Concrete Solution to the Nonsingular Quartic Binary Moment Problem
aa r X i v : . [ m a t h . F A ] D ec CONCRETE SOLUTION TO THE NONSINGULARQUARTIC BINARY MOMENT PROBLEM
RA ´UL E. CURTO AND SEONGUK YOO
Abstract.
Given real numbers β ≡ β (4) : β , β , β , β , β , β , β , β , β , β , β , β , β , β , β , with β >
0, the quartic real momentproblem for β entails finding conditions for the existence of a positive Borelmeasure µ , supported in R , such that β ij = R s i t j dµ (0 ≤ i + j ≤ M (2) be the 6 × β (4) , given by M (2) i , j := β i + j , where i , j ∈ Z and | i | , | j | ≤
2. In this note we find concrete representing measuresfor β (4) when M (2) is nonsingular; moreover, we prove that it is possible toensure that one such representing measure is 6-atomic. Introduction
In this paper we find a direct proof that the nonsingular
Quartic Binary MomentProblem always admits a finitely atomic representing measure with the minimumnumber of atoms, that is, six atoms. We do this in three steps:(i) by normalizing the given moment matrix M (2) to ensure that M (1) is theidentity matrix;(ii) by developing a new rank-reduction tool, which allows us to decompose thenormalized M (2) matrix as the sum of a positive semidefinite moment matrix ^ M (2) of rank 5 and the rank-one moment matrix of the point mass at the origin;and(iii) by proving that when a moment matrix M (2) admits such a decomposition,and ^ M (2) admits a column relation subordinate to a degenerate hyperbola (i.e., apair of intersecting lines), then M (2) admits a 6-atomic representing measure (asopposed to the expected 7-atomic measure).To describe our results in detail, we need some notation and terminology. Givenreal numbers β ≡ β (4) : β , β , β , β , β , β , β , β , β , β , β , β , β , β , β , with β >
0, the
Quartic Real Moment Problem for β entails findingconditions for the existence of a positive Borel measure µ , supported in R , suchthat β ij = R s i t j dµ (0 ≤ i + j ≤ M (2) be the moment matrix for β (4) ,given by M (2) i , j := β i + j , where i , j ∈ Z and | i | , | j | ≤
2; this 6 × M (2) are labeled , X, Y, X , XY, Y . Ina similar way, given a collection of real numbers β (2 n ) one defines the associated Mathematics Subject Classification.
Primary 47A57, 44A60, 42A70, 30A05; Secondary15A15, 15-04, 47N40, 47A20.
Key words and phrases.
Nonsingular quartic binary moment problem, moment matrix exten-sion, flat extensions, rank-one perturbations, invariance under degree-one transformations.The first named author was supported by NSF Grants DMS-0801168 and DMS-1302666. moment matrix by M ( n ) i , j := β i + j , where i , j ∈ Z and | i | , | j | ≤ n .) M (2) ≡ β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β . Assume now that M (2) is nonsingular. A straightforward consequence of Hilbert’sTheorem yields the existence of a finitely atomic representing measure, as follows.Let P be the cone of nonnegative polynomials of degree at most 4 in x, y , regardedas a subset of R . The dual cone is P ∗ := { ξ ∈ R : h ξ, p i ≥ p ∈ P } .If a, b ∈ R and ξ ( a,b ) := (1 , a, b, a , ab, b , a , a b, ab , b , a , a b, a b , ab , b ) ∈ R ,then (cid:10) ξ ( a,b ) , p (cid:11) = p ( a, b ) ≥
