Determinantal Representations and the Hermite Matrix
DDETERMINANTAL REPRESENTATIONS AND THE HERMITE MATRIX
TIM NETZER, DANIEL PLAUMANN, AND ANDREAS THOM
Abstract.
We consider the problem of writing real polynomials as determinants of sym-metric linear matrix polynomials. This problem of algebraic geometry, whose roots go backto the nineteenth century, has recently received new attention from the viewpoint of convexoptimization. We relate the question to sums of squares decompositions of a certain Her-mite matrix. If some power of a polynomial admits a definite determinantal representation,then its Hermite matrix is a sum of squares. Conversely, we show how a determinantalrepresentation can sometimes be constructed from a sums-of-squares decomposition of theHermite matrix. We finally show that definite determinantal representations always exist,if one allows for denominators.
Introduction
A polynomial p ∈ R [ x ] in n variables x = ( x , . . . , x n ) with p (0) = 1 is called a real-zeropolynomial if p has only real zeros along every line through the origin. The terms hyperbolic or real stable polynomial are also common and mean essentially the same, but usually forhomogeneous polynomials. The typical example is a polynomial given by a definite (linearsymmetric) determinantal representation p = det( I + A x + · · · + A n x n ) , where A , . . . , A n are real symmetric matrices and I is the identity. A representation of thisform is a certificate for being a real-zero polynomial. In other words, the fact that p is a real-zero polynomial is apparent from the representation. A definite determinantal representationalso provides a description of the rigidly convex region of p . This is the closed connectedcomponent of the origin in the complement of the zero-set of p . It is always convex, andgiven a definite determinantal representation of p , it coincides with the set of points wherethe matrix polynomial I + A x + · · · + A n x n is positive semidefinite. Date : August 23, 2011.2000
Mathematics Subject Classification.
Primary 11C20, 11E25, 14P10; Secondary 90C22, 90C25, 52B99.
Key words and phrases. determinantal representations, real-zero polynomials, spectrahedra, Hermite matrix,sums of squares.Travel for this project was partially supported by the Forschungsinitiative
Real Algebraic Geometry andEmerging Applications at the University of Konstanz. Daniel Plaumann gratefully acknowledges supportthrough a Feodor Lynen return fellowship from the Alexander von Humboldt Foundation. a r X i v : . [ m a t h . AG ] A ug TIM NETZER, DANIEL PLAUMANN, AND ANDREAS THOM
In recent years, real-zero polynomials and their determinantal representations have beenstudied mostly with a view towards convex optimization, specifically semidefinite and hy-perbolic programming. In general, one would like to answer the following questions:(1) Under what conditions does a real-zero polynomial have a definite determinantalrepresentation?(2) If such a representation exists, what is the minimal matrix dimension and how canthe representation be computed effectively?(3) If no such representation exists, what other certificates for being a real-zero polyno-mial are available?Question (1) is the most immediate and has consequently received the most attention. Itties in with the theory of determinantal hypersurfaces in complex algebraic geometry, whoseroots go back to the nineteenth century. Arguably the most important modern results arethe Helton-Vinnikov theorem in [7], which gives a positive answer for n = 2 , and Brändén’snegative results in higher dimensions in [4]. Since there are various subtle variations of thequestion, it is not always easy to figure out what is known and what is not; we give a verybrief overview after the introduction below.Question (2), which should be of interest for practical purposes, has not been studiedvery systematically so far. Even in the case n = 2 , the classical approach of Dixon forconstructing determinantal representations is quite algorithmic in nature but hard to carryout in practice (see [5], or [14] for a more modern presentation.)One approach to Question (3) is to study the determinantal representability of a suitablepower or multiple of p if no representation for p exists. This is motivated by the GeneralizedLax Conjecture, as described below. On the other hand, the real-zero property does nothave to be expressed