A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX. Thomas Craven and George Csordas

Transcription

1 A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX Thomas Craven and George Csordas Abstract. We establish a sufficient condition for strict total positivity of a matrix. In particular, we show that if the (positive elements of a square matrix grow sufficiently fast as their distance from the diagonal of the matrix increases, then the matrix is strictly totally positive.. Introduction. The importance of total positivity of matrices in several areas of mathematics has been pointed out in an excellent inclusive survey by T. Ando [A]. The authors particular interest in this area of research stems from the applications of total positivity of matrices to the theory of distribution of zeros of entire functions ([CC]. For Toeplitz matrices, that is, matrices of the form T =(a i j n i,j=, a complete characterization of the total positivity, in terms of certain entire functions, has been established in a series of papers by Aissen, Schoenberg and Whitney [ASW], Edrei [E, E2, E3] and Schoenberg [S] (see also Karlin [K]. We recall that (cf. [A], a matrix A is said to be totally positive, if every square submatrix has a nonnegative determinant and A is said to be strictly totally positive, if every square submatrix has a positive determinant. While it is well known that many of the nontrivial examples of totally positive matrices are obtained by restricting certain kernels to appropriate finite subsets of R (see, for example, Ando [A, p. 22] or the exhaustive treatise of Karlin [K], the de facto verification of total positivity is, in general, a very difficult problem. For recent references, see, for example, [B], [GP], [GP2] and [GP3]. The primary purpose of this paper is to provide a new sufficient condition for a matrix to be totally positive. We prove that if M =(a ij, a ij > 0, is an n n matrix with the property that (. a ij a i,j c 0 a i,j a i,j ( i, j n, where c 0 = is a constant defined in Section 2, then M is strictly totally positive (Theorem 2.2. One of the referees kindly pointed out that in [GP2], M. Gasca 99 Mathematics Subject Classification. Primary 5A48, 5A57. Key words and phrases. Hankel matrix, increasing sequence, moment, totally positive matrix. Typeset by AMS-TEX

2 2 THOMAS CRAVEN AND GEORGE CSORDAS and J. M. Peña provide an algorithm of O(n 3 operations to check the total positivity or strict total positivity of an n n matrix. Although our Theorem 2.2 provides only a sufficient condition for strict total positivity, the corresponding computational cost is only O(n 2 operations, making it an attractive alternative when it appears that it may apply. Our principal interest here is in Hankel matrices, that is, matrices which are of the form A =(a ij 2 n i,j=. Corresponding to a sequence of real numbers {λ k} k=0, we can form the matrices A n =(λ ij 2 n i,j= = λ 0 λ... λ n λ λ 2... λ n... λ n λ n... λ 2n (n=0,,2,... and we refer to these matrices as the Hankel matrices associated with the sequence {λ k } k=0 ([G, vol., p. 338]. If det A n > 0forn=0,,2,..., then we will say that the sequence {λ k } k=0 is a positive definite sequence (cf. [W, p ]. In 939, R. P. Boas showed that any sequence {µ k } k=0,where (.2 µ 0 and µ n (nµ n n (n =,2,..., leads to a soluble Stieltjes problem, where the µ n s are the moments of a nondecreasing function µ(t (cf. [W, p. 40]. An example of a sequence satisfying (.2 is µ 0 =, µ n =n nn for n. One consequence (Corollary 2.3 of our main theorem asserts that if {λ k } k=0 is a sequence of positive numbers such that for k =,2,3,..., (.3 λ k λ k cλ 2 k, where c c 0, then the Hankel matrices associated with the sequences {λ k } k=0 and {λ k} k=0 are strictly totally positive. Thus, by the classical results from the theory of moments (cf. [W, p. 36], the sequence {λ k } k=0 leads to a soluble Stieltjes problem, where the λ k s are the moments of a nondecreasing function µ(t. Now if a sequence {λ k } k=0,whereλ 0 =,λ =3, satisfies (.3 with c =9,thenλ n 3 n2. Since these sequences grow slower, by an order of magnitude, than the sequences which satisfy the conditions (.2, Corollary 2.3 is a generalization of the aforementioned theorem of Boas. Other applications of Theorem 2.2 will appear elsewhere. In Section 3, we briefly discuss the role of the constant appearing in inequalities (. and (.3. The question whether or not the constant c 0 = is best possible remains open. The paper concludes with a simple proof of the total positivity of the Hankel matrices associated the sequence {λ k } k=0, where the λ k s satisfy (.3 with a somewhat larger constant, namely, c = (Theorem Main theorem. For any square matrix M, we shall write M(i,...,i r j,...,j r for the submatrix of M with its rows i,...,i r and columns j,...,j r deleted.

