Pacific Journal of Mathematics

Transcription

1 Pacific Journal of Mathematics ON SPACES OF MATRICES CONTAINING A NONZERO MATRIX OF BOUNDED RANK Dmitry Falikman, Shmuel Friedland, and Raphael Loewy Volume 07 No. November 00

2 PACIFIC JOURNAL OF MATHEMATICS Vol. 07, No., 00 ON SPACES OF MATRICES CONTAINING A NONZERO MATRIX OF BOUNDED RANK Dmitry Falikman, Shmuel Friedland, and Raphael Loewy Let M n (R) and S n (R) be the spaces of n n real matrices and real symmetric matrices respectively. We continue to study d(n, n, R): The minimal number l such that every l-dimensional subspace of S n (R) contains a nonzero matrix of rank n or less. We show that d(4,, R) = 5 and obtain some upper bounds and monotonicity properties of d(n, n, R). We give upper bounds for the dimensions of n subspaces (subspaces where every nonzero matrix has rank n ) of M n (R) and S n (R), which are sharp in many cases. We study the subspaces of M n (R) and S n (R) where each nonzero matrix has rank n or n. For a fixed integer q > we find an infinite sequence of n such that any ( q+) dimensional subspace of S n (R) contains a nonzero matrix with an eigenvalue of multiplicity at least q.. Introduction. Let F be a field, M m,n (F) the space of all m n matrices over F and S n (F) the space of all n n symmetric matrices over F. We write M n (F) for M n,n (F). Let V be either M m,n (F) or S n (F), and let k be a positive integer. In the last 80 years there has been interest in the following types of subspaces, all related to the rank function: (a) A subspace of V where each matrix has rank bounded above by k. (b) A subspace of V where each nonzero matrix has rank k. A subspace of this type is said to be a k-subspace. (c) A subspace of V where each nonzero matrix has rank bounded below by k. See the works [3], [], [], [8], [5], [3], [4], [7] and many others. Roughly speaking these problems are divided into two classes depending on whether F is algebraically closed or not. The classical case, which goes back to Radon-Hurwitz, discusses the maximal dimension ρ(n) of an n-subspace U of M n (R) where each 0 A U is an orthogonal matrix times r R. Write n = (a + ) c+4d, where a and d are nonnegative integers, and c 57

3 58 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY {0,,, 3}. Then the Radon-Hurwitz number ρ(n) is defined by (.) ρ(n) = c + 8d. In his famous work Adams [] gave a nonlinear version of the Radon-Hurwitz number by showing that ρ(n) is the maximal number of linearly independent vector fields on the n dimensional sphere S n. (S n R n denotes the Eucledian sphere of radius one centered at the origin.) From this result Adams deduced that the maximal dimension of an n-subspace of M n (R) is exactly ρ(n). Let ρ(x) = 0 if x is not a positive integer. Define now ( n ) (.) ρ s (n) = ρ +. Adams, Lax and Phillips [] showed that the maximal dimension of an n- subspace of S n (R) is exactly ρ s (n). Friedland, Robbin and Sylvester [8] and Berger and Friedland [5] gave further nonlinear versions of the above results by considering odd maps φ from S p to matrices of rank n in M n (R), S n (R) and M n,n+ (R) respectively. In this paper ( 4) we generalize these results to odd maps from S p to rank n matrices in M n (R) and S n (R). The main motivation of this paper is the following quantity studied in [7]. For an integer k, such that k n, let d(n, k, F) be the smallest integer l such that every l dimensional subspace of S n (F) contains a nonzero matrix whose rank is at most k. Note that it is clear that the maximal dimension of a subspace of S n (F) of type (c) above is exactly d(n, k, F). It is our purpose to continue the study of d(n, k, R) started in [7]. Note that any d(n, k, R) dimensional subspace of S n (R) contains a nonzero matrix with an eigenvalue of multiplicity at least n k. (One of the main results of [8] was that any subspace of S n (R) of dimension σ(n) +, where σ(n) = if n 0, ± (mod 8), σ(n) = ρ(4b) if n = 8b, 8b ±, contains a nonzero matrix with a multiple eigenvalue.) The equality ( ) n k + (.3) d(n, k, C) = +, established in [7] is derived straightforwardly from the following dimension computations. Let V k,n (C) (V k,n (R)) be the variety of all matrices in S n (C) (S n (R)) of rank k or less. Then in the projective space PS n (C) the projective variety PV k,n (C) is an irreducible variety of codimension d(n, k, C), which yields (.3). In particular d(n, n, C) =. The results of Adams, Lax and Phillips, cited above, yield that d(n, n, R) = ρ s (n)+. This shows that in general the computation of d(n, k, R) is much more difficult than the computation of d(n, k, C). In [7] we gave a simple condition on n when d(n, k, C) = d(n, k, R) for k n, which trivially holds for

