then A ¼ A11 is nilpotent and B ¼ B22 is invertible. We are going to show that A ¼ 0. Hence, the lemma is satisfied with k ¼ 0. Suppose A 6¼ 0. We may assume that a12 6¼ 0. Let 0 a12 b11 b12 X ¼ and Y ¼ : 00 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 detðBÞ : 2 DðX, Y Þ DðA, BÞ: Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 b11b22 Therefore, D(A, B) has nonempty interior, a contradiction. Similarly, if m ¼ n, then B ¼ 0. Hence, we may assume that 1 m 1 ðam,m bmþ1, mþ1Þ detðA BÞDðX, Y Þ is an elliptical disk in D(A, B), which is impossible. Next, we show that am 1, mþ1 ¼ 0 ¼ bm 1, mþ1. If it is not true, let X, Y2M2 be the principal submatrices of A and B lying in rows and columns m 1 and m þ 1. For any unitary U, V2M2, let ¼ detðÞUXU VYV . Construct U^ (respectively, V^ ) from In by changing the principal submatrix at rows and columns m 1 and m þ 1byU (respectively, V ). Then UA^ U^ is still in upper triangular block form so that its leading ðm 2Þ ðm 2Þ principal submatrix and its trailing ðn m 1Þ ðn m 1Þ principal 120 C.-K. Li et al. submatrix are the same as A. Moreover, since we have shown that am, mþ1 ¼ 0 ¼ bm, mþ1, the principal submatrix of UA^ U^ lying in rows m 1, m, m þ 1 has the form 0 1 B C @ 0 amm 0 A: A similar result is true for VB^ V^ . Hence, detðUA^ U^ VB^ V^ Þ¼ detðA BÞ =ðam 1, m 1 bmþ1, mþ1Þ: As a result, D(A, B) contains the set 1 ðam 1, m 1 bmþ1, mþ1Þ 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 the order of j ¼ m, m 1, ..., 1. Remove the ðm þ 1Þ-th row and column from B and A to get B^ and A^. Then amþ1, mþ1DðA^, B^Þ is a subset of D(A, B) and has no interior point. 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 B11 or 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 ðÞA22 In m and B In ¼ ðÞ B11 Im 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, Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 then we may assume that its leading 2 2 submatrix A0 is non-scalar. But then DðA0, 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 that 4 4 submatrices of A and B lying at rows and columns labeled by m 1, m, m þ 1, m þ 2 have the forms 0 0 0 0 0 A ¼ I2 and B ¼ B B 00 1 2 C 0 0 0 with 2 , a nonzero 2 2 nilpotent matrix B1 and a 2 2 matrix B2 such that B2 6¼ I2. Let 01 P ¼½1 ½1 , V ¼ V1 V2 10 0 t 0 with unitary V1, V2 2M2. Then detðPA P VB V Þ¼ 1 2 with 0 0 1 ¼ detðdiagð ,0Þ V1B1V1Þ and 2 ¼ detðdiagð ,0Þ V2B2V2Þ: Ranks and determinants of the sum of matrices 121 0 Since B2 6¼ I2, by Theorem 3.1, we can choose some unitary V2 such that 2 6¼ 0. Also 0 as B1 is nonzero nilpotent, by Theorem 3.1, one can vary the unitary matrices V1 to get 0 all values 1 in the nondegenerate circular disks Dðdiagð ,0Þ, B1Þ. Hence, 2 1 0 ð bmþ1, mþ1 bmþ2, mþ2Þ 2 detðA BÞ Dðdiagð ,0Þ, B1Þ DðA, BÞ so that D(A, B) also has nonempty interior, which is the desired contradiction. g LEMMA 3.8 Let A; B 2 Mn be normal matrices. Then QðA; BÞ¼f g if and only if one of the following holds. (a) ¼ 0, and A B has an eigenvalue with multiplicity at least n þ 1. (b) 6¼ 0, and one of the matrix A or B is a scalar matrix, and detðA BÞ¼ . 