by a determinantal representation. That a polynomial p in one variablehas only real roots is equivalent to its Hermite matrix being positive semidefinite. This isa symmetric real matrix associated with p , which provides one of the classical methods forroot counting. To treat the multivariate case, we use a parametrized version of the Hermitematrix with polynomial entries. In a typical sums-of-squares-relaxation approach commonin polynomial optimization, we then ask for the parametrized Hermite matrix H ( p ) to be asum of squares, which means that there exists a matrix Q such that H ( p ) = Q T Q . (This iscalled a sum of squares rather than a square, because Q is allowed to be rectangular of anysize). This approach has been used before by Henrion in [9] and by Parrilo (unpublished) asa relaxation for the real-zero property, which is exact in the two-dimensional case.While the Hermite matrix provides a practical way of certifying the real-zero property,having a definite determinantal representation of p is clearly much more desirable, since it ETERMINANTAL REPRESENTATIONS AND THE HERMITE MATRIX 3 also yields a description of the rigidly convex region by a linear matrix inequality. And evenif one is only interested in the real-zero property, the multivariate Hermite matrix is a fairlyunwieldy object compared to the original polynomial, and a sum-of-squares decompositioneven more so.Our main goal is therefore to use a sum-of-squares decomposition of the parametrizedHermite matrix of a polynomial p to construct, as explicitly as possible, a definite determi-nantal representation of p , or at least of some multiple of p . We first show in Section 1 thata definite determinantal representation of some power of p of the correct size always yieldsa sum-of-squares decomposition of H ( p ) (Thm. 1.6). In Section 2, we make an attempt atthe converse. This is partly motivated by our experimental finding that the Hermite matrixof the Vámos polynomial, which is the counterexample of Brändén, is not a sum of squares(Example 1.9). Note also that in the case n = 2 , where every real-zero polynomial possessesa definite determinantal representation by the Helton-Vinnikov theorem, the parametrizedHermite matrix can be reduced to the univariate case. It is therefore a sum of squares ifand only if it is positive semidefinite, by a result of Jakubovič [10]. Given a decomposition H ( p ) = Q T Q , we show that a definite determinantal representation of a multiple of p canbe found if a certain extension problem for linear maps on free graded modules derivedfrom Q has a solution (Thm. 2.5). Given Q , the search for such a solution amounts onlyto solving a system of linear equations. This method can in principle also be applied ifthe sums of squares decomposition uses denominators. Finally, we show that by allowinga sum-of-squares decomposition with denominators, which exists whenever H ( p ) is positivesemidefinite, one can always obtain a determinantal representation with denominators: Theorem.
Let p be a square-free real-zero polynomial with p (0) = 1 . There exists a sym-metric matrix M whose entries are real homogeneous rational functions of degree such that p = det( I + M ) . The precise statement is given in Thm. 3.1.
Acknowledgements.
We would like to thank Didier Henrion, Pablo Parrilo, Rainer Sinn,and Cynthia Vinzant for helpful comments and discussions.
Known results ◦ For n = 2 , every real-zero polynomial of degree d has a real definite determinantalrepresentation of matrix size d by the Helton-Vinnikov theorem [7]. ◦ For n ≥ and d sufficiently large, a simple count of parameters shows that onlyan exceptional set of polynomials can have a real determinantal representation of TIM NETZER, DANIEL PLAUMANN, AND ANDREAS THOM size d . The question whether every real-zero polynomial has a definite determinantalrepresentation of any size became known as the generalized Lax conjecture. ◦ The generalized Lax conjecture was disproven by Brändén who even showed the exis-tence of real-zero polynomials p such that no power p r has a determinantal represen-tation of any size [4]. His smallest counterexample, the so-called Vámos polynomial ,is of degree in variables (see 1.9 below). ◦ Netzer and Thom [11] have proved that only an exceptional set of polynomials canhave a determinantal representation, even if one allows for matrices of arbitrary