A Sufficient condition for strict total positivity of a matrix

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.

Full text

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.
1. Introduction. The importance of total positivity of matrices in several areas of math-
ematics 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=1 , 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 [E1, E2, E3] and Schoenberg [S]
(see also Karlin [K]).
We recall that (cf. [A]), a matrix Ais said to be totally positive, if every square sub-
matrix has a nonnegative determinant and Ais said to be strictly totally positive,ifevery
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. 212] 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], [GP1], [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), aij >0, is an n×nmatrix with the
property that
(1.1) aij ai+1,j+1 ≥c0ai,j +1ai+1,j (1 ≤i, j ≤n−1),
where c0=4.07959562349 ... is a constant defined in Section 2, then Mis strictly totally
positive (Theorem 2.2). One of the referees kindly pointed out that in [GP2], M. Gasca
1991 Mathematics Subject Classification. Primary 15A48, 15A57.
Key words and phrases. Hankel matrix, increasing sequence, moment, totally positive matrix.
Typeset by A
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1

2 THOMAS CRAVEN AND GEORGE CSORDAS
and J. M. Pe˜na provide an algorithm of O(n3) operations to check the total positivity
or strict total positivity of an n×nmatrix. Although our Theorem 2.2 provides only a
sufficient condition for strict total positivity, the corresponding computational cost is only
O(n2) 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
i+j−2
)
n+1
i,j=1 . Corresponding to a sequence of real numbers {λk}∞
k=0, we can form the
matrices
An=(λ
i+j−2
)
n+1
i,j=1 =


λ0λ1... λ
n
λ
1λ
2... λ
n+1
...
λ
nλ
n+1 ... λ
2n


(n=0,1,2,...)
and we refer to these matrices as the Hankel matrices associated with the sequence {λk}∞
k=0
([G, vol. 1, p. 338]). If det An>0forn=0,1,2,..., then we will say that the sequence
{λk}∞
k=0 is a positive definite sequence (cf. [W, p. 132–135]). In 1939, R. P. Boas showed
that any sequence {µk}∞
k=0,where
(1.2) µ0≥1andµ
n
≥(nµn−1)n(n=1,2,...),
leads to a soluble Stieltjes problem, where the µn’s are the moments of a nondecreasing
function µ(t) (cf. [W, p. 140]). An example of a sequence satisfying (1.2) is µ0=1,
µ
n=n
n
nfor n≥1. One consequence (Corollary 2.3) of our main theorem asserts that if
{λk}∞
k=0 is a sequence of positive numbers such that for k=1,2,3,...,
(1.3) λk−1λk+1 ≥cλ2
k,where c≥c0,
then the Hankel matrices associated with the sequences {λk}∞
k=0 and {λk+1}∞
k=0 are strictly
totally positive. Thus, by the classical results from the theory of moments (cf. [W, p. 136]),
the sequence {λk}∞
k=0 leads to a soluble Stieltjes problem, where the λk’s are the moments
of a nondecreasing function µ(t). Now if a sequence {λk}∞
k=0,whereλ
0=1,λ
1=3,
satisfies (1.3) with c=9,thenλ
n≥3
n
2
. Since these sequences grow slower, by an order
of magnitude, than the sequences which satisfy the conditions (1.2), 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 (1.1) and
(1.3). The question whether or not the constant c0=4.079 ... is best possible remains
open. The paper concludes with a simple proof of the total positivity of the Hankel
matrices associated the sequence {λk}∞
k=0, where the λk’s satisfy (1.3) with a somewhat
larger constant, namely, c=4.801 ... (Theorem 3.4).
2. Main theorem. For any square matrix M, we shall write M(i1,...,i
r|j
1,...,j
r)for
the submatrix of Mwith its rows i1,...,i
rand columns j1,...,j
rdeleted.

