INVERSE CLOSED RAY-NONSINGULAR CONES OF MATRICES

Transcription

1 INVERSE CLOSED RAY-NONSINGULAR CONES OF MATRICES CHI-KWONG LI AND LEIBA RODMAN Abstract A description is given of those ray-patterns which will be called inverse closed ray-nonsingular of complex matrices that contain invertible matrices only and are closed under inversion Here two n n matrices are said to belong to the same ray-pattern if in each position either the entries of both matrices are zeros or both entries are nonzero and their quotient is positive Possible Jordan forms of matrices in the inverse closed ray-nonsingular ray-patterns are characterized 1 Introduction In this paper we study ray-patterns of complex matrices with the property that every matrix in the ray-pattern is nonsingular and the inverse of the matrix is again in the same ray-pattern We first introduce some terminology and notation Let Ω = {0} {z C : z = 1} A matrix with entries in Ω will be called a ray-pattern Denote by M n (Ω) the set of n n ray-patterns A pattern A M n (Ω) generates a cone Cone (A) in a natural way: Cone (A) = {X A : X n n matrix with positive entries} where stands for the Hadamard (entrywise) product A ray-pattern A M n (Ω) is called ray-nonsingular if X A is invertible for every n n matrix X with positive entries Ray-patterns in particular ray-nonsingular ray-patterns were studied recently in For a survey of the theory of real ray-nonsingular ray-patterns see the book 1 and references there In applications of matrix analysis to stability (see eg 2 3) one often encounters closed cones of matrices with the property that the set of invertible matrices in the cone is dense and the inverse of every invertible matrix in the cone belongs again to the cone In the context of ray-nonsingular ray-patterns such cones appear as follows We denote by M n (C) the algebra of complex n n matrices Given X = x ij n ij=1 M n (C) define the pattern projection ray (X) of X as follows: ray (X) is the ray-pattern whose (i j)-th entry is 0 if x ij = 0 and is equal to x ij x ij if x ij 0 A ray-nonsingular ray-pattern A is called inverse-closed if ray ((X A) 1 ) = A for every matrix X with positive entries 1991 Mathematics Subject Classification 47L07 15A99 Key words and phrases Ray-nonsingular ray-pattern cone complex matrix The authors gratefully acknowledge the support of the NSF grants DMS and DMS respectively 1

2 2 CHI-KWONG LI AND LEIBA RODMAN Lemma 11 Suppose A is an inverse-closed ray-nonsingular ray-pattern Denote by Cone (A) the cone generated by A and let CCone (A) be the closed cone which is the closure of Cone (A): CCone (A) = {X A : X has nonnegative entries} Then every Y Cone (A) is invertible and every invertible Y CCone (A) has the property that Y 1 CCone (A) Proof Let Y = X A for some X with nonnegative entries and assume that Y is invertible Let X m be a sequence of matrices with positive entries such that X m X as m Then clearly Y m := X m A is invertible for large m and Ym 1 Y 1 By the properties of inverse-closeness we have that Ym 1 Cone (A) Passing to the limit we obtain Y 1 CCone (A) as required In this paper we solve the following problem Problem 12 Describe all inverse-closed ray-nonsingular (in short ICRN) ray-patterns A solution of the problem is known in the real case see 1 Section 74 4 A closely related (but different) problem also in the real case was studied in Another relevant paper is 8 The ICRN ray-patterns transform well with respect to certain similarity transformations Let G n sometimes abbreviated to G if the size n is understood from context be the group of matrices that consists of all products of the form DQ where Q is an n n permutation matrix and D is a unimodular diagonal matrix ie diagonal matrix with unimodular entries on the main diagonal Clearly ga and Ag are ray-patterns for every n n ray-pattern A and every g G n Lemma 13 If a ray-pattern A M n (Ω) is ICRN then so are ±A and ±A t (the transpose of A) and every ray-pattern of the form gag 1 where g G n is arbitrary Furthermore if A 1 A p are ICRN s then so is diag (A 1 A p ) We omit a simple proof In view of Lemma 13 we will describe the ICRN ray-patterns up to a similarity with the similarity matrix in G n The rest of the paper consists of 5 sections Sections 2 and 3 are preparatory and contain descriptions of ICRN ray-patterns without zero entries and irreducible ICRN ray-patterns The main result of the paper Theorem 42 is stated in Section 4 Its rather long proof is delegated to Section 5 In Section 6 we study location of eigenvalues and inertia properties of matrices in the cone Cone (A) for ICRN ray-patterns A In particular we completely describe Jordan forms of matrices in the set A (Cone (A)) where the union is taken over all n n ICRN ray-patterns A

3 INVERSE CLOSED RAY-NONSINGULAR CONES 3 2 ICRN ray-patterns without zero entries Lemma 21 Let A be an n n ray-nonsingular ray-pattern without zero entries such that the ray-pattern projection ray ((X A) 1 ) is independent of any matrix X with positive entries and all entries of ray ((X A) 1 ) are nonzero Then n 2 and there exist g h G n such that gah has the form 1 or 1 1 Proof It is known that there do not exist ray-nonsingular n n ray-patterns without zero entries if n > We will focus on the cases for n = 1 4 and for completeness present also the case n = 5 The hypotheses on A are clearly invariant under pre- and post-multiplication by elements of G n Using a transformation of the form A gah g h G we may assume that the entries of A in the first row and the first column are all one viz A = z ij n ij=1 with z ij = 1 and z ij = 1 whenever 1 {i j} Then the result is trivial if n = 1 Suppose n = 2 Let X = 1 r with r > 0 Then the () entry of X A is z 22 /(rz 22 1) which has the same argument for any r > 0 Thus rz 22 1 has the same argument for all r > 0 and hence z 22 = 1 ie A has the asserted form Consider the case n = 3 Let X = 1 u v 1 1 Then the (3 3) entry and (2 3) entry of (X A) 1 are (21) (z 22 u)/ det(x A) and (v z 23 )/ det(x A) respectively and each of them has a fixed argument for all choices of u v > 0 Dividing the numbers (21) we see that (z 22 u)/(v z 23 ) has a fixed argument for all choices of u v > 0 Fixing v and changing u we see that z 22 = 1; fixing u and changing v we see that z 23 = 1 Now applying the argument to à where à is obtained from A by interchanging its last two rows we see that z 32 = z 33 = 1 But then A = is not ray-nonsingular which is a contradiction Consider the case n = 4 We may assume that z 22 z 3k z 32 z 2k 0 for some k 3; otherwise the (1 4) entry of A 1 is zero which is excluded by the hypotheses We may

