Positivity Properties of Some Special Matrices

Home , Alternant matrix, Pascal matrix

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Li(s)= I + sEi,i 1 and Uj (t)= I + tEj 1,j , − − where 2 i, j n. Matrices of the form Li(s) or Uj (t) are called elementary bidiagonal matrices. For ≤ ≤ a vector (d , d ,...,dn), let diag([di]) denote the diagonal matrix diag(d ,...,dn). An n n matrix is 1 2 1 × totally positive if and only if it can be written as (Ln (lk) Ln 1 (lk 1) L2 (lk n+2)) − − ··· − (Ln (lk n+1) Ln 1 (lk n) L3 (lk 2n+4)) (Ln (l1))D(Un (u1))(Un 1 (u2) Un (u3)) (U2 (uk n+2) − − − ··· −n ··· − ··· − ··· Un 1 (uk 1) Un (uk)), where k = , li,uj > 0 for all i, j 1, 2,...,k and D = diag[(di)] is a di- − − 2 ∈ { } agonal matrix with all di > 0 [12, Corollary 2.2.3]. This particular factorization (with li,uj 0) is called successive elementary bidiagonal (SEB) factorization or Neville factorization. We obtain≥ the 1 SEB factorization for β(i,j) explicitly. Let (x)0 = 1 and forh a positivei integer k, let (x)k = x(x 1)(x 2) (x k +1). The Stirling numbers of first kind s(n, k) [9, p. 213] and the Stirling numbers− of second− ··· kind−S(n, k) [9, p. 207] are respectively defined as n k (x)n = s(n, k)x k X=0 and n n x = S(n, k)(x)k. k X=0 The unsigned Stirling matrix of first kind s = [sij ] is defined as

i j ( 1) − s(i, j) if i j, sij = − ≥ (0 otherwise.

The Stirling matrix of second kind = [ ij ] is defined as S S S(i, j) if i j, ij = ≥ S (0 otherwise. It is a well known fact that s and are totally nonnegative (see for example [13]). We consider the S matrices ss∗ and ∗ and call them symmetrized unsigned Stirling matrix of first kind and symmetrized Stirling matrix ofSS second kind, respectively. By definition, these are positive semidefinite and totally nonnegative. We show that both these matrices are in fact totally positive. Another matrix that we consider is formed by the well known Bell numbers. The sum w(n) = n k=0 S(n, k) is the number of partitions of a set of n objects and is known as the nth Bell number [9, B w n 1 n 1 p. 210]. Consider the matrix = [ (i + j)]i,j−=0. Let X = (xij )i,j−=0 be the lower triangular matrix definedP recursively by

x00 =1, x0j = 0 for j > 0, and xij = xi 1,j 1 + (j + 1)xi 1,j + (j + 1)xi 1,j+1 for i 1. − − − − ≥

(Here xi, 1 = 0 for every i and xin = 0 for i = 0,...,n 2.) This is known as the Bell triangle [8]. − − n 1 − Lemma 2 in [1] gives the Cholesky decomposition of B as LL∗, where L = X diag √i! . It is i=0 B n 1 B h i also shown in [1] that det = i=0− i!. It is a known fact that is totally nonnegative [18]. We show that B is totally positive. We also show that B is inﬁnitely divisible only upto order 4. In Section 2 we give our resultsQ for the beta matrix, namely, its inﬁnite divisibility and total positivity, 1 r its Cholesky decomposition, its determinant, − , and its SEB factorization. We show that ◦ is in B B

2 fact totally positive for all r> 0. We end the discussion on the beta matrix by proving that for positive 1 1 real numbers λ1 < < λn and µ1 < <µn, is an infinitely divisible matrix and ··· ··· β(λi,λj ) β(λi,µj ) is a totally positive matrix. In section 3, we proveh that symmetrizedi Stirling matrices and B areh totallyi positive. For the first kind, we give the SEB factorization for s. For the second kind, we show that is triangular totally positive [12, p. 3]. We also show that B is infinitely divisible if and only if its orderS is less than or equal to 4.

2 The beta matrix

The infinite divisibility and total positivity of are easy consequences of the corresponding results for . For 1 i, j n, let A(i, j) denote the submatrixB of A obtained by deleting ith row and jth column fromP A. Each≤ A≤(i,i) is infinitely divisible, if A is infinitely divisible, and each A(i, j) is totally positive, if A is totally positive.

