Piecewise Certificates of Positivity for matrix polynomials
aa r X i v : . [ m a t h . R A ] J a n PIECEWISE CERTIFICATES OF POSITIVITY FOR MATRIXPOLYNOMIALS
RONAN QUAREZ
Abstract.
We show that any symmetric positive definite homogeneous ma-trix polynomial M ∈ R [ x , . . . , x n ] m × m admits a piecewise semi-certificate,i.e. a collection of identites M ( x ) = P j f i,j ( x ) U i,j ( x ) T U i,j ( x ) where U i,j ( x )is a matrix polynomial and f i,j ( x ) is a non negative polynomial on a semi-algebraic subset S i , where R n = ∪ ri =1 S i . This result generalizes to the settingof biforms.Some examples of certificates are given and among others, we study a vari-ation around the Choi counterexample of a positive semi-definite biquadraticform which is not a sum of squares. As a byproduct we give a representationof the famous non negative sum of squares polynomial x z + z y + y x − x y z as the determinant of a positive semi-definite quadratic matrix poly-nomial. Introduction
The Hilbert 17-th problem asks if a non negative polynomial f ( x ) ∈ R [ x , . . . , x n ]is a sum of squares. The answer given by Artin provides an identity with denomi-nators (or rational functions), namely there are two sums of squares of polynomials q ( x ) and p ( x ) such that q ( x ) f ( x ) = p ( x ).Such an identity is a called Positivestellensatz and can be seen as a certificate which algebraically proves the positivity of the polynomial f ( x ). For a generalpolynomial, the denominator q ( x ) is necessary.One may also consider a ”relative” version, namely when f ( x ) is non negative ona basic semi-algebraic subset S = { x ∈ R n | f ( x ) ≥ , . . . , f r ( x ) ≥ } . This searchof certificates received a lot of contributions especially for those semi-algebraicsubsets S where the Positivestellensatz is denominator free. Essentially it happensto be possible when f ( x ) is positive and when the description of S satisfies somearchimedean property (which can be seen as a “strong” compactedness assumption).Among all the available references, we mention the seminal works of Schm¨udgen[10] and Putinar [8].Another natural and often difficult question is, when it is possible to obtain acertificate of positivity, to study its complexity, i.e. the number of squares that areneeded.In this paper, we are mainly interested in matrix polynomials, namely polyno-mials whose coefficients are matrices. Let x = ( x , . . . , x n ) and R [ x ] m × m be the set Date : November 20, 2018.2000
Mathematics Subject Classification.
14P ; 15A.
Key words and phrases.
Biforms ; Matrix polynomials ; Positive semi-definite ; Positivitycertificate ; Sum of Squares. of all square matrix polynomials of size m with entries in R [ x ]. In that context,by a square we mean a expression of the form U ( x ) T U ( x ) where U ( x ) ∈ R [ x ] m × m ,sometimes we will also use the terminology of hermitian square.Given a symmetric matrix polynomial M ( x ), the natural question is the following: Assume that M ( x ) is positive semi-definite ( psd in short) for all x ∈ R n . Is M ( x )a sum of hermitian squares ? Or in otherwords, do we have a positivity certificatefor M ( x ) ?It is well known that, when n = 1, then M ( x ) is a sum of two squares (see forinstance [4] for a proof), i.e. M ( x ) = U ( x ) T U ( x ) + V ( x ) T V ( x ) where U ( x ) , V ( x ) ∈ R [ x ] m × m .There is also another case where we have a positive answer : the case of bi-quadratic forms in dimension 2. Namely, if m = 2 and if the entries are homoge-neous polynomials of degree 2, then M ( x ) is also a sum of squares.In fact this last case can be seen as a particular case of the homogenized versionof the previous one (see [2] for this fact and a proof).Unfortunaltely these are essentially the only cases where such a certificate existsfor a matrix polynomial. For instance, it becames false even for biquadratic formsin 3 variables and dimension 3 (confer the Choi counterexample in [1]).We may also look at the relative situation : if M ( x ) is positive definite for all x in a semi-algebaric subset S ⊂ R n whose description satisfies an archimedeanproperty, then we get a similar certificate as in the polynomial case. This has beenproved by Hol and Scherer in [6] ; see also [7] for a different and more algebraicproof.In this paper, we are searching for certificates of positivity without denominators for matrix polynomials. We first study certificates which are sums of squares ofmatrix polynomial weighted by a psd (scalar) polynomial. More precisely, given apsd matrix polynomial M ( x ), we wonder if M ( x ) can be written as a finite sum(1) M ( x ) = X i f i ( x ) U i ( x ) T U i ( x )where U i ( x ) ∈ R [ x ] m × m and f i ( x ) is non negative polynomial on R n .We call it a semi-certificate since, roughly speaking, we have a certificate whichis a hermitian sum of squares weighted by some coefficients which are non negativepolynomials.Of course such a certificate may gives some others which will be totally alge-braic by using usual certificates for the psd polynomials f i ( x ). Beware that usingdenominators at this point could break any interest since we may readily get onesuch certificate with denominators simply by using the Gauss algorithm (indeed,it suffices to inverte the m − S whose descip-tion is archimedean would produce a usual denominator free algebraic certificate forthe matrix polynomial on S , using for instance Schm¨udgen or Putinar certificatesfor all the f i,j ’s.Unfortunately such a semi-certificate as in (1) does not exist in general, even fora positive definite matrix polynomial, as it is illustrated in section 3. IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 3
Nevertheless, noticing that one may obtain such certificate locally, we introducethe notion of piecewise semi-certificate, i.e. a collection of semi-certificates as in (1)with respect to a semi-algebraic covering of R n .We show in section 4, that such a certificate exists for any positive definite homogeneous matrix polynomial. We show also that we may translate the notionof semi-certificate and our related results to the case of biforms.Some examples and counterexamples are given and for instance, in section 5, westudy some variations around the Choi counterexample given in [1].As a byproduct we give a representation of the famous psd non sum of squarespolynomial x z + z y + y x − x y z as the determinant of a psd quadraticmatrix polynomial.Of course, a lot a work have to be done to better understand the frameworkof semi-certificate for psd non definite matrix polynomials. Moreover, concerningcomplexity, one may be interested in estimating the lenght of the hermitian sumof squares representation, and also in the number of pieces of the semi-certificate.Since we use a compactedness argument, no bound on the number of pieces of thesemi-algebraic covering can be derived by our method.2. Notations
Matrix polynomials and biforms.
