MINIMAL NORMAL AND COMMUTING COMPLETIONS

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1 INTERNATIONAL JOURNAL OF INFORMATION AND SYSTEMS SCIENCES Volume 4, Number 1, Pages 5 59 c 8 Institute for Scientific Computing and Information MINIMAL NORMAL AND COMMUTING COMPLETIONS DAVID P KIMSEY AND HUGO J WOERDEMAN Abstract We study the minimal normal completion problem: given A A C n n A1, how do we find an n+q n+q normal matri A et := A 1 A of smallest possible size? We will show that this smallest number q of rows and columns we need to add, called the normal defect of A, satisfies nda ma{i AA A A, i + AA A A}, where i ± M denotes the number of positive and negative eigenvalues of the Hermitian matri M counting multiplicities Subsequently, we will show that for some matrices a minimal normal completion can be chosen to be a multiple of a unitary, addressing a conjecture from [H J Woerdeman, Linear and Multilinear Algebra , 59 68] In addition, we study the related question where A C n n and B A C n n A1 are given, and where we look for A et := and B et := A 1 A B B1 such that they commute and are of smallest possible size B 1 B Key Words commuting completions, commuting defect, normal completions, normal defect, inertia, inverse defect, unitary defect 1 Introduction The minimal normal completion problem was introduced in [6] and concerns the following Given A C n n, we wish to find a smallest possible normal matri with A as a principal submatri Recall that a matri A is normal if and only if the commutator of A and its conjugate transpose A, denoted by [A, A ] := AA A A, equals In other words, we would like to find a normal completion of A? :?? C n C q of smallest possible size thus smallest possible q We shall call this smallest number q the normal defect of A, and denote it by nda Clearly, nda = if and A A only if A is normal As observed in [], the matri A is normal, so it follows A that for an n n matri A we have that nda n As was observed in [6], and as we shall see further on, we have in fact that nda n 1 It is also not hard to A come up with the lower bound nda 1 rankaa A A1 A Indeed if A 1 A is of size n + q n + q and normal then AA A A = A 1A 1 A 1 A 1 and thus rankaa A A rank A 1A 1 + rank A 1 A 1 q + q = q C n C q Received by the editors December 8, 6 and, in revised form, May 18, 7 Mathematics Subject Classification 15A4, 15A4, 15A57 5

2 MINIMAL NORMAL AND COMMUTING COMPLETIONS 51 In order to obtain sharper bounds for nda, the so-called unitary defect was introduced in [6] It corresponds to the smallest number of rows and columns we need to add to A such that the completion is a multiple of a unitary matri, and as it turns out we have that 1 uda := rank A A A, where denotes the spectral norm As multiples of unitaries are normal we clearly have that nda uda Formula 1 implies that uda n 1, which yields nda n 1 In order to state a conjecture from [6], let us recall that a matri A C n n is called unitarily reducible if A = U A1 U, with U A unitary and A 1, A of nontrivial size Clearly, with A as above we have that nda nda 1 + nda uda 1 + uda uda As soon as A 1 = A we have that the last inequality in is strict, and thus nda < uda in that case So for a general statement for the case when nda = uda, it is natural to require that A is unitarily irreducible, which by definition means that A is not unitarily reducible An open question from [6] is whether the following conjecture holds Conjecture 1 For a unitarily irreducible matri A we have that nda = uda In this paper we refine some of the estimates for nda and as a result obtain more evidence for this conjecture Let us mention that the separability problem that appears in quantum computation can be seen as a normal completion problem where additional constraints need to be met; see [7] for details In that contet, minimizing the size of the matri corresponds to minimizing the number of states in the separable representation We will end this paper with considering the problem of completing two matrices to make them commute The paper is organized as follows In Section, we obtain an improved lower bound for nda by showing that nda ma{i + [A, A ],i [A, A ]}, where i ± M denotes the number of positive/negative eigenvalues of the Hermitian matri M Using this improved lower bound we are able to show that for some weighted Jordan blocks we have that nda = uda, providing new evidence for Conjecture 1 Net, in Section 3 we eamine matrices for which nda = 1 Finally in Section 4 we eplore the commuting completion problem Main result and normal defect conjecture In this section we shall prove our main result and provide evidence for Conjecture 1 Theorem 1 Given A C n n Then nda ma{i + [A, A ], i [A, A ]} A A1 Proof Let A et := be a normal completion of size n + q n + q A 1 A From the normality of A et we get 3 AA A A = A 1A 1 A 1 A 1 Let us denote the eigenvalues of the Hermitian matrices A 1A 1 A 1 A 1 and A 1A 1 by λ 1 λ n and µ 1 µ n, respectively By the Courant-Fischer

3 5 DP KIMSEY AND HJ WOERDEMAN theorem see eg, Theorem 411 in [3], we get for 1 j n 4 5 λ j = min w 1,w,,w n j C n min w 1,w,,w n j C n ma, C n w 1,w,,w n j ma, C n w 1,w,,w n j A 1A 1 A 1 A 1 A 1A 1 = µ j Since µ n q =, we get λ n q Thus 3 gives i + [A, A ] q Notice that a similar argument can be carried out by looking at the eigenvalues of A 1A 1 A 1 A 1 and A 1 A 1, which will give i [A, A ] q This proves the result Using the well-known connection between normal matrices N and pairs of commuting Hermitian matrices ReN, ImN, where ReN = 1 N + N and ImN = 1 i N N, one can easily deduce the following corollary Corollary 1 Let Hermitian matrices A, B C n n be given If there eist Hermitian matrices A et =, B A B et = of size n + q n + q that commute, then q ma{i + iba AB, i iba AB} Proof Let N = A + ib Calculating NN we get NN = A + iba ib = A iab+iba+b Now if we calculate N N we get N N = A +iab iba+b Thus NN N N = iba AB It then follows from Theorem 1 that q ma{i + iba AB, i iba AB} Net we eplore a class of matrices for which the equality nda = uda is true a 1 Proposition 1 Let A C n n be of the form A := with an 1 either a 1 = = a l > > a n 1 > or < a 1 < < a n l = = a n 1, where 1 l n 1 Then nda = uda = n l Proof Let α := a 1 = = a l = A By Proposition 54 in [6] we have that nda uda Recall that uda = rank A A A Thus α uda = rank a 1 α a n 1 α = rank α a l+1 α a n 1 = n l

4 MINIMAL NORMAL AND COMMUTING COMPLETIONS 53 Thus we have nda uda = n l To achieve a lower bound for nda, we notice that Theorem 1 yields nda ma{i + [A, A ], i [A, A ]} Observe that α [A, A ] = a l+1 a l a n 1 a n a n 1 Therefore i + [A, A ] = 1 and i + [A, A ] = n l Clearly, n l 1 and thus nda n l We conclude that nda = uda = n l The proof for the second part of the statement is similar We observe that the matrices in Proposition 1 are unitarily irreducible, as the following lemma shows Lemma 1 Let A C n n be strictly upper triangular with a i,i+1 for 1 i n 1 Then A is unitarily irreducible Proof Let A = U A1 U with U unitary, A A 1 C p p and A C q q with p + q = n We need to show that p = or q = Notice that n 1 = ranka = ranka 1 + ranka p + q = n So there eist two possibilities, either ranka 1 = p and ranka = q 1 or ranka 1 = p 1 and ranka = q Without loss of generality we assume that ranka 1 = p and ranka = q 1 Thus A 1 is invertible and ranka k 1 = p for all k As = ranka n ranka n 1 = p, it follows that p = 3 When the normal defect equals one Inspired by [4] we consider the case when the normal defect is equal to one For the class of matrices considered before we have the following observation a 1 Proposition Given a matri A of the form A :=, with an 1 a 1,, a n 1 C \ {} Then nda = 1 if and only if a 1 = = a n 1 =: α Furthermore, when n 4 all minimal normal completions have the form a 1 a n 1, with α = a n = a n+1 A et := a n a n+1