4 SPACES OF MATRICES 59 k = n. It was shown by Harris and Tu [0] that the degree of the variety PV k,n (C) is given by the formula (.4) δ k,n := deg PV k,n (C) = n k j=0 ( n+j n k j ( j+ j Then d(n, k, C) = d(n, k, R) if δ k,n is odd. (In this case the result that any d(n, k, R) dimensional subspace of S n (R) contains a nonzero matrix with an eigenvalue of multiplicity at least n k is best possible.) We show that δ n q,n is odd if n > q and ( (.5) n ±q mod log q ). In particular, under the above conditions, ( ) q + (.6) d(n, n q, C) = d(n, n q, R) = +. If n q and n satisfies (.5) then any ( ) q+ subspace of Sn (R) contains a nonzero matrix with an eigenvalue of multiplicity at least q. This statement for q = yields the original Lax s result [] that any 3 dimensional subspace of S n (R) contains a nonzero matrix with a multiple eigenvalue for n (mod 4). (This result and its generalization in [8] is of importance in the study of singularities of hyperbolic systems.) An important part of the paper is devoted to the study of the numbers d(n, n, R). Besides the cases given in (.5-.6) for q = it is easy to see that d(3,, R) = 6 [7]. (Hence the inequality d(n, n, R) σ(n) + established in [7] is not sharp for some n.) Partial results regarding d(n, n, R) were obtained in [7]. In particular, it was shown that for m ) ). (.7) d(4m, 4m, R) 4m +. We obtain here additional results on d(n, n, R). In particular, using numerical and symbolic computations we show that d(4,, R) = 5, which implies that the upper bound in (.7) is sharp for m =. The paper is organized as follows: In Section we show that d(4,, R) = 5. In Section 3 we obtain some upper bounds and some monotonicity properties of d(n, n, R). In Section 4 we consider the existence of continuous and smooth odd maps from the sphere S p to matrices of rank n in M n (R) and S n (R). As a consequence we obtain new upper bounds for the dimension of (n )-subspaces of M n (R) and S n (R). These upper bounds are shown to be sharp in many cases. In Section 5 we discuss the existence of subspaces U S n (R) of dimension n, which do not contain a nonzero matrix of rank less than n. The last section is devoted to remarks and conjectures.

5 60 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY. Computation of d(4,, R). As indicated in the introduction, the first unknown number in the sequence {d(n, n, R)} n=3 is d(4,, R). By (.7) d(4,, R) 5. It is our purpose to prove: Theorem.. d(4,, R) = 5. Proof. It suffices to exhibit a 4-dimensional subspace L of S 4 (R) with the property that every 0 A L has rank 3 or 4. Let L be the subspace of S 4 (R) spanned by the matrices B = B 3 =, B =, B 4 = It is straightforward to check that dim L = 4. Let α, α, α 3, α 4 be indeterminates and let α = (α, α, α 3, α 4 ), and B(α) = 4 i= α ib i. We show that if 0 α R 4 then rank B(α) 3. So let d(α) = det B(α), and for j 4 let d j (α) denote the determinant of the principal submatrix of B(α) obtained by deleting row and column j. We let Maple compute the common zeros of the 5 polynomials d(α), d (α), d (α), d 3 (α), d 4 (α) in C 4. It turns out that there are no common zeros with α = 0 except the trivial solution α = 0. Now we can assume α =. We list the ten solutions found by Maple. They are: (i) The two solutions given by α = (, 0, + x, x), where x is a zero of the polynomial f(z) = z + z +. (ii) The two solutions given by α = (, x, x, 3x ), where x is a zero of the polynomial f(z) = 7z 8z + 3. (iii) The two solutions given by α = (, x, x, 5x ), where x is a zero of the polynomial f(z) = 4z + 4z +.,.

6 SPACES OF MATRICES 6 (iv) The four solutions given by ( α =, x, x + x +, x x ), where x is a zero of the polynomial f(z) = z 4 + 4z 3 + 8z + 4z + 3 = (z + ) 4 + z +. It follows that for any 0 α R 4 either d(α) 0 or j 4 such that d j (α) 0. Thus, the rank of every nonzero matrix in L is at least 3. We explain now why Maple did indeed give us all the common zeros of d(α), d (α), d (α), d 3 (α), d 4 (α). Let L be the subspace of S 4 (C) spanned by B, B, B 3, B 4. We have by (.3) (.) Moreover, by (.4) we get deg P(V,4 (C)) = d(4,, C) = 4. ( 4 )( 5 ) 3 = 0. We also use the known fact that a rank r symmetric matrix with entries in a field has a nonsingular r r principal submatrix. Thus, the ten distinct solutions found by Maple yield ten matrices in L whose rank is at most, and no two of those matrices lie on the same line. In fact, a computation showed that each of those ten matrices has rank. It remains to explain why Maple did not omit other solutions, assuring us it did not miss any real solutions in particular. It follows from (.) that if L is a four dimensional generic subspace of S 4 (C) then P(V,4 (C)) meets P(L ) in at most ten distinct points. Hence, we have to check that L is indeed generic. It suffices to show that P(V,4 (C)) and P(L ) intersect transversally. So let A denote any of the ten matrices in L of rank found by Maple. Let x, y C 4 be linearly independent vectors such that Ax = Ay = 0. A computation showed that the rank of xt B x y t B y x t B y x t B 3 x y t B 3 y x t B 3 y x t B 4 x y t B 4 y x t B 4 y M 3 (C) is 3. This implies that the linear system of three equations x t Bx = 0 y t By = 0 x t By = 0, has no solution B in L such that A, B are linearly independent. This shows that L is generic.