In all the paper, we consider a set ofvariables x = ( x , . . . , x n ) and the polynomial ring R [ x ] whose associated field ofrational functions is R ( x ).A matrix polynomial A ( x ) is just a matrix whose entries are polynomials. Forinstance, for a square matrix polynomial of size m in the variables x , we write A ( x ) ∈ R [ x ] m × m .A form of type ( n, d ) will denote an homogeneous polynomial of degree d in n variables. A biform of type ( n , d ; n , d ) will denote a homogeneous polynomialin n + n variables which is of degree d with respect to a set of n variables andof degree d with respect to the other set of d variables.To a matrix polynomial A ( x ) = ( a i,j ( x )) ∈ R [ x ] m × m whose entries are all formsof type ( n, d ), we may canonically associate a biform of type ( n , d ; m,
2) : f A ( x, y ) = X ≤ i,j ≤ m a i,j ( x ) y i y j . We say that a biform of type ( n , d ; n , d ) is positive definite if f ( x, y ) > x, y ) ∈ S n − × S n − , where S k − denotes the k dimentional unit sphere in R k . Equivalentely, we may alternatively consider the product of projective spacesinstead of unit spheres.If A ( x ) is a symmetric matrix polynomial in R [ x ] m × m whose entries are all formsof type ( n, d ), we say that A ( x ) is positive definite if the associated biform f A ( x, y )is positive definite. We may also note that A ( x ) is psd if and only if f A ( x, y ) is psdon the product of unit spheres.2.2. Sum of squares and Hilbert 17th problem.
Note that sums of hermitiansquares when dealing with matrix polynomials corresponds to usual sum of squaresin the terminology of biforms.
RONAN QUAREZ
Let A ( x ) be a symmetric matrix polynomial in R [ x ] m × m which is positive semi-definite for all substitution of x ∈ R n . Then, the Gauss reduction algorithm gives asolution with denominators to the Hilbert 17-th Problem for a generic matrix poly-nomial : there exist r ∈ N , U ( x ) , . . . , U r ( x ) ∈ R [ x ] m × m and f ( x ) . . . , f r ( x ) , f ( x ) ∈ R [ x ] psd polynomials such that(2) f ( x ) A ( x ) = r X i =1 f i ( x ) U i ( x ) T U i ( x )where f ( x ) can be choosen as the product of the m − A ( x ) whenthey do not identically vanish. Moreover, by multiplying the identity (2) by thedenomiators appearing in a positivity certificate for the f i ’s, we may assume thatthe polynomials f i and f are sum of squares.In our paper, we are mainly interested in positivity certificates without denom-inators, namely when we may choose f ≡
1. Hence, we first look for matrixpolynomials which can be written as in (1) : A ( x ) = r X i =1 f i ( x ) U i ( x ) T U i ( x )where the f i ’s are non-negative polynomials (we cannot ask for them to be sum ofsquares since denominators are already needed for polynomials).One may want to measure the complexity of the certificate given in (1), simplyby counting the number of squares : r . But, one may also prefer another way ofcounting.Noticing that U T U = U T ( E + . . . + E m ) U = ( E U ) T ( E U )+ . . . +( E m U ) T ( E m U )where E i is the diagonal matrix whose only non null entry is 1 at the i -th posi-tion onto the diagonial, we may prefer to count “rank-one” sums of squares, i.e.when U ( x ) has rank one in R ( x ) m × m . The advantage of that count is that it coin-cides with the number of needed squares when we view our matrix polynomial asa biform. 3. Semi-certificates
The following examples shows that it is not possible in general to hope for (1).3.1.
Some examples.Example 3.1.
Let : M = (cid:18) x xyxy x + y (cid:19) Of course M is psd for all ( x, y ) ∈ R .To show that M cannot be written as in (1), we may proceed by applying tothe biform f M which is canonically associated to M ) the technology of cages andGram matrices developped in [3]. But we prefer to produce here an elementary andself-contained argument.Assume that M = P ri =1 f i U Ti U i , where U i ∈ ( R [ x, y ]) × and f i ≥ R . IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 5
By decomposing into rank one sum of squares, we may assume that U Ti U i = (cid:18) a i a i b i a i b i b i (cid:19) , where a i , b i ∈ R [ x, y ] . Identifying the (1 , P ri =1 f i a i = 1 + x . Thus, f i does not dependon y : f i ∈ R [ x ]. Since f i ≥ x ∈ R , we deduce that f i is a sum of twosquares in R [ x ]. Thus, if we increase the integer r and if we change the U i ’s, wemay assume that f i = 1. Identifying the entries, we get P ri =1 a i = 1 + x P ri =1 b i = x + y P ri =1 a i b i = xy By a simple degree consideration, the polynomial b i has necessarily the form b i = α i x + β i y , with α i , β i ∈ R . Likewise, the polynomial a i has necessarily the form a i = γ i + δ i x with γ i , δ i ∈ R . It follows a contradiction with the identity P ri =1 a i b i = xy .The previous pathology is not due to the fact that the matrix M is only psd, asshown by considering the following variation : Example 3.2.
