A Hadamard-type lower bound for symmetric diagonally dominant positive matrices

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1 A Hadamard-type lower boud for symmetric diagoally domiat positive matrices Christopher J. Hillar, Adre Wibisoo Uiversity of Califoria, Berkeley Jauary 7, 205 Abstract We prove a ew lower-boud form of Hadamard s iequality for the determiat of diagoally domiat positive matrices. Keywords: diagoally domiat; positive matrices; determiatal iequality; Hadamard s iequality Subject Classificatios: 5B48; 5A5; 5A45 Itroductio A real matrix J is diagoally domiat if i J) := j i J ij 0, for i =,...,. A particularly iterestig case is whe i J) = 0 for all i; we call such matrices diagoally balaced. Irreducible, diagoally domiat matrices are always ivertible, ad such matrices arise ofte i theory ad applicatios. I this ote we study bouds o the determiat of symmetric diagoally domiat matrices that have positive etries. These matrices are always positive defiite e.g., by Lemma 2.). It is classical that the determiat of a positive semidefiite matrix A is bouded above by the product of its diagoal etries: 0 deta) A ii. This well-kow result is sometimes called Hadamard s iequality [6, Theorem 7.8.]. A lower boud of this form, however, is ot possible without additioal assumptios. Surprisigly, there is such a iequality whe J is diagoally domiat with positive etries. Theorem.. Let 3, ad let J be a symmetric diagoally domiat matrix with off-diagoal etries m J ij l > 0. The, the followig iequality holds: detj) m J + m ) ) exp m + m ) ) as. ii 2 2) l l 2 l l Redwood Ceter for Theoretical Neurosciece, chillar@berkeley.edu; partially supported by NSF grat IIS ad a NSF All-Istitutes Postdoctoral Fellowship admiistered by the Mathematical Scieces Research Istitute through its core grat DMS Departmet of Electrical Egieerig ad Computer Sciece, wibisoo@eecs.berkeley.edu.

2 The result above was discovered i a attempt to prove the followig difficult orm iequality [5]. Let S = 2)I + be the diagoally balaced matrix whose off-diagoal etries are all equal to I is the idetity matrix ad is the -dimesioal colum vector cosistig of all oes). Theorem.2 [5]). Let 3. For ay symmetric diagoally domiat matrix J with J ij l > 0, we have J l S = Moreover, equality is achieved if ad oly if J = ls l 2) ). Here, is the maximum absolute row sum of a matrix, which is the matrix orm iduced by the ifiity orm o vectors i R. The boud i Theorem. depeds o the largest off-diagoal etry of J i a essetial way; see Example 3.3), ad thus is ill-adapted to prove Theorem.2. For istace, combiig Theorem. with Hadamard s iequality applied to the positive defiite J := J detj) the adjugate of J) i the obvious way gives estimates which are worse tha Theorem.2. Nevertheless, Theorem. should be of idepedet iterest, ad we prove it i Sectio 2 usig a block matrix factorizatio. 2 Proof of Theorem. Our argumets for provig Theorem. are ispired by block LU factorizatio ideas i [2]. For i, let J i) be the lower right i+) i+) block of J, so J ) = J ad J ) = J ). Also, for i, let b i) R i be the colum vector such that Jii b ) i) J i) =. b i) J i+) The our block decompositio takes the form, for i, ) ) Ui) si 0 J i) = 0 I i b i) J i+) with s i = b i) J i+) b ) i) ad U i) = b i) J i+). Notice that detj) = J s i, or equivaletly, detj) J = ii s i = b i) J i+) b ) i). ) It remais to boud each factor s i /. We first establish the followig results. Recall the Loewer partial orderig o symmetric matrices: A B meas that A B is positive semidefiite. Lemma 2.. Let J be a symmetric diagoally balaced matrix with 0 < l J ij m for i j. The ls J ms, ad the eigevalues λ λ of J satisfy 2)l λ i 2)m for i ad 2 )l λ 2 )m. Moreover, if J is diagoally domiat, the the lower bouds still hold. 2

