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k k B } ≥ k A ≥ i 1 sdfie as defined is nonnegative o more For . 0 ) > ◦ uhthat such r ,where ), o all for 0 > r 1 . A matrix T = [tij ] ∈ Mn(R), 1 ≤ i, j ≤ n, where tij = 0 for |i − j| > 1 is called tridiagonal. A matrix P = [pij ] ∈ Mn(R), 1 ≤ i, j ≤ n, where pij = 0 for |i − j| > 2 is called pentadiagonal. Let T, P ∈ Mn(R) be symmetric tridiagonal and symmetric pentadiagonal matrices respectively, then

a1 b1 c1 b1 a2 b2 c2 a1 b1   ...... b1 a2 b2 c1 b2 . . .     .. ..  ......  b2 . .  . . . . .      T =  . . .  , P =  . . . . .  , (1)  ......   ......       . .   . . . .   .. .. b   ...... c   n−1  n−2  b a   . . .   n−1 n   ...... b     n−1  c b a   n−2 n−1 n    where ai,bj ,ck ≥ 0 for 1 ≤ i ≤ n, 1 ≤ j ≤ (n − 1), 1 ≤ k ≤ (n − 2). We discuss the infinite divisibility of tridiagonal and a special family of pentadiagonal matrices. A matrix A = [aij ] ∈ Mn(R) is called a of bandwidth d if aij = 0 for |i − j| > d. Hence, the above matrices T and P are band matrices of bandwidth 1 and 2, respectively. A graph G = (V, E), where V = {1,...,n}, is called a band graph of bandwidth d if {i, j} ∈ E if and only if i 6= j and |i − j| ≤ d. Let G = (V, E) be a band graph of bandwidth d, where V = {1,...,n}; ◦r and A = [aij ] ∈ Mn(R) be a PSD matrix such that aij = 0 for i 6= j and (i, j) 6∈ E. Then A is positive semidefinite for r ≥ min(d, n − 2) [6, Cor 3.11]. Hence, if T and P are PSD matrices, which are tridiagonal and pentadiagonal, respectively, then T ◦r and P ◦s are PSD for r ≥ 1 and s ≥ 2, respectively. We show that if T is a tridiagonal PD matrix, then T ◦r is PD for r > 1; and discuss some of its nice consequences. Then, we prove a similar result for a special family of pentadiagonal matrices. In Section 2, we give a characterization for tridiagonal ID matrices. In Section 3, using chain sequences, we show that if T is a tridiagonal PD matrix, then T ◦r is PD for r > 1. We show that if P = [pij ] is a PD (PSD) pentadiagonal matrix, where pij = 0 if |i − j| = 1, and pij > 0 if |i − j| =0 or 2, then P ◦r is PD (PSD) for r> 1. We also give a characterization for such pentadiagonal ID matrices. We end the paper with some important remarks.

2 Tridiagonal infinitely divisible matrices

A symmetric block is PSD (PD,ID) if and only if each block is PSD (PD,ID). Each principal submatrix of A is ID if A is ID. Every PSD matrix of order 2 is ID. The sum of two PSD matrices is PSD.

a1 b1 0 Lemma 2.1. Let A = b a b be a PSD matrix of order 3. Then A is ID if and only if b b =0.  1 2 2  1 2 0 b2 a3   ◦r Proof. Let A be ID and G = limr→0+ A . Then G is a PSD matrix, as it is the limit of a sequence of PSD matrices. Let b1 and b2 be positive. Since A is PSD, a1,a2 and a3 are positive. But then 1 1 0 G = 1 1 1 is not a PSD, we get a contradiction. Hence b1b2 = 0. Conversely, if b1b2 = 0, then 0 1 1 det(A◦r) ≥ 0 for r> 0. Hence A is ID.

We now characterize all tridiagonal ID matrices of order n ≥ 4.

2 Theorem 2.2. Let T be a tridiagonal PSD matrix of order n as given in equation (1). Then, for n ≥ 4, T is ID if and only if bibi+1 = 0 for every i ∈ {1, 2,...,n − 2}, i.e., the sequence {bi}i≥1 has no two consecutive positive entries.

Proof. Let T be ID. Let bkbk+1 > 0, for some 1 ≤ k ≤ (n − 2), then by Lemma 2.1, the principal submatrix ak bk 0 bk ak bk  +1 +1  0 bk+1 ak+2   of T is not ID. Thus T is not ID, which contradicts the hypothesis. Hence, bibi+1 = 0 for every i ∈ {1, 2,...,n − 2}. Conversely, if bibi+1 = 0 for every i ∈ {1, 2,...,n − 2}, then A becomes a block diagonal matrix, where each non-zero diagonal block is a PSD matrix of order 1 or 2, hence T is ID. Remark 2.3. From Theorem 2.2, we have, T is an ID matrix if and only if T is a block diagonal matrix, where each non-zero diagonal block is a PSD matrix of order 1 or 2.