3 A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 3 Lemma 2.. Let c. Assume that M =(a ij, a ij > 0, isann nmatrix with the property that a ij a i,j ca i,j a i,j ( i, j n. If i<kand j < l,thena ij a kl c (l j(k i a il a kj. In particular, if a Hankel matrix satisfies a ij = λ ij 2 and λ k λ k cλ 2 k,thenλ kλ mn k c (n k(m k λ n λ m for all m, n > k 0. Proof. Working with each pair of adjacent rows, the inequality on the entries of M gives a kl a k,l 2 a kj c a k,l c 2 c l j a k,l a k,l a k,l 2 a k,j a k,l c a k,l c 2 a k,l 2 c l ja k,j a k 2,l a k 2,l a k 2,l 2 a k 2,j. a i,l a il c a i,l a i,l c 2 a i,l 2 a i,l 2 Now multiply the rows together to obtain a kl a il c l ja i,j a ij. c (l j(k i a kj a ij as desired. Setting a ij = λ ij 2, n = il 2, m = j k 2 andk=ij 2 yields λ k λ mn k c (n k(m k λ n λ m. The restrictions i<k,j<lonly eliminate the trivially true cases k = m and k = n. We now proceed to the main theorem of the paper. parameters: c 0 = , the unique real root of x 3 5x 2 4x and We begin by introducing two r 0 = 2c 0 c 0 = These numbers arise from the solution of the simultaneous equations 2 r 0 r 0 c 2 0 =, r 0 c 0 r 0 =. Since the proof is long and involved, we begin with a sketch of the main ideas. We shall actually prove a much stronger result than the one stated. We prove strong inequalities on the minors multiplied by the corresponding elements of the matrix: see (2. and (2.2 below. Though the notation is imposing, the message of (2. and (2.2 is simply that the values of the positive expressions a ij det M(i j decrease by a factor of at least r 0 as the position (i, j moves away from the main diagonal along any row or column. The final result follows immediately from these inequalities by expanding the determinant along the last column. To prove (2. and (2.2, we use induction on the size n of the original matrix M. An expression such as a ij det M(i j r 0 a i,j det M(i j is computed by expanding

4 4 THOMAS CRAVEN AND GEORGE CSORDAS both minors along the row in which the matrices differ, giving an expression such as (2.3. The inequality is then proved by rearranging the terms so that the induction hypotheses and the inequalities on the elements of the original matrix (see Theorem 2.2(b below can be applied to each summand to demonstrate positivity. The induction hypothesis applies since the matrices M(i j and M(i j have smaller size. These smaller matrices are not Hankel even if M is, so the proof, even for Hankel matrices, requires the greater generality of the statement of the theorem. The main complications of the proof arise in rearranging these sums to see that they are positive, as it requires looking at many cases, depending on the relative size of the subscripts, whether they are odd or even, and where the entries are located with respect to the main diagonal of the matrix. Theorem 2.2. Let M =(a ij be an n n matrix with the property that (a a ij > 0 ( i, j n and (b a ij a i,j c 0 a i,j a i,j ( i, j n. Then M is strictly totally positive. Proof. For notational simplicity in this proof, we write c and r for c 0 and r 0, respectively. For i<kand j<l, we still have a ij a kl ca il a kj by the preceding lemma, so this relation holds for all submatrices of M. BychoosingMtobeacounterexample of minimum size, we may assume that all proper minors of M are positive. We shall expand det M by minors. In order to estimate the result, we form a much stronger induction hypothesis: (2. a ij det M(i j ra i,j det M(i j 0if2 i j a i,j det M(i j ra ij det M(i j 0ifi>j. Along with this comes the corresponding transpose condition on the columns (2.2 a ij det M(i j ra i,j det M(i j 0if2 j i a i,j det M(i j ra ij det M(i j 0ifj>i. By the symmetry of conditions (a and (b, it suffices to prove the row condition (2., but we shall need both (2. and (2.2 for the induction hypothesis. Inequalities (2. and (2.2 are true for n = 2 by hypothesis (b. If we can prove (2. for arbitrary n, then the expansion of det M along column n gives an alternating sum of a strictly decreasing sequence beginning with a positive term: det M = a nn det M(n n a n,n det M(n na n 2,n det M(n 2 n >0. We prove (2. by induction. Assume that (2. and (2.2 hold for matrices satisfying (a and (b of smaller size. Compute a ij det M(i j ra i,j det M(i j by expanding both