A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 3
Lemma 2.1. Let c≥1. Assume that M=(a
ij ),aij >0,isann×nmatrix with the
property that
aijai+1,j+1 ≥cai,j+1 ai+1,j (1 ≤i, j ≤n−1).
If i<kand j<l,thena
ijakl ≥c(l−j)(k−i)ailakj. In particular, if a Hankel matrix
satisfies aij =λi+j−2and λk−1λk+1 ≥cλ2
k,thenλ
k
λ
m+n−k≥c
(n−k)(m−k)λnλmfor all
m, n > k ≥0.
Proof. Working with each pair of adjacent rows, the inequality on the entries of Mgives
akl
ak−1,l ≥cak,l−1
ak−1,l−1≥c2ak,l−2
ak−1,l−2≥···≥c
l−ja
kj
ak−1,j
ak−1,l
ak−2,l ≥cak−1,l−1
ak−2,l−1≥c2ak−1,l−2
ak−2,l−2≥···≥c
l−ja
k−1,j
ak−2,j
.
.
.
ai+1,l
ail ≥cai+1,l−1
ai,l−1≥c2ai+1,l−2
ai,l−2≥···≥c
l−ja
i+1,j
aij
.
Now multiply the rows together to obtain akl
ail ≥c(l−j)(k−i)akj
aij as desired. Setting aij =
λi+j−2,n=i+l−2, m=j+k−2andk=i+j−2 yields λkλm+n−k≥c(n−k)(m−k)λnλm.
The restrictions i<k,j<lonly eliminate the trivially true cases k=mand k=n.
We now proceed to the main theorem of the paper. We begin by introducing two
parameters:
c0=4.07959562349 ...,
the unique real root of x3−5x2+4x−1and
r
0=2c
0−1
c
0−1=2.3247179572 ....
These numbers arise from the solution of the simultaneous equations
2
r0
+r0
c2
0
=1,r
0
c
0
+1
r
0
=1.
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 (2.1) and (2.2)
below. Though the notation is imposing, the message of (2.1) and (2.2) is simply that the
values of the positive expressions aij det M(i|j) decrease by a factor of at least r0as 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 (2.1) and (2.2), we use induction on the size nof the original matrix
M. An expression such as aij det M(i|j)−r0ai−1,j det M(i−1|j) is computed by expanding

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)andM(i−1|j) have smaller size. These smaller matrices are not
Hankel even if Mis, 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×nmatrix with the property that
(a) aij >0(1≤i, j ≤n)and
(b) aij ai+1,j+1 ≥c0ai,j +1ai+1,j (1 ≤i, j ≤n−1).
Then Mis strictly totally positive.
Proof. For notational simplicity in this proof, we write cand rfor c0and r0, respectively.
For i<kand j<l, we still have aijakl ≥cail akj by the preceding lemma, so this relation
holds for all submatrices of M.BychoosingMto be a counterexample of minimum size,
we may assume that all proper minors of Mare positive. We shall expand det Mby
minors. In order to estimate the result, we form a much stronger induction hypothesis:
(2.1) aij det M(i|j)−rai−1,j det M(i−1|j)≥0if2≤i≤j
a
i−1,j det M(i−1|j)−raij det M(i|j)≥0ifi>j≥1.
Along with this comes the corresponding transpose condition on the columns
(2.2) aij det M(i|j)−rai,j−1det M(i|j−1) ≥0if2≤j≤i
a
i,j−1det M(i|j−1) −raij det M(i|j)≥0ifj>i≥1.
By the symmetry of conditions (a) and (b), it suffices to prove the row condition (2.1),
but we shall need both (2.1) and (2.2) for the induction hypothesis. Inequalities (2.1) and
(2.2) are true for n= 2 by hypothesis (b). If we can prove (2.1) for arbitrary n,then
the expansion of det Malong column ngives an alternating sum of a strictly decreasing
sequence beginning with a positive term:
det M=ann det M(n|n)−an−1,n det M(n−1|n)+a
n−2,n det M(n−2|n)−···>0.
We prove (2.1) by induction. Assume that (2.1) and (2.2) hold for matrices satisfying (a)
and (b) of smaller size. Compute aij det M(i|j)−rai−1,j det M(i−1|j) by expanding both