4 4 CHI-KWONG LI AND LEIBA RODMAN assume that k = 3; otherwise interchange the last two columns of A Let 1 r s 1 X = t 1 with r s t > 0 Then the (4 4) entry and the () entry of (X A) 1 have constant (ie independent of r s t) arguments and so is their quotient: (z 22 z 33 z 23 z 32 ) r(tz 33 z 23 ) + s(tz 32 z 22 ) z 22 z 33 z 44 + z 24 z 32 z 43 + z 42 z 23 z 34 z 42 z 33 z 24 z 32 z 23 z 44 z 43 z 34 z 22 As a result for any t > 0 not equal to z 22 /z 32 or z 23 /z 33 the quantity (z 22 z 33 z 23 z 32 ) r(tz 33 z 23 ) + s(tz 32 z 22 ) has a constant argument for any r s > 0 Hence z 23 tz 33 and tz 32 z 22 have the same argument as z 22 z 33 z 23 z 32 for all t > 0 not equal to z 22 /z 32 or z 23 /z 33 It follows that z 33 = z 23 and z 32 = z 22 But then z 22 z 33 z 23 z 32 = 0 which is a contradiction Now suppose n = 5 Let X = 1 r s t 1 u x 1 v w 1 with r s t u v w x > 0 Denoted by Y (p q) the matrix obtained from the matrix Y by removing its pth row and qth column Let B = (X A)(5 5) Then the (5 5) and () entries of (X A) 1 have constant arguments and so does their quotient which has the form (22) det(b()) r det(b(1 2)) + s det(b(1 3)) t det(b(1 4)) det(x A)() Note that B() and A() depend only on the variable x; B(1 2) B(1 3) B(1 4) depend on the variables u v w x For fixed x > 0 the quotient (22) has constant argument for all choices of r s t u v w > 0 It follows that (23) det(b(1 3)) = u(z 32 z 44 z 42 z 34 ) v(xz 22 z 44 z 42 z 24 ) + w(xz 22 z 34 z 32 z 24 ) is either zero or has a constant argument for all choices of u v w > 0 if x > 0 is fixed Choosing x > 0 such that both xz 22 z 44 z 42 z 24 and xz 22 z 34 z 32 z 24 are nonzero and varying u v w > 0 in (23) we see that xz 22 z 44 z 42 z 24 and xz 22 z 34 z 32 z 24 always have the same arguments Hence they are positive multiples of each other This is true for infinitely many x > 0 It follows that z 44 /z 42 = z 34 /z 32

5 INVERSE CLOSED RAY-NONSINGULAR CONES 5 Interchange the kth row and the third row of A for k = 2 5 and repeat the above argument We conclude that z 44 /z 42 = z k4 /z k2 k = Thus the second column and the fourth column of A() are multiples of each other It follows that det(a()) = 0 which is a contradiction Our proof is complete Proposition 22 An n n ray-pattern A without zero entries is ICRN if and only if n 2 and there exists g G n such that gag 1 has one of the three forms Proof The if part is obvious For the only if part by Lemma 21 we have n 2 The case n = 1 is trivial For n = 2 we may apply a transformation A gag 1 g G and assume that A = v w where v w are unimodular numbers Let X = with x > 0 x 1 and consider the () and (2 1) entries of (X A) 1 We see that w/(w xv) > 0 and 1/(w xv) > 0 for all x > 0 Thus v = w = 1 3 Irreducible ICRN ray-patterns A ray-pattern A is called irreducible if there is no permutation matrix Q such that QAQ 1 A1 = A 21 A 22 where the sizes of the square submatrices A 11 and A 22 are strictly smaller than that of A Proposition 31 Let A be an n n irreducible ray-pattern Then A is ICRN if and only if n {1 2 4} and there exists g G n such that gag 1 has one of the following five forms: (31) Proof The proof follows the argument from 1 Section 74 The if part is clear Assume A is an ICRN ray-pattern First suppose that A is fully indecomposable ie no p q submatrix of A with p + q n is the zero matrix Consider an (n 1) (n 1) submatrix B of A Since the ray-pattern projection ray ((X A) 1 ) is independent of the positive matrix X we have either det (X B) = 0 for every positive (n 1) (n 1)

6 6 CHI-KWONG LI AND LEIBA RODMAN matrix X or det (X B) 0 for every positive (n 1) (n 1) matrix X In the former case all (n 1)! terms in the expression of det B are zeros which implies (as can be seen by induction on the size of B for example) that there exists an r s zero submatrix of B with r + s > n 1 (the Frobenius - König theorem see eg 9) This contradicts the full indecomposability of A Thus the latter case holds which implies that A has no zero entries Now use Proposition 22 Suppose that A is not fully indecomposable We denote by Y α β the α β submatrix of an n n matrix Y defined by the nonempty index set α {1 2 n} of rows and the nonempty index set β {1 2 n} of columns; α stands for the cardinality of a set α Let α and β be such that α + β = n and Aα β = 0 Then for every positive matrix X we have (X A) 1 β α = 0 where α stands for the complement of α in {1 2 n} Since A is ICRN it follows that Aβ α = 0 If α \ β then Aα \ β α \ β = 0 a contradiction with irreducibility of A Thus α \ β = Similarly β \ α = So α = β Since we conclude that n is even and α = n/2 Aα α = 0 Aα α = 0 If Aα α is not fully indecomposable then there exists a γ δ zero submatrix Aγ δ with γ + δ = n and γ δ a contradiction with what was proved above Thus Aα α and similarly Aα α are fully indecomposable Applying a similarity A QAQ 1 with a permutation matrix Q we may assume that A = 0 A1 A 2 0 where A 1 and A 2 are fully indecomposable (n/2) (n/2) ray-patterns Clearly ray ((X A 1 ) 1 ) = A 2 and ray ((X A 2 ) 1 ) = A 1 for every positive matrix X In particular the ray-pattern projections ray ((X A j ) 1 ) are independent of X for j = 1 2 The first paragraph of the proof shows that A j have no zero entries and an application of Lemma 21 completes the proof Lemma 41 Let 4 Reducible ICRN ray-patterns: the main result A = A1 A 21 A 22 be an ICRN n n ray-pattern where A 11 and A 22 are irreducible ICRN ray-patterns Then either A 21 = 0 or there exists g G n such that gag 1 has one of the following forms: (41)