7 6 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY In addition to the computation using Maple, a numerical procedure (using Matlab) was performed to check that there are no nonzero matrices of rank or in L. 3. Upper bounds and monotonicity properties of d(n, n, R). It follows from (.5) and (.6) that if n (mod 4) then d(n, n, R) = d(n, n, C) = 4. It was shown in [7] that (3.) d(n, n, R) 7 for n 4(mod 8). It is our purpose here to get some additional upper bounds and some monotonicity properties of d(n, n, R). Proposition 3.. Let n 3, 5(mod 8). Then d(n, n, R) 7. Proof. It follows from (.4) that for every n 3 ( n n+ )( n+ ) deg P(V n 3,n (C)) = 3)( = (n + )(n + )n (n )(n ) , 5 and this is odd for n 3, 5(mod 8). So, as indicated in the introduction, and using (.3) we have for n 3, 5(mod 8) ( ) 4 d(n, n 3, R) = d(n, n 3, C) = + = 7, and since d(n, n, R) d(n, n 3, R), the proposition follows. It follows from [7] that d(n, n, R) 4 for n 3, 4, 5(mod 8). Thus, we have the exact value for d(n, n, R) whenever n, 6(mod 8), and good upper and lower bounds whenever n 3, 4, 5(mod 8). Proposition 3.. Let k be a fixed integer such that k {3, 4, 5,,, 3}. Then the sequence {d(6m + k, 6m + k, R)} m=0 is a (weakly) monotone increasing sequence, bounded above by 7. There exists M = M(k) such that d(6m + k, 6m + k, R) = d(6m + k, 6M + k, R) for all m M. Proof. It suffices to prove the monotonicity of the given sequence, because the required boundedness follows from (3.) and Proposition 3.. Let L be any 8-dimensional 8-subspace of M 8 (R). Such a subspace exists because ρ(8) = 8 by (.). Let L = {[ 0 A t A 0 ] }, A L. Then L is an 8-dimensional subspace of S 6 (R). Note that every nonzero matrix in L is nonsingular. Let {B i } 8 i= be a basis of L. Let d m,k = d(6m + k, 6m + k, R) and d 0 = d m,k. Let W m,k be any d 0 -dimensional subspace of S 6m+k (R) such that the rank of any

8 SPACES OF MATRICES 63 nonzero matrix in W m,k is at least 6m + k. Note that d 0 7 = 6. Let {C i } d 0 i= be a basis of W m,k. Let W m,k = span{c i B i : i d o }. Then W m,k is a d 0 -dimensional subspace of S 6(m+)+k (R), and clearly every nonzero matrix in it has rank 6 + 6m + k = 6(m + ) + k. This shows that d(6(m + ) + k, 6(m + ) + k, R) d 0 + = d m,k = d(6m + k, 6m + k, R). The following proposition justifies some claims we made in the introduction. Proposition 3.3. Let k n. Then any d(n, k, R) dimensional subspace of S n (R) contains a nonzero matrix which has an eigenvalue of multiplicity at least n k. If d(n, k, R) = d(n, k, C) then this result is sharp. Proof. Let V be a d(n, k, R) dimensional subspace of S n (R). If I n V then is the eigenvalue of I n of multiplicity n n k. Suppose that I n / V. Let ˆV = span{v, I n }. Then ˆV contains 0  such that rank  k.  = A + ai, 0 A V, so a is an eigenvalue of A of multiplicity at least n k. Let Σ n k S n (R) be the variety of all A S n (R) such that A has an eigenvalue of multiplicity at least n k. Clearly Σ n k = {B S n (R) : B = A + ai n for some A V k,n (R) and a R}. Since V k,n (R) is an irreducible variety of codimension d(n, k, C) it follows that Σ n k is an irreducible variety of codimension d(n, k, C). Hence there exists an d(n, k, C) dimensional subspace W S n (R) such that Σ n k W = {0}. 4. Upper bounds for the dimension of (n )-subspaces of M n (R) and S n (R). In this section we obtain upper bounds for the dimension of (n )-subspaces of M n (R) and S n (R). This is done by considering certain odd, smooth maps from S p into M n (R) and S n (R). We also need several known results. Let M n (k) (R) ( S (k) n ) (R) denote the set of all rank k matrices in M n (R)(S n (R)). Given any field F and positive integers m, n, let M 0 m,n(f) denote the set of all matrices in M m,n (F) whose rank is equal to min{m, n}. Lemma 4.. Let C Mn,n+ 0 (F). For j =,,..., n +, let C(j) denote the n n matrix obtained from C by deleting its j-th column. Then the