In general, ID matrices are not closed under addition and multiplication. For example, let X = [xixj ], where x1,...,xn are distinct positive real numbers and Jn be the matrix of order n with each of its entries equals to 1, then X and Jn are both ID, but their sum is not ID, see Theorem 1.1 of [10]. 1 2 The C = [cij ] = i+j , 1 ≤ i, j ≤ 3 is ID [1] but its square C is not ID because 1 1 2 ◦ 4 1 4 h i det(C ) = det i+j < 0. We say that two block diagonal matrices are of same structures if their correspondingh blocks i are square matrices of the same order. The set of block diagonal matrices of the same structure, is closed under addition, multiplication and multiplication by a nonnegative scalar. Hence, by Remark 2.3, we get the following: n k R Corollary 2.4. Let T be a tridiagonal ID matrix of order n ≥ 2 and f(x) = k=0 akx ∈ [x] be a n k polynomial, where ak ≥ 0 for all 0 ≤ k ≤ n, then f(T )= akT is ID. k=0 P Observation 2.5. The results on tridiagonal ID matricesP help us to provide an algorithm to check if a given 3 × 3 PSD matrix is ID or not. Consider the PSD matrix

a1 b1 c1 A = b a b .  1 2 2  c1 b2 a3   Let all entries of A be positive. Then by Corollary 1.6 and Theorem 1.10 of [7], A is ID if and only a1a2 b1b2 log( 2 ) log( ) b1 a2c1 if b1b2 a2a3 is PSD. If c1 = 0, then A becomes tridiagonal and thus by Lemma 2.1, A is log( ) log( 2 ) " a2c1 b2 # ID if and only if b1b2 = 0. In the remaining cases, one can easily observe that A is ID if and only if a1a3 > 0 and b1 =0= b2. Hence, this way we can characterize all ID matrices of order 3.

3 Hadamard powers preserving positivity

To prove our next result we need the following two results related to chain sequences:

n n Theorem 3.1. [3, Theorem 5.7] If {ak}k=1 is a chain sequence and 0 < ck ≤ ak for k ≥ 1, then {ck}k=1 is also a chain sequence.

Let T be a of order n as given in equation (1) with ai,bj > 0 for 1 ≤ i ≤ n, 1 ≤ j ≤ (n − 1). Then we have,

3 Theorem 3.2. [9, Theorem 3.2] The matrix T defined as above is positive definite if and only if 2 n−1 bi aiai is a chain sequence. +1 i=1 n o Let T be a tridiagonal matrix of order n as given in equation (1). If bj = 0 for some 1 ≤ j ≤ (n − 1), then T becomes a block diagonal matrix having two smaller diagonal blocks. Continuing this way with these smaller blocks and keep repeating the process, one can see that T is a block diagonal matrix, where each diagonal block is a tridiagonal matrix with positive entries on its upper and lower diagonals. Moreover, every such block of T is a PD matrix with positive entries on the main, upper and lower diagonals, if T is PD. We are now ready to state and prove our first main result. Theorem 3.3. Let T be a tridiagonal matrix of order n. If T is PD, then T ◦r is PD for r> 1. Proof. Let T be a tridiagonal PD matrix. Then T is a block diagonal matrix, where each diagonal block is a tridiagonal PD matrix with positive entries on its main, upper and lower diagonals. Hence, to show that, T ◦r is PD for r > 1, it’s sufficient to prove the our result for every symmetric tridiagonal matrix with positive entries on its main, upper and lower diagonals. Let T1 be a tridiagonal PD matrix of order n as given in equation (1), where ai,bj > 0 for 1 ≤ i ≤ 2 n−1 bi n, 1 ≤ j ≤ (n−1). By Theorem 3.2, aiai is a chain sequence. Let 1 ≤ i ≤ n−1 and r> 1. Since +1 i=1 2 2 r 2 T is PD, a a − b2 > 0, so bi n< 1, whicho gives us 0 < bi < bi . Thus by Theorem 3.1, 1 i i+1 i aiai+1 aiai+1 aiai+1 2r n−1 bi   ◦r ar ar is also a chain sequence. Hence, by Theorem 3.2, the matrix T1 is PD. This completes i i+1 i=1 then proof.o Corollary 3.4. Let T be a tridiagonal matrix of order n. If T is PSD, then T ◦r is PSD for r> 1.

Proof. Let T be a tridiagonal PSD matrix of order n. There exist a sequence {Tk}k≥1 of tridiagonal PD 1 matrices, where Tk = T + k I for k ≥ 1, such that Tk converges to T entrywise as k → ∞ [8, Page 432]. ◦r ◦r ◦r Let r> 1. By Theorem 3.3, we have, Tk is PD for k ≥ 1. Since T is the limit of a sequence {Tk }k≥1 of PD matrices, T ◦r is PSD. Hence done. The next result is an immediate consequence of Corollary 3.4 and the fact that every (nonnegative) tridiagonal PSD matrix is totally nonnegative [4, page 9]. Corollary 3.5. Let T be a tridiagonal matrix of order n, which is symmetric and totally nonnegative. Then T ◦r is a totally nonnegative for r> 1.