5 54 DP KIMSEY AND HJ WOERDEMAN Proof Observe that a 1 a a 1 [A, A ] =, a n 1 a n a n 1 which yields that ma{i + [A, A ], i [A, A ]} 1, and equality holds if and only if a 1 = = a n 1 Using Theorem 1 it follows that for nda = 1 it is necessary that a 1 = = a n 1 Let now a 1 = = a n 1, and put a 1 β 1 β A et := A β a n 1 =, γ δ β n γ 1 γ γ n δ where β := T β 1 β n and γ := γ1 γ n Let us assume that Aet is normal Then we see that AA + ββ = A A + γ γ, or equivalently [A,A ] = γ γ ββ Recall that a 1 [A, A ] = a n 1 = γ 1 γ 1 γ n γ n β 1 β n β 1 β n γ 1 β 1 γ 1 γ β 1 β γ 1 γ n β 1 β n γ γ 1 β β 1 γ β γ γ n β β n = γ n γ 1 β n β 1 γ n γ β n β γ n β n =: W = w ij n i,j=1 As a 1, a n 1 > we get that γ 1 = a 1 + β 1, β n = a n 1 + γ n From w ii = for i n 1, we see that γ i = β i for i =,, n 1 From w i1 = for i =,, n, we see that γ i γ 1 β i β 1 = This implies that γ i γ 1 β i β 1 = Since γ i = β i for i =,, n 1, we get γ i γ 1 β 1 = So either γ i = β i = or γ 1 = β 1 But γ 1 = β 1 since γ 1 = a 1 + β 1 > β 1 Therefore γ i = β i = for i =,, n 1

6 MINIMAL NORMAL AND COMMUTING COMPLETIONS 55 To find β 1, β n, γ 1 and γ n, we observe the following equation that results from A et being normal, Aγ + βδ = A β + γ δ Rewriting we see that γ a 1 1 β 1 + δ an 1 γ n β n β 1 γ 1 = a 1 + δ a n 1 β n γ n β 1 δ γ 1 δ a 1 β 1 Simplifying we get that + = + a n 1 γ n As n 4 we β n δ γ n δ see that 6 β 1 δ = γ 1 δ, β n δ = γ n δ, 7 a 1 β 1 =, 8 a n 1 γ n = From 7 we get β 1 = since a 1 = From 8 we get γ n = From 6 we get δ = since β 1 = and γ 1 Finally from the equations for w 11 and w nn we see that γ 1 = β n = α In conclusion, for n 4 the normality of A et implies that it is of the form a 1 9 A et := a n 1, a 1 = = a n+1 = a n a n+1 Clearly, for any n, if A et is as in 9, then A et is normal This proves the proposition It is worth mentioning that for n = or n = 3 the above proposition does 3 not describe the full situation For eample, consider A = and à := 1 1 Then A et := and Ãet := 1 also yield minimal normal completions for A and Ã, respectively Net we determine the eigenvalues of the completed normal matri A et from Proposition