9 64 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY solution of the homogeneous linear system Cy = 0 is a line spanned by ( y = detc (), detc (), detc (3),..., ( ) n+ detc (n+)) t (4.). Proof. This is an easy consequence of well-known properties of the determinant. The next theorem appears as a part of Theorems A and F in [8]. See there how it is related to classical results due to Radon, Hurwitz, Adams and Adams-Lax-Phillips. Theorem 4.. Let ϕ : S p ϕ( α) = ϕ(α) α S p. M n (R) be an odd continuous map, i.e., (i) Suppose that ϕ(s p ) M n (n) (R) = GL(n, R). Then, p ρ(n). (ii) Suppose that ϕ(s p ) S n (n) (R). Then, we have p = 0 if n is odd, and p ρ( n ) if n is even. All the inequalities in (i) and (ii) are sharp. The next theorem appears in [5]. Theorem 4.. Let ϕ : S p Mn,n+ 0 (R) be an odd continuous map. Then p max{ρ(n), ρ(n + ) }, and this inequality is sharp. Theorem 4.3. Let n. Let ϕ : S p M n (n ) (R) be n odd smooth map. Then, if n 3, 5, 9 (4.) p max{ρ(n ), ρ(n), ρ(n + ), }. Furthermore, if ϕ(s p ) S n (n ) (R) then p = 0 if n is even, and { ( ) } n (4.3) p max ρ, ρ(n + ) if n is odd. Proof. We prove first (4.). If p (4.) certainly holds, so we may assume p. Hence S p is simply connected. Therefore, we can choose x(α) S n in a well-defined and continuous way such that (4.4) x t (α)ϕ(α) = 0 α S p. For α S p let B(α) be the matrix in M n,n+ (R) obtained from ϕ(α) by augmenting it with the column vector x(α), i.e., B(α) = (ϕ(α), x(α)) M n,n+ (R). Then rank ϕ(α) = n and (4.4) imply that B(α) Mn,n+ 0 (R) for all α S p. Note that the smoothness of ϕ implies that x(α) is also smooth. It is also clear that for each α S p x( α) = ±x(α). Hence, we must have x( α) = x(α) for all α S p, or x( α) = x(α) for all α S p.

10 SPACES OF MATRICES 65 Case. Suppose that x( α) = x(α) for all α S p. Then the map from S p to Mn,n+ 0 (R) defined by α B(α) is an odd continuous map, so by Theorem 4. we have (4.5) and (4.) holds. p max{ρ(n), ρ(n + ) }, Case. We now assume x( α) = x(α) for all α S p. We apply Lemma 4. to B(α) Mn,n+ 0 (R), for each α Sp, and denote the normalized solution ( ) given by (4.) that is, y y by ψ(α). Let There are two subcases now. ψ(α) = (ψ (α), ψ (α),..., ψ n (α), ψ n+ (α)). Case a. Suppose that n is even. It follows from (4.) that We define ψ( α) = ( ψ (α), ψ (α),..., ψ n (α), ψ n+ (α)). C(α) = [ B(α) ψ(α) ] M n+ (R). This matrix is nonsingular because ψ(α) is orthogonal to all the rows of B(α). Let η(α) = (ψ (α), ψ (α),..., ψ n (α)) and let [ ] ϕ(α) D(α) = M η(α) n+,n (R), that is, D(α) is the matrix obtained from C(α) by deleting its last column. Its rank is n for each α S p. Since D( α) = D(α) for all α S p we apply Theorem 4. again and conclude that (4.5) holds, so (4.) holds. Case b. Suppose that n is odd. So n 3. Suppose first that p < n. Since x(α) is smooth, it follows from Sard s theorem that {x(α) : α S p } has measure zero in S n. In particular, ξ S n such that x(α) ±ξ for all α S p. For every α S p let L α = span{x(α), ξ}. Let Q(α) be the unique n n orthogonal matrix satisfying: Q(α) is the identity on L α and its restriction to L α is the rotation in that plane by an angle < π that sends x(α) to ξ. Since x( α) = x(α) we have Q( α) = Q(α). Also, Q(α) is continuous in α (cf. the Proposition in [8]). It follows from (4.4) that (4.6) ξ t (Q t (α)ϕ(α)q(α)) = 0, for all α S p. Let ϕ (α) = Q t (α)ϕ(α)q(α). Then ϕ (α) is an odd continuous function. Without loss of generality we may assume ξ t = (0, 0,..., 0, ). It follows from (4.6) that for each α S p the last row of ϕ (α) is 0, so deleting it