Theorem 2.1. The matrix = 1 is inﬁnitely divisible and totally positive. B β(i,j) h i Proof. By deﬁnition,

1 (i + j 1)! ij(i + j)! 1 i + j = − = = . . (1) β(i, j) (i 1)!(j 1)! (i + j)i!j! 1 + 1 i − − i j

1 Thus = C (1, 1), where C = 1 + 1 is a Cauchy matrix. Both C and (1, 1) are infinitely divisible B ◦P i j P [3]. Since Hadamard product of infinitelyh i divisible matrices is infinitely divisible, we get is infinitely divisible. B Again, note that

1 (i + j 1)! (i + j 1)! (i 1)+ j = − = j − = − j. (2) β(i, j) (i 1)!(j 1)! (i 1)!(j!) i 1 − − − − So is the product of the totally positive matrix (n +1, 1) with the positive diagonal matrix diag([i]). HenceB is totally positive. P B The below remarks were suggested by the anonymous referee.

r r Remark 2.2. Since for each r> 0, C◦ and ◦ are positive definite (see p. 183 of [4]), (1) gives that r n P r ◦ is positive definite. Let = [(i + j)!]i,j=0. Then is a Hankel matrix. Since ◦ is congruent to B r G r n rG G r ◦ via the positive diagonal matrix diag([i! ])i=0, ◦ is positive definite. The matrix ◦ (n +1, 1) = P r r G r G r [(i + j 1)! ] is congruent to ◦ , via the positive diagonal matrix diag [(i 1)! ]. So ◦ (n +1, 1) is − B r − G r also positive definite. Hence ◦ is totally positive, by Theorem 4.4 in [19]. This shows that ◦ is r G P totally positive. By (2), ◦ is also totally positive. B Remark 2.3. Another proof for infinite divisibility of can be given as follows. For positive real B Γ(λi+λj +1) numbers λ , λ ,...,λn, the generalized Pascal matrix is the matrix . The beta matrix 1 2 Γ(λi+1)Γ(λj +1) is congruent to it via a positive diagonal matrix, when λi = i 1h/2. The generalizedi Pascal matrix isB infinitely divisible [3], and hence so is . − B By Theorem 2.1, is positive definite and so, can be written as LL∗. Our next theorem gives L explicitly. B

1 i Theorem 2.4. The matrix [ β(i,j) ] has the Cholesky decomposition LL∗, where L = [ j √j].

3 Proof. We prove the result by the combinatorial method of two way counting. Consider a set of (i+j 1) persons. The number of ways of choosing a committee of j people and a chairman of this committee− is

i + j 1 (i + j 1)! j − = − . · j (i 1)!(j 1!) − − Another way to count the same is to separate (i + j 1) people into two groups of i and (j 1) people each. The number of ways of choosing a committee− of j people from these two groups of− people and i j 1 then a chairman of the committee is j k k j−k , where k varies from 1 to min i, j . Rearranging the terms in this expression, we get · − { } P i j 1 i j 1 j − = j − · k j k k j k k k X − X − i (j 1)! = j − k (j k)!(k 1)! k X − − i j! = k k (j k)!k! k X − i j = k . k k k X

i The last expression is the (i, j)th entry of the matrix LL∗, where L = [ j √j]. Corollary 2.5. The determinant of is equal to n!. B Corollary 2.6. The inverse of the matrix has (i, j)th entry as ( 1)i+j n k k 1 . B − k=1 i j k i i P Proof. By Theorem 2.4, = LL∗, where L = √j . Let Z = [ ] and D′ = diag([√i]). Then B j j n k+j i k 1 i+j i 1 1 1 L = ZD′. Since k=1( 1) k j = δij , weh get thati Z− = [( 1) j ]. So L− = D′− Z− = − − − i+j i 1 1 1 1 1 ( 1) . Thus − = L∗ L− . This gives that the (i, j)th entry of − is − j √i P B B h i n k 1 k 1 ( 1)k+i ( 1)k+j − i √ − j √ k k k X=1 which is equal to n k k 1 ( 1)i+j . − i j k k X=1