7 (42) INVERSE CLOSED RAY-NONSINGULAR CONES z z 1 Conversely all matrices in (41) and (42) are ICRN ray-patterns z Ω Proof The converse statement is verified in a straightforward way Let A be as in the lemma By Proposition 31 each of the matrices A 11 and A 22 has one of the forms (31) We will prove that if at least one of A 11 and A 22 has either the third or the fifth form of (31) then A 21 = 0 Let A 11 = A 22 = 1 1 A 21 = Then for x 11 x 12 x 21 x 22 positive we have x 11 z 11 x 12 z 12 x 21 z 21 x 22 z where The matrix T is computed to be T = z11 z 12 z 21 z 22 A 1 = T 1 T = A 1 x11 z 11 x 12 z A x 21 z 21 x 22 z A 1 22 x11 z 11 + x 21 z 21 + x 12 z 12 + x 22 z 22 x 11 z 11 + x 21 z 21 x 12 z 12 x 22 z 22 x 11 z 11 x 21 z 21 + x 12 z 12 x 22 z 22 x 11 z 11 x 21 z 21 x 12 z 12 + x 22 z 22 Since A is ICRN we must have in particular that each of the entries of T has the same argument (or is zero) irrespectively of the positive values of x 11 x 12 x 21 x 22 It is easy to see that this happens only if at most one number among z 11 z 12 z 21 z 22 is nonzero But if one of those numbers were nonzero we would have ray (T ) A 21 a contradiction with A being ICRN Let where Q = 1 1 Then where 0 Q Q 0 A 11 = Note that Q 1 = 1 Q Let 2 A 22 = Q A 21 = z ij i=12;j=1234 X = x ij i=12;j=1234 x ij > 0 T = A 1 = X A 21 A 22 x13 z Q 13 x 14 z 14 Q x 23 z 23 x 24 z 24 A 1 1 A 1 22 T x11 z Q 11 x 12 z 12 Q x 21 z 21 x 22 z 22

8 8 CHI-KWONG LI AND LEIBA RODMAN As in the preceding case where A 11 and A 22 were both equal to Q it follows (since ray (T ) should be independent of X) that at most one number among z 11 z 12 z 21 z 22 is nonzero and at most one number among z 13 z 14 z 23 z 24 is nonzero Say z 11 0 The right 2 2 submatrix of T is Since A is ICRN we must have ( 1 P x11 z 11 x 11 z 11 x 11 z 11 x 11 z 11 ) x11 z 11 x 11 z 11 = x 11 z 11 x 11 z 11 z13 z 14 z 23 z 24 a contradiction with the property that at least three numbers among z 13 z 14 z 23 z 24 are zeros Thus A 21 = 0 The cases when 0 Q 0 Q A 11 = A 22 = or A Q 0 11 = Q A 22 = Q 0 are treated in a similar manner Let now consider the case 0 Q A 11 = Q 0 Let We have where T = 1 2 A 22 = A 21 = z ij i=12;j=1234 X = x ij i=12;j=1234 x ij > 0 1 A 1 = X A 21 A 22 A 1 1 A 1 22 x23 z 23 + x 24 z 24 x 23 z 23 x 24 z 24 x 21 z 21 + x 22 z 22 x 21 z 21 x 22 z 22 x 13 z 13 + x 14 z 14 x 13 z 13 x 14 z 14 x 11 z 11 + x 12 z 12 x 11 z 11 x 12 z 12 For ray (T ) to be independent of x ij we must have that there is at least one zero in each of the four pairs {z 11 z 12 } {z 13 z 14 } {z 21 z 22 } and {z 23 z 24 } But if one of the z ij is nonzero say z 11 0 then the ray-pattern of the right lower 1 2 corner of T is not equal to z 23 z 24 a contradiction with the ICRN property of A The remaining 0 Q cases of one of the two blocks A 11 and A 22 being equal to either Q or and Q 0 the other block being one of ±1 or can be dealt with similarly Thus leaving aside the case when A 21 = 0 each of the matrices A 11 and A 22 has one of the first second or fourth forms of (31) The rest is elementary For example 1 0 y y1 0 y 0 0 y 5 z 5 y 6 z 6 0 y 4 = y y3 1 y2 1 y 8 z 8 y3 1 y1 1 y 7 z 7 0 y 1 3 y 7 z 7 y 8 z 8 y 3 0 y4 1 y2 1 y 6 z 6 y4 1 y1 1 y 5 z 5 y4 T

9 INVERSE CLOSED RAY-NONSINGULAR CONES 9 where y 1 y 8 > 0 and z 5 z 6 z 7 z 8 Ω Thus the ray-pattern 0 0 A 0 = 0 0 z 5 z 6 z 7 z 8 is ICRN if and only if z 8 = z 5 and z 7 = z 6 Applying a suitable transformation A 0 (h g)a 0 (h g) 1 h g G 2 we reduce A 0 to one of the forms indicated in (42) (if at least one of z 5 z 6 z 7 and z 8 is nonzero) We now state the main result of this paper describing all ICRN ray-patterns up to similarity with a similarity matrix in the group G Theorem 42 Let A A (43) A = 21 A A p1 A p2 A p3 A pp be an ICRN n n ray-pattern where A 11 A pp are irreducible ICRN ray-patterns Then there exists g G n such that gag 1 has a block lower triangular form (44) B := gag 1 C = B ij 1 ij p = C C 2 C 0 3 with the following properties: (α) C 0 is the direct sum of r identical matrices of the form 0 1 identical matrices of the form ; 1 0 (β) C = B 0 C 11 B qq 3 with q = p r s where for j = 1 q: { 1 1 (45) B jj (γ) If the block } ; 1 1 Bii 0 is such that B B ji B ji is a submatrix of C 2 then jj Bii 0 B ji B jj and of s

