THE REAL POSITIVE DEFINITE COMPLETION PROBLEM. WAYNE BARRETT**, CHARLES R. JOHNSONy and PABLO TARAZAGAz

Transcription

1 THE REAL POSITIVE DEFINITE COMPLETION PROBLEM FOR A SIMPLE CYCLE* WAYNE BARRETT**, CHARLES R JOHNSONy and PABLO TARAZAGAz Abstract We consider the question of whether a real partial positive denite matrix (in which the specied o-diagonal entries consist of a full n cycle) has a positive denite completion This lies in contrast to the previously studied chordal case We give two solutions In one, we describe about n independent conditions on angles associated with a normalization of the data that are necessary and sucient The second is more computational and allows presentation of all positive denite completions, as well as answering the existence question Introduction A real partial matrix A is one in which some entries are specied real numbers and the remainder are unspecied, ie, free variables over the real numbers We say that A is partial symmetric if A is square, a ji is specied whenever a ij is, and a ji = a ij We shall assume throughout that the diagonal entries of A are specied An example is () A =?? in which the?'s indicate unspecied entries A partial positive denite matrix is a partial symmetric matrix each of whose specied principal submatrices is positive denite (By a specied portion of a partial matrix we always mean one composed entirely of specied entries) Partial positive semidenite matrices are dened similarly The matrix above is not partial positive denite, but is if the, entry is replaced, for example, by A completion of a partial matrix is a specication of the unspecied entries resulting in a conventional matrix, and the positive denite completion problem is to determine if a positive denite completion exists or to nd a completion of a partial positive denite matrix that is positive denite For example, is a positive denite completion of?? *This manuscript was prepared while the rst two authors were visitors at the Institute for Mathematics and its Applications, Minneapolis, Minnesota **Dept of Mathematics, Brigham Young University, Provo, Utah 80 ydept of Mathematics, The College of William and Mary, Williamsburg, Virginia 8 The work of this author was supported in part by National Science Foundation grant DMS and oce of Naval Research Contract N J-9 zdept of Mathematics, University of Puerto Rico, Mayaguez, Puerto Rico 0009 This author was partially supported by NSF grants RI and EPSCoR of Puerto Rico

2 For convenience, we consider the positive denite and positive semidenite completion problems interchangeably As observed in [GJSW], they are equivalent Not all partial positive semidenite matrices have a positive semidenite completion An example is: () B =???? Let x be the, entry In order that B have a positive semidenite completion we must choose x so that the minors x x ; x x obtained by deleting the last row and column, and by deleting the second row and column are nonnegative But the rst equals jx j which forces x =, while the second equals jx +j which forces x = This conict precludes the possibility of a positive denite completion The existence of a positive denite completion of a partial positive denite matrix A depends on the pattern of specied entries of A and can be most easily described in terms of its undirected graph Definition Let A be a partial positive denite n-by-n matrix The undirected graph G =(N;E) of A has node set N = f; ; ;ng and an edge fi; jg E,i=j,if and only if a ij is specied For example, the undirected graphs of A and B given by () and () above are and () respectively We recall that the graph G is connected if there is a path between any twovertices in N, and that G is chordal if it has no minimal simple circuit of four or more edges Akey result is

3 Theorem [GJSW] Every partial positive denite (semidenite) matrix with graph G has a positive denite (semidenite) completion if and only if G is chordal The graph G in () is the simplest non-chordal graph and the matrix B illustrates that not all partial positive semidenite matrices with pattern G have a positive semidenite completion Our purpose in this paper is to give two solutions to the positive denite completion problem in the case that G is the simple cycle C n =(N;E); E = ff; g; f; g; ;fn ;ng;fn; gg: Thus, we are given () A = satisfying () where a n+ denotes a a b b n b a b? b a? b n b n b n a n ) a i > 0 a i a i+ b i > 0 i =;;n Letting D = diag(= p a ; = p a ; ;= p a n ), the congruence C = DAD gives C = b p a a p b a a bn p a an p b a a? p b a a? bn p a an bn p an an bn p an an