size.This is true for n ≥ and d sufficiently large, or d ≥ and n sufficiently large. Theyalso show that if p is a real-zero polynomial of degree , then there exists r ≥ suchthat p r has a determinantal representation. On the other hand, there exists such p where one cannot take r = 1 . ◦ Another result of Helton, McCullough and Vinnikov [8] (see also Quarez [13]) saysthat every real polynomial has a real symmetric determinantal representation, thoughnot necessarily a definite one. This means that the constant term in the matrixpolynomial cannot be chosen to be the identity matrix in their result. ◦ The most general form of the Lax conjecture says that every rigidly convex set isa spectrahedron. In terms of determinantal representations, this amounts to thefollowing: For every real-zero polynomial p there exists another real-zero polynomial q such pq has a real definite determinantal representation and such that q is non-negative on the rigidly convex set of p . This conjecture is still wide open, even withoutthe additional positivity condition on q . Note that if pq has a definite determinantalrepresentation, then q is automatically a real-zero polynomial.1. The Hermite Matrix
In this section we introduce the parametrized Hermite matrix H ( p ) of a polynomial. Itis positive semidefinite at each point if and only if p is a real zero polynomial. If some powerof p admits a determinantal representation of the correct size, then H ( p ) even turns out tobe a sum of squares of polynomial matrices.Let p = t d + p t d − + · · · + p d − t + p d ∈ R [ t ] be a monic univariate polynomial of degree d and let λ , . . . λ d be the complex zeros of p . Then N k ( p ) = d (cid:88) i =1 λ ki ETERMINANTAL REPRESENTATIONS AND THE HERMITE MATRIX 5 is called the k -th Newton sum of p . The Newton sums are symmetric functions in the roots,and can thus be expressed as polynomials in the coefficients p i of p . The Hermite matrix of p is the symmetric d × d matrix H ( p ) := ( N i + j − ( p )) i,j =1 ,...d . It is a Hankel matrix whose entries are polynomial expressions in the coefficients of p . Notethat H ( p ) = V T V , where V is the Vandermonde matrix with coefficients λ , . . . , λ d .The following well-known fact goes back to Hermite. For a proof, see for exampleTheorem 4.59 in Basu, Pollack and Roy [1]. Theorem 1.1.
Let p ∈ R [ t ] be a monic polynomial. The rank of H ( p ) is equal to the numberof distinct zeros of p in C . The signature of the Hermite matrix H ( p ) is equal to the numberof distinct real zeros of p .In particular, H ( p ) is positive definite if and only if all zeros of p are real and distinct,and H ( p ) is positive semidefinite if an only of all zeros are real. (cid:3) Now let p ∈ R [ x ] be a polynomial of degree d in n variables x = ( x , . . . , x n ) . Thepolynomial p is called a real-zero polynomial (with respect to the origin) if p (0) = 1 andfor every a ∈ R n , the univariate polynomial p ( ta ) ∈ R [ t ] has only real zeros. We want toexpress this condition in terms of a Hermite matrix. Write p = (cid:80) di =0 p i with p i homogeneousof degree i , and let P ( x, t ) = (cid:80) di =0 p i t d − i be the homogenization of p with respect to anadditional variable t . We consider P as a monic univariate polynomial in t and call theHermite matrix H ( P ) the parametrized Hermite matrix of p , denoted H ( p ) . Its entries arepolynomials in the homogeneous parts p i of p . The ( i, j ) -entry is a homogeneous polynomialin x of degree i + j − . Corollary 1.2.
A polynomial p ∈ R [ x ] with p (0) = 1 is a real-zero polynomial if and only ifthe matrix H ( p )( a ) is positive semidefinite for all a ∈ R n .Proof. By Theorem 1.1, H ( p )( a ) is positive semidefinite for a ∈ R n if and only if the uni-variate polynomial t d p ( a t − , . . . , a n t − has only real zeros. Substituting t − for t , we seethat this is equivalent to p ( ta ) having only real zeros. (cid:3) The following is Proposition 2.1 in Netzer and Thom [11].
Proposition 1.3.
Let M = x M + · · · + x n M n be a symmetric linear matrix polynomial, andlet p = det( I − M ) . Then for each a ∈ R n , the nonzero eigenvalues of M ( a ) are in one toone correspondence with the zeros of the univariate polynomial p ( ta ) , counting multiplicities.The correspondence is given by the rule λ (cid:55)→ λ . (cid:3) TIM NETZER, DANIEL PLAUMANN, AND ANDREAS THOM
Lemma 1.4.