0, for all p ∈ P . Thus, ξ ( a,b ) ∈ P ∗ for all a, b ∈ R ,and ξ ( a,b ) is also an extreme point. Consider now an arbitrary moment sequence β (4) with a nonsingular moment matrix M (2). Regarded as a point in R , β (4) is in the interior of P ∗ , since every p ∈ P is a sum of squares of polynomials.By the Krein-Milman Theorem and Carath´eodory’s Theorem, the Riesz functionalΛ β (4) is a convex combination of evaluations ξ ( a,b ) ; that is, β (4) admits a finitelyatomic representing measure, with at most 15 atoms. (In recent related work,L.A. Fialkow and J. Nie [FiNi] have obtained this result as a consequence of a moregeneral result on moment problems.)In this note we obtain a concrete 6-atomic representing measure for M (2). TheQuartic Real Binary Moment Problem admits an equivalent formulation in termsof complex numbers and representing measures supported in the complex plane C ,as follows. Given complex numbers γ ≡ γ (4) : γ , γ , γ , γ , γ , γ , γ , γ , γ , γ , γ , γ , γ , γ , γ , with γ ij = ¯ γ ji , one seeks necessary and sufficientconditions for the existence of a positive Borel measure µ , supported in C , suchthat γ ij = Z ¯ z i z j dµ (0 ≤ i + j ≤ . Just as in the real case, the Quartic Complex Moment Problem has an associatedmoment matrix M (2), whose columns are conveniently labeled 1 , Z, ¯ Z, Z , ¯ ZZ, ¯ Z .The most interesting case of the Singular Quartic Binary Moment Problem ariseswhen the rank of M (2) is 5, and the sixth column of M (2), labeled ¯ Z , is a linearcombination of the remaining five columns. Depending on the coefficients in thelinear combination, four subcases arise in terms of the associated conic C [CuFi3,Section 5]: (i) C is a parabola; (ii) C is a nondegenerate hyperbola; (iii) C is a pairof intersecting lines; and (iv) C is a circle. In subcase (iii), it is possible to provethat the number of atoms in a representing measure (if it exists) may be 6 [CuFi3,Proposition 5.5 and Example 5.6]; that is, in some soluble cases the rank of M (2)may be strictly smaller than the number of atoms in any representing measure. Proposition 1.1. ( [CuFi3, Proposition 5.5] ) If M (2) ≥ , if rank M (2) = 5 , andif XY = 0 in the column space of M (2) , then M (2) admits a representing measure µ with card supp µ ≤ . When combined with previous work on truncated moment problems, Proposi-tion 1.1 led to the following solution to the truncated moment problem on pla-nar curves of degree ≤
2. Given a moment matrix M ( n ) and a polynomial ONCRETE SOLUTION TO THE NONSINGULAR QUARTIC BINARY MOMENT PROBLEM3 p ( x, y ) ≡ P p ij x i y j , we let p ( X, Y ) := P p ij X i Y j . A column relation in M ( n ) istherefore always described as p ( X, Y ) = 0 for some polynomial p , with deg p ≤ n .We say that M ( n ) is recursively generated if for every p with p ( X, Y ) = 0 andevery q such that deg pq ≤ n one has ( pq )( X, Y ) = 0. In what follows, v denotesthe cardinality of the associated algebraic variety, defined as the intersection of thezero sets of all polynomials which describe the column relations in M ( n ). Theorem 1.2. ( [CuFi4, Theorem 2.1] , [Fia, Theorem 1.2] ) Let p ∈ R [ x, y ] , with deg p ( x, y ) ≤ . Then β (2 n ) has a representing measure supported in the curve p ( x, y ) = 0 if and only if M ( n ) has a column dependence relation p ( X, Y ) = 0 , M ( n ) ≥ , M ( n ) is recursively generated, and r ≤ v . The proof of Theorem 1.2 made use of affine planar transformations to reduce ageneric quadratic column relation to one of four canonical types: Y = X , XY = 1, XY = 0 and X + Y = 1; each of these cases required an independent result. Weshall have occasion to use the affine planar transformation approach in Section 3.To date, most of the existing theory of truncated moment problems is foundedon the presence of nontrivial column relations in the moment matrix M ( n ). Onone hand, when all columns labeled by monomials of degree n can be expressed aslinear combinations of columns labeled by monomials of lower degree, the matrix M ( n ) is flat, and the moment problem has a unique representing measure, whichis finitely atomic, with exactly rank M ( n −
1) atoms [CuFi1, Theorem 1.1]. Asa straightforward consequence, we conclude that for n = 1, an invertible M ( n )always admits a flat extension, while that is not the case for n ≥
