Vandermonde Matrices for Intersection Points of Curves

This preprint is in final form. It appeared in: Jaén J. Approx. (009) 1, 67 81 Vandermonde Matrices for Intersection Points of Curves Hakop Hakopian, Kurt Jetter and Georg Zimmermann Abstract We reconsider and reprove Cayley-Bacharach type theorems, as discussed in [1], Theorems CB1-CB5. Our proof naturally uses Max Noether s famous basis theorem for finitely generated polynomial ideals, but in addition only envolves rank properties of Vandermonde matrices. In this way, we even arrive at a slight extension of Theorem CB5. Applications to poised or non-poised problems of multivariate polynomial interpolation are also discussed. Keywords: Cayley-Bacharach, Max Noether s basis theorem, Hilbert function, polynomial interpolation, poised, multivariate Mathematics Subject Classification (000). Primary: 41A05, 41A63; Secondary: 14H50. Introduction The purpose of this short note is a new proof and a slight extension of the Cayley- Bacharach type theorems as stated, e.g., in [1], Theorems CB1-CB5. Our analysis is based, naturally, on Max Noether s famous AF + BG -theorem, but in addition will only use rank properties of Vandermonde matrices, and properties of point sets which will be derived from these. We will not rely on the famous Riemann-Roch theorem, nor do we focus on an extension to more abstract versions of Cayley-Bacharach (as paper [1] does). Instead, we will be satisfied with a rather straightforward understanding of these problems in algebraic geometry from a classical point of view. For simplicity, we will deal only with bivariate problems. There are rather early versions of Cayley-Bacharach theorems, one of them the classical Pascal s Theorem: If a hexagon is inscribed in a conic, then the three opposite sides meet in collinear points. A special case of this, when the conic degenerates to a set of two lines, is the famous antique Pappus Theorem. Let us give a version of these results in our notation which implies both Pascal s Theorem and its converse statement. 1

Hakop Hakopian, Kurt Jetter and Georg Zimmermann Theorem. Assume that two cubics, p and q, intersect at nine points, and let X = {x 1,x,...,x 9 } R, X = U U c with U c = X \ U and #U = 6. Then the six points of U lie on a conic if and only if the remaining three points of U c are collinear. Figure 1 illustrates this result: On the left, the hexagon ABCDEF is inscribed in the dotted ellipse. The sides span the lines l 1 = l(a,b), l = l(b,c),..., l 6 = l(f,a), and pairs of opposite lines intersect in the points A = l(a,b) l(d,e), B = l(b,c) l(e,f) and C = l(c,d) l(f,a), respectively. These three points lie on the dashed line. On the right, however, point D has been moved off the ellipse; and consequently, A, B, and C are no longer collinear. l 6 l 1 l 5 l 3 A F C l 6 l 1 l 5 l 3 A F C l B B C l B B C l 4 A E D l 4 A E D Figure 1: Pascal s Theorem with p = l l 4 l 6 and q = l 1 l 3 l 5. A generalization of this result, for the case that X consists of N = m n different points which are the common points of intersection of two algebraic curves, of degree m and n, respectively, will be given in Theorem 8 and Theorem 9 below. To this end, we will elaborate on properties of such point sets X and of subsets thereof, such as being independent, or complete, or basic with respect to the space of algebraic polynomials of a certain degree k. These properties can be analyzed using properties of the ideal of all polynomials vanishing on X, and of its Hilbert function, and we introduce these notations and develop the needed results in our first two sections. The main part of the paper is section 3 where we present and prove a series of propositions and the two main theorems. In this way, we arrive at a version of Cayley- Bacharach which is slightly more general than Theorem CB5 in [1]. This extension makes heavy use of formula (13), which was apparently missed in reference [1].