4 in which all o-diagonal entries have modulus less than one Since A has a positive denite completion if and only if C has, it suces to consider the question: When does the matrix () C = c c n c c? c admit a positive denite completion? c n c n c n ; jc ij < ; i N It is helpful to rst consider the case n = separately Then () C = and its graph G is c? c c c?? c c c? c : If the, and, entries of C can be chosen so that C is still partial positive denite, G becomes the chordal graph It then follows from theorem that the, and, entries can be chosen to make C positive denite Therefore, C has a positive denite completion if and only if there exists a real x for which c x c c x c These inequalities may be rewritten > 0 and x c x c c c > 0: ( c )( c ) (c c x) > 0; and ( c )( c ) (c c x) > 0;

5 which in turn may be written (8) jx c c j < q ( c )( c ); jx c c j < q ( c )( c ): There is an x satisfying the inequalities (8) if and only if q (9) jc c c c j < q( c )( c )+ ( c )( c ): Squaring and simplifying gives (0) i= c i < +c c c c + Y i=! ( c ) i : We record this as: Proposition This matrix C given by () has a positive denite completion if and only if the inequality (0) holds Parameterization of completable cycles Fiedler [F] has already given necessary and sucient conditions for the matrix C given by () to have a positive denite completion However, his result is not phrased in the language of partial positive denite matrices and completion problems, and to the best of our knowledge predates other work on matrix completion problems It is geometrical in nature and is obtained by viewing the matrix C as a Gram matrix of n unit vectors Our purpose is to give an alternate view and a combinatorial extension to his results and to discuss its implications We express each c i in () as c i = cos i, i (0;), i = ;;n, or i [0;], i = ;;n when considering the positive semidenite completion problem Note that the i are uniquely determined This parameterization of the c i enables one to state criteria for existence of a positive denite completion of C in terms of linear inequalities on the i We will need the following result concerning -by- positive semidenite matrices, which is of independent interest Proposition Let 0, ; Then the matrix C = is positive semidenite if and only if cos cos cos cos cos cos + ; + ; + ; and + + :

6 Furthermore, C is singular if and only if one of these inequalities is an equality () Proof We have the following chain of equivalences: () () () () () () () () () C is positive semidenite det C 0 + cos cos cos cos cos cos 0 cos cos + cos cos cos cos cos cos + cos cos ( cos )( cos ) (cos cos cos ) sin sin jcos sin sin cos cos cos cos cos j cos cos sin sin sin sin cos cos cos + sin sin cos( + ) cos cos( ) cos( + ) cos cos j j Now 0 + and 0, j j Since cos x is decreasing on [0;] and increasing on [;], () holds if and only if () j j + which is equivalent to the four inequalities: or in other words ; ; + ; and + () + ; + ; + ; and + + : Finally, C is singular if and only if one of the inequalities in () is an equality, ie, if and only if one of the equalities in () is an equality

7 The forward implication in Proposition is intuitively clear if one thinks of C as the Gram matrix of three unit vectors u ;u ;u in R, and ; ; as the angles between the pairs of vectors fu ;u g, fu ;u g, and fu ;u g respectively Then C is singular if and only if u ;u and u are coplanar, that is if and only if one of the inequalities in () is an equality Proposition makes it trivial to construct singular -by- positive semidenite matrices For instance, = = = gives and =, =, = gives p p 0 0 We also note that proposition parameterizes all real positive semidenite matrices up to diagonal congruence We know of no result analogous to proposition for complex positive semidenite matrices, which is the reason we have restricted ourselves to real matrices We now give criteria for a partial positive semidenite matrix whose graph is a simple cycle to have a positive semidenite completion The corresponding results for positive denite matrices are obtained by replacing all inequalities by strict inequalities We begin with n = : Theorem Let 0 ; ; ; Then the matrix C = cos? cos cos cos?? cos cos cos? cos has a positive semidenite completion if and only if () k i= i + k k=;; Proof As in the argument preceding proposition, C has a positive semidenite completion if and only if there is a [0;] such that when cos is substituted for the, and, entries, the principal submatrices cos cos cos cos cos cos ; cos cos cos cos cos cos