3; that is, thereexist examples of positive and invertible M (3) without a representing measure (cf.[CuFi1, Section 4]).When n = 2, the idea is to extend the 6 × M (2) to a bigger10 ×
10 moment matrix M (3) by adding so-called B and C blocks, as follows: M (3) ≡ (cid:18) M (2) B (3) B (3) ∗ C (3) (cid:19) . A result of J.L. Smul’jan [Smu] states that M (3) ≥ M (2) ≥ B (3) = M (2) W for some W ; and (iii) C (3) ≥ W ∗ M (2) W . Moreover, M (3) is a flat extension of M (2) (i.e., rank M (3) = rank M (2)) if and only if C (3) = W ∗ M (2) W . Further, when M (2) is invertible, one easily obtains W = M (2) − B (3), so in the flat extension case C (3) can be written as B (3) ∗ M (2) − B (3).However, writing a general formula for M (3) is nontrivial, even with the aid of Mathematica , because of the complexity of ( M (2)) − and the new moments con-tributed by the block B (3). On the other hand, if only one column relation ispresent (given by p ( X, Y ) = 0), then v = + ∞ , and the condition r ≤ v , while nec-essary, will not suffice. One knows that the support of a representing measure mustlie in the zero set of p , but this does not provide enough information to decipherthe block B (3). The situation is much more intriguing when no column relationsare present; this is the nonsingular case, for which very little is known.2. Statement of the Main Result
Theorem 2.1.
Assume M (2) is positive and invertible. Then M (2) admits arepresenting measure, with exactly atoms; that is, M (2) actually admits a flatextension M (3) . RA´UL E. CURTO AND SEONGUK YOO
The proof of Theorem 2.1 is constructive, in that we first prove that it is alwayspossible to switch from the invertible M (2) to a related singular matrix ^ M (2), withrank ^ M (2) = 5, for which Theorem 1.2 applies. Since singular positive semidefinitematrices M (2) always admit representing measures with 6 atoms or less, we canthen conclude that an invertible positive M (2) admits a representing measure withat most 7 atoms. While this would already represent a significant improvementon the upper bound given by Carath´eodory’s Theorem (15 atoms), we have beenable to establish that all positive invertible M (2)’s actually have flat extensions,and therefore their representing measures can have exactly 6 atoms.3. A New Tool
We begin this section with a result that will allow us to convert a given momentproblem into a simpler, equivalent, moment problem. One of the consequencesof this result is the equivalence of the real and complex moment problems, via thetransformation x := Re[ z ] and y := Im[ z ]; this equivalence has been exploited amplyin the theory of truncated moment problems. For us, however, this simplificationwill allow us to assume that the submatrix M (1) is the identity matrix.We adapt the notation in [CuFi3] to the real case. For a, b, c, d, e, f ∈ R , bf − ce = 0, let Ψ( x, y ) ≡ (Ψ ( x, y ) , Ψ ( x, y )) := ( a + bx + cy, d + ex + f y ) ( x, y ∈ R ). Given β (2 n ) , define ˜ β (2 n ) by ˜ β ij := L β (Ψ i Ψ j ) (0 ≤ i + j ≤ n ), where L β denotes the Riesz functional associated with β . (For p ( x, y ) ≡ P p ij x i y j , the Rieszfunctional is given by L β ( p ) := p ( β ) ≡ P p ij β ij .) It is straightforward to verifythat L ˜ β ( p ) = L β ( p ◦ Ψ) for every p of degree at most n . Proposition 3.1. (Invariance under degree-one transformations; [CuFi3])