Let A be a of order n. For α = {α1,...,αk}⊆{1, 2,...,n}, where α1 < α2 < ··· < αk, let A[α] denotes the principal submatrix of A obtained by picking rows and columns indexed by α. A is PD if and only if all its leading principal minors are positive. If A is a PD matrix then all its principal submatrices are PD. Let Perm(i1,...,in), 1 ≤ i1,...,in ≤ n be the ′ ′ of order n, whose k th row is ikth row of the of order n. If A is a PD (PSD) matrix of order n, then XAX∗ is PD (PSD) for any square matrix X of order n. We give our second main result as follows:

Theorem 3.6. Let P = [pij ] be a pentadiagonal matrix of order n as given in equation (1), where pij =0 for |i − j| =1, and pij > 0 for |i − j| =0 and 2. We have, 1. If P is PD (PSD), then P ◦r is PD (PSD) for r> 1.

2. P is ID if and only if the sequences {c2i}i≥1 and {c2i−1}i≥1 have no two consecutive positive entries.

4 Proof. Let P = [pij ] be a pentadiagonal matrix of order n as given in equation (1), where pij = 0 ∗ ∗∗ for |i − j| = 1, and pij > 0 for |i − j| = 0 and 2. Let Al = P [β] and Am = P [γ], where β = {1, 3,..., (2l − 1)},γ = {2, 4,..., 2m} for 1 ≤ l,m ≤ k if n = 2k and 1 ≤ l ≤ (k + 1), 1 ≤ m ≤ k if ∗ ∗∗ n =2k +1. From the definition of P, one can observe that the principal submatrices Al and Am of P (n−2) are tridiagonal with the upper and lower diagonal entries from the set {ci}i=1 , and the main diagonal n entries from the set {ai}i=1, as given below:

a1 c1 ··· 0 0 a2 c2 ··· 0 0 c1 a3 c3 0 0 c2 a4 c4 0 0  . . . .   . . . . .  A∗ = . c .. .. . and A∗∗ = ...... l  3  m    .   .   0 0 .. a c   0 0 .. a c   2l−3 2l−3   2m−2 2m−2  0 0 ··· c a   0 0 ··· c a   2l−3 2l−1l×l  2m−2 2m m×m     Again, from the definition of P , it can be observed that for every r > 0, P ◦r is congruent to the M ◦r via a permutation matrix X of order n, i.e., M ◦r = XP ◦rX∗ for r> 0, where

∗ Ak 0 ∗∗ , if n =2k, " 0 Ak # Perm(1, 3,..., (2k − 1), 2, 4,..., 2k), if n =2k  X = ,M =  Perm(1, 3,..., (2k + 1), 2, 4,..., 2k), if n =2k +1  (  ∗  Ak+1 0 ∗∗ , if n =2k +1. " 0 Ak #   We prove the required results for the case when n is even (the case when n is odd can be proved analogously). Let r > 1 and n = 2k for some positive integer k ≥ 1. If P is PD (PSD), then by ∗ ∗ ∗∗ M = XPX ,M is PD (PSD), so Ak and Ak are PD (PSD) matrices. Hence, by Theorem 3.3 ∗ ◦r ∗∗ ◦r ◦r ◦r (Corollary 3.4), (Ak) and (Ak ) are PD (PSD), which gives M is PD (PSD), so P is PD (PSD), by P ◦r = X−1M ◦r(X−1)∗. This proves Part (1). ◦r ◦r ∗ ∗ Since M = XP X for every r > 0, P is ID if and only if M is ID. Hence P is ID if and only if Ak ∗∗ and Ak are ID. Thus Part (2) follows by Theorem 2.2 (b).

Remark 3.7. The results analogous to Theorem 3.3 and Corollary 3.4 are not true for all band matrices. For example, let

234 50 357 90 K = 4 7 10 13 0 , 5 9 13 17 0   000 01     then K is a band matrix of bandwidth 3, which is PSD, but K◦r is not PSD for 1

Remark 3.8. In Theorem 3.3, the lower bound 1 for the Hadamard power r is sharp. Let 0 < r0 < 1. 1 1 0 1 + r Consider the tridiagonal PD matrix A(ǫ) = 1 (2+ ǫ) 1 , where ǫ ∈ R . For every ǫ < (2 0 − 2), 0 1 1 det(A(ǫ)◦r0 )=(2+ ǫ)r0 − 2 < 0, so A(ǫ)◦r0 is not PD. 

5 References

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