7 56 DP KIMSEY AND HJ WOERDEMAN Lemma For the normal matri A et := α e iθ 1 α e iθ n 1 α e iθn α e iθn+1 all eigenvalues of A et are eactly λ k = α e i ψ+πk n+1 where ψ = θ θ n+1 and k =,, n Proof Computing the characteristic polynomial of A et we see that λ n+1 = α n+1 e iψ, where ψ = θ 1 + +θ n+1 Thus λ k = α e i ψ+πk n+1 with k =,, n λ n n k= λ λ k λ λ k λ n λ n+1 α n+1 e iψ = In [4] the following problem was considered Let λ,, λ n and µ 1,, µ n be two sequences of comple numbers When can one find a n + 1 n + 1 normal matri with eigenvalues λ,, λ n whose n n principal submatri has eigenvalues µ 1,, µ n? In [4] the author obtained the following result Define n j=1 λ := λ µ j n k= λ λ k Then there eists a normal matri A with spectrum λ,, λ n and with an n n principal submatri with spectrum µ 1,, µ n if and only if the rational function has only simple poles and Res λk λ, k =,, n If we take A et in Lemma we observe that A et has spectrum λ k = α e i ψ+πk n+1, where ψ = θ θ n+1 and k =,, n When we remove the last row and last column in A et the remaining matri has eigenvalues µ i =, i = 1,, n λ Thus by the result in [4] we must have that Res λk n is satisfied This is indeed true, since Res λk = lim λ n+1 α n+1 e iψ λ λk λ lim n +λ λ k nλ n 1 λ λk n+1λ = 1 n n+1, where we used L Hôpital s rule in the second equality 4 Commuting defect In [1] the following minimal commuting completion problem was introduced Given A 1,, A d C n n, how do we find A 1 et,, A d et C n+q n+q of smallest possible size with Ai A i et = such that [A i et, A j et ] =, i j We will restrict our investigation to completing only two matrices: given A, B C n n how do we find A et, B et of smallest possible size, with A A1 1 A et = A 1 A, B et = B B1 B 1 B such that [A et, B et ] = We shall call the smallest possible number q the commuting defect, and denote it cda, B Clearly, cda, B = if and only if [A, B] = As shown in [1], cda, B 1 rank[a, B] Indeed if [A et, B et ] = then AB BA = B 1 A 1 A 1 B 1 and thus rank[a, B] rankb 1 A 1 + ranka 1 B 1 q + q = q The results in [1] also show that cda, B n One easily sees this by taking A B B A A et = and B B A et = A B One useful observation for this problem is that if two square matrices C and D satisfy CD = αi for some α, then automatically CD = DC With this in,

8 MINIMAL NORMAL AND COMMUTING COMPLETIONS 57 mind we introduce the minimal inverse completion problem: Given A, B C n n, how do we find A et, B et C n+q n+q as in 1 of smallest possible size such that A et B et = αi n+q, for some α We shall call this smallest number q the inverse defect and denote it by ida, B The inverse defect of a pair of matrices is easily determined as the following theorem shows Theorem For A, B C n n, suppose α is the nonzero eigenvalue of AB with the highest geometric multiplicity Then ida, B = rankαi n AB Proof Let A et A? B? := and B?? et := eist such that A?? et B et = αi n+q, where α We notice that A et B et = αi n+q if and only if αin+q A rank et = n + q, B et I n+q αin which is obviously bigger than or equal to rank B A As the latter equals I n n + rankαi n AB by a Schur complement argument, we conclude that n + q n + rankαi n AB or q rankαi n AB as required Now let α be a nonzero eigenvalue of AB with highest geometric multiplicity or equivalently, the α so that rankαi n AB is minimal, and set αi n A? q = rankαi n AB We define W := αi q?? B? I n After a permutation similarity we arrive at W := αi n A??? I q αi q??? B I n By inspection we see?? I q that W is a partial banded matri with a pattern J, say Therefore we can apply Theorem 11 from [5] We now have that min rank W = ma T J min rank W T, where W T is the partial matri obtained from W by only keeping the known entries { that lie in the triangular } subpattern T This gives us that min rank W = αin A ma n + q, rank = n + q, by the choice of q Thus B I has a completion n of rank n + q and consequently we can find A et and B et such that A et B et = αi n+q Thus ida, B rankαi n AB This proves the proposition Let us outline how to find a minimal inverse completion Algorithm 1 Let A, B and α be as in Theorem 41 and put q = ida, B and p = n q First determine an invertible matri S so that SABS 1 αip P = for Q C11 C some P and Q of sizes p q and q q, respectively Write now SA = 1, C 1 C BS 1 D11 D = 1 with C D 1 D 11, D 11 of size p p and C, D of size q q Choose X and Y to be invertible matrices of size q q and let D et = D 11 D 1 D 11 C 1 + D 1 C Y 1 D 1 D D 1 C 1 + D C αiy 1 X XC Y 1