Let M ( n ) and ˜ M ( n ) be the moment matrices associated with β and ˜ β , and let J ˆ p := [ p ◦ Ψ . Then the following statements hold. (i) ˜ M ( n ) = J ∗ M ( n ) J . (ii) J is invertible. (iii) ˜ M ( n ) ≥ ⇔ M ( n ) ≥ . (iv) rank ˜ M ( n ) = rank M ( n ) . (v) M ( n ) admits a flat extension if and only if ˜ M ( n ) admits a flat extension. We are now ready to put M (2) in “normalized form.” Without loss of generality,we always assume that β = 1. Let d i denote the leading principal minors of M (2);in particular, d = − β + β d = − β β + 2 β β β − β − β β + β β . Consider now the degree-one transformationΨ( x, y ) ≡ ( a + bx + cy, d + ex + f y ) , where a := β β − β β √ d d , b := β − β β √ d d , c := − q d d , d := − β √ d , e := √ d ,and f := 0. Note that bf − ce = − q d = 0. Using this transformation, and astraightforward calculation, we can prove that any positive definite moment matrix ONCRETE SOLUTION TO THE NONSINGULAR QUARTIC BINARY MOMENT PROBLEM5 M (2) can be transformed into the moment matrix β ˜ β ˜ β β ˜ β ˜ β β ˜ β ˜ β ˜ β ˜ β β ˜ β ˜ β ˜ β ˜ β β ˜ β ˜ β ˜ β ˜ β . Thus, without loss of generality, we can always assume that M (1) is the identitymatrix. We will now introduce a new tool in the study of moment matrices: thedecomposition of an invertible M (2) as a sum of a moment matrix of rank 5 and arank-one moment matrix.Assume now that M (2) is invertible and that the submatrix M (1) is the identitymatrix. For u ∈ R decompose M (2) as follows: M (2) = − u β β β β β β β β β β β β β β β β β β β β β + u . Denote the first summand by \ M (2) and the second summand by P . It is clearthat P is positive semidefinite and has rank 1 if and only if u >
0, and in that case P is the moment matrix of the 1-atomic measure uδ (0 , , where δ (0 , is the pointmass at the origin. Proposition 3.2.
Let M (2) , \ M (2) and P be as above, and let u := det M (2) R ,where R is the (1 , entry in the positive matrix R := ( M (2)) − . Then, withthis nonnegative value of u , we have (i) \ M (2) ≥ ; (ii) rank \ M (2) = 5 ; and (iii) \ M (2) is recursively generated. Moreover, u is the only value of u for which \ M (2) satisfies (i)–(iii). For the proof of Proposition 3.2 we will need to following auxiliary result, whichis an easy consequence of the multilinearity of the determinant.
Lemma 3.3.
Let M be an n × n invertible matrix of real numbers, let E be therank-one matrix with (1 , -entry equal to and all other entries are equal to zero,and let u ∈ R . Then det( M − uE ) = det M − u det M { , , ··· ,n } , where M { , , ··· ,n } denotes the ( n − × ( n − compression of M to the last n − rows and columns.In particular, if u = det M ( M − ) , then det( M − uE ) = 0 .Proof of Proposition 3.2. (ii) Observe that 6 = rank M (2) ≤ rank \ M (2)+rank P =rank \ M (2) + 1, so rank \ M (2) ≥ . Since det \ M (2) = 0, we have rank \ M (2) = 5.(i) Using the Nested Determinant Test starting at the lower right-hand cornerof \ M (2), we know that \ M (2) is positive semidefinite since the nested determinantscorresponding to principal minors of size 1, 2, 3, 4 and 5 are all positive, and therank of \ M (2) is 5. This also implies that 1 − u ≥
0. We now claim that 1 − u is strictly positive. If 1 − u = 0, then the positive semidefiniteness of \ M (2) would RA´UL E. CURTO AND SEONGUK YOO force all entries in the first row to be zero. Since this is evidently false, we concludethat 1 − u > \ M (2) are linearlyindependent. Consider the third leading principal minor of \ M (2), which equals 1 − u , and is therefore positive. Thus, there is no linear dependence in this submatrix,and as a result the same holds in \ M (2).Finally, the uniqueness of u as the only value satisfying (i)–(iii) is clear. (cid:3) Proof of the Main Result