Remark 2.7. The above theorems hold true if i, j are replaced by λi, λj , where 0 < λ < < λn are 1 ··· 1 positive integers. For r > 0, the matrix r is totally positive because it is a submatrix of the β(λi,λj ) r 1 1 λn λn matrix ◦ . So is alsoh inﬁnitelyi divisible. We have = LL∗, where L is the × B β(λi,λj ) β(λi,λj ) λi n λn matrix [ √j].h The proofi is same as for Theorem 2.4. h i × j The next theorem gives the SEB factorization for the matrix , which also gives another proof for to be totally positive. B B

4 Theorem 2.8. The matrix = 1 can be written as B β(i,j) n n 1h i n 3 n Ln Ln 1 − L2 (2) Ln L3 ... Ln D n 1 − n 2 ··· n 1 ··· 2 n 1 − − − − n 3 n n 1 n Un ... U3 ...Un U2 (2) Un 1 − Un , n 1 2 n 1 ··· − n 2 n 1 − − − − where D = diag([i]). To prove this theorem, we ﬁrst need a lemma. (k) Lemma 2.9. For 1 k n 1, let Yk = y be the n n lower triangular matrix where ≤ ≤ − ij × i i h(n k)i j j−(n−k) if n k j

Then  n n k +1 n n k +2 n Yk = Ln Ln (k 1) − Ln Ln (k 2) − Ln . n 1 ··· − − n k n 1 ··· − − n k +1 ··· n 1 − − − − (3)−

n Proof. For k = 1, the right hand side is Ln n 1 , which is same as Y1. We show below that for − 1 k n 2, ≤ ≤ − ( n n 1 n k)Yk = Yk+1, (4) L L − ···L − p where p denotes Lp p 1 . This will show that (3) is true for k =1, 2,...,n 1. We also keep note L − − of the fact that multiplying p on the left of a matrix A is applying the elementary row operation p L row p row p + p 1 row (p 1) on A, which we will use in the cases 2, 3 and 4. We also note that → − × − for k = n 2, only cases 1 and 4 are relevant. − Case 1: Let 1 i j n. Since p and Yk are lower triangular matrices with diagonal entries 1, ≤ ≤ ≤ L so is ( n n 1 n k)Yk. So the (i, j)th entry of both the matrices in (4) is same. L L − ···L − (k) Case 2: Let 1 j n k 2 and j +1 i n k 1. In this case, y = 0. Since multiplying Yk ≤ ≤ − − ≤ ≤ − − ij on the left by n n 1 n k will keep its rows j +1,...,n k 1 unchanged, we get that the (i, j)th L L − ···L − − − entry of ( n n 1 n k)Yk is zero. L L − ···L − Case3: Let 1 j n k 2 and n k i n. Multiplying Yk on the left by n k, n (k 1),..., n ≤ ≤ − − − ≤ ≤ L − L − − L successively, we get that the (i, j)th entry of ( n n 1 n k)Yk is given by L L − ···L − (k) i (k) i 1 (k) n k +1 (k) n k (k) yij + yi 1,j + − yi 2,j + + − yn k,j + − yn k 1,j i 1 − i 2 − ··· n k − n k 1 − − ··· − − − − − (k) i (k) i (k) i (k) = yij + yi 1,j + yi 2,j + + yn k 1,j . (5) i 1 − i 2 − ··· n k 1 − − − − − − (k) Now ypj = 0 for all 1 j n k 2 and p = j. So the (i, j)th entry of ( n n 1 n k)Yk is zero. ≤ ≤ − − 6 L L − ···L − Case 4: Let i > j n k 1. Again, the (i, j)th entry of ( n n 1 n k)Yk is given by (k) i (k) i (k≥) − − i (k) i (k) L L − ···L −(k) yij + i 1 yi 1,j + i 2 yi 2,j + + j yjj + + n k 1 yn k 1,j . For j = n k 1, ypj = 0 for p = j. − − − − ··· ··· − − − − − − 6 i (k) So the (i,n k 1)th entry of ( n n 1 n k)Yk is . For j n k, ypj = 0 for p < j and − − n k 1 (k) p p n−+k − L L ···L − − ≥ − ypj = j j−n+k for p j. So we obtain that the (i, j)th entry of ( n n 1 n k)Yk is − ≥ L L − ···L − i i n + k i i 1 (i 1) n + k i i 2 (i 2) n + k i = − + − − − + − − − + + j j n + k i 1 · j j n + k i 2 · j j n + k ··· j − − − − − i i p n + k = − . (6) j j n + k p=j X −