A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 5
matrices along the row in which they differ.
(2.3)
aij det M(i|j)−rai−1,j det M(i−1|j)
=aij X
k<j
(−1)k+i−1ai−1,k det M(i−1,i|k, j)
+X
k>j
(−1)k+iai−1,k det M(i−1,i|j, k)!
−rai−1,j X
k<j
(−1)k+i−1aik det M(i−1,i|k, j)
+X
k>j
(−1)k+iaik det M(i−1,i|j, k)!
=
j−1
X
k=1
(−1)k+i−1[aijai−1,k −rai−1,j aik ]detM(i−1,i|k, j)
+
n
X
k=j+1
(−1)k+i[aij ai−1,k −rai−1,jaik]detM(i−1,i|j, k).
We continue (2.3) in several cases. First assume that i≤j,iis even and jis odd. We
begin with the “generic” case (i≥4,j≥i+ 3). Then, rearranging terms, we have
aij det M(i|j)−rai−1,j det M(i−1|j)
=(a
ij ai−1,1−rai−1,jai,1)detM(i−1,i|1,j)(k=1)
+rai−1,jai,2det M(i−1,i|2,j)
+a
ij 1
2ai−1,3det M(i−1,i|3,j)−a
i−1,2det M(i−1,i|2,j)

+1
2a
ij ai−1,3−rai−1,jai,3det M(i−1,i|3,j)(k=2,3)
+···
+rai−1,jai,i−2det M(i−1,i|i−2,j)
+a
ij 1
rai−1,i−1det M(i−1,i|i−1,j)−a
i−1,i−2det M(i−1,i|i−2,j)

+a
ij 1
rai−1,i−1det M(i−1,i|i−1,j)−a
i−1,i det M(i−1,i|i, j)
+r
c2aij ai−1,i−1−rai−1,jai,i−1det M(i−1,i|i−1,j)
+rai−1,jai,i det M(i−1,i|i, j)(k=i−2,i−1,i)
+

r
c
a
ij ai−1,i+1 −rai−1,jai,i+1det M(i−1,i|i+1,j)

6 THOMAS CRAVEN AND GEORGE CSORDAS
+aij 1
rai−1,i+1 det M(i−1,i|i+1,j)−a
i−1,i+2 det M(i−1,i|i+2,j)

+rai−1,jai,i+2 det M(i−1,i|i+2,j)(k=i+1,i+2)
+···
+1
2a
ij ai−1,j−2−rai−1,jai,j−2det M(i−1,i|j−2,j)
+a
ij 1
2ai−1,j−2det M(i−1,i|j−2,j)−a
i−1,j−1det M(i−1,i|j−1,j)

+rai−1,j(ai,j−1det M(i−1,i|j−1,j)−a
i,j+1 det M(i−1,i|j, j +1))
+a
ij ai−1,j+1 det M(i−1,i|j, j +1)
(k=j−2,j−1,j+1)
+((r−1)ai−1,jai,j+2 −aij ai−1,j+2)detM(i−1,i|j, j +2)
+a
i−1,j(ai,j+2 det M(i−1,i|j, j +2)−rai,j+3 det M(i−1,i|j, j +3))
+a
ij ai−1,j+3 det M(i−1,i|j, j +3) (k=j+2,j+3)
+···
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−1|j). In groupings with a common
determinant, we have used Lemma 2.1 to ensure that the first term is at least c2times the
second one. Note that it does not matter whether nis odd or even. This works also in the
special case i+1=j(ieven) with slight modifications: k=j=i+ 1 does not occur, so
the (k=i+1,i+ 2) grouping is changed to
···+a
ijai−1,i+2 det M(i−1,i|j, i +2)
−rai−1,jai,i+2 det M(i−1,i|j, i +2)
+···.
Here the new negative term −rai−1,jai,i+2 det M(i−1,i|j, i+2) = −rai−1,jai,j+1 det M(i−
1,i|j, j + 1) must now be paired with the unused positive term rai−1,jaii det M(i−1,i|i, j )
from k=iand we use the induction hypothesis on M(i−1|j). For the special case i=2
(j≥3), the groupings for k=1,2,3 must be changed to
1
2a2ja11 −ra1ja21det M(1,2|1,j)
+ra1ja22 det M(1,2|2,j)
+a
2j1
2a
11 det M(1,2|1,j)−a
12 det M(1,2|2,j)

+(a
2ja
13 −ra1ja23)detM(1,2|3,j)
+···.