10 10 CHI-KWONG LI AND LEIBA RODMAN has one of the following forms: (a) ± 0 ±1 (d) for some w z Ω (b) 0 0 z z ±1 ± z 1 (e) (c) ± 0 z z w z z w Conversely every ray-pattern of the form (44) with the properties (α) - (γ) is ICRN The cases when one or more integers among q r s are zeros with the obvious interpretation are not excluded in Theorem 42 If q = 1 then the property (β) is interpreted in the sense that B 11 has one of the forms as in (45) For convenience of reference we define the following five types of block forms: (I) 1 (II) 1 (III) (46) (IV) 1 1 (V) The rather long proof of the theorem will be given in the next section It will be seen in the proof of Theorem 42 that when the proof is specialized to real ray-patterns A the following real analogue of Theorem 42 is obtained Let G n (r) be the group of n n matrices of the form DQ where Q is a permutation matrix and D is a diagonal matrix with diagonal entries ±1 Theorem 43 Let A A (47) A = 21 A A p1 A p2 A p3 A pp be an ICRN n n ray-pattern with entries 0 ±1 where A 11 A pp are irreducible ICRN ray-patterns Then there exists g G n (r) such that gag 1 has a block diagonal form (44) with the properties (α) (β) and (γ) where z w {0 1 1} in (b) - (e) Conversely every real ray-pattern of the form as described is ICRN Theorem 43 is essentially a reformulation of 1 Theorem 746

11 INVERSE CLOSED RAY-NONSINGULAR CONES 11 5 Proof of Theorem 42 We prove first the direct statement of the theorem We use induction on p For p 2 the result is established in Proposition 31 and Lemma 41 Let now p 3 and assume that Theorem 42 is proved for all smaller values of p Let A be given as in the Theorem In what follows we formally put A ij = 0 if i < j By Proposition 31 we may assume that each diagonal block A jj has one of the forms (I) - (V) Step 1 Suppose that one of the diagonal blocks say A j0 j 0 where j 0 < p has the form (IV) or (V) Applying the induction hypothesis to A jk p 1 jk=1 we may assume that A kj0 = 0 for k = j p 1 and A j0 k = 0 for k = 1 j 0 1 (if it happens that A pp has one of the forms (IV) or (V) we apply the induction hypothesis to A jk p jk=2 and argue similarly) Partition the matrix A jk p jk=j 0 as follows: (51) A jk p jk=j 0 = A j 0 j C 0 A pj0 D A pp where C = A jk p 1 jk=j 0 +1 D = A pj 0 +1 A pj0 +2 A pp 1 Since A is ICRN it is easy to see that the matrix A jk p jk=j 0 is also ICRN For any positive matrix X = X uv 3 uv=1 of appropriate size partitioned conformably with (51) we have ( ) X Ajk p 1 jk=j 0 = (X 11 A j0 j 0 ) 0 0 (X 22 C) Q (X 33 A pp ) 1 where and Q = (X 33 A pp ) 1 (X 31 A pj0 )(X 11 A j0 j 0 ) 1 ray (Q) = A pj0 On the other hand ( ) 1 X11 X 13 Aj0 j 0 0 (X11 A = j0 j 0 ) X 31 X 33 A pj0 A pp Q (X 33 A pp ) 1 Aj0 j and therefore the ray-pattern 0 0 is ICRN By Lemma 41 A A pj0 A pj0 = 0 Permuting the j 0 -th and p-th block rows and columns of A we obtain a block diagonal pp matrix 0 0 A j0 j 0 and an application of the induction hypothesis completes the proof in this case

12 12 CHI-KWONG LI AND LEIBA RODMAN Step 2 Thus we assume from now on in the proof that the blocks A jj have forms (I) (II) (III) and therefore q = p in our notation Using the induction hypothesis we assume also that A A 21 A A p 11 A p 12 A p 13 A p 1p 1 = B B 21 B B p 11 B p 12 B p 13 B p 1p 1 where the B jk s have the properties described in the theorem We prove the property (β) first As an intermediate step the following statement will be proved: (A) For every index i 2 i p 1 either the blocks A i1 A ii 1 are all zeros or the blocks A i+11 A pi are all zeros or both Arguing by contradiction suppose there exist an index i 0 2 i 0 p 1 such that not all blocks A i0 1 A i0 i 0 1 are zeros and not all blocks A i0 +1i 0 A pi0 are zeros We select the smallest i 0 with these properties Because of the induction hypothesis made above (52) A i0 +1i 0 = 0 A p 1i0 = 0 A pi0 0 The matrix A jk p jk=i 0 can be partitioned as follows: (53) A jk p jk=i 0 = A i 0 i C 0 A pi0 D A pp analogously to (51) and as in the proof of Step 1 we conclude that the submatrix Ai0 i 0 0 A pi0 A pp is ICRN Thus by Lemma 41 the cases when A i0 i 0 = A pp = ±1 cannot occur Denoting by n j (n j {1 2}) the size of the block A jj and applying a similarity transformation A (h g)a(h g) 1 where h G ni0 g G np we may assume in view of the same Lemma 41 in addition to the assumptions already made that the block A pi0 has the following structure: (α ) if A i0 i 0 = A pp = ±1 then A pi0 = 1; (β 1 ) if A i0 i 0 = ±1 and A pp = then A pi0 = ; 1 (γ ) if A i0 i 0 = and A pp = 1 then A pi0 = 1 ± 1; (δ 1 z ) if A i0 i 0 = A pp = then A pi0 = for some z Ω z 1

13 INVERSE CLOSED RAY-NONSINGULAR CONES 13 Let X = X ij p ij=1 be a positive matrix partitioned conformably with the partition of A Partition (X A) 1 again conformably with that of A: Q Q (X A) 1 = 21 Q Q p1 Q p2 Q p3 Q pp We have Q p1 = (X pp A pp ) 1 (X p1 A p1 )(X 11 A 11 ) 1 + (X pp A pp ) 1 X p2 A p2 X pp 1 A pp 1 X 21 A 21 X ij A ij X p A 31 ij=2 X p 11 A p 11 (X 11 A 11 ) 1 Since A 21 = = A i0 11 = 0 the above expression for Q p1 can be rewritten in the form (54) Because of (52) (55) Q p1 = (X pp A pp ) 1 (X p1 A p1 )(X 11 A 11 ) 1 + (X pp A pp ) 1 X pi0 A pi0 X pp 1 A pp 1 X i0 1 A i0 1 X i0 +11 A i0 +11 X ij A ij p 1 ij=i 0 1 X p 11 A p 11 (X 11 A 11 ) 1 Q p1 = (X pp A pp ) 1 (X pi0 A pi0 )(X i0 i 0 A i0 i 0 ) 1 (X i0 1 A i0 1)(X 11 A 11 ) 1 We let + {terms that are independent of X pi0 and of X i0 1} W = (X pp A pp ) 1 (X pi0 A pi0 )(X i0 i 0 A i0 i 0 ) 1 (X i0 1 A i0 1)(X 11 A 11 ) 1 Step 3 Consider the case when all entries of W are nonzero for some choice of positive matrices (56) X pp X pi0 X i0 i 0 X i0 1 and X 11 In view of the conditions (α ) - (γ ) and (a) - (e) (the latter conditions with B ij replaced by A ij are satisfied by the blocks of the matrix A ij p 1 ij=1 ) this case does not occur only if (57) A 11 = A i0 i 0 = A pp =