11 66 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY results in an (n ) n matrix of rank n. Applying Theorem 4. again we conclude that p max{ρ(n ), ρ(n) } so (4.) holds. Now suppose that p n. We may consider S n as contained in S p, and then consider the restriction of the given ϕ(α) and x(α) to S n. We repeat the proof given for p < n and conclude in the same way that n max{ρ(n ), ρ(n) } = ρ(n ). Since this cannot happen by (.) when n 3, 5, 9 we have completed the proof of (4.). Suppose now that ϕ(s p ) S n (n ) (R). Recall that the inertia of A S n (R) is the triple (π, ν, δ), where π is the number of positive eigenvalues of A, ν is the number of negative eigenvalues of A and δ is the number of zero eigenvalues of A. Let n be even and suppose that p > 0. Let α S p, and consider a path in S p from α to α. Let J denote the image of this path under ϕ. Since every matrix in J has rank n, it follows that the inertia of each matrix on this path is equal to inertia (ϕ(α)). In particular, inertia (ϕ( α)) = inertia(ϕ(α)), but this is impossible since ϕ( α) = ϕ(α). Hence p = 0. Let n be odd. Suppose first that n = 3. We show that p in this case, so (4.3) holds. Suppose to the contrary that p. It is clear that ϕ(α) has the same inertia for each α S p. In particular, given any α S p, ϕ(α) and ϕ( α) = ϕ(α) have the same inertia. Hence each ϕ(α) has one positive, one negative and one zero eigenvalue. So for each α S p, the eigenvalues of ϕ(α) are pairwise distinct, contradicting Theorem B of [8]. Note that while we have shown that there is no odd continuous map from S to S () 3 (R), there is an odd continuous map from S to M () 3 (R). For example, consider the map that sends (x, y, z) S to 0 x y x 0 z We assume now n 5. Since max{ρ( n ), ρ(n + ) }, (4.3) holds if p. Hence we may assume that p. We go along the proof of the nonsymmetric part of the theorem. If x(α) is odd, then (4.5) holds, and since ρ(n) = for odd n, we get { ( ) } n p ρ(n + ) max ρ, ρ(n + ), and (4.3) holds. y z 0.

12 SPACES OF MATRICES 67 So we may assume that x(α) is even. Suppose also that p < n. Then we get (4.6) again, where ϕ (α) = Q t (α)ϕ(α)q(α) is an odd continuous function, and we may assume ξ t = (0, 0,..., 0, ). It follows that for each α S p the matrix obtained from ϕ (α) by deleting its last row and column is in S (n ) n (R). Hence, by part (ii) of Theorem 4. we get ( ) n p ρ, so (4.3) holds. If p n we consider S n as contained in S p, and then consider the restriction of ϕ(α), x(α) to S n. A repetition of the argument for p < n yields n ρ( n ), which is impossible for n odd, n 5. This completes the proof. Corollary 4.. Let n and let L be an (n )-subspace of M n (R). (a) If n 3, 5, 9, then (4.7) dim L max{ρ(n ), ρ(n), ρ(n + ), 3}. (b) If L S n (R) (that is, L is a subspace of S n (R)), then { if n is even, (4.8) dim L max { ρ ( ) } n +, ρ(n + ) if n is odd. Proof. (a) Let d = dim L and let A, A,..., A d be a basis of V. For A V we write A = d i= α ia i and let α = (α, α,..., α d ). Then we may view the map α A, restricted to α such that α =, as an odd smooth map from S d to M n (n ) (R). Then d is bounded by the right-hand side of (4.), so (4.7) follows. (b) The proof is similar to the previous case, using now (4.3) instead of (4.). Consider next the sharpness of the inequalities (4.), (4.3), (4.7) and (4.8). For that purpose we have the following lemma. Lemma 4.. (i) There exists a ρ(n ) dimensional (n )-subspace V of M n (R). (ii) There exists an (n )-subspace V of S n (R) such that { if n is even, dim V = ρ ( ) n +, if n is odd. Proof. (i) As indicated in the introduction, there exists an ρ(n ) dimensional (n )-subspace V of M n (R). Then we let V be the subspace of M n (R) obtained from V by appending a row and column of 0 s to each matrix in V.

13 68 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY (ii) The claim is trivial if n is even, so suppose n is odd. As indicated in the introduction, there exists an (n )-subspace V of S n (R) of dimension ρ ( ) n +. Now V is obtained from V as in Part (i). Corollary 4.. Let n (mod 4). Then the inequalities (4.3) and (4.8) are sharp, and the inequalities (4.) and (4.7) are sharp provided that n 5, 9. Proof. For n (mod 4) the maximum of the right-hand sides of (4.) and (4.7) are ρ(n ) and ρ(n ), respectively. Let V be the ρ(n ) dimensional (n )-subspace defined in the Proof of Lemma 4., Part (i). This shows the sharpness of (4.7) provided that n 5, 9. The sharpness of (4.) is obtained by considering an odd smooth map from S ρ(n ) to M n (n ) (R) as in the proof of Corollary 4. Part (a). The sharpness of (4.3) and (4.8) is proved similarly, using the ρ ( ) n + dimensional (n )-subspace V of S n (R) in Part (ii) of Lemma 4.. Remark. In Theorem 4.3 it is possible to replace the assumption that ϕ is an odd smooth map by the assumption that ϕ is an odd continuous map. The proof of the remark is achieved as follows: First approximate ϕ(α) arbitrarily by a smooth odd ϕ(α) (which is in S n (R) if ϕ(s p ) S n (n ) (R)). We can assume that for each α S p, ϕ(α) has a simple eigenvalue λ(α) which is the smallest in absolute value among all eigenvalues of ϕ(α) (and is also real if ϕ(s p ) S n (n ) (R)). Clearly λ( α) = λ(α) α S p. Let ψ(α) be the projection of ϕ(α) corresponding to λ(α). Then ψ(α) is a smooth odd function of rank at most, and it follows that ϕ(α) ψ(α) satisfies the conditions of Theorem 4.3, so (4.) and (4.3) hold respectively. 5. Existence of certain n dimensional subspaces of S n (R). Let n 4 be an even integer. It follows from [7] and this paper that d(n, n, R) n +, with equality for n = 4. Now suppose that n 5 is an odd integer. We do not know if a similar result holds, although it seems plausible. In this section we show that if there exists an n + dimensional subspace L of S n (R) such that each nonzero matrix in L has rank n or n, then there is a severe restriction on the possible inertias attained in L. We also derive here additional results on n + dimensional subspaces of n n matrices where each nonzero matrix has rank n or n. Let F be any field and let V be an n + dimensional subspace of M n (F) spanned by A, A,..., A n+. Sometimes we find it convenient to identify PF n with F n+ and PV with PF n. That is, we identify α PF n and à Ṽ := PV with lines spanned by 0 α = (α, α,..., α n+ ) t and 0 A in F n+ and V respectively. Let A(α) = n+ i= α ia i. We identify α with Ã(α).