5 A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 5 matrices along the row in which they differ. a ij det M(i j ra i,j det M(i j ( = a ij k<j( ki a i,k det M(i,i k, j k>j( ki a i,k det M(i,i j, k (2.3 ra i,j ( k<j( ki a ik det M(i,i k, j k>j( ki a ik det M(i,i j, k j = ( ki [a ij a i,k ra i,j a ik ]detm(i,i k, j k= n k=j ( ki [a ij a i,k ra i,j a ik ]detm(i,i j, k. We continue (2.3 in several cases. First assume that i j, i is even and j is odd. We begin with the generic case (i 4, j i 3. Then, rearranging terms, we have a ij det M(i j ra i,j det M(i j =(a ij a i, ra i,j a i, detm(i,i,j (k = ra i,j a i,2 det M(i,i 2,j a ij 2 a i,3 det M(i,i 3,j a i,2 det M(i,i 2,j 2 a ija i,3 ra i,j a i,3 det M(i,i 3,j (k =2, 3 ra i,j a i,i 2 det M(i,i i 2,j a ij r a i,i det M(i,i i,j a i,i 2 det M(i,i i 2,j a ij r a i,i det M(i,i i,j a i,i det M(i,i i, j ( r c 2 a ija i,i ra i,j a i,i det M(i,i i,j ra i,j a i,i det M(i,i i, j ( r c a ija i,i ra i,j a i,i det M(i,i i,j (k = i 2, i,i

6 6 THOMAS CRAVEN AND GEORGE CSORDAS a ij r a i,i det M(i,i i,j a i,i2 det M(i,i i2,j ra i,j a i,i2 det M(i,i i2,j (k = i,i2 2 a ija i,j 2 ra i,j a i,j 2 det M(i,i j 2,j a ij 2 a i,j 2 det M(i,i j 2,j a i,j det M(i,i j,j ra i,j (a i,j det M(i,i j,j a i,j det M(i,i j, j a ij a i,j det M(i,i j, j (k=j 2,j,j ((r a i,j a i,j2 a ij a i,j2 detm(i,i j, j 2 a i,j (a i,j2 det M(i,i j, j 2 ra i,j3 det M(i,i j, j 3 a ij a i,j3 det M(i,i j, j 3 (k=j2,j3 in which all the terms are nonnegative. In groupings with two different determinants, we have used the induction hypothesis (2.2 on the matrix M(i j in all instances except the last two, where it is applied to the matrix M(i j. In groupings with a common determinant, we have used Lemma 2. to ensure that the first term is at least c 2 times the second one. Note that it does not matter whether n is odd or even. This works also in the special case i =j (ieven with slight modifications: k = j = i does not occur, so the (k = i,i 2 grouping is changed to a ij a i,i2 det M(i,i j, i 2 ra i,j a i,i2 det M(i,i j, i 2. Here the new negative term ra i,j a i,i2 det M(i,i j, i2 = ra i,j a i,j det M(i,i j, j must now be paired with the unused positive term ra i,j a ii det M(i,i i, j from k = i and we use the induction hypothesis on M(i j. For the special case i =2 (j 3, the groupings for k =,2,3 must be changed to 2 a 2ja ra j a 2 det M(, 2,j ra j a 22 det M(, 2 2,j a 2j 2 a det M(, 2,j a 2 det M(, 2 2,j (a 2j a 3 ra j a 23 detm(, 2 3,j.