Vandermonde Matrices for Intersection Points of Curves 3 Cayley-Bacharach theorems have interesting applications in problems on multivariate polynomial interpolations, a subject from Approximation Theory where the present authors are originally interested in; see [] and the references there. We are optimistic that further applications to actual problems in our subject will follow, and that our approach will be helpful in further investigations. 1. Notation and elementary facts We are going to deal with a finite set of points, X = {x 1,x,...,x N } R or C, (1) and with spaces Π and Π k, respectively, of all algebraic polynomials, and of polynomials of total degree less or equal k. Let P X = {p Π : p X = 0} and P k,x = P X Π k () be the ideal of all polynomials vanishing on X and, respectively, its restriction to polynomials of degree at most k. The Hilbert function of the ideal is then defined as { dim Π k dim P k,x, k = 0,1,..., H(k,X) = (3) 0, k = 1,,.... For k 0, it is the dimension of the factor space Π k /P k,x of equivalence classes of polynomials which take the same values on X. When identifying P k,x as the set of solutions of a homogeneous system of linear equations, the following interpretation of the Hilbert function is immediate. Let B k (x) = {x α : α = (α 1,α ) N 0, α := α 1 + α k}, k = 0,1,..., (4) be the monomial basis of Π k ; here we use multi-index notation. The Vandermonde matrix B k (x 1 ) B k (x ) V k (X) = (5). B k (x N ) is an N N k -matrix, with N = #X the number of rows, and ( ) k + N k = dim Π k = the number of columns, and we have H(k,X) = rank V k (X). (6) Hence, the Hilbert function is monotonely non-decreasing, if the number k increases, or if the set X is enlarged.

4 Hakop Hakopian, Kurt Jetter and Georg Zimmermann 1.1 Independent points If V k (X) has full row rank, i.e., if rank V k (X) = H(k,X) = #X, we say that the points X are Π k -independent. Otherwise, X has positive Π k -excess, where the Π k - excess as a non-negative integer is defined by E k (X) := #X rank V k (X). (7) Π k -independence of X is equivalent to the fact that the interpolation problem p Π k and p(x) = c x, x X, is always solvable, for any chosen data (c x ) x X. However, the solution is not unique unless #X = dim Π k (see below). The following results are clear: Lemma 1. For given k = 0, 1,... the following hold true: (i) A point set X with positive Π k -excess can be reduced to a subset X = X \ {x}, for appropriate x X, such that H(k,X ) = H(k,X) and E k (X ) = E k (X) 1. (ii) A Π k -independent point set X with #X < dim Π k can be enlarged to a superset X = X {x } which is also Π k -independent, whence satisfies H(k,X ) = H(k,X) + 1. In particular, by applying these statements several times, we see that, for given k, any set X with positive Π k -excess can be reduced to a Π k -independent subset without changing the value of the Hilbert function. And likewise, a Π k -independent set X with #X < dim Π k can be enlarged to a superset Y with #Y = dim Π k, such that the Vandermonde matrix V k,y is (square and) regular. It is this situation where and the interpolation problem rank V k (Y ) = #Y = dim Π k, p Π k and p(y) = c y, y Y, is always uniquely solvable, for any real (or complex) data c y, y Y. Such interpolation problems will be called to be Π k -poised.

Vandermonde Matrices for Intersection Points of Curves 5 1. Complete and basic sets of points If X and k, and hence H(k,X) are fixed, we can look at values of H(k,U), for subsets U X. If H(k,U) = H(k,X) for U X, (8) we call U to be Π k -complete in X, and if in addition U is Π k -independent, we say that U is Π k -basic in X. In terms of the corresponding ideals, we have that P k,x P k,u for U X ; here, stands for the relation of being a linear subspace. Completeness means that or in other words, P k,x = P k,u for U X, p Π k and p U = 0 = p X = 0, and basic sets are minimal subsets of X with this property. In terms of the Vandermonde matrices, complete (and basic) subsets refer to an appropriate selection of a (minimal) number of rows in the matrix which span the row space, i.e., rank V k (U) = rank V k (X) for U X. For later reference, we state some facts about subsets U of X. Lemma. Given a set X as in (1), the following hold true for k = 0,1,...: (i) There are Π k -basic subsets U X, and each has cardinality #U = H(k,X). (ii) If U is Π k -independent, then #U H(k,X). In case of strict inequality, U can be enlarged to a Π k -basic subset of X; in case of equality, U is Π k -basic in X. (iii) If U is Π k -complete in X, then #U H(k,X). In case of strict inequality, U can be reduced to a Π k -basic subset of X; in case of equality, U is Π k -basic in X. Complementary to the notion of Π k -excess for the set X, we call D k (U,X) := dim P k,u /P k,x = dim P k,u dim P k,x = H(k,X) H(k,U) (9) the Π k -defect of U modulo X, for U X. Theorem 10 below will refer to this terminology.