14 SPACES OF MATRICES 69 (5.) Given any 0 y F n, let L y = {A V : Ay = 0}. Since dim V n + it is clear that dim L y for every y 0. Let (5.) M(y) = [A y, A y,..., A n+ y] M n,n+ (F). ( n+ ) For α = (α, α,..., α n+ ) t it is clear that i= α ia i y = 0 if and only if M(y)α = 0. It follows that dim L y = if and only if rank M(y) = n. Observe that if rank M(y) = n then, by Lemma 4., ker M(y) is spanned by ( α(y) = det M(y) (), det M(y) (),..., ( ) n+ det M(y) (n+)) t (5.3). Let (5.4) T = { ỹ PF n : α(y) = 0 }. Hence, T consists of those ỹ for which rank M(y) < n. Given α = (α, α,..., α n+ ) we consider the homogeneous polynomial ( n+ ) ψ(α) = det α i A i, and let (5.5) i= Z(ψ) = the zero set of ψ(α), considered as a subset of PF n. Lemma 5.. Let V be an n + dimensional subspace of M n (C), spanned by A, A,..., A n+, and such that: (i) A V such that det A 0. (ii) B V such that rank B = n and ker B is spanned by a vector 0 u C n with dim L u =. Then the following holds: Let H PC n be the irreducible component of the hypersurface Z(ψ) passing through B. Then there exists a birational map defined as follows: ϕ : H PC n, θ = ϕ : PC n H (a) For each α H with rank C = n, where C = n+ i= α ia i, ϕ( α) = ỹ, where y C n is the basis of ker C. (b) For each ỹ PC n such that dim L y =, θ(ỹ) = α, where α = α(y).

15 70 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY Proof. The maps ϕ and θ are rational. For y in the neighborhood of u, dim L y =. Hence θ is holomorphic in the neighborhood of ũ. Let B = n+ i= β ia i. The assumptions (ii) imply that ϕ is holomorphic in the neighborhood of β and ϕ θ is the identity map (as a rational map). Hence θ : PC n H. Remark. As θ : PC n H is a rational map, it is not holomorphic on a variety of codimension at least (cf. [9, Chapter 4, Section ]). Hence codim T. Lemma 5.. Let V be an n + dimensional subspace of M n (R), such that each nonzero matrix in V has rank n or n. Then: (i) A V such that det A 0. (ii) B V such that rank B = n and ker B is spanned by a vector 0 y R n with dim L y =. Proof. By Theorem of [3], V cannot be an (n )-subspace or an n- subspace. Hence V contains nonsingular matrices and singular matrices. Let (5.6) Let Ṽ s = {Ã Ṽ : det A = 0}. η : Ṽs PR n be defined as follows: Suppose that B V is any singular matrix, and suppose 0 y R n satisfies By = 0. Then let η( B) = ỹ. It is known that the set of singular points of Ṽs is a proper subvariety of Ṽ s. As η(ṽs) = PR n, it follows that there exists a regular point B of Ṽs such that Dη( B) has the rank n. Let y S n be such that By = 0. Suppose that dim L y. Then it is clear that dim ker Dη( B). It follows that the dimension of the tangent space to B in Ṽs has dimension n, implying that dim Ṽs n, a contradiction. Lemma 5.3. Let n be odd, n 3, and let V be an n + dimensional subspace of S n (R) such that each nonzero matrix in V has rank n or n. Suppose that V = {B S n (R) : tr (AB) = 0 A V } does not contain rank one matrices. Let Ṽs be as in (5.6). Then, Ṽs is a smooth connected variety in PR n. ( Proof. Let A k = ( n+ ) det i= α ia i a (k) ij ) n i,j=, k =,,..., n+ be a basis of V. Let ψ(α) =. For α Ṽs we have ψ(α) = 0, so A(α) = n+ i= α ia i has