14 14 CHI-KWONG LI AND LEIBA RODMAN and we will consider separately the case when (57) holds true Let the matrices (56) be fixed so that all entries of W are nonzero By keeping all other diagonal blocks of X fixed and choosing all other nondiagonal blocks of X sufficiently small we have by virtue of (55) that (58) Q p1 W < ɛ for any prescribed ɛ > 0 On the other hand since A is ICRN we have that ray (Q p1 ) = A p1 is independent on X and since all entries of W are nonzero (55) and (58) give (59) ray (W ) = A p1 (provided W has no zero entries) (Use here the fact that the ray-pattern projection P is continuous on the set of matrices with no zero entries) In particular (510) the matrix A p1 has no zero entries and since ray (Q p1 ) = A p1 it follows that Q p1 has no zero entries either Returning to formulas (54) and (55) write (511) Q p1 = (X pp A pp ) 1 (X p1 A p1 )(X 11 A 11 ) 1 + W + {terms that are independent of X pi0 of X i0 1 and of X p1 } Keeping in this formula X p1 X 11 and X pp fixed keeping fixed all other diagonal blocks of X and letting all other nondiagonal blocks of X tend to zero we obtain Q p1 ( (X pp A pp ) 1 (X p1 A p1 )(X 11 A 11 ) 1) < ɛ for any prescribed ɛ > 0 Note that since A p1 has no zero entries and since A 11 and A pp have the forms (I) (II) or (III) it is easy to see that the matrix U := (X pp A pp ) 1 (X p1 A p1 )(X 11 A 11 ) 1 has no zero entries Using again the continuity of the ray-pattern projection on the set of matrices with no zero entries with U playing the role of W in the above argument it follows (letting ɛ tend to zero) that (512) ray (U) = ray (Q p1 ) = A p1 As a result we obtain that the matrix A1 (513) A p1 A pp is ICRN We now consider several situations that may occur (S1) A 11 = ±1 Then A i0 i 0 ±1 (or else the block A i0 1 would have been zero by (a) - (e)) and if A i0 i 0 = 1 then A pp 1 (for a similar reason in view of (α ) - (δ )) Also by (510) and (513) A pp ±1 Thus we have the following situations: (S11) A 11 = ±1 A i0 i 0 = 1 A pp = ;

15 INVERSE CLOSED RAY-NONSINGULAR CONES 15 (S12) A 11 = ±1 A i0 i 0 = A pp = 1; (S13) A 11 = ±1 A i0 i 0 = A pp = (S2) A 11 = Then A i0 i 0 = A pp = ±1 is impossible and (57) is excluded (so far) so we have the following situations: (S21) A 11 = A i0 i 0 = ±1 A pp = 1; (S22) A 11 = A i0 i 0 = ±1 A pp = ; (S23) A 11 = A i0 i 0 = A pp = ±1 We now consider each of the six situations (S11) - (S13) (S21) - (S23) separately In these considerations it will be assumed that the matrices X 11 X i0 i 0 and X pp are matrices of all 1 s Assume (S11) and let X 1 11 X i0 1 X i0 i 0 = x 2 1 x X p1 X pi0 X 4 x 6 x 2 x 7 > 0 pp x 5 x 7 By virtue of (a) - (e) (α ) - (δ ) (510) and (513) we have 1 z A pi0 = A ±1 i0 1 = 1 A p1 = z = 1 z A computation shows that W = x7 x 2 U = x 6 x 2 x 4 z x 4 z We have ray (W ) ray (U) a contradiction with (59) and (512) Assume (S12) and let X 1 11 X i0 1 X i0 i 0 = x 2 x X p1 X pi0 X 3 x 2 x 10 > 0 pp x 8 x 9 x 1 Also 1 A pi0 = 1 ± 1 A i0 1 = 1 A p1 = z z = 1

16 16 CHI-KWONG LI AND LEIBA RODMAN A computation shows that W = ±x 10 x 2 x 9 x 8 Thus ray (W ) is not constant a contradiction with (59) Assume (S13) and let X 1 11 x 2 X i0 1 X i0 i 0 = x 3 X p1 X pi0 X pp x 8 x 10 x 1 1 x 2 x 13 > 0 x 9 x 12 x 13 Also we have 1 u A pi0 = u 1 1 A i0 1 = 1 for some u = 1 or A pi0 = A p1 = A computation shows that x (514) W = 12 ux 3 x 13 x 2 x 10 x 3 ± x 11 ux 2 x 1 1 depending on the form of A pi0 and U = z z x 9 z x 8 z for some z = 1 or W = x 12 x 3 ±x 11 x 2 If W has the second form in (514) then clearly ray (W ) ray (U) a contradiction If W has the first form in (514) then ray (W ) is not constant (which happens if u 1 in the case of upper signs or if u 1 in the case of lower signs) or ray (W ) is constant and ray (W ) ray (U) (which happens if the signs are lower and u = 1) or ray (W ) is constant and ray (W ) = ray (U) (which happens if the signs are upper and u = 1) but 1 then ray (W ) = a contradiction in all cases; in the latter case a contradiction 1 is obtained with ray (W ) = A p1 The cases (S21) and (S23) can be reduced to (S11) and (S13) respectively by taking transposed matrices and then permuting rows and columns appropriately to get lower triangular forms Thus we obtain a contradiction in the cases (S21) and (S23) Assume (S22) and let X 11 X i0 1 X i0 i 0 = x 5 x 6 1 X p1 X pi0 X pp x 8 x 9 x 12 x 5 x 13 > 0 x 10 x 11 x 13 Here A i0 1 = 1 1 A pi0 = 1 1