8 are positive semidenite By proposition this occurs if and only if or equivalently 9 >= >; 8 >< >: + + which is equivalent to the following inequalities Eight of these inequalities hold automatically The others are: () Therefore C has a positive semidenite completion if and only if the inequalities () hold 8

9 For example,???? ( = = = = ) has a positive semidenite completion, but???? ( = = =0; =) does not Theorem seems much more understandable than the equivalent result, proposition We note that proposition (n = ) and theorem (n = ) are similar in form although only the latter is concerned with a completion problem Apparently, there will be n linear constraints on the i in general, and if the inequalities are presented so that the coecient ofeach i is, there will be an odd number of i on the left hand side of each inequality The following gives necessary and sucient conditions for general n matrix Theorem Let n, let N = f; ; ;ng, and 0 ; ;; n Then the C = cos cos n cos cos? cos? cos n cos n cos n has a positive semidenite completion if and only if for each S N with jsj odd, () is i (jsj ) + is c i : Proof The proof is by induction on n If n=, the inequalities () agree with () and the result follows by theorem Suppose that the theorem holds for n We examine the situation which occurs when cos is substituted for the ;n and n ; entries The graph is then the simple cycle C n with the edge f;n g added 9

10 () Now (;;;;n ;) is a cycle in this graph and C[f;n ;ng] is a fully specied principal submatrix Thus, in order that the original partial matrix C have a positive semidenite completion, it is necessary (by the inheritance property for principal submatrices) that there exist a [0; ] such that B = cos cos cos?? cos n cos cos n has a positive semidenite completion, and E = cos cos n cos cos n cos n cos n is positive semidenite Conversely, if Bhas a positive semidenite completion e B and E is positive semidenite, substitution of the entries from the completion e B into their corresponding positions in C gives a partial positive semidenite matrix C e whose graph is obtained from () by adding all missing edges with vertices in f; ;n g It is easy to see that this is a chordal graph and so by theorem, C, e and hence C, has a positive semidenite completion Thus, the matrix C has a positive semidenite completion if and only if B has a positive semidenite completion and E is positive semidenite 0

11 By the induction hypothesis and theorem, these occur if and only if (8) (9) is i (jsj ) + is c i + for every S f;;n g;jsj odd ; i + jtj+ i it itc for every T f;;n g;jtj even ; (0) () () () + n n n + n + n n + n + n Equivalently, () i i (jsj ) is is c for every S f;;n g;jsj odd () () () n n n n i i + jt j it c it for every T f;;n g;jtj even ; (8) (9) n + n n n

12 Equations (){(9) are equivalent to the 9 inequalities (0) () () () () () () () (8) is is is i i i For every S f;;n g;jsj odd and for every T f;;n g;jtj even i (jsj ) is c is c i (jsj ) n + n i i + jt j it c it i (jsj ) n n is c n n i + jt j it c i n n n + n it n n n n n n n n n + n i i + jt j it c it n n n n Inequalities (), (), () and (8) hold automatically written The inequalities (0) can be (9) is i + it i (jsj + jt j ) + is c i + it c i for every S f;;n g;jsj odd and every T f;;n g;jtj even : Fix S and T Since jsj is odd and jt j is even, S = T Therefore, either S \ T c = or S c \ T = Assume S \ T c = and let k S \ T c Then k occurs in both sums, P is i and P it c i Subtracting k from both sides of (9), (0) is fkg i + it i (jsj + jt j ) + is c i + Since there are jsj + jt j terms i on the left hand side of (0) and each is bounded above by, the inequality (0) holds automatically If S c \ T =, the same type of argument applies It follows that the inequalities (9), and hence (0), hold automatically Therefore, A has a positive semidenite completion if and only if the inequalities (), it c fkg i :

13 (), () and () hold These are: () () () () is is it it i (jsj ) + is c i + n + n for every S f;;n i + n g;jsj odd + n (jsj +)+ is c i for every S f;;n i + n jtj+ for every T f;;n i + n jtj+ But (){() are exactly equivalent to and this completes the proof it c i + n it c i + n for every T f;;n is i (jsj ) + is c i for every S N;jSj odd g;jsj odd g;jtj even g;jtj even Since the inequalities () are unchanged under an arbitrary permutation of ; ; n, wemay assume that the i 's are relabeled so that Then C has the form () C = n : cos i cos in cos i cos i? cos i? cos in cos in cos in ; and is i (jsj ) + is c i

14 holds for all S with jsj = k if and only if Thus, we have k i= i (k ) + Corollary Let n and 0 n n n i=k+ i : Then the matrix C in () has a positive semidenite completion if and only if for k odd, k N, () k i= i (k ) + n i=k+ Thus, the n inequalities in () of theorem can be replaced by about n inequalities, although ; ; n are no longer completely symmetric in () In certain cases it suces that the rst inequality in () hold For example, if P P P then k i = + k i + k (k ) for k In this case, C has i= i= = k+ i= a positive semidenite completion if and only if convenience Corollary Let n, 0 ; ; n ; ; n is greater than Then C = i : n P i= cos cos n cos cos? cos? cos n cos n cos n has a positive semidenite completion if and only if Corollary Let () C = max kn k c c n c c? c? c n c n c n i We restate this slightly for and assume that at most one of n k= k : ; jc ij; in:

15 If two or more of the c i are 0, then C has a positive semidenite completion Proof Writing c i = cos P i, i =;;n,wehave j = k = for some j; k N, j = k P Let S N If j; k S, i = + is P i +(jsj ) =(jsj ) If is fj;kg P j S and k S c, or vice versa, i P +(jsj ) (jsj ) + i Finally, if is is P c j; k S c, i jsj=(jsj ) + (jsj ) + i The result follows from is is c theorem Corollary may also be proved directly as in [F, p ] We now derive Fiedler's main result Corollary The matrix C given by () has a positive semidenite completion if and only if max kn arccos jc k j knarccos jc k j for c c c n >0 and kn arccos jc k j for c c c n 0: Proof P If two or more of the c k are 0, then C has a positive semidenite completion and arccos jc k j = kn So assume at most one c k = 0 Since P T CP has a positive semidenite completion for any cyclic permutation P if and only if C does, we may assume that c ;c ;;c n are nonzero Let D = diag(; sign c ; sign(c c ); ;sign(c c n )) Then C has a positive semidenite completion if and only if DCD = jc j "jc n j jc j jc j? jc j? jc n j "jc n j jc n j " = sign(c c c n ), has one Take " =0ifc n =0 If " > 0, by corollary P, C has a positive semidenite completion if and only if max kn arccos jc k j arccos jc k j If " 0, then arccos( jc n j)= arccos jc n j and, kn again by corollary, C has a positive semidenite completion if and only if np P arccos jc k j,or arccos jc k j This completes the proof k= kn ; arccos jc n j Of course, Corollary gives a complete characterization for the positive semidenite completion problem in the case that the undirected graph is a simple cycle, as does theorem

16 However, we believe that theorem provides additional insight not available from corollary It is also dicult to see how theorem might be proved from corollary We give two more corollaries to illustrate the utility of theorem and its corollaries Corollary Let n and jcj Then C = c c c c? c? c c c has a positive semidenite completion for all c [ ; ] if n is even and for c cos n n if n is odd Proof Write c = cos By corollary, C has a positive semidenite completion if and only if k (k ) +(n k) for k odd, k N Equivalently, (k n)(k ), k odd, k N, or k for k odd, k N, and k >n Since (k )=(k n) is k n decreasing, this is n for n odd n for n even Thus C has a positive semidenite completion if and only if c [ c cos n for n odd n In particular, the partial matrix?? has a positive semidenite completion, namely, if n is even and no positive semidenite completion if n is odd We end with the case of a partial positive semidenite Toeplitz matrix ; ] for n even and