A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 7
If, instead of being odd, we make jeven, this chain of summands changes, beginning at
k=j−1; just prior to this point we have been balancing pairs of summands with kfirst
odd, then even (as in k=i+1,i+ 2 above). Now j−1 is odd, but the even k=jdoes
not occur. Thus we continue
···+(a
ij ai−1,j−1−rai−1,jai,j−1)detM(i−1,i|j−1,j)
+(a
i−1,jai,j+1 −aij ai−1,j+1 )detM(i−1,i|j, j +1)
+a
i−1,j((r−1)ai,j +1 det M(i−1,i|j, j +1)−rai,j+2 det M(i−1,i|j, j +2))
+a
ij ai−1,j+2 det M(i−1,i|j, j +2)
(k=j−1,j+1,j+2)
+···
and then continue balancing pairs with kfirst odd, then even, as in the previous case where
jis odd. Here we have used the hypothesis (2.2) on M(i−1|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−1, we group terms as follows
···+rai−1,iai,i−2det M(i−1,i|i−2,i)
+a
ii 1
rai−1,i−1det M(i−1,i|i−1,i)−a
i−1,i−2det M(i−1,i|i−2,i)

+r
ca
iiai−1,i−1−rai−1,iai,i−1det M(i−1,i|i−1,i)
+···.
For k=i+1,..., the signs are reversed from the jodd case, so the summands are balanced
in pairs as in k=j+2,j+3above:
···+((r−1)ai−1,iai,i+1 −aiiai−1,i+1 )detM(i−1,i|i, i +1)
+a
i−1,i(ai,i+1 det M(i−1,i|i, i +1)−rai,i+2 det M(i−1,i|i, i +2))
+a
ij ai−1,i+2 det M(i−1,i|i+2,j)
+···.
When iis odd and 2 <i≤j, the situation is nearly the same. The computation begins
aij det M(i|j)−rai−1,j det M(i−1|j)
=rai−1,jai,1det M(i−1,i|1,j)
+a
ij 1
2ai−1,2det M(i−1,i|2,j)−a
i−1,1det M(i−1,i|1,j)

+1
2a
ij ai−1,2−rai−1,jai,2det M(i−1,i|2,j)(k=1,2)
+···

8 THOMAS CRAVEN AND GEORGE CSORDAS
and continues as in the case of ieven without the first term; that is, the same pattern
of pairing values of kis followed with kshifted by 1. The grouping of terms follows the
cases above depending on whether iand jhave the same or different parity modulo two.
In particular, the difficult case of k=i−2,k=i−1,k=iwith i≤j+ 1 is identical to
the earlier computation.
We now turn our attention to the second expression of (2.1), in which i>j. As before,
we expand both matrices along the row in which they differ, obtaining
(2.4)
ai−1,j det M(i−1|j)−raij det M(i|j)
=ai−1,j X
k<j
(−1)k+i−1aik det M(i−1,i|k, j)
+X
k>j
(−1)k+iaik det M(i−1,i|j, k)!
−raij X
k<j
(−1)k+i−1ai−1,k det M(i−1,i|k, j)
+X
k>j
(−1)k+iai−1,k det M(i−1,i|j, k)!
=
j−1
X
k=1
(−1)k+i−1[ai−1,jaik −raij ai−1,k]detM(i−1,i|k, j)
+
n
X
k=j+1
(−1)k+i[ai−1,jaik −raijai−1,k ]detM(i−1,i|j, k).
We again begin by assuming that iis even and greater than 2 and that jis odd and
j≤i−3. In this case, the terms of (2.4) can be regrouped as
ai−1,j det M(i−1|j)−raij det M(i|j)
=ai−1,jai,1det M(i−1,i|1,j)
+a
ij((r−1)ai−1,2det M(i−1,i|2,j)−rai−1,1det M(i−1,i|1,j))
+(a
ijai−1,2−ai−1,j ai,2)detM(i−1,i|2,j)(k=1,2)
+···
+(a
i−1,jai,j+1 −raijai−1,j+1 )detM(i−1,i|j, j +1) (k=j+1)
+raijai−1,j +2 det M(i−1,i|j, j +2)
+a
i−1,j 1
2ai,j+3 det M(i−1,i|j, j +3)−a
i,j+2 det M(i−1,i|j, j +2)

+1
2a
i−1,jai,j+3 −raij ai−1,j+3det M(i−1,i|j, j +3)
(k=j+2,j+3)