17 and the property that the matrix zero entries imply that INVERSE CLOSED RAY-NONSINGULAR CONES 17 A1 is ICRN together with A A p1 A p1 not having pp A p1 = u v v u for some u v with u = v = 1 A computation shows that ±x13 x W = 6 x 13 x 5 x11 u x and U = 10 v x 12 x 6 ±x 12 x 5 x 9 v x 8 u Clearly ray (W ) ray (U) a contradiction again Step 4 Consider the so far excluded case when (57) holds true Applying a transformation A (h I)A(h I) 1 h G 2 the block A i0 1 may be transformed to the form 1 z (515) z Ω z 1 Thus we assume that A i0 1 and A pi0 have the form (515) Assume first that at least one of A i0 1 and A pi0 has the form (515) with z = 1 Then for some choice of the matrices X ij i j {1 i 0 p} i j still keeping the matrices X ii i {1 i 0 p} all 1 s the matrix W has no zero entries and we may repeat the arguments of Step 3 Thus the properties obtained in Step 3 are valid In particular the matrix A1 A p1 A pp is ICRN and therefore (also because A p1 has no zero entries) we have u v A p1 = u = v = 1 v u Let A i0 1 = 1 z z 1 z Ω A pi0 = 1 w w 1 w Ω where not both z and w are zero As in Step 3 we compute W and U and obtain a contradiction with one of the properties of W and U We let X 11 X i0 1 X i0 i 0 x = 5 x 6 x X p1 X pi0 X x 7 x 8 5 x 20 > 0 pp x 13 x 14 x 17 x 18 x 15 x 16 x 19 x 2 1

18 18 CHI-KWONG LI AND LEIBA RODMAN A computation shows that x20 x (516) W = 6 z + x 19 wx 8 x 20 x 5 + x 19 wx 7 z x 18 wx 6 z x 17 x 8 x 18 wx 5 x 17 x 7 z U = x 16 u x 14 v x 15 v x 13 u If ray (W ) is not constant (ie independent of x j as long as W has no zero entries) we obtain a contradiction with (59) If ray (W ) is constant then its off diagonal entries must be both negative whereas the off diagonal entries of U are negative multiples of each other (or are equal to zero if v = 0) Thus ray (W ) ray (U) a contradiction again with (59) and (512) Step 5 We take up the remaining case when (57) holds true and A i0 1 = A pi0 = 0 1 Partition the positive matrix X = X ij p ij=1 and (X A) 1 = Q ij p ij=1 conformably with (43) and fix X 11 X i0 i 0 and X pp to be the matrices of all 1 s Then Q p1 takes the form (cf (511)): (517) where Let X i0 1 = Q p1 = ( (X p1 A p1 ) ) + W + {terms that are independent of X pi0 of X i0 1 and of X p1 } W = (X pi0 A pi0 ) (X i0 1 A i0 1) x1 x 2 x5 x X x 3 x p1 = 6 x9 x X 4 x 7 x pi0 = 10 z1 z A 8 x 11 x p1 = 2 12 z 3 z 4 A computation shows that the matrix Q p1 has the form x (518) Q p1 = 8 z 4 x 7 z 3 x 12 x 1 + { } x 6 z 2 x 9 x 4 x 5 z 1 where the ellipse stands for terms that are independent of x 1 x 12 Let us analyze (518) Regardless of the value of z 2 the (2 1) entry of x 8 z 4 x 7 z 3 x 12 x 1 + { } x 6 z 2 x 9 x 4 x 5 z 1 is nonzero for some x 6 x 9 x 4 Fix these values and (for a given ɛ > 0) select the blocks of X other than X ij i j {1 i 0 p} so that x8 z (519) Q p1 4 x 12 x 2 x 7 z 3 < ɛ x 6 z 2 x 5 z 1 x 9 x 3 Since ray (Q p1 ) = A p1 passing to the limit in (519) as ɛ 0 we obtain that x 6 z 2 x 9 x 4 = cz 3

19 INVERSE CLOSED RAY-NONSINGULAR CONES 19 for some positive c as long as x 6 z 2 x 9 x 4 0 Thus (520) z 2 = z 3 = 1 Applying a similar argument to the (1 2) entry of Q p1 we obtain z 3 = z 2 = 1 a contradiction with (520) This completes the proof of the statement (A) for the matrix A (after a similarity A gag 1 for some g G) Property (β) is now deduced without difficulties Still using the induction hypothesis assume that A = A jk p 1 jk=1 = B jk p 1 jk=1 where the B jk s have the properties as in Theorem 42 and let t be such that each of A ij t ij=1 and A ij p 1 ij=t+1 is a direct sum of matrices of forms (I) (II) and (III) If for some index s t + 1 it happens that A ps 0 then by the already proven part we have A 1s = 0 A s 1s = 0 If s = t + 1 then the block A ij t+1 ij=1 is a direct sum of matrices of forms (I) (II) and (III) If s > t + 1 then we interchange in A the s-th and (t + 1)-th block rows and the s-th and (t + 1)-th block columns resulting in a matrix whose (t + 1) (t + 1) block upper left corner is a direct sum of matrices of forms (I) (II) and (III) Continuing this process we eventually obtain a matrix B with property (β) Property (γ) will follow by elementary considerations from Property (β) (cf proof of Lemma 41) once we show that all ray-patterns Bii 0 1 i t < j p B ji B jj are ICRN To this end let X be a positive matrix of the same size as B and partition: (X B) 1 = Y km p km=1 It is clear from property (β) that for fixed i j (1 i t < j p) we have Y ji = (X jj B jj ) 1 (X ji B ji )(X ii B ii ) 1 and by the ICRN property of B it follows that (521) ray (Y ji ) = B ji On the other hand Y ji is also the off-diagonal block of the matrix ( ) 1 Xii X ij Bii 0 X ji X jj B ji B jj Bii 0 and the ICRN property of follows from (521) B ji B jj Finally the converse part of Theorem 42 is verified in a straightforward way taking into account that the ray-pattern B given by (44) with the property (β) is ICRN if the