Let N = (cid:18) x + ǫ ( x + y ) xyxy ǫ (1 + x ) + x + y (cid:19) where ǫ >
0. This is a positive definite matrix polynomial on all R (and evenat infinity, i.e. the associated homogeneized matrix polynomial remains positivedefinite).Assume that N can be written as in (1) : N = r X i =1 f i ( x, y ) (cid:18) A i A i B i A i B i B i (cid:19) + r X j =1 g j ( x, y ) (cid:18) U j U j V j U j V j V j (cid:19) + P rk =1 (cid:18) R k R k S k R k S k S k (cid:19) where the f i ’s are polynomials of degree at most 4, the g j , R k , S k ’s are polyno-mials of degree at most 2, the U j , V j ’s are polynomials of degree at most 1 andthe A i , B i ’s are constant. Moreover, the polynomials f i , g j are assumed to be nonnegative on R . Since their degrees are at most 4 and the number of variables is 2,we know that they are sum of squares of polynomials. Hence, we may assume that(3) N = r X i =1 (cid:18) R i R i S i R i S i S i (cid:19) where R i = a i + b i x + c i x + d i y + e i y + f i xy,S i = α i + β i x + γ i x + δ i y + ν i y + µ i xy, ( a i , b i , c i , d i , e i , f i , α i , β i , γ i , δ i , ν i , µ i ) ∈ R . RONAN QUAREZ
Let us consider the following vectors in the usual euclidien space R r :¯ a = ( a i ) , ¯ b = ( b i ) , ¯ c = ( c i ) , ¯ d = ( d i ) , ¯ e = ( e i ) , ¯ f = ( f i ) , ¯ α = ( α i ) , ¯ β = ( β i ) , ¯ γ = ( γ i ) , ¯ δ = ( δ i ) , ¯ ν = ( ν i ) , ¯ µ = ( µ i ) . Identifying the non diagonal entries in (3), we get(4) ¯ b · ¯ δ + ¯ β · ¯ d + ¯ a · ¯ µ + ¯ f · ¯ α = 1where · denotes the usual inner product on R r .By identifying the diagonal entries in (3) and by the Cauchy-Schwarz inequality,we get ¯ b = 1 − a · ¯ c ≤ √ ¯ a √ ¯ c ≤ √ ǫ ¯ δ = − α · ¯ ν ≤ √ ¯ α √ ¯ ν ≤ √ ǫ ¯ β = 1 − α · ¯ γ ≤ ǫ ¯ d = − a · ¯ c ≤ √ ǫ ¯ a = 1¯ µ = − γ · ¯ ν ≤ √ ǫ ¯ f = − c · ¯ e ≤ √ ¯ c √ ¯ e ≤ ǫ ¯ α = ǫ Again, by the Cauchy-Schwarz inequality, we have : ¯ b · ¯ δ ≤ p √ ǫ ) √ ǫ ¯ β · ¯ d ≤ p ǫ ) √ ǫ ¯ a · ¯ µ ≤ p √ ǫ ¯ f · ¯ α ≤ √ ǫ Thus, if we take ǫ small enough, then we get a contradiction to (4). Namely, N does not admit a certificate as in (1).3.2. Semi-algebraic covering.
Thus, even for positive definite matrix polynomi-als, a certificate (1) does not exist in general. Although, one may remark that sucha certificate exists locally in our examples.Look at example 3.1. When | x | ≥ | y | , then consider the identity (cid:18) x xyxy x + y (cid:19) = (cid:18) x xyxy y (cid:19) + (cid:18) (cid:19) + ( x − y + y ) (cid:18) (cid:19) whereas when | y | ≥ | x | , then consider (cid:18) x xyxy x + y (cid:19) = (cid:18) xyxy x y (cid:19) + x (cid:18) (cid:19) + ( y − x y ) (cid:18) (cid:19) Hence we are going to change our definition by considering a semi-algebraiccovering of our space.From now on, we will focus on forms , namely homogeneous polynomials.
Definition 3.3.
Let f ( x, y ) be a biform of type ( n, d ; m, d ) . A semi-certificate ofpositivity is the data of a semi-algebraic covering R n = S ∪ . . . ∪ S r such that ∀ x ∈ S i , f ( x, y ) = r i X j =1 f i,j ( x )( g j,i ( x, y )) where each f i,j ( x ) is a form of degree e i,j which is non negative on S i and g i,j ( x, y ) is a biform of type ( n, d − e i,j ; m, d ) . Roughly speaking, the semi-certificate is a piecewise identity which is a sum ofsquares with respect to one set of variables y and only psd with respect to the otherset of variables x .For convenience, we may formulate what happens when we are dealing withmatrix polynomials. The symmetric psd matrix polynomial M ( x ) ∈ R [ x ] m × m admits a semi-certificate if there exists a (finite) semi-algebraic covering ( S i ) ≤ i ≤ r of R n and( ∀ i )( ∃ r i ∈ N )( ∀ j ∈ { , . . . , s i } )( ∃ f i,j ( x ) ∈ R [ x ])( ∃ U i,j ( x ) ∈ R [ x ] m × m )such that(5) M ( x ) = r i X j =1 f i,j ( x ) U Ti,j ( x ) U i,j ( x ) and ( ∀ x ∈ S i , f i,j ( x ) ≥ R necessarily have morethan one piece. Remark . Note that if we restrict ourselves to certificates where the U i,j ( x ) areconstant matrices, then we a strictly smaller class of certificates. For instance,consider the matrix polynomial (cid:18) x xyxy y (cid:19) which is psd on all R , which is obvi-ously psd on a neighbourhood V of the point (1 ,
1) ( we may proceed likewise theneighbourhood of any given point).Let us deshomogenize by setting y = 1, set for simplicity V = [1 , ǫ [, ǫ > (cid:18) x xx (cid:19) = X i f i ( x ) (cid:18) α i α i α i (cid:19) + X j g j ( x ) (cid:18) (cid:19) where the f i ’s and g j ’s are univariate polynomials psd on V and α i ∈ R .Let (cid:26) f i ( x ) = a i + b i ( x −
1) + c i ( x − g j ( x ) = d j + e j ( x −
1) + f j ( x − Then x = P i α i f i ( x ) + P j g j ( x ) x = P i α i f i ( x )1 = P i f i Combining these three identities we get X i ( α i − f i ( x ) + X j g j ( x ) = ( x − By assumption, we have a i ≥ d j ≥
0. Then, for all j we have d j = 0 and e j = 0. Furthermore, if α i = 1, then a i = 0 and also b i = 0. But substracting thelast two equalities of the previous system yields x − X i ( α i − f i ( x ) RONAN QUAREZ a contradiction. 4.
Main results
There is one case when semi-certificates exist : when the matrix polynomial ispositive definite. Remind that it means that it is positive definite on the unit sphereor the associated projective space.