Let p ∈ R [ x ] be a real-zero polynomial of degree d , and assume that p r =det( I − M ) is a symmetric determinantal representation of size k , for some r > . Then H ( p ) i,j = 1 r · (cid:18) tr (cid:0) M i + j − (cid:1)(cid:19) , except possibly for ( i, j ) = (1 , , where H ( p ) , = d and tr (cid:0) M (cid:1) = k .Proof. For each a ∈ R n , the trace of M ( a ) s is the s -power sum of the nonzero eigenvaluesof M ( a ) . These eigenvalues are the inverses of the zeros of p ( ta ) , by Proposition 1.3, buteach such zero gives rise to r many eigenvalues. Since the zeros of p ( ta ) correspond to theinverses of the zeros of t d p ( t − a ) , the trace of M ( a ) s equals the s -power sum of the zeros of t d p ( t − a ) multiplied with r . This proves the claim. (cid:3) Definition 1.5.
Let
H ∈
Sym d ( R [ x ]) be a symmetric matrix with polynomial entries. H isa sum of squares , if there is a d (cid:48) × d -matrix Q with polynomial entries, such that H = Q T Q .This is equivalent to the existence of d (cid:48) many d -vectors Q i with polynomial entries, suchthat H = (cid:80) ki =1 Q i Q Ti . Theorem 1.6.
Let p ∈ R [ x ] be a real-zero polynomial of degree d . If a power p r admitsa definite determinantal representation of size r · d , for some r > , then the parametrizedHermite matrix H ( p ) is a sum of squares.Proof. Let p r = det( I − M ) with M of size k = rd , and denote by q ( s ) (cid:96)m the ( (cid:96), m ) -entry of M s . Put Q (cid:96)m = (cid:16) q (0) (cid:96)m , . . . , q ( d − (cid:96)m (cid:17) T ∈ R [ x ] d . Then we find k (cid:88) (cid:96),m =1 Q (cid:96)m Q T(cid:96)m = (cid:32) k (cid:88) (cid:96),m =1 q ( i − (cid:96)m q ( j − (cid:96)m (cid:33) i,j =1 ,...,d = (cid:0) tr( M i − M j − ) (cid:1) i,j =1 ,...,d = r H ( p ) , by Lemma 1.4. (cid:3) Remarks . (1) If the determinantal representation of p r is of size k > rd , then H ( p ) becomes a sum of squares after increasing the (1 , -entry from d to k/r . This is clear fromthe above proof.(2) It was shown in Netzer and Thom [11] that if a polynomial p admits a definitedeterminantal representation, then it admits one of size dn , where d is the degree of p and n is the number of variables. So if any power p r admits a determinantal representation of any size, then H ( p ) is a sum of squares, after increasing the (1 , -entry from d to dn . Notethat this is independent of r. (3) The determinant of H ( p ) is the discriminant of t d p ( t − x ) in t . If H ( p ) = Q T Q , itfollows from the Cauchy-Binet formula that the determinant of H ( p ) is a sum of squares in ETERMINANTAL REPRESENTATIONS AND THE HERMITE MATRIX 7 R [ x ] . Thus, by the above theorem, the discriminant of det( tI + M ) in t is a sum of squares,a fact that has long been known, at least since Borchardt’s work from 1846 [3].(4) The sums-of-squares decomposition of H ( p ) obtained by Thm. 1.6 from a deter-minantal representation p r = det( I − M ) is extremely special. In principle, it is possibleto characterize the decompositions of H ( p ) coming from a determinantal representation bya recurrence relation that they must satisfy. But this does not appear to be a promisingapproach for finding determinantal representations. Example . It was shown in Netzer and Thom [11] that if p is quadratic, a high enoughpower admits a definite determinantal representation of the correct size. Thus H ( p ) is a sumof squares in this case. This can also be shown directly. Write p = x T Ax + b T x + 1 with A ∈ Sym n ( R ) and b ∈ R n . Then p is a real-zero polynomial if and only if bb T − A (cid:23) ,as is easily checked. We find t p ( t − x ) = x T Ax + b T x · t + t , and so we compute H ( p ) = (cid:32) − b T x − b T x x T ( bb T − A ) x (cid:33) . Write bb T − A = (cid:80) ni =1 v i v Ti as a sum of squares of column vectors v i ∈ R n . Set Q = − b T x v T x ... ... v Tn x . Then H ( p ) = 2 · Q T Q . (cid:3) Example . We consider Brändén’s example from [4]. It is constructed from the Vámoscube as shown in Figure 1. Its set of bases B consists of all four element subsets of { , . . . , } that do not lie in one of the five affine hyperplanes. Define q := (cid:88) B ∈B (cid:89) i ∈ B x i , a degree four polynomial in R [ x , . . . , x ] . It contains as its terms the product of any choiceof four pairwisely different variables, except for the following five: x x x x , x x x x , x x x x , x x x x , x x x x . Now p = q ( x + 1 , . . . , x + 1) turns out to be a real-zero polynomial, of which Brändén hasshown that no power has a determinantal representation. TIM NETZER, DANIEL PLAUMANN, AND ANDREAS THOM
Figure 1.