7 A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 7 If, instead of being odd, we make j even, this chain of summands changes, beginning at k = j ; just prior to this point we have been balancing pairs of summands with k first odd, then even (as in k = i, i 2 above. Now j is odd, but the even k = j does not occur. Thus we continue (a ij a i,j ra i,j a i,j detm(i,i j,j (a i,j a i,j a ij a i,j detm(i,i j, j a i,j ((r a i,j det M(i,i j, j ra i,j2 det M(i,i j, j 2 a ij a i,j2 det M(i,i j, j 2 (k=j,j,j2 and then continue balancing pairs with k first odd, then even, as in the previous case where j is odd. Here we have used the hypothesis (2.2 on M(i j. Now the special case is i = j, where summands are balanced in pairs as in k =2, 3uptotheoddk=i 3. For k = i 2, i, we group terms as follows ra i,i a i,i 2 det M(i,i i 2,i a ii r a i,i det M(i,i i,i a i,i 2 det M(i,i i 2,i ( r c a iia i,i ra i,i a i,i det M(i,i i,i. For k = i,..., the signs are reversed from the j odd case, so the summands are balanced in pairs as in k = j 2,j3above: ((r a i,i a i,i a ii a i,i detm(i,i i, i a i,i (a i,i det M(i,i i, i ra i,i2 det M(i,i i, i 2 a ij a i,i2 det M(i,i i2,j. When i is odd and 2 <i j, the situation is nearly the same. The computation begins a ij det M(i j ra i,j det M(i j = ra i,j a i, det M(i,i,j a ij 2 a i,2 det M(i,i 2,j a i, det M(i,i,j 2 a ija i,2 ra i,j a i,2 det M(i,i 2,j (k =, 2

8 8 THOMAS CRAVEN AND GEORGE CSORDAS and continues as in the case of i even without the first term; that is, the same pattern of pairing values of k is followed with k shifted by. The grouping of terms follows the cases above depending on whether i and j have the same or different parity modulo two. In particular, the difficult case of k = i 2, k=i,k=iwith i j is identical to the earlier computation. We now turn our attention to the second expression of (2., in which i>j. As before, we expand both matrices along the row in which they differ, obtaining a i,j det M(i j ra ij det M(i j = a i,j ( k<j( ki a ik det M(i,i k, j k>j( ki a ik det M(i,i j, k (2.4 ra ij ( k<j( ki a i,k det M(i,i k, j k>j( ki a i,k det M(i,i j, k j = ( ki [a i,j a ik ra ij a i,k ]detm(i,i k, j k= n k=j ( ki [a i,j a ik ra ij a i,k ]detm(i,i j, k. We again begin by assuming that i is even and greater than 2 and that j is odd and j i 3. In this case, the terms of (2.4 can be regrouped as a i,j det M(i j ra ij det M(i j = a i,j a i, det M(i,i,j a ij ((r a i,2 det M(i,i 2,j ra i, det M(i,i,j (a ij a i,2 a i,j a i,2 detm(i,i 2,j (k =, 2 (a i,j a i,j ra ij a i,j detm(i,i j, j (k=j ra ij a i,j2 det M(i,i j, j 2 a i,j 2 a i,j3 det M(i,i j, j 3 a i,j2 det M(i,i j, j 2 2 a i,ja i,j3 ra ij a i,j3 det M(i,i j, j 3 (k=j2,j3

9 A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 9 ra ij a i,i det M(i,i j, i ( r c 2 a i,ja ii ra ij a i,i det M(i,i j, i a i,j r a ii det M(i,i j, i a i,i det M(i,i j, i a i,j r a ii det M(i,i j, i a i,i det M(i,i j, i ra ij a i,i det M(i,i j, i (k=i, i, i ( r c a i,ja i,i2 ra ij a i,i2 det M(i,i j, i 2 a i,j r a i,i2 det M(i,i j, i 2 a i,i3 det M(i,i j, i 3 ra ij a i,i3 det M(i,i j, i 3. (k=i2,i3 As before, we need a modification for j = i. Since k = j = i is no longer in the sum, the k = i, i grouping is now ( r c a i,ja ii ra ij a i,i det M(i,i j, i a i,j r a ii det M(i,i j, i a i,i det M(i,i j, i ra ij a i,i det M(i,i j, i For the special case i = 2, the terms for k = j = do not occur, so there is no problem at the beginning. If j is even (j i 3, the grouping is as above for k =,2,...,j 2, and continues as a i,j a i,j det M(i,i j,j ra ij (a i,j det M(i,i j, j a i,j det M(i,i j,j a i,j 2 a i,j2 det M(i,i j, j 2 a i,j det M(i,i j, j 2 a i,ja i,j2 ra ij a i,j2 det M(i,i j, j 2 (k=j,j,j2 (r a ij a i,i det M(i,i j, i a ij (a i,i det M(i,i j, i ra i,i det M(i,i j, i