6 Hakop Hakopian, Kurt Jetter and Georg Zimmermann. The Hilbert function for intersection points It is well known that, in general, the Hilbert function for the ideal P X is not known a priori. A different situation occurs when the ideal has a basis of two elements. Here, we consider the special case that X = X 1 X = {x 1,x,...,x N } with N = m n, (10) where X 1 and X are the zero sets of two polynomials F m and G n, of degree m and n, respectively, i.e., we have F m Π m, X 1 = {x C : F m (x) = 0} and G n Π n, X = {x C : G n (x) = 0}. In this case, the Hilbert function is well-known, see [1], section 1. or [3], where the Hilbert function is used for describing extremal signatures in problems of multivariate Chebyshev approximation. And the reason for this is the fact that the ideal P X can be represented according to Max Noether s fundamental theorem: Lemma 3. For k = 0,1,... we have p k P k,x if and only if it can be represented as p k = A k m F m + B k n G n with A k m Π k m and B k n Π k n. In case k m + n 1, this representation is unique. Here, A k m = 0 for k < m and B k n = 0 for k < n. The representation allows a counting argument to determine the dimension of P k,x : ( ) k m + dim P k,x = + ( k n + ) ( ) k n m +, (11) with the usual convention that ( ν ) = 0 for an integer ν <. The value of the Hilbert function thus follows from (3). We state the result for later use. Lemma 4. Let X = {x 1,x,...,x N }, with N = m n, be the intersection points of two curves of degree m and n respectively, and assume that k m+n 1. Then, the value of the Hilbert function is given by and we have the identity ( ) k + H(k,X) = ( k m + ) ( ) k n +, (1) H(k,X) + H(m + n k 3,X) = m n. (13)

Vandermonde Matrices for Intersection Points of Curves 7 k = 1 k = 3 k = 6 k = 8 k = 9 Figure : Illustration of equation (13) for #X = m n = 10 4. Eq. (13) is a binomial identity which can be verified in a straightforward way by a simple, but lengthy case study. However, the following interpretation of the Hilbert function makes the identity immediately transparent. The number ( ) k+ equals the number of points in a triangular array of points having k + 1 points at the horizontal and at the vertical side. From this triangular array we cut off two smaller triangular arrays, from the top and from the right, for k m or k n, resulting with the number of points given by (1). If the complementary value of the Hilbert function in (13) is interpreted in the same way, we see that the set of m n points is divided into two complementary sets as illustrated in Figure : The number of points below and above the respective lines refers to the two complementary values of the Hilbert function. For bigger values of k, the Hilbert function is constant. If we take k = m + n, we find ( ) ( ) ( ) m + n n m H(m + n,x) = = m n = #X. This shows that, for any k m + n, the Vandermonde matrix V k (X) has full row rank, hence X is Π k -independent, hence H(k,X) = #X. In particular, eq. (13) holds true for any k, in view of the notation in (3). We are interested in properties of subsets U X, and would like to relate these to properties of their complementary sets (as was done in our version of Pascal s theorem). In what follows, we will use the following general assumption: Assumption A: X = {x 1,x,...,x N } with N = m n is the set of intersection points of two curves, of degree m and n, respectively, and X = U U c with U X and U c = X \ U.