The Vámos Cube1 7 25 64 38We can apply the sums-of-squares-test to the Hermite matrix H ( p ) here. Unfortunately,the matrix is too complicated to do the computations by hand. When using a numericalsums-of-squares-plugin for matlab, such as Yalmip, the result however indicates that H ( p ) isnot a sum of squares. In view of Theorem 1.6 this shows again that no power of p admits adeterminantal representation. Note that if some power p r has a determinantal representation,then it has one of size r . This was proven by Brändén or follows more generally from Netzerand Thom, Theorem 2.7 [11].Finally, we can apply the sums-of-squares-test also to small perturbations of Brändén’spolynomial. For example, p can be approximated as closely as desired by real-zero poly-nomials, which have only simple roots on each line through the origin (in other words, theHermite matrix is positive definite at each point a (cid:54) = 0 ). Such a smoothening procedure isfor example describe in Nuij [12]. Still, Yalmip reports that the Hermite matrix is not a sumof squares, if the approximation is close enough. This is exactly what one expects, since thecone of sums of squares of polynomial matrices is closed, and the Hermite matrix dependscontinuously on the polynomial.2. A general construction method
In this section we are interested in the converse of the above result. Namely, can a sums-of-squares decomposition of H ( p ) be used to produce a definite determinantal representationof p or some multiple? We describe a method to do this, which amounts to only solving asystem of linear equations.Let p = 1 + p + · · · + p d ∈ R [ x ] be a real-zero polynomial of degree d . Since the matrix H ( p ) is everywhere positive semidefinite, it can be expressed as a sum of squares if one allowsdenominators in R [ x ] . This generalization of Artin’s solution to Hilbert’s 17th problem was ETERMINANTAL REPRESENTATIONS AND THE HERMITE MATRIX 9 first proved by Gondard and Ribenboim in [6]. We need to make a slight adjustment to oursituation.
Lemma 2.1.
There exist a matrix polynomial
Q ∈
Mat k × d (cid:0) R [ x ] (cid:1) , for some k > , and ahomogeneous non-zero polynomial q ∈ R [ x ] such that q H ( p ) = Q T Q . Proof.
By the original result of Gondard and Ribenboim [6] there is some non-zero polyno-mial q ∈ R [ x ] such that q H ( p ) = Q T Q for some Q ∈
Mat k × d ( R [ x ]) . We want to make q homogeneous.Write q = q r + q r +1 + · · · + q R , where each q i is homogeneous of degree i , and q r (cid:54) = 0 , q R (cid:54) = 0 . Since the i -th diagonal entry in H ( p ) is homogeneous of degree i − , each entryin the i -th column of Q has homogeneous parts of degree between r + i − and R + i − . Let Q min be the matrix one obtains from Q by choosing only the homogeneous part of degree r + i − of each entry in each i -th column. Put (cid:101) Q = Q − Q min and note that all entries in the i -th column of (cid:101) Q have non-zero homogeneous parts only in degrees at least r + i . We nowcompute q H ( p ) = Q T min Q min + Q T min (cid:101) Q + (cid:101) Q T Q min + (cid:101) Q T (cid:101) Q , compare degrees on both sides,and find q r H ( p ) = Q T min Q min , as desired. (cid:3) We will now describe the setup that we are going to use for the rest of this section. Wefix a representation of q H ( p ) = Q T Q as in Lemma 2.1. As before, let P = t d · p ( t − x ) = t d + p t d − + · · · + p d ∈ R [ x, t ] , and consider the free R [ x ] -module A = R [ x, t ] / ( P ) ∼ = d − (cid:77) i =0 R [ x ] · t i ∼ = R [ x ] d . Since P is homogeneous, the standard grading induces a grading on A . We shift this gradingby r , the degree of q , and obtain a grading with deg( t i ) = r + i for i = 0 , . . . , d − .This turns A into a graded R [ x ] -module, where R [ x ] is equipped with the standard grading.Furthermore, we equip A with a symmetric R [ x ] -bilinear and R [ x ] -valued map (cid:104)· , ·(cid:105) p definedby (cid:104) f, g (cid:105) p := f T (cid:0) q H ( p ) (cid:1) g, for f = ( f , . . . , f d ) T and g = ( g , . . . , g d ) T in A .Next, consider the map L t : A → A given by multiplication with t . This is an R [ x ] -linearmap which we can compute with respect to our chosen basis: L t : ( f , . . . , f d ) T (cid:55)→ ( − p d f d , f − p d − f d , . . . , f d − − p f d ) T . Note that L t is of degree with respect to the grading, i.e. deg (cid:0) L t ( f ) (cid:1) = deg( f ) + 1 . Weidentify L t with the matrix that represents it, so that L t = − p d − p d − . . . ... · · · − p , which is exactly the companion matrix of P , viewed as a univariate polynomial in t . It iswell known and easy to see that P is the characteristic polynomial of L t , so that det ( I − L t ) = p Lemma 2.2.
The linear map L t is self-adjoint with respect to (cid:104)· , ·(cid:105) p , i.e. (cid:104)L t f, g (cid:105) p = (cid:104) f, L t g (cid:105) p holds for all f, g ∈ A. Proof.
We may divide by q on both sides and hence assume that q = 1 . It is enough toshow (cid:104)L t e i , e j (cid:105) p = (cid:104) e i , L t e j (cid:105) p for all i, j , where e i is the i -th unit vector. For i, j < d , thisfollows from the fact that H ( p ) is a Hankel matrix. For i = j = d , it is clear from symmetry.So assume j < i = d . We find (cid:104)L t e d , e j (cid:105) p = − d (cid:88) i =1 p d − i +1 e i H ( p ) e j = − d (cid:88) i =1 p d − i +1 N i + j − , where N k is the k -th Newton sum of P . On the other hand, we compute (cid:104) e d , L t e j (cid:105) p = (cid:104) e d , e j +1 (cid:105) p = N d + j − . In conclusion, we have to show that d (cid:88) i =0 p d − i N i + j − = 0 , where we have set p = 1 . This statement is equivalent to (cid:80) di =0 p i N k − i = 0 , where k = d + j − ≥ d . This last equation, however, follows immediately from the Newton identity kp k + (cid:80) k − i =0 p i N k − i = 0 , where we let p k = 0 for k > d . (cid:3) Let B = R [ x ] k . The k × d -matrix Q in the decomposition of H ( p ) describes an R [ x ] -linear map A = R [ x ] d → B , f (cid:55)→ Q f . From the degree structure of H ( p ) , we see that eachentry in the i -th column of Q is homogeneous of degree r + i − . So Q is of degree withrespect to the canonical grading on B . Lemma 2.3.
ETERMINANTAL REPRESENTATIONS AND THE HERMITE MATRIX 11 (1) If p is square-free, then Q : A → B is injective.(2) We have (cid:104) f, g (cid:105) p = (cid:104)Q f, Q g (cid:105) for all f, g ∈ A . In other words, Q is an isometry, taking (cid:104) , (cid:105) p to the canonicalbilinear form (cid:104) , (cid:105) on B .Proof. (2) is immediate from the fact that q H ( p ) = Q T Q . (1) If Q f = 0 , then (cid:104)Q f, Q f (cid:105) = (cid:104) f, f (cid:105) p = q · f T H ( p ) f. For each a ∈ R n for which p ( ta ) has only distinct roots, the matrix H ( p )( a ) is positivedefinite. So f ( a ) = 0 for generic a , and thus f = 0 . (cid:3) Time for a brief summary of what we have done so far.