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; BÞ¼f g. If both A and B are not scalar matrices, then A has at least two distinctQ eigenvalues, say a1, a2 and B alsoQ has two distinct eigenvalues, say b1; b2. n n Then ¼ i¼1ðai biÞ¼ða1 b2Þða2 b1Þ i¼3ðai biÞ implies that ¼ 0. Now we claim that condition (a) holds. Suppose not, then every eigenvalue of A B has multiplicity at most n. For k ¼ 1; 2; ...; n, let Sk ¼fi : bi 6¼ akg. Suppose m 1 k1 < k2 < < km n. Then i 62[j¼1Skj if and only if bi ¼ ak1 ¼ ak2 ¼ ¼akm . m m Therefore, there are at most n minot in [j¼1Skj . Hence, [j¼1Skj contains at least m elements. By the theoremQ of P. Hall [19], there exist ik 2 Sk, k ¼ 1; ...; n, with 0 n 0 g ik 6¼ ik for k 6¼ k . Thus, k¼1ðak bik Þ 6¼ 0, which contradicts the fact that ¼ 0. LEMMA 3.9 Suppose A, B2Mn are normal matrices such that A has at least three distinct 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 eigenvalues a1, a2, a3 of A satisfying the following condition. For any eigenvalue b of B with b2f= a1, a2, a3g, there exist eigenvaluesQ b1, b2, b3 of B with n b1 2f= b2, b3g, b3 ¼ b, and the remaining eigenvalues can be labeled so that j¼4ðaj bjÞ 6¼ 0. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 Moreover, if A has more than three distinct eigenvalues, and B has exactly two distinct eigenvalues, then we can replace a3 by any eigenvalue of A different from a1, a2, a3, and get the same conclusion. Proof Let A and B have k distinct common eigenvalues 1, 2, ..., k so that j has multiplicity mj in the matrix A B for j ¼ 1, ..., k, with m1 mk. By our assumption, n 1 m1. The choices for ai and bi depend on k. We illustrate the different cases in the following table. k ¼ 0 k ¼ 1 k ¼ 2 k 3 a1 * 1 1 1 a2 ** 2 2 a3 *** 3 b1 6¼b 1 1 1 b2 6¼b1 6¼b1 2 2 b3 bbbb 122 C.-K. Li et al. 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 b2f= a1, a2, a3g, set b3 ¼ b and choose b1 ¼ 1 if k 1 and b1 to be any eigenvalue of B not equal to b. Since the multiplicity of b1 is 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 matrices A^ ¼ diagða4, ..., anÞ and B^ 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 A^ and B^ then the multiplicity of in the matrix A^ B^ is at most n 3. To verify our claim, let be a common eigenvalue of A^ and B^. Then ¼ r with r2f1, ..., 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 A B^ equals mr 2 n 3. If r ¼ 3, then b3 6¼ i for i ¼ 1, 2, 3. Thus, m3 ðm1 þ m2 þ m3Þ=3 ð½ 2n 1Þ=3 n 2, where ½t is the integral part of the real number t. Since one of the entries ^ ^ in ða1, a2, a3, b1, b2, b3Þ equals 3, we see that the multiplicity of 3 in A B equals m3 1 n 3. Suppose r ¼ 4. If n ¼ 4 then ða1, a2Þ¼ðb1, b2Þ¼ð 1, 2Þ, a3 ¼ 3 and b3 ¼ 4 by our construction. Thus, a4 b4 ¼ 4 3 6¼ 0. If n 5, then the multiplicity of r in A^ B^ is at most m4 ½ 2n=4 n 3. ^ If r 5, then n r 5 and the multiplicity of r in A B^ 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 of the 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 3 t detðUAU BÞ¼detðAÞþð 1Þ detðBÞ ðb1, b2, b3ÞSUð0, 0, aÞ t þðb2b3, b1b3, b1b2ÞSUða,1,0Þ : Let t t C ¼ða,1,0Þ ðb2b3, b1b3, b1b2Þ ð0, 0, aÞ ðb1, b2, b3Þ: Then 3 detðUAU BÞ¼detðAÞþð 1Þ detðBÞþtr ðCSUÞ: Ranks and determinants of the sum of matrices 123 It follows that the set ÈÉ 2 R¼ tr C jursj : 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 0 1 1 1 1 B þ þð Þ C 0 1 B 3 3 3 C B C 1 B 1 1 1 C B C S ¼ Sð , Þ¼B þ ð Þ C ¼ S0 þ @ 1 Að Þ: B 3 3 3 C @ A 1 1 1 0 3 3 3 Since rffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 1 1 1 1 2, 2, ð Þ2 , 15 9 9 9 3 2 by the result in [1], there is a unitary ðursÞ such that ðjursj Þ¼S. Direct calculation shows that trðCSÞ¼trðCS0Þþðb1 b2Þ½ ðab3 þ aÞ ðb3 þ aÞ : The set R having empty interior implies that ðab3 þ aÞ and ðb3 þ aÞ are linearly independent over reals, which is possible only when b3 is real. Thus, fa1, a2, a3, b3g R and the result follows. g Proof of Theorem 3.3 The implication (b) ) (a) is clear. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 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 1 full rank and the eigenvalues of ðB InÞ are collinear. We may replace (A, B)by 1 ðA In, B InÞ and assume that ¼ 0. Since B is normal with collinear eigenva- lues, the numerical range WðB 1Þ of B 1 is a line segment. Let U2Mn be unitary, and b11 U BU ¼ B22 1 with B22 2Mn 1. Then detðB22Þ is the (1, 1) entry of detðBÞU B U. Hence, 1 detðB22Þ=detðBÞ2WðB Þ. Thus, n 1 n detðUAU BÞ¼detðA U BUÞ¼að 1Þ detðB22Þþð 1Þ detðU BUÞ 2 að 1Þn 1 detðBÞWðB 1Þþð 1Þn detðBÞ: 124 C.-K. Li et al. If (c.3) holds, then D(A, B) is the line segment joining 0 and ð 1Þn k detðA~Þ detðB~Þ by 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 a singleton, neither A nor B is a scalar matrix. Suppose A and B have eigenvalues 0 0 a1, ..., an and b1, ..., bn. Let A ¼ diagða1, ..., anÞ and B ¼ diagðb1, ..., bnÞ. By Lemma 3.5, DðA0, B0Þ¼DðA, BÞ and hence DðA0, B0Þ is not singleton. It then follows from 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 j has multiplicity mj in the matrix A B for j ¼ 1, ..., k, with m1 mk. Since 0 0 QðA, BÞ¼QðA , B Þ 6¼f0g, 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 k ¼ 0ormj 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 roles of 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 the leading 2 2 principal submatrix of A. We can also assume that B is upper triangular with diagonal b1, ..., bn, where b1 6¼ b2 satisfies the following additional assumptions: (1) If fa1, a2g\f 1, ..., kg¼6 0, then b1 and b2 are chosen so that j 2fb1, b2g for 1 j minfk,2g. (2) If fa1, a2g\f 1, ..., kg 6¼ 6 0, then b1 and b2 are chosen so that j 2fa1, a2, b1, b2g for 1 j minfk,3g. Q n Then b3, ..., bn can be arranged so that p ¼ j¼3 aj bj 6¼ 0. It follows from Theorem 3.1 that fp : 2DðA1, diagðb1, b2ÞÞg is a nondegenerate elliptical disk in D(A, B), which is a contradiction. Now, suppose both A and B are normal, and assume that k ¼ 0ormj n 1 for all 1 j k. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 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 that A ¼ diagða,0,...,0Þþa2In. 