9 58 DP KIMSEY AND HJ WOERDEMAN and put B et = D et S I q, A et = αb 1 et Then one can check that A et and B et have the required form and obviously A et B et = αi Let us try the algorithm out on an eample 1 1 Eample 1 Let A = and B = Then AB = 1 eigenvalues ± Let us choose α = and S =, SA = Thus we get D et = 1 1 A et = 4 y, B et = 1 y y y y y has Then BS 1 = y y y 1 1 and As observed before, we have that cda, B ida, B In general there is no equality For instance, when A = and B = 1 we have cda, B = and ida, B = 1 That such a simple eample eists seems to be due to the fact that we are ecluding the possibility α = in the definition of ida, B; we do this since CD = does not imply DC = But now one can ask what happens when A and B are nonsingular Even in that case one can in general improve upon the estimate cda, B ida, B by doing the following Suppose there is an invertible matri S so that A 1 B 1 11 SAS 1 =, SBS 1 =, A d B d where A i and B i, i = 1,, d, are square matrices of the same nontrivial size Then completing A i and B i to A i et and B i et that commute for all i {1,, d} yields completions A et and B et for A and B, respectively, that commute Thus cda, B d i=1 cda i, B i d i=1 ida i, B i We are now led to the following question: Let A and B be nonsingular matrices so that for no invertible S we have that 11 holds with d Is it then true that cda, B = ida, B? The questions in this section may also be pursued in the class of real symmetric matrices In other words, let A and B be real symmetric and look for A et and B et that are also real symmetric As a comple symmetric matri N is normal if and only if the real symmetric matrices A = Re N and B = Im N commute, Corollary 1 applies The real symmetric case is of interest in deriving multivariable quadrature formulas; see [1] In their setting A and B have a tridiagonal block form and A et and B et are required to have this form as well For this reason it may not be optimal to look for A et and B et with A et B et = αi n+q We hope to further pursue the real symmetric case in a future publication Acknowledgments The authors are very grateful for the helpful suggestions that were provided by the anonymous referee In addition, we wish to thank Professor Ilya Spitkovsky for suggesting a more compact proof of Theorem 1 Both authors were partially supported by NSF grant DMS-5678 David P Kimsey performed the research as part of an REU project

10 MINIMAL NORMAL AND COMMUTING COMPLETIONS 59 References [1] I Degani, J Schiff, and D J Tannor Commuting etensions and cubature formulae Numer Math, 113:479 5, 5 [] P R Halmos Subnormal suboperators and the subdiscrete topology In Anniversary volume on approimation theory and functional analysis Oberwolfach, 1983, volume 65 of Internat Schriftenreihe Numer Math, pages Birkhäuser, Basel, 1984 [3] R A Horn and C R Johnson Matri analysis Cambridge University Press, Cambridge, 199 Corrected reprint of the 1985 original [4] S M Malamud Inverse spectral problem for normal matrices and the Gauss-Lucas theorem Trans Amer Math Soc, 3571: electronic, 5 [5] H J Woerdeman Minimal rank completions of partial banded matrices Linear and Multilinear Algebra, 361:59 68, 1993 [6] H J Woerdeman Hermitian and normal completions Linear and Multilinear Algebra, 43:39 8, 1997 [7] H J Woerdeman The separability problem and normal completions Linear Algebra Appl, 376:85 95, 4 Department of Mathematics, Dreel University, Philadelphia, PA 1914, USA dpk7@dreeledu and hugo@mathdreeledu