10 0 THOMAS CRAVEN AND GEORGE CSORDAS a i,j 2 a ii det M(i,i j, i a i,i det M(i,i j, i a i,j 2 a ii det M(i,i j, i a i,i det M(i,i j, i ra ij a i,i det M(i,i j, i (k=i, i, i ( r c a i,ja i,i2 ra ij a i,i2 det M(i,i j, i 2 a i,j r a i,i2 det M(i,i j, i 2 a i,i3 det M(i,i j, i 3 ra ij a i,i3 det M(i,i j, i 3. (k=i2,i3 In the special case where j = i 2 is the deleted value of k, this should be interpreted as using the spare term (r a ij a i,i det M(i,i j, i from the case k = i to dominate ra ij a i,j det M(i,i j,j from the case k = i 3. The final remaining case is where i > j and i is odd. In this case, the first term (k = is (ra ij a i, a i,j a i, detm(i,i,j>0. The remaining terms can be grouped as in the case where i is even. This completes the proof of (2.. As noted above, condition (2.2 simply replaces rows by columns, so it holds by a symmetric argument. Here we confine our attention to a single application of Theorem 2.2 and provide the following generalization of a result of Boas [W, p. 40] mentioned in the introduction. Corollary 2.3. Let {λ k } k=0 be a sequence of positive numbers satisfying λ kλ k c 0 λ 2 k. Then, for each positive integer n, the Hankel matrices A = {λ ij 2} n i,j= and A = {λ ij } n i,j=, associated with {λ k} k=0 and {λ k} k=0, respectively, are strictly totally positive. Moreover, there is a nondecreasing function µ(t with infinitely many points of increase such that (2.5 λ n = 0 t n dµ(t (n =0,,2,... Proof. Fix a positive integer n and set a ij = λ ij 2.Then a ij a i,j c 0 a i,j a i,j = λ ij 2 λ ij c 0 λ 2 ij 0 ( i, j n. Therefore, by Theorem 2.2, both matrices A and A are strictly totally positive and a fortiori the sequences {λ k } k=0 and {λ k} k=0 are positive definite sequences. But then by a well-known theorem from the theory of moments [W, p. 38], this is equivalent to the existence of a function µ(t with the stated properties such that (2.5 holds.

11 A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 3. Remarks on the growth constant. An open question is whether the constant c 0 in Theorem 2.2 is merely an artifact of the proof or actually best possible. We note that the crucial conditions on r 0 and c 0 come from the i = j case when k = i 2, i and from when k = i, iin the j odd case. The i = j case is truly a problem. If we could get r to be 3 in this case, the proof would go through for c = 4. However, the following 4 4 example shows that the largest we can hope for r with c =4is 8 3. Example 3.. Let 4 x M = 4 x x 2 4 x x 2 y x x 2 4y y 2 x ( 2 Computing a 44 det M(4 4 ra 34 det M(3 4 yields 8 3r 32 x 256 x 2 y 2. Since y may be made arbitrarily large, the coefficient of y 2 must be positive. Within this, x can be arbitrarily large, so that we must have 8 3r 0. For special sequences, we can certainly do better for the value of c. For the prototypical sequence {λ k2 } k=0, with λ>, any constant c λ2 works. In fact, we can compute the determinant directly. Example 3.2. The Hankel matrices associated with the sequence {λ k2 } k=0, λ>, are all strictly totally positive. For the (n (n matrix, one can factor all the powers of λ out of the rows and columns to see that the determinant is... (λ λ 4 λ n2 2 λ 2... λ 2n =(λ λ 4 λ n2 2 (λ 2j λ 2i.... λ 2n... λ 2n2 0 i<j n The Vandermonde matrix is strictly totally positive by [PS, vol. II, Part V, No. 48]. For an arbitrary 3 3 matrix based on a sequence {λ k } k=0 satisfying λ k λ k cλ 2 k for k, one can check that any constant greater than c = 2 will always work. With a careful choice of larger matrices, we can get larger lower bounds for c asshowninthenext example. Example 3.3. Consider the matrix c x M = c x x 2 c x x 2 y x x 2 y cy 2 x 2,