Theorem . Let A ( x ) ∈ ( R [ x ]) m × m be a positive definite marix polynomial. Then,there is a finite semi-algebraic covering of R n = ∪ ri =1 S i , some forms f i,j ( x ) ∈ R [ x ] ,some matrix polynomials A i,j ( x ) ∈ ( R [ x ]) m × m such that, ∀ x ∈ S i , A ( x ) = r i X j =1 f i,j ( x ) A Ti,j ( x ) A i,j ( x ) with the condition that f i,j ( x ) ≥ for all x ∈ S i .Proof. Assume that the entries of A ( x ) are d -forms ( d is even).We start with a lemma Lemma . Let a ( x ) be a positive definite d -form and c ( x ) be a positive semi-definite d -form. Then, there exists ǫ > such that a ( x ) − ǫc ( x ) remains positivedefinite.Proof. The d -form e c ( x ) = c ( x ) + x d + . . . + x dn is positive definite and the subset S ˜ c = { x | ˜ c ( x ) = 1 } is compact. Since a ( x ) > S ˜ c we have for all x ∈ S ˜ c , a ( x ) ≥ m >
0. Hence, for all x , a x d p ˜ c ( x ) , x d p ˜ c ( x ) ! ≥ m > . Then a ( x ) − m ˜ c ( x ) is positive definite, which concludes the proof since ˜ c ( x ) ≥ c ( x ). (cid:3) Let A ( x ) = ( a i,j ( x )) ≤ i,j ≤ m . For a given x ∈ R n \ { } , the matrix A ( x ) ispositive definite in a semi-algebraic subset U x which is open in the unit sphere S n − of R n .Let U be the square matrix of size m whose all entries are equal to 1. By 4.2 andup to resizing U x , we may assume that B ( x ) = A ( x ) − ǫ ( x d + . . . + x dn ) U remainspositive definite in U x for ǫ small enough, and moreover that all entries of B ( x )are non-zero.Let B ( x ) = ( b i,j ( x )) ≤ i,j ≤ m . Assume that b , ( x ) >
0. We may write B ( x ) = b , ( x ) C where C ∈ R m × m is positive definite. Let h x ( x ) be a d -form satisfying h x ( x ) = 1. Then, by 4.2, b , ( x ) C − ǫ h x ( x )Id m remains positive definite for some small ǫ > B ( x ) = ( b , ( x ) C − ǫ h x ( x )Id m ) + e B ( x )Since e B ( x ) = ǫ Id m , the matrix e B ( x ) is positive definite in an open semi-algebraicneighbourhood of x which we still denote by U x .Then, e B ( x )) = (˜ b i,j ( x )) ≤ i,j ≤ m is such that ˜ b , ( x ) = 0. IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 9
Of course, we proceed likewise when b , ( x ) <
0, writing B ( x ) = − b , ( x ) C where C ∈ R m × m is positive definite. Thus, B ( x ) = ( − b , ( x ) C − ǫ h x ( x )Id m ) + e B ( x )Then, we repeat the same process to get rid off all the b i,j ( x )’s such that i = j and reduce, up to resize U x , to the case where B ( x ) is diagonal. Namely, there isan open semi-algebraic subset U x ⊂ S n − such that for all x ∈ U x ,(6) A ( x ) = X k f k ( x ) U Tk U k where U k ⊂ R m × m is a constant matrix and f k ( x ) is a d -form which is psd on U x .To conclude the proof, we extract a finite semi-algebraic covering by compact-edness of S n − .By homogeneity, we extend the identities (6) to all x ∈ R n . Moreover, we maymanage in order that the polynomials describing the U x ’s are forms, and hencethe identity (6) is true on S x , the cone in R n with origin 0 and basis U x . Thus,we get a finite covering of R n . (cid:3) Note that the proof gives an open semi-algebraic covering, and that the U k ’s areconstant matrices.Likewise, we have an analogeous result for biforms : Theorem . Let f ( x, y ) ∈ R [ x, y ] be a positive definite biform of type ( n, d ; m, d ) .Then, there is a finite semi-algebraic covering of R n = ∪ ri =1 S i , some forms f i,j ( x ) ∈ R [ x ] , some biforms g i,j ( x ) ∈ R [ x, y ] such that, ∀ x ∈ S i , f ( x, y ) = r i X j =1 f i,j ( x )( g i,j ( x, y )) with the condition that f i,j ( x ) ≥ for all x ∈ S i .Proof. We proceed as in the proof of Theorem 4.1. We get rid of any monomialappearing in f with some odd power with respect to at least one indeterminate. (cid:3) Semi-certificates on orthant-neighbourhoods
We say that V is an orthant-neighbourhood of x if it contains the intersection ofa neighbourhood of x and an orthant O x centered at x (i.e. a subset defined byan open condition x = x + ( X , . . . , X n ) ∈ O x if ǫ X > , . . . , ǫ n X n > ǫ , . . . , ǫ n ) ∈ {− , +1 } n ). We will write that O x is an orthant-neighbourhood of( x , ǫ )The orthant-neighbourhoods fit naturally with the use of Taylor expansion for-mula. Let us recall it relatively to an orthant-neigbourhood ( x , ǫ ) for a matrixpolynomial A ( x ) whose entries are not necessarily homogeneous polynomials : A ( x + ǫX ) = X α ( ǫX ) α α ! A ( α ) ( x ) where we use the standart multi-index symbol for products. Under a conditionof domination by the constant term, we will see how to derive some certificate onorthant-neighbourhoods. Definition . Let A ( x ) be a matrix polynomial. Denote by Γ x the set of all multi-indexes β which are minimal (for the lexicographic ordering) such that A ( β ) ( x ) = 0 .We say that A ( x ) satisfies the domination condition at x if for all multi-index α there is some multi-index β ∈ Γ x and a non negative real number r α,β such that β ≤ α and r α,β A ( β ) ( x ) ± A ( α ) ( x ) is psd. Note that Γ x = { } when A ( x ) = 0 which simplifies the domination condition.Note also that if β ∈ Γ x then necessarily all its coordinates are even integers.Finally, mention that to check positivity via the Taylor formula, it would beenough to have the domination condition for any multi-index α whose at least onecomponent is odd.A typical use of this domination condition appears in the following situation : Proposition . Let A ( x ) be a psd matrix polynomial on a neighbourhood of x .Then, A ( x ) satisfies the domination condition at x if and only if it admits a semi-certificate at x where the U i,j ( x ) ’s appearing in (5) are constant matrices.Proof. Assume that A ( x ) satisfies the domination condition at x . Let V be anorthant-neighbourhood of ( x , ǫ ). We have A ( x + ǫX ) = P β ∈ Γ x P α ∈ ∆ β ∪{ β } ( ǫX ) α α ! A ( α ) ( x )where ∆ β is a subset of all multi-index α such that α > β (we have to be carefullthat one multi-index α may be greater than several elements of Γ x ). A ( x + ǫX ) = P β ∈ Γ x (cid:16)(cid:16) ǫ β X β β ! A ( β ) ( x ) (cid:17) (cid:16) − P ∆ β ǫ β β ! α ! X α − β r α,β (cid:17) + P ∆ β X α α ! (cid:0) ǫ α A ( α ) ( x ) + r α,β A ( β ) ( x ) (cid:1)(cid:17) Since ǫ β = 1, we obtain a semi-certificate on the orthant-neighbourhood V .Conversely, assume that A ( x ) = X i f i ( x ) V i where each V i is a constant psd matrix and each f i is non negative on a neighbour-hood of x . We write the Taylor expansion of f i at x : f i ( x + X ) V i = X α X α α ! f ( α ) i ( x ) ! × V i For the matrix polynomial f i ( x ) V i , consider the set Γ x . If β ∈ Γ x , then f ( β ) i ( x ) > α > β , we clearly have the existence of a positive realnumber r α,β such that r α,β f ( β ) i ( x ) ± f ( α ) i ( x ) ≥
0. This is the domination condi-tion. (cid:3)
Remark . Since positive definite matrices obviously satisfy the domination con-dition, we may recover a version of Theorem 4.1 with orthant-neighbourhoods. But,maybe (highly heuristic !) it will produce a lot more pieces for the covering, since
IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 11 (again roughly speaking) we need 2 n − orthant-neighourhoods to recover a usualneighbourhood.5.1. Semi-certificates relative to a semi-algebraic subset.