17 Corollary Let n and ; [0;] Then C = cos cos cos cos? cos? cos cos cos has a positive semidenite completion if and only if (n ) (n ) + for n even and (n ) (n ) for n odd : Proof Note that () holds automatically if occurs on both sides of the inequality Thus, we need only consider the cases in which S has minimal or maximal cardinality If jsj =, the required inequality is (n ) If n is even and jsj = n, we need (n ) (n ) +, while if n is odd we require (n ) + (n ) This completes the proof An alternative solution As we saw in the introduction, the partial matrix C corresponds to the graph C n Even though the way we labeled the nodes in C n is natural, it is not the only way We relabel the nodes to produce an alternate pattern for the matrix associated with a simple cycle Given the cycle, if n is even we can label the nodes as follows:

18 For n odd the labeling is Now the associated matrix is (8) D = 0 d d d? d d??? d? : :?? d n?? d n d n? d n d n d n C A ; jd i j < ;i N and its pentadiagonal structure is clear The unspecied entries inside the pentadiagonal structure are the only ones that concern us; call them x i, i =; ;n We use Theorem [GJSW] to decide if D has a positive denite completion and to see how tochoose the x i, i =; ;n in this event First consider the case n =, which illustrates all important features of the problem The corresponding matrix is (9) D = 0 We proceed as we did to obtain Proposition d d?? d x d? d x x d? d x d?? d d From Theorem [GJSW] this matrix D has a positive denite completion if and only if there exist x and x such that jx j < ; jx j < and such that the three -by- 8 C A

19 contiguous principal minors are positive For then, the matrix is partial positive denite and its associated graph is chordal We can transform the conditions given in the statementaboveby computing the -by- determinants The rst one is given by det d d d x d x and from (8) we obtain the inequality (0) jx d d j < We will denote by I the set of solutions of (0) The second determinant down the diagonal is det x d x x d x and the associated inequality is = det d x d d x d q ( d )( d ): =+d x x x x d which may be written as the quadratic form x +x d x x < d ; () (x ;x ) d d x x < d : The last determinant is det x d x d d d = det d x d d x d and again from (8) we obtain the inequality q () jx d d j < ( d )( d ): and we denote by I the interval of solutions Now we can rewrite our statement as the following result 9

20 Proposition The matrix D given by (9) has a positive denite completion if and only if there exists a solution for the quadratic system () 8 >< >: jx d d j < p ( d )( d ) (x ;x ) d d x x < d jx d d j <p ( d )( d ): At this point, we make some observation about the inequalities that remain true for the general case Clearly the rst and the last inequalities are similar and the solution sets are intervals contained in [-,], in other words d d d )( d) <x <d d + ( q d )( d) q( d d q( d )( d) <x <d d + ( d )( d) The quadratic inequality in the middle has an associated quadratic form For the corresponding matrix +d and d are the eigenvalues and the eigenvectors are p p (= ; = ) T and p p (= ; = ) T Therefore the admissible solutions to the inequality are those (x ;x ) lying inside an ellipse: q The equation of this ellipse is x + x d x x = d : 0