12 2 THOMAS CRAVEN AND GEORGE CSORDAS where as in Example 3., we may make x and then y arbitrarily large. The determinant is (c 2 3c y 2,sothatc It is interesting that a vastly simpler proof is available for the application of our main theorem to Hankel matrices if we allow a somewhat higher constant. The following, though a weaker version of Corollary 2.3, is interesting for the simplicity of its proof. Theorem 3.4. Let c be the number (approximately such that k=0 ( c k2 = 3/2 and let {λ k } k=0 be a sequence of positive numbers satisfying λ k λ k cλ 2 k. Then the Hankel matrices associated with {λ k } k=0 are all strictly totally positive. Proof. We first prove that the (n (nhankelmatrixaassociated with {λ k } k=0 is positive definite. Let D be the diagonal matrix with diagonal entries λ0, λ2, λ4,..., λ2n. Then it suffices to show that the matrix DAD is positive definite. Indeed, if DAD is positive definite, then x T Ax =(D x T (DAD(D x and therefore A is also positive λ definite. Now the (i, j entryofdad is ij 2 and so the diagonal entries of λ2i 2 λ2j 2 DAD are all. Since λ k λ k cλ 2 k, we can invoke Lemma 2. with m = n = i j 2 and k =2i 2 to conclude that λ 2i 2 λ 2j 2 c (j i2 λ 2 ij 2. Using this inequality, we can estimate the sum of the entries in the ith row of DAD by n j= λ ij 2 λ2i 2 λ2j 2 < < 2 n j= c (j i2 /2 ( i c k2 ( c k2 k=0 k= ( c k2. k=0 By our hypothesis on c, the row sum is less than 2 and thus the matrix DAD is strictly diagonally dominant. Hence by the Gerschgorin circle theorem, the eigenvalues of DAD are all positive (and in fact lie in the open interval (0, 2. Hence DAD and therefore A is positive definite. Similarly, the n n Hankel matrix A associated with the sequence λ,λ 2,λ 3,... is also positive definite. By a classical theorem of Fekete (see, for example, [A, Theorem 2.5], in order to verify the strict total positivity of A, it suffices to check the signs of all those minors of A which are the determinants of submatrices with consecutive rows and columns. But each of these minors is a principal minor of either A or A,and thus is positive. Therefore it follows that A is strictly totally positive.

13 A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 3 Acknowledgement. The authors wish to thank the referees for several helpful suggestions. References [A] T. Ando, Totally positive matrices, Linear Algebra Appl. 90 (987, [ASW] M. Aissen, I. J. Schoenberg and A. M. Whitney, On generating functions of totally positive sequences I, J. Anal. Math. 2(952, [B] F. Brenti, Unimodal, Log-Concave and Pólya Frequency Sequences in Combinatorics, Memoirs of the Amer. Math. Soc., vol. 8, No. 43, Providence, RI, 989. [CC] T. Craven and G. Csordas, Complex zero decreasing sequences, Methods Appl. Anal. 2 (995, [E] A. Edrei, On the generating functions of totally positive sequences II, J. Anal. Math. 2 (952, [E2] [E3] [G] [GP] A. Edrei, Proof of a conjecture of Schoenberg on the generating function of a totally positive sequence, Canad. J. Math. 5 (953, A. Edrei, On the generating function of a doubly infinite totally positive sequence, Trans.Amer. Math. Soc. 74 (953, F. R. Gantmacher, The Theory of Matrices, vols. and 2, Chelsea Publishing Co., New York, 959. M. Gasca and J. M. Peña, On the characterization of totally positive matrices, Approximation Theory, Spline Functions and Applications (Maratea, 99, , NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., vol. 356, Kluwer Acad. Publ., Dordrecht, 992. [GP2] M. Gasca and J. M. Peña, Total positivity and Neville elimination, Linear Algebra Appl. 65 (992, [GP3] M. Gasca and J. M. Peña, On factorizations of totally positive matrices, Total Positivity and its Applications (Jaca, 994, 09 30, Kluwer Acad. Publ., Dordrecht, 996. [K] S. Karlin, Total Positivity, vol., Stanford Univ. Press, Stanford, Calif., 968. [PS] G. Pólya and G. Szegö, Problems and Theorems in Analysis, vols. I and II, Springer-Verlag, New York, 976. [W] D. V. Widder, The Laplace Transform, Princeton Univ. Press, Princeton, 94. [S] I. J. Schoenberg, On Pólya frequency functions I. The totally positive functions and their Laplace transforms, J. Anal. Math. (95, Department of Mathematics, University of Hawaii, Honolulu, HI 96822

14 4 THOMAS CRAVEN AND GEORGE CSORDAS address: Department of Mathematics, University of Hawaii, Honolulu, HI address: