Zeros of real polynomials

Zeros of Real Polynomials Richard Aron, Raquel Gonzalo∗, and Andriy Zagorodnyuk†

Abstract We study vector subspaces of the set of zeros of real valued symmetric odd polynomials and of real homogeneous polynomials of low degree.

Zeros of complex polynomials have been an object of interest for many years, and techniques from many areas of mathematics have been developed for this study (see, e.g., [4] and [5]). In recent years, there has been increased interest in the existence of subspaces contained in the set of zeros of a polynomial which vanishes at the origin. Typically, these results have been along the following lines: Given any two positive integers n and d, there is a positive integer m = m(n, d) such that for any complex polynomial P : Cm → C of degree d, there is a vector subspace XP ⊂ Cm of dimension n such that P |XP = P (0) (see, e.g., [9] and [1], and for further background on related problems [6] and [7]). For instance, it is not difficult to see that P2n m(n, 2) = 2n + 1, and for example the polynomial P (z1 , ..., z2n ) ≡ j=1 zj2 vanishes on the n−dimensional subspace generated by {e1 + ie2 , ..., e2n−1 + ie2n }. Our interest here is in the real case. Our results will be for three special situations. We first prove a general result for real, symmetric, homogeneous polynomials of odd degree. We next study real 2−homogeneous polynomials P , relating the size of the subspace of zeros of P with the signature of the associated matrix. Finally, we prove a general result concerning the zeros of real 3−homogeneous polynomials. It will be useful to recall that a d−homogeneous polynomial P : X → Y between Banach spaces X and Y has the defining property that there is a unique associated symmetric continuous d−linear mapping A : X × ... × X → Y such that P (x) = A(x, ..., x) for every x ∈ X. We will denote the space of real-valued d−homogeneous polynomials P : X → IR by P(d X). For background material, the reader is referred to [2]. Proposition 1. Let P be a symmetric homogeneous polynomial P : IRm → IR of odd degree. Then there is an [m/2]-dimensional subspace contained in P −1 (0). Proof. Let P P be a d-homogeneous polynomial on IRm where d is an odd integer. It is known that the m polynomials { i=1 xri | r ∈ IN} form an algebraic basis for the symmetric polynomials on IRm (see, e.g., [8, p. 79]). Therefore there is a polynomial Q such that Ãm ! Ãm ! m X X X d P xi ei = Q xi , . . . , xi . i=1

Now, P

Ãm X i=1

! xi ei

=

i=1

X

i=1

à ,...,ks αik11,...,i s

m X i=1

i1 ,...,is ; i1 k1 +i2 k2 +...+is ks =d

!k1 xii1

à ...

m X

!ks xiis

.

i=1

Note that in each expression i1 k1 +. . .+is ks = d, some of the i0j s must be odd. Consider the [m/2]-dimensional subspace H = [e1 − e2 , . . . , e2[m/2]−1 − e2[m/2] ]; ∗ Research supported in part by the Kent State University-Universidad Complutense de Madrid Cooperative Exchange Agreement and by CGYCIT grant PB96-0607 † Research supported by a COBASE Grant of the National Research Council

1

clearly, if x ∈ H then P (x) = 0. So, H ⊂ P −1 (0). 2 The same works in the infinite dimensional case: Proposition 2. Let X be a Banach space with a symmetric basis and let P be a homogeneous symmetric polynomial of odd degree. Then P −1 (0) contains an infinite dimensional subspace. Proof. Let P be a d−homogeneous polynomial on X where d is an odd integer, and let {en } be the symmetric basis of X. By the representation of symmetric P∞ polynomials given in [3], either P = 0 or there exists an integer N ≥ 1 such that the set of polynomials { i=1 xri | r ≥ N } is an algebraic basis for the space of symmetric polynomials on X. Therefore, if d < N we have P = 0; otherwise, there is a polynomial Q : IRd−N +1 → IR such that Ã∞ ! Ã∞ ! ∞ X X X N d P xi ei = Q xi , . . . , xi i=1

Then, P

̰ X i=1

! xi ei

=

i=1

i=1

Ã

X

,...,ks αik11,...,i s

∞ X

!k1 xii1

à ...

i=1

i1 ,...,is ; i1 k1 +i2 k2 +...+is ks =d

∞ X

!ks xiis

.

i=1

Consider now the infinite dimensional subspace: H = [e1 − e2 , e2 − e3 , . . . , e2n−1 − e2n , . . . , ]. Clearly, H is an infinite dimensional subspace contained in P −1 (0), as we required. 2 Corollary 1. Let X be a Banach space with a symmetric basis and let P be a homogeneous symmetric polynomial operator of odd degree from X to a Banach space Y . Then, P −1 (0) contains an infinite dimensional subspace. Proof. For each y 0 ∈ Y ∗ , y 0 ◦ P : X → IR is a scalar-valued homogeneous symmetric polynomial of odd degree. Applying Proposition 2, ker y 0 ◦ P ⊃ H, where H = [e1 − e2 , e2 − e3 , . . . , e2n−1 − e2n , . . . , ]. Thus, ker P ⊃ H. 2 We turn now to searching for subspaces contained in thePzero set of 2−homogeneous real polynomials. 2n The example of the symmetric polynomial P (x1 , ..., x2n ) ≡ j=1 x2j shows that there is no hope of either generalizing Proposition 1 or of attempting to complexify. Nevertheless, we are able to obtain a simple, general result, which depends only on the signs of the eigenvalues of the quadratic form associated to the polynomial. We recall that a quadratic form Q on IRk (or, equivalently, the associated symmetric matrix or bilinear form A) is said to be positive definite if Q(x) > 0 (equivalently, A(x, x) > 0) for all x ∈ IRk , x 6= 0. In the same way, Q will be negative definite whenever −Q is positive definite. We denote by p(Q) (resp. n(Q), z(Q)) the number of positive (resp. negative, zero) eigenvalues with their multiplicity. Proposition 3. Let Q ∈ P(2 IRk ). Then, if r = min{p(Q), n(Q)} + z(Q) there is an r-dimensional subspace Y such that Y ⊂ Q−1 (0). Proof. Consider a basis {w1 , . . . , wk } with respect to which Q is diagonal. Then, Ã k ! k X X Q yi wi = µi yi2 , i=1

i=1

2

where without loss of generality, we may assume that µi = ±1 or 0. We may also assume that these eigenvalues are written so that µ1 = 1, µ2 = −1, µ3 = 1, ..., µ2s−1 = 1, µ2s = −1, where s = min{p(Q), n(Q)}, and that µk−z(Q)+1 = ... = µk = 0. It is easy to verify that the r-dimensional subspace Y = [w1 + w2 , . . . , w2s−1 + w2s , wk−z(Q)+1 , . . . , wk ] is contained in Q−1 (0). 2 The rest of this note will be devoted to proving that every 3−homogeneous polynomial P : IRk → IR vanishes on a subspace whose dimension n depends only on k, where n → ∞ as k → ∞. For this, we will prove some preliminary results which are of independent interest. Lemma 1. Let Ψ : [0, 1] → P(2 IRk ) be an analytic mapping such that the quadratic form associated to Ψ(0) is positive definite and the quadratic form associated to Ψ(1) is negative definite. Then there is 0 < t < 1 k such that Pt = Ψ(t) vanishes on a [ k+1 2 ]-dimensional subspace of IR . Proof. For each t ∈ [0, 1], consider the triple (p(t), z(t), n(t)), consisting of the number of positive, negative and zero eigenvalues of At , respectively, where At is the symmetric matrix associated with Pt ∈ P(2 IRk ). Observe that t → det At is analytic, and thus this function can have only a finite number of zeros in the interval [0, 1]. In other words, we can break the interval [0, 1] up into disjoint open intervals Ij such that for t ∈ Ij , z(t) ≡ 0. Fix t0 ∈ (0, 1) with, say, p(t0 ) = m. We claim that there is a δ > 0 such that if |t − t0 | < δ, then p(t) ≥ m. Indeed, there is an m−dimensional subspace H ⊂ IRk such that At0 |H is positive definite. Therefore, 0 < min{At0 (x, x)|x ∈ H, kxk = 1} = R, say. By continuity of Ψ, we may choose δ > 0 such that if |t − t0 | < δ, then kΨ(t) − Ψ(t0 )k < R/2. Consequently, At |H is positive definite, which implies that p(t) ≥ m. The same argument applies to n(t). It follows that p(t) and n(t) are constant on each Ij . Note that (p(0), z(0), n(0)) = (k, 0, 0) and also (p(1), z(1), n(1)) = (0, 0, k). Therefore, it must be the case that on some subinterval Ij ⊂ [0, 1], the triple takes the constant value (p(t), n(t), z(t)) = (q, 0, k − q) where q > k − q, and on the ‘next’ subinterval Ij+1 ⊂ [0, 1], the triple takes the constant value (p(t), n(t), z(t)) = (`, 0, k − `), where ` ≤ k − `. Let b be the common endpoint of the two intervals. If ` = k − `, then we apply Proposition 3 to Ψ(t) for t ∈ Ij+1 . If ` < k − `, then the above argument shows that for ² > 0 sufficiently small, p(b) ≤ p(b + ²) = ` < k2 and n(b) ≤ n(b − ²) = k − q < k2 . Since p(b) + z(b) + n(b) = k, we have min{p(b), n(b)} + z(b) = k − max{p(b), n(b)} > k2 . The conclusion follows by applying Proposition 3 to Ψ(b). 2 Corollary 2. Let P ∈ (n IRk ) be a polynomial with associated n−linear symmetric form A, where k ≥ 2 and n ≥ 3 is odd. Then there is y ∈ IRk , y 6= 0, such that the quadratic polynomial Qy (x) ≡ A(y, . . . , y, x, x) k vanishes on a [ k+1 3 ]-dimensional subspace of IR . Proof. Let x0 ∈ IRk , x0 6= 0. Then Qx0 (x, x) = −Q−x0 (x, x). With the notation of Proposition 3, if k+1 r = min{p(Qx0 ), n(Qx0 )}+z(Qx0 ) ≥ [ k+1 3 ], we are done. Otherwise, suppose that r < [ 3 ] and, without loss of generality, that n(Qx0 ) = min{p(Qx0 ), n(Qx0 )}. Therefore, there is a p = p(Qx0 ) (≥ [2k/3])− dimensional subspace on which Qx0 is positive definite. By restricting to this subspace, we may assume without loss of generality that Qx0 is positive definite. Let γ : [0, 1] → IRk be an analytic curve such that γ(0) = x0 , γ(1) = −x0 and 0 6∈ γ(t). Then Ψ(t) = Qγ(t) (x, x) is an analytic function from [0, 1] to P(2 IRp ), such that Qγ(0) (x, x) > 0 and Qγ(1) (x, x) < 0. From Lemma 1 it follows that there is t, 0 < t < 1, such that the quadratic polynomial Qγ(t) (x, x) vanishes on a [ p+1 2 ]-dimensional subspace. Since [