A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 9
+···
+raijai−1,i−1det M(i−1,i|j, i −1)
+r
c2ai−1,jaii −raij ai−1,idet M(i−1,i|j, i)
+ai−1,j 1
raii det M(i−1,i|j, i)−ai,i−1det M(i−1,i|j, i −1)
+ai−1,j 1
raii det M(i−1,i|j, i)−ai,i+1 det M(i−1,i|j, i +1)

+raijai−1,i+1 det M(i−1,i|j, i +1) (k=i−1,i,i+1)
+r
ca
i−1,jai,i+2 −raij ai−1,i+2 det M(i−1,i|j, i +2)
+a
i−1,j 1
rai,i+2 det M(i−1,i|j, i +2)−a
i,i+3 det M(i−1,i|j, i +3)

+raijai−1,i+3 det M(i−1,i|j, i +3) (k=i+2,i+3)
+···.
As before, we need a modification for j=i−1. Since k=j=i−1 is no longer in the
sum, the k=i, i + 1 grouping is now
···+r
ca
i−1,j aii −raij ai−1,idet M(i−1,i|j, i)
+ai−1,j 1
raii det M(i−1,i|j, i)−ai,i+1 det M(i−1,i|j, i +1)

+raijai−1,i+1 det M(i−1,i|j, i +1)
+···
For the special case i= 2, the terms for k=j= 1 do not occur, so there is no problem at
the beginning.
If jis even (j≤i−3), the grouping is as above for k=1,2,...,j−2, and continues as
+···
+a
i−1,j ai,j −1det M(i−1,i|j−1,j)
+raij (ai−1,j+1 det M(i−1,i|j, j +1)−a
i−1,j−1det M(i−1,i|j−1,j))
+ai−1,j 1
2ai,j+2 det M(i−1,i|j, j +2)−a
i,j+1 det M(i−1,i|j, j +1)

+1
2a
i−1,jai,j+2 −raij ai−1,j+2det M(i−1,i|j, j +2)
(k=j−1,j+1,j+2)
+···
+(r−1)aij ai−1,i−1det M(i−1,i|j, i −1)
+aij (ai−1,i−1det M(i−1,i|j, i −1) −rai−1,i det M(i−1,i|j, i))

10 THOMAS CRAVEN AND GEORGE CSORDAS
+ai−1,j 1
2aii det M(i−1,i|j, i)−ai,i−1det M(i−1,i|j, i −1)
+ai−1,j 1
2aii det M(i−1,i|j, i)−ai,i+1 det M(i−1,i|j, i +1)

+raij ai−1,i+1 det M(i−1,i|j, i +1) (k=i−1,i,i+1)
+r
ca
i−1,jai,i+2 −raij ai−1,i+2det M(i−1,i|j, i +2)
+a
i−1,j 1
rai,i+2 det M(i−1,i|j, i +2)−a
i,i+3 det M(i−1,i|j, i +3)

+raij ai−1,i+3 det M(i−1,i|j, i +3) (k=i+2,i+3)
+···.
In the special case where j=i−2 is the deleted value of k, this should be interpreted
as using the spare term (r−1)aijai−1,i−1det M(i−1,i|j, i −1) from the case k=i−1
to dominate −raij ai−1,j−1det M(i−1,i|j−1,j) from the case k=i−3. The final
remaining case is where i>jand iis odd. In this case, the first term (k=1)is
(raijai−1,1−ai−1,j ai,1)detM(i−1,i|1,j)>0. The remaining terms can be grouped as in
the case where iis even.
This completes the proof of (2.1). 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. 140] mentioned in the introduction.
Corollary 2.3. Let {λk}∞
k=0 be a sequence of positive numbers satisfying λk+1λk−1≥
c0λ2
k. Then, for each positive integer n, the Hankel matrices A={λi+j−2}n
i,j=1 and
A1={λi+j−1}n
i,j=1 , associated with {λk}∞
k=0 and {λk+1}∞
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) λn=Z∞
0
tndµ(t)(n=0,1,2,...).
Proof. Fix a positive integer nand set aij =λi+j−2.Then
a
ij ai+1,j+1 −c0ai,j +1ai+1,j =λi+j−2λi+j−c0λ2
i+j−1≥0(1≤i, j ≤n−1).
Therefore, by Theorem 2.2, both matrices Aand A1are strictly totally positive and a
fortiori the sequences {λk}∞
k=0 and {λk+1}∞
k=0 are positive definite sequences. But then by
a well-known theorem from the theory of moments [W, p. 138], this is equivalent to the
existence of a function µ(t) with the stated properties such that (2.5) holds. 