We first observe that by combining Proposition 3.2 with Theorem 1.2, it sufficesto consider the case when \ M (2) has a column relation corresponding to a pair ofintersecting lines. For, in all other cases, there exists a representing measure for \ M (2) with exactly five atoms; when combined with the additional atom comingfrom the matrix P , we see that M (2) admits a 6-atomic representing measure.We thus focus on the case when \ M (2) is subordinate to a degenerate hyper-bola. After applying an additional degree-one transformation, we can assume, asin Proposition 1.1, that the column relation XY = 0 is present in \ M (2). However,we may not continue to assume that the submatrix \ M (1) is the identity matrix,since the degree-one transformation that produces the column relation XY = 0will, in general, change the low-order moments. That is, \ M (2) is of the form \ M (2) = a b c da c e b d fc e g d f h . In this case, the original moment matrix M (2) is written as M (2) = \ M (2) + u (cid:0) p q p p q q (cid:1) T (cid:0) p q p p q q (cid:1) , for some u > p q = 0. That is, M (2) is the sum of a moment matrix of rank 5with column relation XY = 0 and a positive scalar multiple of the moment matrixassociated with the point mass at ( p, q ), with pq = 0. Without loss of generality,we can assume that p = q = 1 (this requires an obvious degree-one transformation,i.e., ˜ x := xp , ˜ y := yq ). As a result, the form of M (2) is now as follows: M (2) = u a + u b + u c + u u d + ua + u c + u u e + u u ub + u u d + u u u f + uc + u e + u u g + u u uu u u u u ud + u u f + u u u h + u We will show that M (2) admits a flat extension, and that will readily imply that itadmits a rank M (2)-atomic (that is, 6-atomic) representing measure. The B (3)-block in an extension M (3) can be generated by letting β = β = β = β = u , ONCRETE SOLUTION TO THE NONSINGULAR QUARTIC BINARY MOMENT PROBLEM7 so that B (3) can thus be written as e + u u u f + ug + u u u uu u u h + uβ u u uu u u uu u u β . As usual, let W := M (2) − B (3) and let C (3) ≡ ( C ij ) := W ∗ M (2) W . Note that if C (3) turns out to be Hankel, then M (3) is a flat extension of M (2). Since C (3) issymmetric, to ensure that C (3) is Hankel (and therefore M (3) is a moment matrix)we only need to solve the following system of equations: E := C − C = 0 E := C − C = 0 E := C − C = 0 . (4.1)This is a system of equations involving quadratic polynomials with 2 unknownvariables (the new moments β and β ). A straightforward calculation showsthat E = 0, E = 0, and that E = 0 ⇐⇒ ( c − ae )( d − bf ) β β + ( c − ae )( f − df h + bh − d u + bf u ) β +( d − bf )( e − ceg + ag − c u + aeu ) β +( e − ceg + ag − c u + aeu )( f − df h + bh − d u + bf u ) = 0 ⇐⇒ κλβ β + κµβ + λνβ + νµ = 0 , where κ , λ , µ and ν have the obvious definitions. If κ, λ = 0, then β = − µν + κµβ κλβ + λν (for β = − νκ ), which readily implies that E = 0 admits infinitely many solutions.When κ = 0 and λ = 0, we see that E = λνβ + µν , from which it follows thata solution always exists (and it is unique when ν = 0). A similar argument showsthat κ = 0 and λ = 0 also yields a solution (which is unique when µ = 0). Weare thus left with the case when both κ ≡ c − ae and λ ≡ d − bf are equal tozero. Since c and d are in the diagonal of a positive semidefinite matrix, they mustbe positive. Thus, all of a, b , e , and f are nonzero and we can set e := c /a and f := d /b . In this case, the moment matrix is M (2) = u a + u b + u c + u u d + ua + u c + u u c a + u u ub + u u d + u u u d b + uc + u c a + u u g + u u uu u u u u ud + u u d b + u u u h + u Let k := det M (2) / det M (2) { , , , , } . As in the proof of Proposition 3.2, we seethat k = − b c − a d + cdcd > RA´UL E. CURTO AND SEONGUK YOO M (2) has rank 5 and is positive semidefinite (note that the (1 , u − k ): M (2) = b c + a d + cducd a + u b + u c + u u d + ua + u c + u u c + aua u ub + u u d + u u u d + bub c + u c + aua u g + u u uu u u u u ud + u