20 20 CHI-KWONG LI AND LEIBA RODMAN and only if each of the 2 2 blocks (see the preceding paragraph) Bii 0 where 1 i t < j p is ICRN B ji B jj 6 Inertias and Jordan forms of matrices with ICRN ray-patterns We start with inertia considerations Let Y M n (C) and assume that Y has no eigenvalues on the imaginary axis We define the inertia In Y = {i (Y ) i + (Y )} where i (Y ) (resp i + (Y )) is the number of eigenvalues of Y (counted with multiplicities) in the open left (resp right) halfplane Thus i (Y ) + i + (Y ) = n Lemma 61 Let A be an ICRN ray-pattern and let Cone (A) be the cone generated by A Then the spectrum of every Y Cone (A) does not intersect the imaginary axis Moreover there exist two nonnegative integers i ± (A) depending on A only that sum up to n such that In Y = {i (A) i + (A)} for every Y Cone (A) Proof Let Y Cone (A) and arguing by contradiction assume Y x = iλx for some nonzero x and λ R \ {0} Then Y 1 x = i 1 λ x Now ( ) 1 1 Y + λ Y x = 0 λ which is impossible because 1 λ Y + λ Y 1 Cone (A) and therefore 1 λ Y + λ Y 1 is invertible in view of the inverse closeness property of A The second statement of the lemma follows easily from the first using a standard argument that involves arcwise connectedness of Cone (A) and continuity of eigenvalues of Y We define the inertia of an ICRN ray-pattern A as the numbers i ± (A) introduced in Lemma 61 The inertia of ICRN ray-patterns are described as follows Theorem 62 If A is an n n ICRN ray-pattern then trace A is an integer of the same parity as n trace A n and i ± (A) = n ± trace A 2 In particular all matrices in Cone (A) are stable (ie all eigenvalues have negative real parts) if and only if A = I

21 INVERSE CLOSED RAY-NONSINGULAR CONES 21 Proof In view of Theorem 42 we need only to prove the result for each of the five forms (31) For the first four forms this is immediate for the form (V) it follows from Lemma 63 below (which will be also used in the proof of the next theorem) Next we turn to the eigenvalues and Jordan forms of matrices in Cone (A) We first establish two lemmas Lemma 63 Let Then: Y = 0 0 a b 0 0 c d e f 0 0 g h 0 0 a b c d e f g h > 0 (1) Y has no eigenvalues with zero real part; (2) if λ is an eigenvalue of Y then so is λ and the algebraic multiplicities of λ and of λ are the same; (3) if λ = µ + iν C µ ν R is an eigenvalue of Y then ν < µ Conversely if (λ 1 λ 2 λ 3 λ 4 ) is a quadruple of not necessarily distinct complex numbers which is closed under negation and complex conjugation and satisfies Im λ j < Re λ j j = then there exist a b c d e f g h > 0 such that (61) σ(y ) = {λ 1 λ 2 λ 3 λ 4 } Proof (1) follows from Lemma 61 because A := is an ICRN The characteristic polynomial of Y is computed to be λ 4 + λ 2 ( dh cf ae bg) + (aedh + adfg + behc + bcfg) Thus (2) follows Moreover Re (λ 2 ) > 0 for every eigenvalue λ of Y Property (3) now follows For the converse statement observe that by letting f = d = h = c and e = b = g = a the characteristic polynomial of Y takes the form λ 4 + λ 2 ( 2c 2 2a 2 ) + 4a 2 c 2 = (λ 2 2a 2 )(λ 2 c 2 ) and therefore for every quadruple of nonzero real numbers (λ 1 λ 2 λ 3 λ 4 ) which is closed under negation there exist a b c d e f g h > 0 such that (61) holds Next fix µ > 0 and let d = h = c = f = a = e = µ g = µ b

22 22 CHI-KWONG LI AND LEIBA RODMAN Then the characteristic polynomial of Y is λ 4 + λ 2 ( 4µ) + (2µ 2 + µ µ(b + µ b )) and therefore for eigenvalues λ 0 of Y we have λ 2 0 = 2µ ± i µ µ(b + µ b ) 2µ2 Clearly by a suitable choice of µ > 0 and b > 0 λ 2 0 can be made equal to any nonreal complex number z = α + iβ with positive real part α Indeed let µ = α/2 and b a positive solution of the quadratic equation Lemma 63 is proved b 2 2 µb + µ β2 µ µ = 0 Lemma 64 (a) Suppose B 1 = B ij q ij=1 is a block lower triangular matrix in the form (44) satisfying the conditions (1) (2) of Theorem 42 Let X be a matrix with positive entries such that λ R \ {0} is an eigenvalue of X B Then X B is similar to a matrix of the form T U where λi α λi β (62) T = 0 λi x R 1 λi γ 0 0 λi x S 1 λi δ 0 R 2 R 3 0 λi y S 2 S 3 0 λi y and none of ±λ is an eigenvalue of U (b) If T has the form (62) with λ 0 for some matrices R j S j (j = 1 2 3) of suitable sizes then the Jordan form of T consists of blocks of the forms: λ 1 λ 1 λ λ 0 λ 0 λ with multiplicities r 1 s 1 r 2 s 2 respectively such that (63) r 1 + 2r 2 s 2 and s 1 + 2s 2 r 2 Proof Note that X B has the form C1 (64) C 21 C 22 such that each of C 11 and C 22 is a direct sum of 1 1 and 2 2 matrices Applying a block permutation similarity we can assume that C 11 = U 1 λi α λi β X and C 22 = λi γ λi δ Y U 2 such that none of ±λ is an eigenvalue of U 1 or U 2 and each of X and Y is a direct sum of 2 2 real matrices with zero diagonal and eigenvalues ±λ Thus X and Y are similar to the matrices λ(i x I x ) and λ(i y I y ) respectively We may apply a similarity transformation to C ij 2 ij=1 so that X and Y are changed to λ(i x I x ) and λ(i y I y ) respectively

23 INVERSE CLOSED RAY-NONSINGULAR CONES 23 We now write (64) after a transformation as indicated above as the block lower triangular matrix V = V ij 10 ij=1 such that and V ij 5 ij=1 = U 1 λi α λi β λi x λi x V ij 10 ij=6 = λi γ λi δ λi y λi y U 2 By property (β) of Theorem 42 we see that V 62 and V 73 are zero blocks Next we show that one can apply a sequence of block permutation similarity transformations to the matrix V ij 10 ij=1 that convert all V ij to zero except for V 64 V 75 V 82 V 84 V 93 V 95 V 101 and keep the blocks V 62 and V 73 zeros Then by a block similarity transformation the resulting matrix will be similar to T U with U U = V 101 U 2 and with T given by (62) In the following all block matrices having the same size as V are partitioned according to the block form of V Suppose 1 i 5 < j 10 are such that (i j) / {(6 2) (7 3) (10 1)} and V ii and V jj have no common eigenvalue Let W = W ij 10 ij=1 be obtained from I (having the same size as V ) by changing its (j i) block W ji = V ji (V jj V ii ) 1 Here note that V ii or V jj is a scalar matrix Then the (j i) block of W 1 V W is zero and this transformation will not change other blocks in the matrix V Hence we can apply a number of such similarity transformations until we get all the desired zero blocks For part (b) let T be given by (62) with λ 0 Then the Jordan form of T has Jordan blocks of size at most 2 This follows from a general fact that the Jordan form of a matrix λi 0 Z = Y λi consists of blocks of size at most 2 and the number of Jordan blocks of size 2 is equal to the rank of Y (This fact is easily proven by using a rank decomposition Irank Y = W Y 0 1 W where W 1 W 2 are invertible) Now to prove (63) note that r 1 + 2r 2 = α + γ + x + y and s 1 + 2s 2 = β + δ + x + y; moreover r 2 and s 2 are equal to the ranks of the matrices α x 0 R1 γ R 3 y R 2 and β x 0 S1 δ S 3 y S 2