16 SPACES OF MATRICES 7 rank n. Hence the (n )th compound of A(α), C n (A(α)), has rank. So there exists 0 x(α) = (x (α), x (α),..., x n (α)) t and ɛ = ± such that C n (A(α)) = ɛx(α)x t (α). Let C n (α) = (c ij (α)) n i,j=. Observe that ψ(α) n = a (k) ij α c ij(α) = ɛx t (α)a k x(α), k =,,..., n +. k i,j= Since V does not contain a rank one matrix, we deduce that ψ(α) 0. Thus each α Ṽs is a smooth point and the local dimension of Ṽs at α is n. Let V C be the complexification of V. Let T C, T R be defined by (5.5) for F = C, R, respectively. By the remark following Lemma 5., codim C T C. Since T R = T C PR n, we have codim R T R. Let Y r = PR n \T R. Then Y r is connected. Using (ii) of Lemma 5. we can define H and θ as in Lemma 5.. Let H R = H PR n. Then θ(y r ) is connected in H R, and this implies that θ(y r ) = H R is connected. To finish the proof it suffices to show that Ṽs = H R. Since Y r = PR n it follows that for any ỹ PR n there exists β H R such that A(β)y = 0. Let γ be an arbitrary point in Ṽs. Then 0 u R n such that A(γ)u = 0. If dim L u =, then γ H R. So suppose dim L u. Then β H R such that A(β) L u. Hence β, γ L u, and clearly β and γ are connected in L u Ṽs. As Ṽs is a smooth variety, it follows that γ H R. Hence Ṽ s = H R. Recall that the standard inner product in S n (R) is given by A, B = tr(ab), so with respect to this inner product a matrix A is normalized if and only if tr(a ) =. Corollary 5.. Let n and V satisfy the assumptions of Lemma 5.3. Let V ν = {A V : tr (A ) = } and V νs = {A V ν : det A = 0}. Then V νs is a smooth connected hypersurface in V ν. Proof. Assume that the basis A, A,..., A n+ is orthonormal with respect to the standard inner product in S n (R). Let Ṽs, T R and Y r be as in Lemma 5.3. Identify Ṽs with V νs /{±I}. It follows from Lemma 5.3 that V νs is smooth. To show that V νs is connected, it suffices to show we can connect some A V νs to A in V νs. Take y S n such that ỹ Y r. Connect y to y by a path J (in S n ) such that J Y r. For z S n such that z / T R let ˆα(z) = α(z), α(z)

17 7 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY where α(z) is given by (5.3). Since α is odd, A(ˆα(z)), z J, connects A(ˆα(y)) to A(ˆα( y)) = A(ˆα(y)). Theorem 5.. Let n be odd, n 3, and let V be an n + dimensional subspace of S n (R) such that each nonzero matrix in V has rank n or n. Then each nonzero matrix in V has at least n positive and negative eigenvalues. Proof. Let A, A,..., A n+ be a basis of V. We distinguish two cases. Case. Suppose that V does not contain a matrix of rank one. Let V νs be as in Corollary 5.. Then, each matrix in V νs has rank n, and by the connectedness of V νs it follows that A and A have the same inertia for each A V νs. It follows that every singular matrix in V \{0} has n positive n eigenvalues, negative eigenvalues, and zero eigenvalue. Continuity yields immediately that each 0 A V has at least n positive and negative eigenvalues. Case. Suppose now that V does contain a rank one matrix. Note that we have dim P(V ) = n(n+) (n + ), while dim PV,n (R) = n. Hence, for any ɛ > 0 there exist A,ɛ, A,ɛ,..., A n+,ɛ S n (R) which satisfy the following three conditions: (i) A i,ɛ A i < ɛ, i =,,..., n +. (ii) V ɛ = span{a,ɛ, A,ɛ,..., A n+,ɛ } is n + dimensional. (iii) Vɛ does not contain a rank one matrix. For ɛ > 0 small enough, we can apply Case and conclude that any nonzero matrix in V ɛ has at least n positive and negative eigenvalues. Let ɛ 0 and use continuity to finish the proof. 6. Concluding remarks and conjectures. Theorem 6.. Let q Z. Then deg PV n q,n (C) is odd if n > q and ( (6.) n ±q mod log q ). Proof. Let 0 q Z. Let µ(q) = log q and 0 ν(q) Z be the largest integer such that ν(q) divides q. Then ν(q) + µ(q). Clearly if l m 0 (mod µ ) for some µ Z then ν(l) = ν(m). Let δ k,n := deg PV k,n (C). We claim that for n satisfying (6.) (6.) q ν(δ n q,n ) = ν j=0 )) ((ˆq + j ν q j (( j + j )), ˆq = ±q.

18 SPACES OF MATRICES 73 Here for any real x and 0 k Z we let ( ) x x(x )... (x k + ) =. k k! In particular (6.3) ( ) x k ( ) x + k = ( ) k, k =,.... k Observe that for j q ( ) n + j (n + j)(n + j )... (n q + j + ) =. q j... (q j) For n satisfying (6.) it follows that n + l ±q + l 0 (mod µ(q) ) if q + l q. Hence (6.) holds. Our theorem is equivalent to the statement that ν(δ n q,n ) = 0 if (6.) holds. We first consider the case n q (mod µ(q) ). Then q ν(δ n q,n ) = ν j=0 q = ν j=0 q = ν j=0 (( q + j q j )) ν (( )) j + ( ( )) (q j) ( ) q j q j (( )) (q j) q j j q (( )) j + ν j j=0 q (( )) j + ν = 0. j We now consider the case n q (mod µ(q) ). In this case it is enough to show the identity ) (6.4) q j=0 ( q+j q j ( j+ j j=0 ) =, q =,.... We prove the above identity by induction on q. For q = (6.4) trivially holds. Assume that (6.4) holds for q = m. Let q = m +. Use the identity ( ) m + + j = m + + j m + j m + j to deduce (6.4) for q = m + from q = m. ( ) m + j, j = 0,,..., m, m j Corollary 6.. Let n > q and let (6.) hold. Then ( ) q + d(n, n q, R) = d(n, n q, C) = +.