One may natu-rally want to extend the framework of semi-certificates relatively to a basic closedsemi-algebraic subset S . The problem is that the result given in this section doesnot take into account the equations describing S . We mainly use the underlyingsemi-algebraic set rather than the preordering or the quadratic module generatedby the equations of S as it is desired for a relative Positivestellensatz. In fact, thissection concerns more the study of local semi-certificates rather than relative ’s ones. Theorem . Let A ( x ) ∈ R [ x ] m × m be a homogeneous symmetric matrix polynomial.Assume that A ( x ) is positive definite for all x ∈ S , where S is a closed semi-algebraically subset of R n defined by homogeneous polynomials. Then, there is afinite semi-algebraic covering of S = ∪ ri =1 S i , some forms p i,j ( x ) ∈ R [ x ] , somehomogeneous matrix polynomials A i,j ( x ) ∈ R [ x ] m × m such that, ∀ x ∈ S i , A ( x ) = r i X j =1 p i,j ( x ) A Ti,j ( x ) A i,j ( x ) with the condition that p i,j ( x ) ≥ for all x ∈ S i .Proof. We may perform the same proof as in Theorem 4.1. (cid:3)
As an example of a case where A ( x ) is not definite, we consider a general matrixpolynomial of degree at most 2 in a single (non homogeneous) variable x . Example . Assume that A ( x ) is psd in the neighbourhood of 0 + .We assume moreover that A (0) is not positive definite otherwise we are doneby 5.4 or by a domination argument. Up to a base change, we may assume that A (0) = (cid:18) (cid:19) . Then, let us write A ( x ) = (cid:18) (cid:19) + x (cid:18) a b b c (cid:19) + x (cid:18) a b b c (cid:19) = (cid:18) a x + a x b x + b x b x + b x c x + c x (cid:19) Since A ( x ) is psd at 0 + , we have c ≥ c >
0, then the domination is satisfied and we get the following certificate A ( x ) = (1 − µx − αx ) (cid:18) (cid:19) + x (cid:18) a + µ b b c (cid:19) + x (cid:18) a + α b b c + βc (cid:19) where α, β, µ are positive real numbers chosen such that the constant ma-trices of the identity are psd. Note that (1 − µx − αx ) is obviously positiveon a neighbourhood of 0 + .* If c = 0, then A ( x ) is psd on 0 + if c − b ≥
0. If c − b >
0, then wehave the certificate A ( x ) = (1 − αǫx ) (cid:18) (cid:19) + − ǫ b xb x ( b x ) − ǫ ! + x a + αǫ b b c − b − ǫ ! where ǫ and α are positive real numbers such that the last constant matrixin the identify is positive definite. * Now, consider the case when c = 0 and c − b = 0. By positivity of A ( x )we have a c − b b ≥ b = 0 is trivial since we must have b = 0 and thecertificate follows. If b = 0, then we get the condition a b − b ≥ a b − b >
0. We may write A ( x ) = (cid:18) x (cid:16) a − b b (cid:17) + x (cid:18) a − (cid:16) b b (cid:17) (cid:19)(cid:19) (cid:18) (cid:19) + (cid:16) b b x (cid:17) ( b x ) (cid:16) b b x (cid:17) ( b x ) (cid:16) b b x (cid:17) ( b x ) * In the remaining case when c = 0, c − b = 0, b = 0, a b − b = 0,the positivity of A ( x ) at 0 + says that a b − b ≥
0. The desired certificatefollows from the identity given in the previous case.This inspection of the most elementary situation leads to conjecture that anymatrix polynomial in a single variable admits a local certificate of positivity.We give a proof of this fact, although we do not give explicit formulas dependingon the entries as in the previous worked example.