5 n m+k m+n+1 i i n+k+1 Since m = m+1 , the expression in (6) is equal to j j−n+k+1 . In all the above four cases, k=0 − the (i,P j)th entry of ( n n 1 n k)Yk is the same as that of Yk+1. Hence we are done. L L − ···L − Proof of Theorem 2.8. By Theorem 2.4 we have that = i D i ∗. So it is enough to show B j j that h i h i i ( n n 1 2)( n n 1 3) ( n n 1)( n)= . (7) L L − ···L L L − ···L ··· L L − L j This is easily obtained by putting k = n 1 in (3). −

Remark 2.10. Let p0,...,pn 1 be functions from a set X to a ﬁeld and λ1,...,λm X. Then − ∈ the m n matrix deﬁned by [pj 1(λi)]1 i m,1 j n is called an alternant matrix [2, p. 112]. Let × − ≤ ≤ ≤ ≤ x n = x(x + 1) (x + n 1). For positive integers 0 < λ < < λn, we have h i ··· − 1 ···

1 (λi + λj 1)! λj λ = − = h i i . β(λi, λj ) (λi 1)!(λj 1)! (λi 1)! − − −

λj x 1 Thus with pj (x)= h i , is an alternant matrix. 1 (x 1)! β(λi,λj ) − − h i 1 We now consider the more general matrix for positive real numbers λ , λ ,...,λn. The β(λi,λj ) 1 2 (i, j)th entry of this matrix is given by Γ(λi+λj )h. The proofi for inﬁnite divisibility of generalized Pascal Γ(λi)Γ(λj ) matrix Γ(λi+λj +1) is given in [3]. Inﬁnite divisibility of 1 follows by a similar argument. Γ(λi+1)Γ(λj +1) β(λi,λj )

h i 1 Γ(λi+λjh+1) i 1 Alternatively, one can also observe that = 1 1 and deduce its infinite β(λi,λj ) Γ(λi+1)Γ(λj +1) + ◦ λi λj divisibility. This is also same as saying thath 1 i is anh infinitely divisiblei kernel [15] on R+ R+. β( , ) × 1 · · R+ R+ We observe that β( , ) is also a totally positive kernel [17] on . For that we first show that · · × [Γ(λi + µj )] is totally positive, where 0 < λ1 < < λn and 0 < µ1 < <µn. The proof of this was guided to us by Abdelmalek Abdesselam and Mateusz··· Kwa´snicki 1. ···

Theorem 2.11. Let 0 < λ < < λn and 0 < µ < < µn be positive real numbers. Then 1 ··· 1 ··· [Γ(λi + µj )] is totally positive.

Proof. Since all the minors of [Γ(λi + µj )] are also of the same form, it is enough to show that y x + det ([Γ(λi + µj )]) > 0. Let K (x, y)= x and K (x, y)= y . For any λ, µ R , 1 2 ∈

∞ t λ+µ 1 Γ(λ + µ) = e− t − dt

Z0 ∞ e t = tλ+µ − dt t Z0 ∞ e t = tλtµσ(dt), where σ(dt)= − dt t Z0 ∞ = K2(λ, t)K1(t,µ)σ(dt). (8) Z0 1https://mathoverﬂow.net/questions/306366/

6 Let 0 0.

Corollary 2.12. For positive real numbers 0 < λ < λ < < λn and 0 < µ < µ < < µn, the 1 2 ··· 1 2 ··· matrix 1 is totally positive. β(λi,µj ) h i 1 1 Proof. Since = diag [Γ(λi + µj )] diag , we obtain the required result. B Γ(λi) Γ(µi) h i h i 3 Combinatorial matrices