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, for any unitary matrix U, let u be the first column of U, and let U~ be obtained from U by removing u. Then detðA U BU Þ¼ð 1Þn detðBÞ a detðU~ BU~ Þ : Note that detðU~ BU~ Þ= detðBÞ is the (1, 1) entry of ðU BUÞ 1, and equals u B 1u. So, ÈÉ DðA, BÞ¼ 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. Ranks and determinants of the sum of matrices 125 Case 2 Suppose both A and B have two distinct eigenvalues and each eigenvalue of A and 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 b2 is not on the circle passing through a1, a2 and b1. Then there is a factional linear transform f(z) such that f(A) and f(B) has eigenvalues 1, 0 and a , b, respectively, where a2R nf1, 0g and b2= R. By Lemma 3.4, D(A, B) has empty interior if and only if Dð fðAÞ, fðBÞÞ has empty interior. We may replace (A, B)byð fðAÞ, fðBÞÞ and assume that A ¼ diagð1, 0, 1, 0Þ A2, B ¼ diagða, b, a, bÞ B2 with detðA2 B2Þ 6¼ 0. By Theorem 3.1, ÈÉ ð ð Þ ð ÞÞ ¼ ð Þ ð Þþ ð Þ 2½ D diag 1, 0 , diag a, b ÈÉ1 s a b 1 sb a 1 : s 0, 1 ¼ aðb 1Þþsða bÞ: s 2½0, 1 : Hence, D(A, B) contains the set ÈÉ R¼ detðA2 B2Þðaðb 1Þþsða bÞÞðaðb 1Þþtða bÞÞ: s, t 2½0, 1 () ! a b a b 2 ¼ detðA þ B Þðaðb 1ÞÞ2 1 þðs þ tÞ þ st : s, t 2½0, 1 : 2 2 aðb 1Þ aðb 1Þ Note that fðst, s þ tÞ: s, t2½0, 1 g has nonempty interior. Let r ¼ða bÞ=aðb 1Þ. Then ðar þ 1Þb ¼ að1 þ rÞ and so r cannot be real. Therefore, the complex numbers r and r2 are linearly independent over reals. Hence the mapping ðu, vÞ ° 1 þ ur þ vr2 is an invertible map from R2 to C. Thus, the set R DðA, BÞ has nonempty interior, which is a contradiction. Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 Case 3 Suppose each eigenvalue of A and B has multiplicity at most n 2 and one of the matrices has at least three distinct eigenvalues. Assume that A has at least three distinct eigenvalues. Otherwise, interchange the roles of A and B. By Lemma 3.8, there are three distinct eigenvalues of A, say, a1, a2, a3, such that 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 B with b2f= a1, a2, a3g we can get b1; b2 and b3 ¼Qb satisfying the conclusion of n Lemma 3.8. Therefore, D(A, B) contains the set f j¼4ðaj bjÞ: 2Dðdiagða1, a2, a3Þ, diagðb1, b2, b3ÞÞg. Since D(A, B) has empty interior, Lemma 3.9 ensures that a1, a2, a3 and b are collinear. Therefore, all eigenvalues of B lie on the line L passing through a1, a2, a3. Suppose B has three distinct eigenvalues. We can interchange the roles of A and B and conclude that the eigenvalues of A lie on the same straight line L. Suppose B has exactly two eigenvalues, and a is an eigenvalue of A with a2f= a1, a2, a3g such that a is not an eigenvalue of B. By Lemma 3.8, we may replace a3 by a and show that 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 126 C.