Theorem . Let M ∈ R [ x ] n × n be a symmetric matrix polynomial whose entriesare polynomials in a single variable x . Assume that M is psd on a neighbourhoodof + .Then, M admits a semi-certificate of positivity at + .Proof. If M (0) is invertible, then we are done by 4.1. Hence, from now on, weassume that det( M (0)) = 0.Let us consider the Smith normal form of M : M = EDF where E and F are invertible in R [ x ] n × n and D = diag( d , . . . , d n ) is diagonal in R [ x ] n × n with diagonal entries ( d , . . . , d n ). By changing M to ( F T ) − M F − , wemay assume that M = ED . Moreover, if d i ≡ M , hence we will assume in the following that d i ≥ + and vanishesonly at 0.The matrix M has the form M = d e , d e , . . . d n e ,n d e , d e , . . . d n e ,n ... ... d n e ,n d n e ,n . . . d n e n,n Hence the matrix E has the form E = e , e , . . . e ,nd d e , e , . . . e ,n ... . . . . . . ...... . . . . . . ... d n d e ,n d n d e ,n . . . d n d n − e ,n e n,n First, let us introduce some notations.Define by induction the following sequence of integers :
IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 13
Let k = 1 and set k i to be the first integer j > k i − such that d j d j − (0) = 0. Ityields an increasing sequence of integers :1 = k < k < . . . < k r ≤ n = k r +1 − . We divide the matrix M = ( m k,l ) ≤ k,l ≤ n into block matrices : M = ( M i,j ) ≤ i,j ≤ r ,where we set M i,j = ( m ( i,j ) k,l ) ≤ k ≤ k i +1 − k i , ≤ l ≤ k j +1 − k j with m ( i,j ) k,l = m k + k i − ,l + k j − .Likewise, we divide E = ( E i,j ) ≤ i,j ≤ r into similar block matrices.Let us define the block matrix M i = ( R k,l ) ≤ k,l ≤ r such that R k,l = M k,l when(1 ≤ k ≤ i and l = i ) or (1 ≤ l ≤ i and k = i ) and R k,l = 0 otherwise. Namely M i = . . . M ,i . . . . . . M i − ,i ... ... M i, . . . M i,i − M i,i ... ...0 . . . . . . . . . . . . . . . . . . . . . . . . . Let us write now M = d k M + d k M + . . . + d k r M r . Observe then by construction that E (0) is block upper triangular. Since E (0) isinvertible we get that E i,i (0) is invertible for each i = 0 , . . . , r .Moreover, we have(7) det( M i,i ) = det( E i,i ) × k i +1 − Y j = k i d j By definition of the k i ’s, it implies that det( d ki M i,i ) does not vanish at 0 . Hencewe may derive a kind of domination condition for the matrix M at 0 + . Namely, letus write M = d k M − d k × µ × Id k ,k − . . . − d k r × µ r × Id k ,k + d k ( M + µ Id k ,k ) − d k × µ × Id k ,k − . . . − d k r × µ r × Id k ,k ...+ d k r ( M r + µ r Id k ,k r )= d k N + d k N + . . . + d k r N r where the µ i ’s are positive real numbers and Id p,q is the diagonal matrix in R n × n whose i -th diagonal entry is +1 if p ≤ i ≤ q − µ i ’s such that the matrix polynomials N i are psdon a neighbourhood of 0 + . Then, using the same argument as in Theorem 4.1, we complete the constructionof a semi-certificate. (cid:3)
The result is no more true with more than one variable :
Proposition . Let M ( x, y ) = (cid:18) x − y − x − x y (cid:19) . Then, M ( x, y ) is psd for x ≥ y ≥ and x ≤ y ≤ althought it does not admita certificate in a neighbourhood of (1 + , + ) .Proof. First note that det( M ( x, y )) = ( y − x − y ) to conclude to the positivitydomain of M ( x, y ).For convenience, let us consider the following change of variables y = 1 + Y and x = Y + H . We also perform a base change to “simplify” the expression of M (0 ,
0) = (cid:18) − − (cid:19) . Let P = (cid:18) −
11 1 (cid:19) and Q = (cid:18) − (cid:19) . Then, P × M (0 , × Q = (cid:18) (cid:19) and thus let us introduce the following matrix N ( X, Y ) = P × M ( x, y ) × Q = (cid:18) H + 4 Y + H + Y + 2 HY H − Y + H − Y + 2 HY H − Y + H − Y + 2 HY H + Y + 2 HY (cid:19) The matrix N ( X, Y ) is psd at an orthant-neighbourhood (0 + , + ) with respect tothe variables ( H, Y ). Let us assume that it has a semi-certificate of positivity.Namely :(8) N ( X, Y ) = X i F i (cid:18) ǫ i ǫ i ǫ i (cid:19) + X j (cid:18) U j U j V j U j V j V j (cid:19) + G (cid:18) (cid:19) where all F i ’s and G are psd on an orthant-neighbourhood (0 + , + ). The degreesof the (2 , f i has only monomials of degree 2 and each V j has monomials of degree 1. Hence : F i = a i H + b i Y + c i HYU j = γ j + µ j H + ν j YV j = α j H + β j YG = a ′ H + b ′ Y + c ′ Y H + d ′ H + e ′ Y + f ′ Identifying the (2 , X i a i + X j α j = 1 , X i b i + X j β j = 1 X i c i + 2 X j α j β j = 2Identifying the coefficients in H and Y of the (1 , , IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 15 (10) X j α j γ j = 2 X j β j γ j = − X j γ j + f ′ = 4Let us introduce ¯ α = ( α j ), ¯ β = ( β j ) and ¯ γ = ( γ j ), vectors of the standarteuclidean space (whose dimension is equal to the number of indexes j ).After (10) and (9), we get¯ γ ≤ , ¯ α ≤ α · ¯ γ = 2 . By the Cauchy- Schwartz case of equality, we get ¯ γ = 4 and ¯ α = 1. Thus, f ′ = 0 and P i a i = 0 which means that for all i , a i = 0. Moreover, the vectors ¯ α and ¯ γ must be colinear and hence ¯ α = ¯ γ .Likewise β = − γ and P i c i = 4. Moreover, for all i we have b i = 0. And hencewe may assume that there is only one index i and we may write(11) N ( X, Y ) = 4 HY (cid:18) ǫ ǫǫ (cid:19) + X j (cid:18) U j U j V j U j V j V j (cid:19) + G (cid:18) (cid:19) . Identifying the (1 , µ = ( µ j ) and ¯ ν = ( ν j ) we have(12)
12 ¯ γ · ¯ µ = 112 ¯ γ · ¯ ν = 1¯ γ · (¯ µ − ¯ ν ) + 4 ǫ = 2We readily deduce that ǫ = . Identifying the (1 , γ · ¯ µ + d ′ = 4 , γ · ¯ ν + e ′ = 4 , µ · ¯ ν + c ′ + 1 = 2 , ¯ µ + a ′ = 1 , ¯ ν + b ′ = 1 , By (12) and (13), we immediately have d ′ = e ′ = 0. Moreover, we get also¯ γ · ¯ µ = 2 , ¯ µ ≤ γ = 4 . By the Cauchy-Schwartz case of equality, we deduce that a ′ = 0 and ¯ µ = ¯ γ .Likewise b ′ = 0 and ¯ ν = ¯ γ .Then, G = c ′ HY where necessarily c ′ ≥
0. To conclude, it suffices to note that2 ¯ µ · ¯ ν + c ′ + 1 = 2 gives 2 + c ′ + 1 = 2, a contradiction. (cid:3) Remember that any psd biquadratic form in dimension 2 is a sum of squares,and hence admits a semi-certificate. This example shows that the local counterpartis no more true.
In general we may mention the following result. Although completely elementary,it looks very much like the ones we can find in [7] or [8] for hermitian squares ormore classical sums of squares.
Proposition . Let A ( x ) ∈ R [ x ] m × m be a symmetric matrix polynomial. Then, A ( x ) is psd if and only if for all real ǫ > the matrix polynomial A ǫ ( x ) = A ( x ) + ǫ n X i =1 x i ! Id m admits a semi-certificate. Of course the complexity of the semi-certificate may increaes as ǫ goes to zero. Remark . The complexity of a semi-certificate shall be measured by the size ofthe semi-algebraic partitions and also by the number of squares. Here the degreesof the polynomials appearing in a (homogeneous) semi-certificate are bounded bythe degree of the given matrix polynomial.