[ 2k ] + 1 2k + 3 k+1 p+1 ]≥[ 3 ]=[ ]=[ ], 2 2 6 3

y = γ(t) will be the required element from IRk . 2 3

We can now prove the main result: Theorem 1. Let P : IRk → IR be a 3-homogeneous polynomial and k≥

3n (6n + 5) − 1 . 4

for some n > 0. Then, there is an [ n2 ]-dimensional subspace contained in P −1 (0). Proof. Let A be the symmetric 3−linear form associated to the 3−homogeneous polynomial P. We will show that there is a set of n linearly independent vectors x1 , ..., xn such that for any i 6= j, A(xi , xi , xj ) = 0 and such that for any i, j, and l, not all equal,A(xi , xj , xl ) = 0. As a consequence, a term like A(xi − xj , xi − xj , xi − xj ) will be equal to P (xi ) − P (xj ). By Corollary 2, we can choose x1 ∈ IRk , x1 6= 0, such that A(x1 , x, x) vanishes on some subspace S1 ⊂ IRk , k+1 dim S1 ≥ [ k+1 3 ]. Let V1 = S1 ∩ ker A(x1 , x1 , x). Then dim V1 ≥ [ 3 ] − 1. Now, we want to judiciously select the next vector x2 ∈ V1 to be linearly independent of x1 . If x1 ∈ / V1 , then of course there is no problem, and we’ll let H1 = V1 . Otherwise, let H1 = V1 \[x1 ] be the vector space complement of the subspace [x1 ] of V1 . In any case, dim H1 ≥ [ k+1 3 ] − 2, which must necessarily be ≥ 1. Next, choose x2 ∈ H1 , x2 6= 0, such that the restriction of the function x → A(x2 , x, x) to H1 vanishes on [( k+1 ]−2)+1

some subspace S2 ⊂ H1 where dim S2 ≥ [ 3 3 ]. Put V2 = S2 ∩ ker A(x1 , x2 , x) ∩ ker A(x2 , x2 , x). As above, if x2 ∈ / V2 , let H2 = V2 . Otherwise, let H2 = V2 \[x2 ] be the vector space complement of [x2 ] ⊂ V2 , so [( k+1 ]−2)+1

that dim H2 ≥ [ 3 3 ] − 3 which must be ≥ 1. We continue this process, obtaining at the nth step a linearly independent set {x1 , . . . , xn } and subspaces IRk = H0 ⊃ H1 ⊃ H2 ⊃ ... ⊃ Hn such that xi+1 ∈ Hi and such that A(xi , xj , xl ) = 0 provided i 6= j, i 6= l, or d1 −1 j 6= l. Let us now estimate the dimension of each Hi . Set d1 = [ k+1 3 ], so that dim H1 ≥ d1 − 2. Let d2 = 3 , dm−1 −(m−1) so that dim H2 is an integer which is ≥ d2 − 3. In general, if we let dm = , then dim H is an m 3 integer ≥ dm − (m + 1), which must be at least 1. A routine calculation shows that for every n, d1 ≥ 2 · 3n−1 +

n X

j · 3j−1 .