u d + bub u u h + u + k The only column relation in the first summand is(4.2) XY = cd − bc − ad + cd − ad − bc − ad + cd X − bc − bc − ad + cd Y =: ξ − ηX − θY. Unless ηθ = − ξ , the conic that represents this column relation is a nondegeneratehyperbola, and therefore the moment sequence associated to the moment matrixhas a 5-atomic measure, by Theorem 1.2. In the case when the conic in (4.2) is apair of intersecting lines (i.e., ( x + θ )( y + η ) = 0), we must have c = a or d = b .Thus, the remaining two specific cases to cover are M (2) with c = a or d = b .Since M (2) is invertible, for any B (3) block we will be able to find W such that M (2) W = B (3). We propose to use a B (3) block with new moments β = β = β = 0, and to then extend M (2) to M (3) using Smul’jan’s Lemma, that is, wewill define C (3) := W ∗ B (3). The goal is to establish that C (3) is a Hankel matrix,and that requires verification of (4.1). Before we begin our detailed analysis, weneed to make a few observations.Let d i denote the principal minor of M (2) for i = 1 , . . . ,
6; since M (2) is positiveand invertible, we know that these minors are all positive. Then d = − (cid:0) b c + a d − cd (cid:1) (cid:0) − c + a g (cid:1) ua and d = d (cid:0) − d + b h (cid:1) b , which implies (cid:0) b c + a d − cd (cid:1) (cid:0) − c + a g (cid:1) < − d + b h > . (4.3)Next, we use Mathematica to solve E = 0 for β and E = 0 for β , and weobtain β = 1 a c ( b c + a d − cd ) ( − d + b h ) u ( α β + α β + α ) ,β = 1 b d ( b c + a d − cd ) ( c − a g ) ( α β + α ) , where the α ij ’s are polynomials in a, b, c, d, g, h, and u . Since a, b = 0, c, d > β and β above are well defined. We nowsubstitute these values in E and check that E is a quadratic polynomial in β ;indeed, we can readily show that the leading coefficient of E is nonzero if c = a or d = b . Thus, if the discriminant ∆ of this quadratic polynomial is nonnegative,then (4.1) has at least one solution. We are now ready to deal with the two specialcases: c = a and d = b . If c = a , then∆ = a u ( a − g ) (cid:0) − d + b h (cid:1) F ( a, b, d, h ) b d , where F ( a, b, d, h ) = ( − a ) b h + 2 b d (cid:0) b − d + 3 ad (cid:1) h − d (cid:0) b − d + 4 ad (cid:1) ONCRETE SOLUTION TO THE NONSINGULAR QUARTIC BINARY MOMENT PROBLEM9 is a concave upward quadratic polynomial in h . Notice that ∆ ≥ F ≥
0, which means that the discriminant of F , ∆ := 16 b d (cid:0) b − d + ad (cid:1) ,needs to be zero or negative. In this case, we observe that c = a > ,d = − ab + ad − a d + au − a u − b u + du − adu > ,d = − d ( a − g ) > ⇒ a − g < ,d = a (cid:0) b − d + ad (cid:1) ( a − g ) u > , which leads to b − d + ad <
0. Therefore, ∆ < > d = b , then∆ = (cid:0) − c + a g (cid:1) ( b − h ) u F ( a, b, c, h ) a , where F ( a, b, d, h ) = ( − b ) h c + 2 a ( − b ) (cid:0) − b + 3 b h + ah − bh (cid:1) c + a (cid:0) − ab + b + 6 ab h − b h + a h − abh + b h (cid:1) is a concave upward quadratic polynomial in c . The discriminant of F is ∆ :=16 a ( − b ) b ( b − h ) ; we observe that d = b > d = − d ( b − h ) >
0, whichleads to b − h <
0. Therefore, ∆ < >
0, which completes the proof. (cid:3)
Acknowledgments . The authors are deeply grateful to the referee for manysuggestions that led to significant improvements in the presentation. Many of theexamples, and portions of the proofs of some results in this paper were obtainedusing calculations with the software tool
Mathematica [Wol].
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Department of Mathematics, The University of Iowa, Iowa City, Iowa 52242
E-mail address : [email protected] Department of Mathematics, Seoul National University, Seoul 151-742, Korea
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