Remark . By [6] with respect to the (compact) unit sphere { P ni =1 x i = 1 } , thenwe obtain for M a Positivestellensatz which is no more homogeneous and wherethe degrees are no more bounded. Remark . Consider a semi-certificate, and a compact semi-algebraic subset S i of the given partition. If we assume that the desciption of S i is archimedean, thenwe may deduce a “true” algebraic certificate without denominators for the matrixpolynomial on S i , using for instance Schm¨udgen or Putinar certificates.6. Around the Choi counterexample
Let us consider the counterexample of a biqudratic psd non sum of squares givenin [1] : x + 2 z − xy − xz − xy y + 2 x − yz − xz − yz z + 2 y Altough it is not a sum of squares, it admits a semi-certificate. Indeed, when | x | ≥ | z | , it can be decomposed as x − xy − xz − xy y yz − xz yz z + 2 z x − yz − yz y = ( − x, y, z ) T ( − x, y, z ) + 2(0 , − z, y ) T (0 , − z, y )+2 z + 2( x − z ) And by symmetry, we deduce an analogeous certificate when | z | ≥ | x | .Let us consider a variation around this example ; in the following proposition,the Choi counterexample is just M : IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 17
Proposition . Let M λ = x + ( λ + 1) z − xy − xz − xy y + ( λ + 1) x − yz − xz − yz z + ( λ + 1) y where λ ∈ R .Then, M λ is never a sum of squares and it is psd if and only if λ ∈ [0 , + ∞ [ .Moreover, det( M ) is a psd non sos polynomial and M λ admits a semi-certificateof positivity for any λ ∈ ]0 , + ∞ [ .Proof. The exact argument as in [1] works to show that any M λ is not a sum ofsquares. We may reproduce it for convenience. We translate our ramewok into thelanguage of biforms, setting f M λ = ( x +( λ +1) z ) S − xyST − xzSU +( y +( λ +1) x ) T − yzT U +( z +( λ +1) y ) U Assume that f M λ = P f i where f i is a bilinear form. Since there is no monomials x U , y S , z T in f M λ , there is no such monomials in each f i , and hence nomonomials xU , yS nor zT in each f i . We may write f i = g i + h i were g i dependsonly in the monomials xS , yT , zU and h i depends only in the monomials xT , yU , zS .Then, f M λ = P ( g i + h i ) shows that X i g i = x S + y T + z U − xyST + yzT U + xzSU ) , a contradiction since the right hand side of the equality takes negative values for x = y = z = S = T = U = 1.The fact that M is psd and not a sum of squares may also be shown directly byusing the following consequence of the Cauchy-Binet formula. Lemma . Let M ( x ) be a symmetric matrix polynomial. If M is a sum of squares,then its determinant is also a sum of squares. .Proof. We first state the Cauchy-Binet formula.Given matrices A ∈ R m × s and B ∈ R s × m , the Cauchy-Binet formula states thatdet( AB ) = X S det( A S ) det( B S )where S ranges over all the subsets of { , . . . s } with m elements, and A S (re-spectively B S ) denotes the matrix in R m × m whose columns are the columns of A (respectively whose rows are the rows of B ) with index from S .If M ( x ) is a sum of squares, then it can be written M ( x ) = A T ( x ) A ( x ) for somematrix polynomial A ( x ) ∈ R [ x ] s × m . Then,det( M ( x )) = P S det( A ( x ) T ) S det( A ( x ) S )= P S det( M S ( x ) T ) det( M S )= P S ( det( M S )) . (cid:3) The Cauchy-Binet formula shows even more : if M is a sum of squares, then all its principalminors are sums of squares. Note that M has as a determinant :det( M ) = x y + y z + z x − x y z a variation of the celebrated Motzkin polynomial which is psd and not a sum ofsquares !Thus, the matrix polynomial M , which is clearly psd since all its principalminors are psd, cannot be a sum of squares since its determinant is not !Thus, M (and thus all M λ for λ ≥
0) are psd matrix polynomial. Moreover,det( M λ (1 , , λ ( λ + 3) which shows that M λ is not psd for small λ < λ < λ >
0, the set of real singular points of det( M λ ) have projective coordinatesis [1 : 0 : 0], [0 : 1 : 0] and [0 : 0 : 1] (they correspond to the points where the nonnegative polynomial det( M λ ) vanishes). The identity M λ = x − xy − xz − xy y yz − xz yz z + λz z λ − yz − yz λy + 1 λ ( λx − z ) gives a semi-certificate at the neighbourhood of [1 : 0 : 0]. We may also proceedlikewise at the neighbourhood of [0 : 1 : 0] and [0 : 0 : 1].Elsewhere the matrix M λ is positive definite so that we can argue as in theproof of Theorem 4.1 and obtain a finite semi-algebraic covering, and hence a semi-certificate.Whereas, the matrice M has a lot more singular points, namely : [1 : 0 : 0],[0 : 1 : 0], [0 : 0 : 1], [1 : 1 : 1], [1 : 1 : − − − M at thepoints [1 : 1 : 1], [1 : 1 : − − − M = x + z − xy − xz − xy y + x − yz − xz − yz z + y = y − yz − yz y + y − xy − xy x
00 0 0 + x + z − y − xz − xz y But, setting z = 1 in the last matrix, we get a matrix polynomial which is psdat an orthant-neighbourhood of (1 + , + ) with respect to the variables ( x, y ), butwhich does not admit any semi-certificate as shown by Proposition 5.7.Nevertheless, let us see in the following how it is possible to get a semi-certificatein some orthant-neighbourhood of [1 : 1 : 1], and hence a partial result. IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 19
Let z = 1 and x = 1 + X , y = 1 + Y . Write the Taylor formula at (1 ,
1) : M = A + XA + X A + Y B + Y B + XY C Consider the following matrices : A X = 512 A + XA + X A = + 2 X + X − − X − − X − − X + 2 X + X − − − X −
512 56 A Y = 512 A + Y B + Y B = − − Y − −
512 56 + 2 Y + Y − − Y − − − Y + 2 Y + Y A XY = 212 A + XY C = − − XY − − − XY − − −
16 13
One may check that C X = A X − A (cid:18) X − X (cid:19) = X +9 X +56 − X − X − − X − X − − X − X −
512 18 X +9 X +56 − X +3 X − − X − X − − X +3 X −
512 12 X − X +56 is positive semi-definite on all R . Since it is a psd matrix polynomial with respectto the single variable X , it is a sum of squares.Likewise, one may check that C Y = A Y − A (cid:18) − Y − Y (cid:19) = Y +3 Y +56 − Y − Y − − Y − Y − − Y − Y −
512 18 Y +15 Y +56 − Y − Y − − Y − Y − − Y − Y −
512 18 Y +15 Y +56 is positive semi-definite on all R . Since it is a psd matrix polynomial with respectto the single variable Y , it is a sum of squares.Finally, write A XY = (cid:18) XY )12 A (cid:19) − XY ( A − C )Where we check that ( A − C ) is a psd constant matrix.If we sum up all these informations, we obtain a desired certificate of positivity M = A X + A Y + A XY = A (cid:18) X − X (cid:19) + C X + A (cid:18) − Y − Y (cid:19) + C Y + A XY with respect to the open orthant-neighbourhood of (0 ,
0) defined by
X >
Y <
X <
Y >
XY > (cid:3) Concluding remarks
Open questions about semi-certificates.