21 We can solve, for example, for x the quadratic equation x (d x )x +(x +d )=0 obtaining x = d x q ( d )( x ): For x = we obtain the tangency points, (d ; ) and ( d ; ), between the ellipse and the lines x = : A similar analysis gives us the other tangency points (;d ) and ( ; d )between the ellipse and the lines x = Observe that in the picture we assumed d > 0; for d =0we obtain a unit circle and for d < 0 the result is an ellipse rotated 90 degrees Finally note that in the general case each variable only appears in two inequalities This fact suggests an easy algorithm to decide if the system has a solution and to compute solutions To solve the system means to nd x I and x I such that the point (x ;x ) is inside the ellipse We describe an algorithm that can be extended to the general case Given an interval I [ ; ] we dene as follows (I) =fx : there exist x I such that (x ;x ) satises ()g: It is easy to verify that for any interval I [ for >0 small enough ([ ; ]) [ ; ] ; ], (I) is a nonemptyinterval Moreover Even though we will not do it here, it is not a dicult task to give an explicit denition of Applying this solvability condition to the system () we have an alternate version of Proposition Proposition The matrix D has a positive denite completion if and only if (I ) \ I = ;: Proof We have that x (I ) \ I if and only if x I and there exists x I such that (x ;x ) satises () if and only if (x ;x ) is a solution of system () Consider an example We will denote by ^D the matrix ^D = 0 : :?? : x :? : x x :? : x :?? : : C A

22 For this example the system () becomes () 8 >< >: jx :j <: x + x x x <: jx :j <: For every x ( ; ) we dene e(x )=fx : (x ;x ) satises the second inequality of ()g: Observe that for this example I = I =( :;) It is easy to see that for each x ( :; ), e(x ) \ ( :; ) = ; which clearly implies ( :; ) \ ( :; ) = ;; which says that there exists a positive denite completion Also, we may easily display the set of all solutions, i e, every pair (x ;x ) that allows a positive denite completion Given x ( :; ) and x e(x ) \ ( :; ), then (x ;x ) allows a positive denite completion In order to see the relation with Theorem, we can rewrite our matrix ^D to generate ^C dened by 0 :?? : : :?? ^C = B? : C?? : :A : :?? : Because cos( )=:, we have = = = For our example N = f; ; ;g and given S N we have to consider S such that jsj is equal to, or Since every i is equal the set of inequalities () is very simple jsj =: jsj=: jsj=: < <+ < The existence of a positive denite completion is assured, but specic information about the solution set was readily obtained in the alternate approach Now we have all the elements to make a straightforward generalization about a positive denite completion of the matrix D, but rst we will generalize the notation The unspeci- ed entries in the position (; ); (; ); ;(n ;n ) will be denoted by x ;x ; ;x n

23 Now wehaven contiguous principal minors of size that have to be positive The rst and the last one yield linear inequalities, and those in the middle yield quadratic inequalities like the second one given in system (); in other words we obtain () (x i ;x i+ ) di+ d i+ xi x i+ < d i+ i =; ;n : Then these positive minors are equivalent to the following system () 8 >< >: jx d d j < p ( d )( d ) [x i ;x i+ ] d i+ d i+ jx n d n d n j< xi x i+ q ( d n )( d n): < d i+ i =; ;n : Associated with each quadratic inequality we have a -function dened as follows For every I [ ; ] i+ = fx i+ : there exist x i I such that (x i ;x i+ ) satises ()g for every i =; ;n Because every i+ ; i =; ;n satises the properties pointed out after the denition of,we can apply them sequentially since i+ (I) i =; ;n is always an interval contained in [-,] We now take I to be the interval dened by the third equation in () We can now establish the main result of this section Theorem Given the matrix D, the following statements are equivalent i) D has a positive denite completion ii) The system () has a solution iii) n (( (I ))) \ I = ; The proof of this theorem follows the proofs of propositions and Acknowledgement We thank Miroslav Fiedler for making us aware of his paper and for several helpful discussions concerning it We also appreciate some useful conversations with Michael Lundquist REFERENCES [F] Miroslav Fiedler, Matrix Inequalities, Numerische Mathematik 9 (9), pp 09{9 [GJSW] R Grone, CR Johnson, E Sa, and H Wolkowicz, Positive Denite Completions of Partial Hermitian Matrices, Linear Algebra and its Applications 8 (98), pp 09{