A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 11
3. Remarks on the growth constant. An open question is whether the constant c0in
Theorem 2.2 is merely an artifact of the proof or actually best possible. We note that the
crucial conditions on r0and c0come from the i=jcase when k=i−2,i−1 and from
when k=i−1,iin the jodd case. The i=jcase 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 rwith c=4is 8
3
.
Example 3.1. Let
M=



11 4 x
14 xx
2
4xx
2y
xx
2y4y
2
x
2




Computing a44 det M(4|4) −ra34 det M(3|4) yields 8−3r+32
x−256
x2y2+···.Since
ymay be made arbitrarily large, the coefficient of y2must be positive. Within this, xcan
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 {λk2}∞
k=0, with λ>1, any constant c≤λ2works. In
fact, we can compute the determinant directly.
Example 3.2. The Hankel matrices associated with the sequence {λk2}∞
k=0,λ>1, are
all strictly totally positive. For the (n+1)×(n+ 1) matrix, one can factor all the powers
of λout of the rows and columns to see that the determinant is
(λ·λ4···λ
n
2)
2






11... 1
1λ
2... λ
2n
...
1λ
2n... λ
2n
2







=(λ·λ
4···λ
n
2)
2Y
0≤i<j≤n
(λ2j−λ2i).
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 {λk}∞
k=0 satisfying λk−1λk+1 ≥cλ2
k
for k≥1, 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 casshowninthenext
example.
Example 3.3. Consider the matrix
M=



11 cx
1cxx
2
cxx
2y
xx
2ycy2
x2




,

12 THOMAS CRAVEN AND GEORGE CSORDAS
where as in Example 3.1, we may make xand then yarbitrarily large. The determinant
is (c2−3c+ 1)y
2+···,sothatc≥3+√
5
2≈2.618.
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 cbe the number (approximately 4.8105828) such that P∞
k=0(√c)−k2=
3/2and let {λk}∞
k=0 be a sequence of positive numbers satisfying λk−1λk+1 ≥cλ2
k.Then
the Hankel matrices associated with {λk}∞
k=0 are all strictly totally positive.
Proof. We first prove that the (n+1)×(n+1)HankelmatrixAassociated with {λk}∞
k=0
is positive definite. Let Dbe the diagonal matrix with diagonal entries
1
√λ0
,1
√λ2
,1
√λ4
, ..., 1
√λ
2n
.
Then it suffices to show that the matrix DAD is positive definite. Indeed, if DAD is
positive definite, then xTAx =(D
−1
x)
T
(DAD)(D−1x) and therefore Ais also positive
definite. Now the (i, j)entryofDAD is λi+j−2
√λ2i−2√λ2j−2
and so the diagonal entries of
DAD are all 1. Since λk−1λk+1 ≥cλ2
k, we can invoke Lemma 2.1 with m=n=i+j−2
and k=2i−2 to conclude that λ2i−2λ2j−2≥c(j−i)2λ2
i+j−2. Using this inequality, we can
estimate the sum of the entries in the ith row of DAD by
n
X
j=1
λi+j−2
pλ2i−2pλ2j−2≤
n
X
j=1
1
c(j−i)2/2
<
∞
X
k=0
(√c)−k2+
i−1
X
k=1
(√c)−k2
<2
∞
X
k=0
(√c)−k2−1.
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×nHankel matrix A1associated with the sequence
λ1,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 Awhich are the determinants of submatrices with consecutive
rows and columns. But each of these minors is a principal minor of either Aor A1,and
thus is positive. Therefore it follows that Ais strictly totally positive. 

A SUFFICIENT CONDITION FOR STRICT TOTAL POSITIVITY OF A MATRIX 13
Acknowledgement. The authors wish to thank the referees for several helpful sugges-
tions.
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Department of Mathematics, University of Hawaii, Honolulu, HI 96822

14 THOMAS CRAVEN AND GEORGE CSORDAS
E-mail address:tom@math.hawaii.edu
Department of Mathematics, University of Hawaii, Honolulu, HI 96822
E-mail address:george@math.hawaii.edu