The unsigned Stirling matrix of ﬁrst kind s is totally nonnegative as well as invertible. Theorem 2.2.2 in [12] says that every invertible totally nonnegative matrix can be written as (Ln (lk) Ln 1 (lk 1) L2 (lk n+2)) − − ··· − (Ln (lk n+1) Ln 1 (lk n) L3 (lk 2n+4)) (Ln (l1))D(Un (u1))(Un 1 (u2) Un (u3)) (U2 (uk n+2) − − − ··· −n ··· − ··· − ··· Un 1 (uk 1) Un (uk)), where k = , li,uj 0 for all i, j 1, 2,...,k and D = diag([di]) is a diagonal − − 2 ≥ ∈{ } matrix with all di > 0. The below proposition gives that uj = 0 for s, which is not surprising in view of [16, Theorem 7]. Proposition 3.1. The n n unsigned Stirling matrix of ﬁrst kind s can be factorized as ×

s = (Ln(n 1)Ln 1(n 2) L2(1))(Ln(n 2)Ln 1(n 3) L3(1)) (Ln(2)Ln 1(1))(Ln(1)). (9) − − − ··· − − − ··· ··· − 1 Proof. Since (Li(s))− = Li( s), so it is enough to show that −

(Ln( 1)) (Ln 1( 1)Ln( 2)) (L3( 1) Ln 1( (n 3))Ln( (n 2))) (L2( 1) − − − − ··· − ··· − − − − − − ··· Ln 1( (n 2))Ln( (n 1))) s = In,(10) − − − − − where In is the n n identity matrix. We shall prove (10) by induction on n. For clarity, we shall denote × 1 0 the n n matrices s and Li(s), by sn and Li(s) , respectively. For n = 2, s = , which is clearly × (n) 2 1 1 equal to L2(1)(2). Let us assume that (10) holds for n. We have the following recurrence relation [14, p. 166] for sij :

s00 =1, s0j =0, si0 =0

si+1,j = si,j 1 + i sij . (11) −

So for 1 k n, multiplying sn+1 on the left by (Lk+1( k))(n+1) replaces its (k + 1)th row by the row whose ﬁrst≤ element≤ is 0 and jth element is the (j 1)th− element of the previous row. So we get − 1 0 L2( 1)(n+1) Ln( (n 1))(n+1)Ln+1( n)(n+1) sn+1 = . (12) − ··· − − − 0 sn It is easy to see that 1 0 Li(s)(n+1) = . 0 Li 1(s)(n) − Hence

Ln ( 1) Ln( 1) Ln ( 2) L ( 1) Ln( (n 2)) Ln ( (n 1)) +1 − (n+1) − (n+1) +1 − (n+1) ··· 3 − (n+1) ··· − − (n+1) +1 − − (n+1) 1 0 = . 0 Ln( 1)(n) Ln 1( 1)(n)Ln( 2)(n) L2( 1)(n) Ln 1( (n 2))(n)Ln( (n 1))(n) − − − − ··· − ··· − − − − − 7 Using induction hypothesis and (12), we obtain

Ln ( 1) Ln( 1) Ln ( 2) L ( 1) Ln( (n 2)) Ln ( (n 1)) +1 − (n+1) − (n+1) +1 − (n+1) ··· 3 − (n+1) ··· − − (n+1) +1 − − (n+1) L ( 1) Ln( (n 1)) Ln ( n) sn = In . 2 − (n+1) ··· − − (n+1) +1 − (n+1) +1 +1

As an immediate consequence, we obtain the following.

Theorem 3.2. The symmetrized unsigned Stirling matrix of ﬁrst kind ss∗ is totally positive. Proof. This follows from the above Proposition 3.1 and Corollary 2.2.3 of [12].

We now show that symmetrized Stirling matrix of second kind ∗ is totally positive. For that SS we ﬁrst prove that is triangular totally positive. For α = α ,...,αp , γ = γ ,...,γp with S { 1 } { 1 } 1 α < < αp n and 1 γ < <γp n, let A[α, γ] denotes the submatrix of A obtained by ≤ 1 ··· ≤ ≤ 1 ··· ≤ picking rows α1,...,αp and columns γ1,...,γp of A. The dispersion of α, denoted by d(α), is deﬁned as d(α)= αp α1 (p 1). Note that d(α) = 0 if and only if α1, α2,...,αp are consecutive p numbers. − − − (p) We denote by α′ the set α +1, α +1,...,αp +1 , and by σ the set 1,...,p . { 1 2 } { } Proposition 3.3. The Stirling matrix of second kind is triangular totally positive. S Proof. By Theorem 3.1 of [10], is triangular totally positive if and only if det( [α, σ(p)]) > 0 for all S S 1 p n and for all α = α1,...,αp satisfying 1 α1 < < αp n and d(α) = 0. For p = n, this ≤ ≤ { } ≤ ··· ≤ p (p) (p) is obviously true. Let 1 p 0. ≤ (p) (p) S Next we show that if det( [α, σ ]) > 0, where d(α) = 0, then det( [α′, σ ]) > 0 (and d(α′) = 0). S S i if i = j

Let T = [tij ] be deﬁned as tij = 1 if j i =1. We prove that  −  0 otherwise

 (p) (p) [α′, σ ]= [α, σ ]T for 1 p < n. (13) S S ≤ (p) The (i, j)th entry of [α, σ ]T is jS(αi, j)+ S(αi, j 1). The Stirling numbers of second kind satisfy the following recurrenceS relation: − S(0, 0) = 1; for ℓ,m 1, S(0,m)=0= S(ℓ, 0), S(ℓ,m)= mS(ℓ 1,m)+ S(ℓ 1,m 1). ≥ − − − (p) (p) Thus the (i, j)th entry of [α, σ ]T is = S(αi +1, j), which is also the (i, j)th entry of [α′, σ ]. S (p) (p) S Hence (13) holds, which gives that det( [α′, σ ]) = p! det( [α, σ ]) > 0. S S Theorem 3.4. The symmetrized Stirling matrix of second kind ∗ is totally positive. SS Proof. This follows from Proposition 3.3 and Corollary 2.4.2 of [12]. B w n 2 Next, we show that = [ (i + j)] is totally positive. Let Y = (yij )i,j−=0 be the lower triangular matrix defined recursively by y00 = 1, y0j = 0 for j > 0, and yij = yi 1,j 1 + (j + 2)yi 1,j + (j + − − − 1)yi 1,j+1 for i 1, where yi, 1 = 0 for every i and yin =0 for 0 i n 3. − ≥ − ≤ ≤ − Theorem 3.5. The matrix B = [w(i + j)] is totally positive. B w n 2 B Proof. Let (n, 1) = [ (i + j + 1)]i,j−=0 be the matrix obtained from by deleting its first column n 1 − and nth row. From the proof of the Theorem in [1], B = LL∗, where L = X diag √i! , and i=0 n 2 − h i B(n, 1) = L′L′∗, where L′ = Y diag i!) . Hence B and B(n, 1) are positive semidefinite. The i=0 B B Theorem in [1] also shows that bothhpandi (n, 1) are nonsingular, hence they are positive definite. Since B is a Hankel matrix, the result now follows from Theorem 4.4 of [19].

8 Now we show that B is infinitely divisible only upto order 4. For A = [aij ], let log A = [log(aij )]. n 1 Let ∆A denote the (n 1) (n 1) matrix [aij + ai ,j ai ,j ai,j ] − . − × − +1 +1 − +1 − +1 i,j=1 Theorem 3.6. The n n matrix B is infinitely divisible if and only if n 4. × ≤ Proof. We denote the n n matrix B by Bn. Since Bn is a principal submatrix of Bn , it is enough × +1 to show that B4 is infinitely divisible but B5 is not infinitely divisible. By Corollary 1.6 and Theorem 1.10 of [15], to prove infinite divisibility of B4, it is enough to prove that ∆ log B4 is positive definite. Now 11 2 5 0 0 log 2 log 5 1 2 5 15 0 log 2 log 5 log 15 B = , log B = , 4 2 5 15 52  4 log 2 log 5 log 15 log 52  5 15 52 203 log 5 log 15 log 52 log 203       log 2 log(5 /4) log(6/5)  and ∆ log B4 = log(5/4) log(6/5) log(52/45) . log(6/5) log(52/45) log(3045/2704)   Since all the leading principal minors of ∆ log B4 are positive, we get that ∆ log B4 is positive 1 ( 4 ) 9 definite. Hence B is infinitely divisible. Since det B◦ = 1.62352 10− < 0, B is not infinitely 4 5 − × 5 divisible. Acknowledgement The possibility of studying these problems was suggested to us by Rajendra Bhatia. It is a pleasure to record our thanks to him. We are also very thankful to the anonymous referee for several helpful comments.

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