8 Hakop Hakopian, Kurt Jetter and Georg Zimmermann 3. Cayley-Bacharach type theorems A first Cayley-Bacharach type theorem is the following extension of Chasles theorem, see [1], Theorems CB3 and CB4. This is case k = m+n 3, where the Hilbert function takes the value H(m + n 3,X) = m n 1 = #X 1. Proposition 5. For the set X of intersection points and any x X, we have P m+n 3,X = P m+n 3,X\{x}. In other words: Any polynomial of degree at most m + n 3 vanishing at all but one points of X, must necessarily vanish at the remaining point, too. Or again: If all but one points of X are contained in a curve of degree at most m + n 3, then also the last point is on the curve. Proof. For completeness, we present a proof using our notation. Given x X, we have to show that p Π m+n 3 and p X\{x} = 0 = p(x) = 0. (14) We are going to use the same notation l for a linear polynomial and for its zero set, which is a line. Fix x X, and choose any line l such that X l = {x}. Under the assumption in the left-hand side of (14), we see that p l P m+n,x and p l = A n F m + B m G n (15) by Lemma 3. Now, l divides both, A n and B m. Namely, l has m common zeros (in C ) with F m, and by selection of l only one of these zeros, namely x, is in X. From the representation, the remaining m 1 zeros must be zeros of B m, which shows that l divides B m. Likewise, l divides A n, and performing this division by l in (15), we see that p P X. Next, we look at dual interpolation problems, refering to subsets U X and to points y U c. For 0 k m + n 3, these are the following two interpolation problems: p Π k, p U = 0 and p(y) = 1, (16) and its dual problem: q Π m+n k 3, q U c \{y} = 0 and q(y) = 1. (17) Each of these problems might or might not be solvable, uniquely or not uniquely. But we have the following general statement.

Vandermonde Matrices for Intersection Points of Curves 9 Proposition 6. The two dual interpolation problems are never simultaneously solvable. For otherwise, the polynomial r = p q satisfies r Π m+n 3, r X\{y} = 0 and r(y) = 1, which contradicts Proposition 5. In terms of Vandermonde matrices, this proposition can be interpreted as follows; here, we refer to the notation in (4) and (5), and we point to the fact that these Vandermonde matrices need not be square: If B k (y) is not in the linear span of the row vectors of V k (U), then B m+n k 3 (y) must be a linear combination of the remaining row vectors of V m+n k 3 (U c ); and vice versa. In particular, H(k,U {y}) = H(k,U) + 1 implies H(m + n k 3,U c \ {y}) = H(m + n k 3,U c ). Proposition 7. If, for some 0 k m + n 3, the set U is not Π k -complete in X, then the complementary set U c is not Π m+n k 3 -independent. Equivalently, P k,x P k,u = #U c > rank V m+n k 3 (U c ). Proof. By assumption, we find a polynomial p Π k vanishing on U, but not vanishing on the entire set X. Thus, for an appropriate y U c, the interpolation problem (16) is solvable. If the complementary set U c were Π m+n k 3 -independent, the dual problem (17) would be solvable. But this contradicts Proposition 6. Combining our findings we arrive at our two main results. The first one deals with subsets U of X, with cardinality H(k,X). The second one shows that the converse of Proposition 7 is also true. Theorem 8. Given Assumption A, suppose that Then the following hold true: #U = H(k,X) for some 0 k m + n 3. (i) #U c = H(m + n k 3,X). (ii) U is Π k -independent if and only if U c is Π m+n k 3 -independent. (iii) If the sets are independent, they are both basic in X. In particular, U is Π k -basic in X if and only if U c is Π m+n k 3 -basic in X. Theorem 9. Given Assumption A, the following are equivalent for 0 k m+n 3:

10 Hakop Hakopian, Kurt Jetter and Georg Zimmermann (i) U is Π k -complete in X. (ii) The complementary set U c is Π m+n k 3 -independent. Proof of Theorem 8. The first statement follows from (13), and the last statement follows from the second one via Lemma. Also, in order to prove the second statement, it is sufficient to show one direction only, since the reverse implication is the dual statement to the direct implication with U replaced by U c and k replaced by m + n k 3. Now, assume that U c is Π m+n k 3 -independent. Then, by Proposition 7, the set U is Π k -complete in X, hence Π k -basic in X by Lemma. In particular, U is Π k -independent. Proof of Theorem 9. Due to Proposition 7 only the implication (i) = (ii) has to be verified. If U is Π k -complete in X then, by Lemma, U can be reduced to a Π k -basic set U U. By Theorem 8, the complementary set (U ) c is Π m+n k 3 -independent, and U c (U ) c is Π m+n k 3 -independent as well. Let us show that Theorem CB5 of [1] follows from Theorem 9 as a corollary. Here, we use the notions introduced in (7) and (9). Theorem 10. [1] For 0 k m + n 3, the Π k -defect of a set U X equals the Π m+n k 3 -excess of the complementary set U c : D k (U,X) = E m+n k 3 (U c ). Proof. If D k (U,X) = dim P k,u dim P k,x > 0, we find a polynomial p Π k and a point y U c such that p U = 0 and p(y) = 1, whence in view of the remarks after Proposition 6, D k (U {y},x) = D k (U,X) 1 and E m+n k 3 (U c \ {y}) = E m+n k 3 (U c ) 1. These equations show that, by adding y U c to U, the Π k -defect of U and the Π m+n k 3 -excess of the complementary set are both reduced by one. Proceeding in this way, by adding D k (U,X) points from U c to U, step by step, we arrive at a superset Ū X of U which is Π k-complete in X, and the excess of its complementary set reduces to 0 E m+n k 3 ((Ū)c ) = E m+n k 3 (U c ) D k (U,X). But by Theorem 9, if the set Ū is Π k-complete in X, then the complementary set (Ū)c is Π m+n k 3 -independent, thus This proves the theorem. E m+n k 3 ((Ū)c ) = 0.