Setup 2.4. ◦ Let p ∈ R [ x ] be a real-zero polynomial of degree d with p (0) = 1 , and let H ( p ) beits parametrized Hermite matrix. Fix a decomposition q H ( p ) = Q T Q , where q ishomogeneous of degree r and Q is a matrix of size k × d with entries in R [ x ] . ◦ We have equipped the free module A = R [ x ] d with a particular grading and with abilinear form (cid:104) , (cid:105) p : A × A → A . ◦ Let B = R [ x ] k be equipped with the canonical bilinear form and the canonical grad-ing. ◦ The map Q : A → B is an isometry and of degree . ◦ Let L t be the companion matrix of t d p ( t − x ) with respect to t , so that det( I − L t ) = p. The map L t : A → A is self-adjoint with respect to (cid:104)· , ·(cid:105) p and of degree .The following is our main result. Theorem 2.5.
Let p ∈ R [ x ] be a square-free real-zero polynomial of degree d with p (0) = 1 .Assume that there exists a homogeneous symmetric linear matrix polynomial M of size k × k such that the following diagram commutes: R [ x ] d = A Q (cid:47) (cid:47) L t (cid:15) (cid:15) B = R [ x ] k M (cid:15) (cid:15) R [ x ] d = A Q (cid:47) (cid:47) B = R [ x ] k Then p divides det( I − M ) . Remark . Note that the above described setup exactly means that we can hope for sucha linear symmetric M to exist. Indeed the "strange" symmetry of L t is transformed into thestandard symmetry by Q , and the "strange" grading is translated to the standard grading. Proof.
For generic a ∈ R n , the map Q ( a ) is injective by Lemma 2.3. Therefore, all eigenvaluesof L t ( a ) are also eigenvalues of M ( a ) . The eigenvalues of L t ( a ) are precisely the zeros of P ( t, a ) , i.e. the inverses of the zeros of p ( ta ) . So q = det( I − M ) vanishes on the zero set of p , by Proposition 1.3. Since p is a square-free real zero polynomial, the ideal ( p ) generatedby p in R [ x ] is real-radical (see Bochnak, Coste and Roy [2], Theorem 4.5.1(v)). It followsthat q is contained in ( p ) , in other words p divides q . (cid:3) Remark . Whether there exists such M can be decided by solving a system of linearequations. Indeed, set M = x M + · · · + x n M n , where the M i are symmetric matrices withindeterminate entries. The equation MQ = QL t of matrix polynomials can be consideredentrywise, and comparison of the coefficients in x gives rise to a system of linear equationsin the entries of the M i . Example . Let p ∈ R [ x ] be quadratic. Write p = x T Ax + b T x + 1 with A ∈ Sym n ( R ) and b ∈ R n . We have seen in Example 1.8 that H ( p ) admits a sums of squares decomposition if p is a real-zero polynomial, given by the matrix Q = √ · − b T x v T x ... ... v Tn x if bb T − A = (cid:80) ni =1 v i v Ti . It is now easy to find a homogeneous linear matrix polynomial M that makes the diagram in Theorem 2.5 commute, namely we can take M = 12 · − b T x v T x · · · v Tn xv T x − b T x ... . . . v Tn x − b T x . The resulting determinantal representation is det ( I − M ) = (cid:18) · b T x (cid:19) n − · p. To give an explicit example, consider p = ( x + √ − x − x − x − x , which itselfdoes not admit a determinantal representation (by Netzer and Thom [11]). The procedure ETERMINANTAL REPRESENTATIONS AND THE HERMITE MATRIX 13 just described now gives rise to the linear matrix polynomial M = −√ x x x x x x x −√ x x −√ x x −√ x x −√ x x −√ x and finally det( I − M ) = (1 + √ x ) · p. Example . There are also examples where no suitable M exists. We are grateful toRainer Sinn and Cynthia Vinzant for helping us find this example. Consider the plane cubic p = ( x − ( x + 1) − x . One computes H ( p ) = x x + 2 x x x + 2 x x + 3 x x x + 2 x x + 3 x x x + 8 x x + 2 x = Q T Q , where Q = x ax x − x bx x √ √ x √ x + x )1 − x x and a = ( √ , b = ( √ − . The equation MQ = QL t has 12 entries, each ofwhich gives rise to several linear equations by comparing coefficients in x. One can checkthat already the equations obtained from the first two rows of MQ = QL t are unsolvable.3. Rational representations of degree one
There is always a way to make the diagram from the last section commute, if one allowsfor rational linear matrix polynomials. This will lead to rational determinantal representa-tions, as described now.Let p be a square-free real-zero polynomial. Since the parametrized Hermite matrix H ( p ) evaluated at a point a ∈ R n is positive definite for generic a , the matrix polynomial H ( p ) is invertible over the function field R ( x ) . Recall that the degree of a rational function f /g ∈ R ( x ) is defined as deg( f ) − deg( g ) . Furthermore, we say that f /g is homogeneousif both f and g are homogeneous, not necessarily of the same degree. Equivalently, f /g ishomogeneous of degree d if and only if ( f /g )( λa ) = λ d ( f /g )( a ) holds for all a ∈ R n with g ( a ) (cid:54) = 0 . Theorem 3.1.