j=1

Differentiating the function f (t) = so that

1−tn+1 1−t

=

Pn j=0

k≥ Therefore

tj and evaluating at t = 3 then yields d1 ≥

3n−1 (6n+5)+1 , 4

3n (6n + 5) − 1 . 4

n n n n n X X X X X P( ai xi ) = A( ai xi , ai x i , ai xi ) = a3i P (xi ) i=1

i=1

i=1

i=1

i=1

for all a1 , . . . , an ∈ IR, since all the ‘mixed’ terms are 0. By multiplying by a non-zero scalar, we may assume that P (xi ) = 1 for i = 1, . . . v for some v ≤ n, and that P (xi ) = 0 for i = v + 1, . . . , n. Then the subspace [x1 − x2 , . . . , x2[v/2]−1 − x2[v/2] , xv+1 , . . . xn ] is a subspace of dimension greater than or equal to [ n2 ] which is contained in P −1 (0), and the proof is complete. 2 It is easy to see that every 3−homogeneous polynomial in at least k = 2 variables vanishes on a one dimensional subspace. The estimate provided by the Theorem in this case is for k to be no larger than 38. If a 2−dimensional subspace is required, so that n = 4, then the Theorem gives the estimate that any 3−homogeneous polynomial in k = 587 variables has such a subspace. We remark that more, and better, information seems to be known for the analogous problem for complex polynomials. For example, it is not hard to see that every homogeneous complex polynomial (of any degree) in two or more variables vanishes on 4

a complex line. In fact, one can show that any 3−homogeneous complex polynomial in 2n−1 (n + 1) variables vanishes on an n−dimensional subspace [1], although here too it is apparently unknown if this estimate is even of the correct order. It would be interesting to get a better estimate and, of course, to extend this result to real polynomials having higher homogeneity. The authors would like to thank the Referee for several helpful suggestions which improved the presentation of this manuscript; in particular, the Referee’s help in obtaining a better estimate in Lemma 1 (and therefore in Corollary 2 and Theorem 1) is gratefully acknowledged. In addition, the second and third authors thank the Department of Mathematics of Kent State University for its hospitality during April May, 1998, when this work was initiated.

References [1] Aron, R. M. and Rueda, M.P., A problem concerning zero-subspaces of homogeneous polynomials, Linear Topological Spaces and Complex Analysis 3 (1997), 20-23. [2] Dineen, S., Complex analysis in locally convex spaces, North-Holland Mathematics Studies, 57 (1981). [3] Gonzalez, M., Gonzalo, R. and Jaramillo, J. A., Symmetric polynomials on rearrangment invariant function spaces, Jour. London Math. Soc. (to appear). [4] Gunning, R. and Rossi, H. Analytic functions of several complex variables, Prentice-Hall (1965). [5] Harris, J. Algebraic Geometry, Grad. Texts in Math. 113, Springer-Verlag (1992). [6] Pestov, V., Two 1935 questions of Mazur about polynomials in Banach spaces: A counter-example, Questiones Mathematicae (to appear). [7] Plichko, A. and Zagorodnyuk, A., On automatic continuity and three problems of “The Scottish Book” concerning the boundedness of polynomials functionals, J. Math. Anal. App., 220 (1998), 477-494. [8] van der Waerden, B. L., Modern Algebra, Ungar (1964). [9] Zagorodnyuk, A., The zero-subspaces of symmetric homogeneous polynomials, preprint. Department of Mathematics, Kent State University, Kent, OH 44242,

USA, ([email protected])

Departamento de Matematica Aplicada, Facultad de Informatica, Campus Montegancedo s/n, Boadilla del Monte, 28660 Madrid, Spain, ([email protected]) Inst. for Applied Problems of Mechanics and Mathematics, Ukrainian Academy of Sciences, 3 b, Naukova str., Lviv, Ukraine, 290601, ([email protected])

5