Vandermonde Matrices for Intersection Points of Curves 11 4. Applications to interpolation The case H(k,X) = dim Π k and H(m + n k 3,X) = dim Π m+n k 3 is of special interest, since it deals with square Vandermonde matrices. It is this case, where these matrices refer to poised or non-poised problems of polynomial interpolation. This generates a series of extensions and modifications of Pascal s Theorem and of its converse statement. Example 1. This is case m = n and k = n 1. Here, dim Π k = dim Π n 1 = ( ) n+1 and dim Πk 1 = dim Π n = ( n ). Since ( n + 1 ) + ( ) n = n = m n, we may choose ( ) n + 1 U X with #U = and #U c = ( ) n. Both Vandermonde matrices V n 1 (U) and V n (U c ) are square, and Theorem 8 tells that they are either both regular or both singular. In conclusion, the two interpolation problems are either both poised or both non-poised. Pascal s theorem is the special case when m = n = 3, #U = 6 and #U c = 3. Figure 3 illustrates case m = n = 4, with #U = 10 = dim Π 3 and #U c = 6 = dim Π. Figure 3: Generalization of Pascal s Theorem, case m = n = 4.

1 Hakop Hakopian, Kurt Jetter and Georg Zimmermann Example. In this case Since we may choose m = n + 1 and k = n 1. ( ) n + 1 = (n + 1) n = m n, ( ) n + 1 #U = #U c = = dim Π n 1 in order to have square Vandermonde matrices V n 1 (U) and V n 1 (U c ). Again, the matrices are either both regular or both singular. In this case the two polynomial interpolation problems deal with polynomials of the same degree. Figure 4 illustrates case n = 3, i.e., interpolation at 6 points using quadratic polynomials. Figure 4: Modification of Pascal s Theorem, case n = 3, m = 4. Acknowledgements Part of this work was done while the first author was visiting the University of Hohenheim. Support from the DAAD, grant # A/07/04610, is gratefully acknowledged.

Vandermonde Matrices for Intersection Points of Curves 13 Bibliography [1] D. Eisenbud, M. Green, and J. Harris, Cayley-Bacharach theorems and conjectures, Bull.Amer.Math.Soc. 33 (1996), 95 34. [] H. Hakopian, The multivariate fundamental theorem of Algebra, Bezout s theorem and Nullstellensatz, in: Approximation Theory (D. K. Dimitrov et al., eds.), pp. 73 97, Marin Drinov Acad. Publ. House, Sofia, 004. [3] H. M. Möller, Linear abhängige Punktfunktionale bei zweidimensionalen Interpolations- und Approximationsproblemen, Math. Z. 173 (1980), 35 49. Hakop Hakopian Department of Informatics and Applied Mathematics Yerevan State University A. Manookian str. 1 Yerevan 005, Armenia E-mail: hakop@ysu.am Kurt Jetter and Georg Zimmermann Institut für Angewandte Mathematik und Statistik Universität Hohenheim D 70593 Stuttgart, Germany E-mail: kjetter@uni-hohenheim.de, Georg.Zimmermann@uni-hohenheim.de