We have introduced the notionof piecewise semi-certificate of positivity for matrix polynomials. For the moment,the only general result is that all positive definite matrix polynomial admit such acertificate. A lot of things remain to be studied. • We shall better understand the set of all psd matrix polynomials whichadmit a semi-certifiacte. Beginning with biquadratic forms for instance ? • We shall developp some effective algorithm to produce the certificates. • About the complexity of certificates : how can we bound the number ofsquares, and also the number of pieces of the semi-algebraic partition ?On the other hand, one may also rise some questions about the familly of psd poly-nomials which have a quadratic determinantal representation. This is the object ofthe last subsection :7.2.
Semi-definite quadratic determinatal representations.
In the spirit ofwhat happens in Proposition 6.1, one may be interested in determining what psdpolynomial can be written as the determinant of a psd quadratic matrix polynomial(whose entries are homogeneous polynomials of degree 2). Namely,
Question . Let f ( x ) ∈ R [ x ] be a form of degree 2 d which is suppose to be nonnegative. Does-there exists a quadratic matrix polynomial M ( x ) ∈ R [ x ] d × d whichis psd for all x ∈ R n and such that(14) f ( x ) = det( M ( x ))?One motivation for such a question is that it appears as the quadratic analo-geaous of very classical linear determinantal representations which have been stud-ied for a very long time. The specifically real considerations being more recents andesentially du to Vinnikov (see for instance [11] and all related papers). Let us re-call that these have a lot of applications for instance to Linear Matrix Inequalities,convex modelling, etc...On the other hand, biquadratic forms are a quite common object found at variousareas in ingenering applications. For instance, any determinantal representation asin (14) of a psd non sum of squares polynomial provides, by Cauchy-Binet formula,another example of a psd biquadratic form which is not a sum of squares.Of course, a simple count on the number of parameters shows Question 7.1 hasa negative answer in general.We can be even more precise : if f is a form of degree 4 in 4 variables, sucha representation never exists when f is psd non sum of squares. Indeed, if f = IECEWISE CERTIFICATES OF POSITIVITY FOR MATRIX POLYNOMIALS 21 det( M ( x )) with M ( x ) ∈ R [ x ] × , then M ( x ) is a psd biquadratic form which is asum of squares (in dimension 2), hence det( M ) is a sum of squares of polynomialsby Cauchy-Binet formula.We may also look at the trivial case when n = 2, which we may deshomogenizefor simplicity. The answer to Question 7.1 is clearly positive by an elementaryargument. Indeed, let us write f ( x ) = d X i =0 a i x i = Y j ( x − α j ) × Y k ( x + β k ) + γ k )Then, the result follows obviously from the multiplicative property of the determi-nant and the fact that the ( x − α i ) ’s and the ( x + β k ) + γ k )’s are psd quadraticpolynomials.We may even give a more algorithmic construction (a polynomial time algorithmwith respect to the size of the coefficients of the polynomial) using arrows matricesand following [5] and [9].Back to Question 7.1, and before expecting general results, one may wonder iffor instance the celebrated Motzkin and Robinson polynomials Mo = z + x y + x y − x y z Ro = x + y + z − x y − x z − y x − y z − z x − z y + 3 x y z can be written as in (14) ?It is quite easy, using for instance the linear well-know determinantal representa-tions of cubics curves, to produce quadratic determinantal representations for thesetwo polynomials, but unfortunately none such representation yields a psd quadraticmatrix polynomial. References [1] M.D. Choi,
Positive Semidefinite Biquadratic Forms Linear Algebra Appl.
12 (1975), 95-100[2] M.D. Choi, T.Y. Lam, B. Reznick,
Real zeros of positive semidefinite forms. I.
Math. Z.171 (1980), no. 1, 1–26.[3] M.D. Choi, T.Y. Lam, B. Reznick,
Sums of squares of real polynomials. , K -theory and al-gebraic geometry: connections with quadratic forms and division algebras (Santa Barbara,CA, 1992), 103–126, Proc. Sympos. Pure Math., 58, Part 2, Amer. Math. Soc., Providence,RI, 1995.[4] D. Z. Djokovic, Hermitian Matrices over Polynomial Rings , Journal of Algebra, 359-374,1976[5] M. Fiedler,
Expressing a polynomial as the characteristic polynomial of a symmetric ma-trix , Linear Algebra Appl. 141 (1990), 265-270.[6] C.W.J. Hol, C.W. Scherer,
Matrix sum-of-squares relaxations for robust semi-definite pro-grams , Math. Program. 107 (2006), no. 1-2, Ser. B, 189-211[7] I. Klep, M. Schweighofer,
Pures States, Positive matrix polynomial and sums of HermitianSquares , arxiv[8] M. Putinar,
Positive polynomials on compact semi-algebraic sets.
Indiana Univ. Math. J.42 (1993), no. 3, 969-984.[9] R. Quarez,
Repr´esentations D´eterminantales des Polynˆomes Univari´es par les MatricesFl`eches
Pr´epublication IRMAR(2008). [10] K. Schm¨udgen,
The K -moment problem for compact semi-algebraic sets. , Math. Ann. 289(1991), no. 2, 203-206.[11] V. Vinnikov, Self-adjoint determinantal representations of real plane curves
Math. Ann.296 (1993), pp. 453-479.
IRMAR (CNRS, URA 305), Universit´e de Rennes 1, Campus de Beaulieu, 35042 RennesCedex, France
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