24 24 CHI-KWONG LI AND LEIBA RODMAN respectively Now r 2 = rank R1 + rank R R 2 R 3 rank R 3 3 x + y rank R 3 x + y + β + δ = s 1 + 2s 2 Similarly one can prove that s 2 r 1 + 2r 2 We are now ready to describe the Jordan form of matrices X A where X is entrywise positive and A is an ICRN ray-pattern Theorem 65 Let K be an n n matrix in the Jordan form Then K is similar to a matrix X A where X is entrywise positive and A is an ICRN ray-pattern if and only if K has blocks of the following types only (perhaps after a permutation of the Jordan blocks of K): (1) λ λ R \ {0}; (2) λ 1 λ λ R \ {0}; 0 λ (3) λ 1 0 λ (4) λ 1 0 λ (5) λ 0 0 λ λ 1 0 λ λ 1 0 λ λ 0 0 λ λ 1 0 λ λ R \ {0}; λ R \ {0} λ = µ + iν 0 < ν < µ Proof The only if part By Theorem 42 and Lemma 63 we see that the Jordan form corresponding to nonreal eigenvalues has the form (5) For a real eigenvalue λ we can use Theorem 42 and Lemma 64 to conclude that the Jordan blocks corresponding to λ have the forms: (a) λ (b) λ (c) λ 1 0 λ (d) λ 1 0 λ with multiplicities r 1 s 1 r 2 s 2 respectively such that r 1 + 2r 2 s 2 and s 1 + 2s 2 r 2 Since s 1 + 2s 2 r 2 we can use construct matrices of the form (4) and (2) until we use up all the Jordan blocks of the form (c) If we also used up the Jordan blocks of the form (d) then we are left with Jordan blocks of the forms (a) and (b) and we are done So suppose that we have not used up all the Jordan blocks of the form (d) Converting one matrix of the form in (4) to two matrices of the form (3) we can use up more Jordan blocks of the form (d) If we are not able to use up all the Jordan

25 INVERSE CLOSED RAY-NONSINGULAR CONES 25 blocks of the form (d) after converting all type (4) matrices to type (3) matrices then construct matrices of the form λ 1 (2 ) λ 0 λ which is a type (2) matrix with the roles of λ and λ interchanged If this still do not exhaust all the Jordan blocks of the form (d) convert type (3) matrices to λ 1 λ 1 λ 1 (4 ) 0 λ 0 λ 0 λ which is a type (4) matrix with the roles of λ and λ interchanged Since r 1 +2r 2 s 2 we will exhaust all the Jordan blocks of the form (d) through such conversions Now we see that all the Jordan blocks associated with λ and λ can be put in the forms (1) (2) (3) (4) (3 ) (4 ) and we are done The if part For each of the matrices of the form (1) (4) by Theorem 42 one can find a positive matrix X and an ICRN ray-pattern A such that X A have the following forms: (1) λ (2) λ (3) λ (4) λ and Lemma 63 shows that one can construct X A for case (5) A direct sum of the above matrices will give rise to a matrix with the desired Jordan form structure As it follows from the proof of Theorem 65 conditions (1) - (4) of the theorem can be expressed also as follows: The Jordan blocks corresponding to real eigenvalues of a matrix of the form X A where X is entrywise positive and A is an ICRN ray-pattern are of size at most two and if r 1 (λ) and r 2 (λ) are the numbers of the Jordan blocks of X A of sizes 1 and 2 respectively corresponding to the real eigenvalue λ then (65) r 1 (λ) + 2r 2 (λ) r 2 ( λ) Note that inequality (65) implies that if λ is an eigenvalue of X A but λ is not then there are no Jordan blocks of size 2 corresponding to λ; indeed apply (65) with λ replaced by λ and interpret r 1 ( λ) and r 2 ( λ) as zeros References 1 R A Brualdi and B L Shader Matrices of sign-solvable linear systems Cambridge University Press N Cohen and I Lewkowicz Convex invertible cones of state space systems Math Control Signals Systems 10 (1997)

26 26 CHI-KWONG LI AND LEIBA RODMAN 3 N Cohen and I Lewkowicz Convex invertible cones and the Lyapunov equation Linear Algebra Appl 250 (1997) C Eschenbach F J Hall and C R Johnson Self-inverse sign patterns In: Combinatorial and graph-theoretical problems in linear algebra IMA Vol Math Appl 50 Springer New York C A Eschenbach F J Hall and Z Li From real to complex sign pattern matrices Bull Austral Math Soc 57 (1998) G Y Lee J J McDonald B L Shader and M J Tsatsomeros Extremal properties of raynonsingular matrices Discrete Math 216 (2000) C K Li T Milligan and B L Shader Non-existence of five by five full ray-nonsingular matrices Electronic Linear Algebra 11 (2004) J J McDonald D D Olesky M J Tsatsomeros and P van den Driessche Ray patterns of matrices and nonsingularity Linear Algebra Appl 267 (1997) H Minc Permanents Addison-Wesley Reading Mass J Y Shao On digraphs and forbidden configurations of strong sign nonsingular matrices Linear Algebra Appl 282 (1998) C Thomassen Sign-nonsingular matrices and even cycles in directed graphs Linear Algebra Appl 75 (1986) Erratum Linear Algebra Appl 240 (1996) C Thomassen When the sign pattern of a square matrix determines uniquely the sign pattern of its inverse Linear Algebra Appl 119 (1989) Department of Mathematics William & Mary Williamsburg VA address: ckli@mathwmedu Department of Mathematics William & Mary Williamsburg VA address: lxrodm@mathwmedu