-K. Li et al. 3.2. Sharp points A boundary point of a compact set S in C is a sharp point if there exists d>0 and 0 t1 It was shown [3, Theorem 2] that for two normal matrices A, B2Mn such that the union of the spectra of A and B has 2n distinct elements, a nonzero sharp point of D(A, B)is a -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 BÞ is a -point as follows. For simplicity, assume U ¼ In so that detðA BÞ is a sharp point of D(A, B). For each Hermitian H2Mn, consider the following one parameter curve in D(A, B): ÈÉ ° det A e i HBei H ¼ detðA BÞ 1 þ i tr ððA BÞ 1½H, B Þ þ O 2 , where ½X, Y ¼XY YX. Since detðA BÞ is a sharp point, 0 ¼ trðA BÞ 1½H, B ¼tr H½B, ðA BÞ 1 for all Hermitian H, and hence 0 ¼½B, ðA BÞ 1 ¼BðA BÞ 1 ðA BÞ 1B: Consequently, 0 ¼ðA BÞB BðA BÞ, equivalently, AB ¼ BA. Thus, there exists a unitary V such that both VAV* and VBV* are in triangular form. As a result, detðA BÞ¼detðVðA BÞV Þ is a -point. Next, we refine the previous argument to treat the case when detðA BÞ¼0is a sharp point. If the spectra of A and B overlap, then 0 is a -point. So, we assume Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 that A and B has disjoint spectra. By Theorem 2.7, there is unitary U such that UAU B has rank n 1. Assume U ¼ In so that A B has rank n 1 and detðA BÞ¼0 is a sharp point. Then for any Hermitian H and 1 k n, det A e i HBei H ¼ det A B þ i ðHB BHÞþ 2M ! Xn 2 ¼ detðA BÞþi rkjsjk þ O , j¼1 Pwhere adjðA BÞ¼R ¼ðrpqÞ and HB BH ¼ S ¼ðspqÞ. Thus, for k ¼ 1, ..., n we have n j¼1 rkjsjk ¼ 0, and hence 0 ¼ tr RS ¼ trðadjðA BÞðHB BHÞÞ ¼ tr H ½BadjðA BÞ adjðA BÞB for every Hermitian H. So, B and adjðA BÞ commute. Since A B has rank n 1, the matrix adjðA BÞ has rank 1, and equals uvt for some nonzero vectors u , v. Comparing the columns of the matrices on left and right sides of the equality Buvt ¼ uvtB, we see that Bu ¼ bu for some b2C. Similarly, we have Auvt ¼ uvtA and hence Au ¼ au for Ranks and determinants of the sum of matrices 127 some a2C. Consequently, 0 ¼ðA BÞadjðA BÞ¼ðA BÞuvt ¼ða 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 that Marcus–Oliveira conjecture holds if D(A, B) has empty interior. COROLLARY 3.11 Let A, B2Mn.IfD(A, B) has empty interior, then D(A, B) equals the convex hull of Q(A, B). By Theorem 2.9, 02DðA, BÞ if for every 2C, the singular values of A In and B In do not lie in two disjoint closed intervals in ½0, 1Þ. Following is a sufficient condition for A, B2Mn to have 0 as a sharp point of D(A, B) in terms of W(A) and W(B). PROPOSITION 3.12 Let A, B2Mn be such that 02DðA, BÞ and ÈÉ WðAÞ[Wð BÞ rei : r 0, 2ð =ð2nÞ, =ð2nÞÞ : Then ÈÉ DðA, BÞ rei : r 0, 2ð =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 i is in upper triangular form. Hence, app bqq ¼ rpe Qp with rp 0 and p 2ð =ð2nÞ, n i =ð2nÞÞ for p ¼ 1, ..., n. So, detðUAU VBV Þ¼ p¼1ðapp bppÞ¼re with r 0 and 2ð =2, =2Þ. g 4. Further extensions Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 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, ..., Ak 2Mn for k 2to ÈÉ max kX þ þX k: X 2UðA Þ, j ¼ 1, ..., k 1()k j j Xk Xk ¼ min kAj jInk: j 2 C, j ¼ 1, ..., k, j ¼ 0 : j¼1 j¼1 However, we are not able to extend the maximum rank result in section 2 to k matrices C with k>2P at this point. In any event, it is easy to show that for any 1, ..., k 2 k satisfying j¼1 j ¼ 0, ÈÉÈÉ min rankðX1 þ þXkÞ: Xj 2UðAjÞ, j ¼ 1, ..., k max rankðAj jInÞ : j ¼ 1, ..., k : 128 C.