Let p be a square-free real-zero polynomial. Write q H ( p ) = Q T Q with q homogeneous as in Lemma 2.1 and let M := q − QL t H ( p ) − Q T . The matrix M is symmetric with entries in R ( x ) homogeneous of degree , and satisfies det( I − M ) = p. Proof.
Abbreviate H ( p ) by H and L t by L . By Sylvesters determinant theorem, we have det( I k − AB ) = det( I d − BA ) for any matrix polynomials A of size k × d and B of size d × k .In our situation, this yields det( I k − M ) = det( I k − q − QLH − Q T ) = det( I d − q − LH − Q T Q ) = det( I d − L ) = p. We find M T = q − Q ( H − ) T L T Q T = q − QLH − Q T = M , where we have used L T H = H T L , which is Lemma 2.2. Thus M is symmetric.Let r be the degree of q . By examining the degree structure of q H , we find Q ( λa ) = Q ( a ) · diag( λ r , λ r +1 , . . . , λ r + d − ) H ( λa ) = diag( λ , . . . , λ d − ) · H ( a ) · diag( λ , . . . , λ d − ) L ( λa ) = diag( λ d , . . . , λ ) · L ( a ) · diag( λ − d +1 , λ − d +2 . . . , λ ) for all a ∈ R n and λ (cid:54) = 0 . Hence for all a ∈ R n for which H ( a ) is invertible and q ( a ) (cid:54) = 0 ,and all λ (cid:54) = 0 , we have M ( λa ) = λ − r q ( a ) − Q ( a ) · diag( λ r , . . . , λ r + d − ) · diag( λ d , . . . , λ ) L ( a ) · diag( λ − d +1 , . . . , λ ) · diag( λ , . . . , λ − d +1 ) · H − ( a ) · diag( λ , . . . , λ − d +1 ) · diag( λ r , . . . , λ r + d − ) · Q T ( a )= λ − r q ( a ) − Q ( a ) λ r + d L ( a ) λ − d +1 H − ( a ) λ r Q T ( a )= λ · M ( a ) . (cid:3) Remark . Note that a representation p = det ( I − M ) as in Theorem 3.1 gives an alge-braic certificate for p being a real-zero polynomial. Since p ( ta ) = det ( I − t M ( a )) , usinghomogeneity, the zeros of p ( ta ) are just the inverses of the eigenvalues of M ( a ) . Since M is symmetric, all of these zeros are real. Theorem 3.1 now states that such an algebraiccertificate exists for each real-zero polynomial p . ETERMINANTAL REPRESENTATIONS AND THE HERMITE MATRIX 15
Example . Consider the quadratic polynomial p = ( x + 1) − x − x − x . We have H = (cid:18) − x − x x + x + x + x ) (cid:19) = Q T Q with Q T = (cid:18) √ −√ x √ x √ x √ x (cid:19) , which results in M = − x x x x x − x x x + x + x − x x x x + x + x − x x x x + x + x x − x x x x + x + x − x x x + x + x − x x x x + x + x x − x x x x + x + x − x x x x + x + x − x x x + x + x . References [1] S. Basu and R. Pollack and M.-F. Roy,
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A note on hyperbolic polynomials , Math. Scand. (1968), 69–72 (1969). ↑ Symmetric Determinantal Representation of Polynomials , Preprint. ↑ Complete description of determinantal representations of smooth irreducible curves , LinearAlgebra Appl. (1989), 103–140. ↑ Tim Netzer, Universität Leipzig, Germany
E-mail address : [email protected] Daniel Plaumann, Universität Konstanz, Germany
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Andreas Thom, Universität Leipzig, Germany
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