19 74 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY Corollary ( 6.. Let n q and assume that (6.) holds. Then any q+ ) dimensional subspace U of Sn (R) contains a nonzero matrix with an eigenvalue of multiplicity q at least. Furthermore, there exists an ( ) q+ dimensional subspace V S n (R) such that each eigenvalue of 0 A V has a multiplicity less than q. Proof. For n = q U = S n (R) and the eigenvalue of I q has multiplicity q. Assume that V S q (R) does not contain I q. Then each eigenvalue of 0 A V has multiplicity less than q. Assume that q < n. Then the claim of the corollary follows from Corollary 6. and Proposition 3.3. Clearly δ n,n = n is odd if and only if n is odd. As d(n, n, R) = ρ s (n) + it follows that d(n, n, R) > d(n, n, C) = if n is even. Conjecture 6.. Let n > q and assume that (6.) does not hold. Then δ n q,n is even. Furthermore d(n, n q, R) > d(n, n q, C). For q =, 3, 4, 5 it is straightforward to check that δ n q,n is odd if and only if (6.) holds. Finally, we return to the sequence {d(n, n, R)} n=3 and state two questions. The first unknown number in this sequence is d(5, 3, R). As we have seen in Section 3, d(5,, R) = 7, so d(5, 3, R) 7. Question. Is d(5, 3, R) = 6? We think that the answer is yes. In [6] it is shown that d(5, 3, R) 6. Let L be the 5 dimensional subspace of S 5 (R) spanned by the matrices B =, B =, B 3 = B 5 =, B 4 = A local minimization procedure using Matlab seems to indicate that every nonzero matrix in L has rank 4 or 5. This has yet to be confirmed by other means, but if it is correct, then d(5, 3, R) = 6..,

20 SPACES OF MATRICES 75 (6.5) Note that for every field F we have d(n,, F) d(n,, F) d(n, n, F) d(n, n, F). Question. Is there strict inequality everywhere in (6.5) if F = R? Acknowledgement. We would like to thank H. Tracy Hall for his help in the computations using Maple associated with the example that is discussed in Theorem.. References [] J.F. Adams, Vector fields on spheres, Ann. of Math., 75 (96), , MR 5 #64, Zbl [] J.F. Adams, P.D. Lax and R.S. Phillips, On matrices whose real linear combinations are nonsingular, Proc. Amer. Math. Soc., 6 (965), 38-3, MR 3 #343, Zbl [3] L.B. Beasley and T.J. Laffey, Linear operators on matrices: The invariance of rank-k matrices, Linear Algebra Appl., 33 (990), 75-84, MR 9f:5008, Zbl [4] L.B. Beasley and R. Loewy, Rank preservers on spaces of symmetric matrices, Linear and Multilinear Algebra, 43 (997), 63-86, MR 98k:500, Zbl [5] M.A. Berger and S. Friedland, The generalized Radon-Hurwitz numbers, Compositio Mathematica, 59 (986), 3-46, MR 88a:035, Zbl [6] S. Friedland and A. Libgober, Generalizations of the odd degree theorem and applications, to appear in Israel J. Math.. [7] S. Friedland and R. Loewy, Spaces of symmetric matrices containing a nonzero matrix of bounded rank, Linear Algebra Appl., 87 (999), 6-70, MR 99j:504, Zbl [8] S. Friedland, J.W. Robbin and J.H. Sylvester, On the crossing rule, Comm. on Pure and Appl. Math., 37 (984), 9-37, MR 85b:500, Zbl [9] P. Griffiths and J. Harris, Principles of Algebraic Geometry, John Wiley & Sons, 978, MR 80b:400, Zbl [0] J. Harris and L.W. Tu, On symmetric and skew-symmetric determinantal varieties, Topology, 3 (984), 7-84, MR 85c:403, Zbl [] A. Hurwitz, Über die komposition der quadratischen formen, Math. Ann., 88 (93), -5. [] P.D. Lax, The multiplicity of eigenvalues, Bull. Amer. Math. Soc., 6 (98), 3-5, MR 83a:5009, Zbl [3] J. Radon, Lineare Scharen orthogonaler matrizen, Abh. Math. Sem. Univ. Hamburg, (93), -4. Received March 5, 00 and revised August 6, 00 Department of Industrial Engineering Technion Israel Institute of Technology address: dmfalikm@ie.technion.ac.il Department of Mathematics, Statistics and Computer Science University of Illinois at Chicago Chicago, IL

21 76 D. FALIKMAN, S. FRIEDLAND, AND R. LOEWY address: Department of Mathematics Technion Israel Institute of Technology address: The second author was a Lady Davis Visiting Professor at the Technion in the Fall of 000. The research of the third author was supported by the Fund for the Promotion of Research at the Technion.