-K. Li et al. It is challenging to determine all the possible rank values of matrices in UðA1Þþ þUðAkÞ. For Hermitian matrices A1, ..., Ak, there is a complete description of the eigenvalues of the matrices in UðA1Þþ þUðAkÞ; see [16]. Evidently, the set () ! Xk ðA1, ..., AkÞ¼ det Xj : Xj 2UðAjÞ, j ¼ 1, ..., k j¼1 is a real line segment. When k ¼ 2, the end points of the line segment have the form detðX1 þ X2Þ for some diagonal matrices X1 2UðA1Þ and X2 2UðA2Þ; see [15]. However, this is not true if k>2 as shown in the following example. Example 4.1 Let 30 31 1 1 A ¼ , B ¼ , C ¼ : 01 11 15 Then for any unitary U, V, W2M2, the matrix UAU þ VBV þ WCW is positive definite with eigenvalues 7 þ d and 7 d with d2½0, 7Þ. Hence detðUAU þ VBV þ WCW Þ 72 ¼ detðA þ B þ CÞ: Thus, the right end point of the set ðA, B, CÞ is not of the form ða1 þ b ð1Þ þ c ð1ÞÞða2 þ b ð2Þ þ c ð2ÞÞ for permutations and of (1, 2). It is interesting to determine the set ðA1, ..., AkÞ for Hermitian, normal, or general matrices A1, ..., Ak 2Mn. Inspired by the Example 4.1, we have the following observations. (1) Suppose A1, ..., Ak are positive semi-definite matrices. If there are unitary U1, ..., Uk such that Xk Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 UjAUj ¼ In ð4:1Þ j¼1 n for some scalar , then max ðA1, ..., AkÞ¼ . The necessary and sufficient conditions for (4.1) to hold can be found in [15]. (2) Let A1, A2, A3 be Hermitian matrices such that detðA1 þ A2 þ A3Þ¼ max ðA1, A2, A3Þ. Then there exist unitary U and V such that UA1U and VðA2 þ A3ÞV are diagonal and detðUA1U þ VðA2 þ A3ÞV Þ¼detðA1 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 matrix P such that detðD1 þ P diagð 1, ..., nÞP Þ detðA1 þðA2 þ A3ÞÞ: Ranks and determinants of the sum of matrices 129 So if V is unitary such that VðA2 þ A3ÞV ¼ P diagð 1, ..., nÞP , then detðA1 þ A2 þ A3Þ¼max 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 group actions. For example, we can consider the usual similarity orbit of A2Mn ÈÉ 1 SðAÞ¼ SAS : S 2 Mn is invertible ; the unitary equivalence orbit of A2Mn VðAÞ¼fUAV: U, V 2 Mn are unitaryg; the unitary congruence orbit of A2Mn ÈÉ t t U ðAÞ¼ UAU : U 2 Mn is unitary : It is interesting to note that for any A, B2Mn, maxfrankðUAU þ VBV Þ: U, V 2 Mn are unitaryg ÈÉ 1 1 max rank SAS þ TBT : S, T 2 Mn are invertible minfrankðA þ InÞþrankðB 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; Downloaded By: [Li, Chi-Kwong] At: 00:07 20 March 2009 [17,22,28]. One may also consider similar problems for matrices over arbitrary fields or rings. Some problems are relatively easy. For example, the set fdetðSAS 1 þ TBT 1Þ: S, T are invertibleg is either a singleton or C. But some of them seem very hard. For example, it is difficult to determine when ÈÉ 1 1 1 0 2 det S1A1S1 þ S2A2S2 þ S3A3S3 : 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 like to thank the referee for some helpful comments. In particular, in an earlier version of the article, the implication (a) ) (b) in Theorem 3.3 was only a conjecture. 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