3 Proof. We first show that if P 0 is a symmetric diagoally domiat matrix, the P 0. For ay x R, x P x = P ij x i x j P ij P ij x i x j = P ij x i + x j ) 2 0. j i i<j P ii x 2 i + 2 i<j x 2 i + 2 i<j Sice the matrices P = J ls ad Q = ms J are symmetric ad diagoally balaced with oegative etries, it follows that P, Q 0 by the discussio above, which meas ls J ms. The eigevalues of S are { 2,..., 2, 2 )}, so the result follows by a applicatio of [6, Corollary 7.7.4]. If J is diagoally domiat, the ls J, ad hece the lower bouds still hold. Lemma 2.2. Let J be a symmetric diagoally balaced matrix with 0 < l J ij m for i j. For each i, let J i) be the lower right i + ) i + ) block of J as defied above, ad suppose the eigevalues of J i) are λ λ i+. The 2)l λ j 2)m for j i ad 2 i )l λ i+ 2 i )m. Moreover, if J is diagoally domiat, the the lower bouds still hold. Proof. Write J i) = H +D, where H is the i+) i+) diagoally balaced matrix ad D is diagoal with oegative etries. Note that i )li D i )mi, so i )li + H J i) i )mi + H. Thus by [6, Corollary 7.7.4] ad by applyig Lemma 2. to H, we get, for j i, ad for j = i +, 2)l = i )l + i )l λ j i )m + i )m = 2)m, 2 i )l = 2 i)l + i )l λ i+ 2 i)m + i )m = 2 i )m. If J is diagoally domiat, the i )li + H J i) ad hece the lower bouds still hold. Proof of Theorem.: Suppose J is diagoally domiat. For each i we have j i J ij b i) i, ad by Lemma 2.2, the maximum eigevalue of J i+) is at most 2)l. Thus, b i) J i+) b i) b i) b i) 2)l 2)l b i) b i) i + ) m b i) 2)l b i) b i) b i). Sice each etry of b i) is bouded by l ad m, the reverse Cauchy-Schwarz iequality [8, Ch. 5] gives us b i) J i+) b i) i + ) m 2)l Substitutig this iequality ito ) gives us the desired boud. 3 Examples We close with several examples. l + m 2 lm i + ) = m 2 2) l + m ). l Example 3.. The matrix S = 2)I + has eigevalues { 2,..., 2, 2 )}, so dets) S = 2 2) ) ii ) = 2 ) 2 e as. 3

4 Example 3.2. Whe J is strictly diagoally domiat, the ratio detj)/ ca be arbitrarily close to. For istace, cosider J = αi + with α 2, which has eigevalues { + α), α,..., α} so detj) + α)α J = ii α + ) as α. Example 3.3. The followig example demostrates that we eed a upper boud o the etries of J i Theorem. a). Let = 2k for some k N, ad cosider the matrix J i the followig block form: ) A B J =, A = km + kl 2l)I B A k + l k k, B = m k k. By the determiat block formula sice A ad B commute), we have detj) = deta 2 B 2 ) [ ] = det km + kl 2l) 2 I k + 2klm + 3kl 2 4l 2 km 2 ) k k = 4lk )km + kl l) km + kl 2l) 2k 2, where the last equality is obtaied by cosiderig the eigevalues of A 2 B 2. The detj) 4lk )km + kl l) km + kl 2l)2k 2 J = ) 2k 4l ii km + kl l l + m exp 2l ) l + m as k. Note that the last quatity above teds to 0 as m/l. Upo submissio of this paper, we also cojectured the followig. We thak Mighua Li for allowig us to iclude his proof [7] of this cojecture. Cojecture 3.4. For a positive, diagoally balaced symmetric J, we have the boud: detj) J dets) ii ) = 2 ) 2 e. Without loss of geerality, we may assume = for all i. The we ca write J = I + B, where B is a symmetric stochastic matrix with B ii = 0 for all i. Recall that a row) stochastic matrix is a square matrix of oegative real umbers with each row summig to. Theorem 3.5 Mighua Li). Let B be a symmetric stochastic matrix with B ii = 0 for all i. The deti + B) 2 ). 2) Moreover, this iequality is sharp. We start with some lemmas that are eeded i the proof. Lemma 3.6. If B is a symmetric stochastic matrix with B ii = 0 for all i, the tr B 2. Equality holds if ad oly if B ij = for all i j. Proof. By the Cauchy-Schwarz iequality, 2 ) Bij 2 B ij i j i j 2 = 2, so tr B 2 = i j B 2 ij. The equality case is trivial. 4

5 Lemma 3.7. For a > 0, the fuctio ft) = + at) t/a) a2, 0 t a is decreasig. Proof. It suffices to show ft) = log ft) is decreasig for 0 < t < a. Observig that the coclusio follows. f t) = a + at a t/a = a + a2 )t + at)a t) < 0, The key to the proof of Theorem 3.5 is the followig lemma. Lemma 3.8. [] or [3, Eq..2)]) Let A be a positive semidefiite matrix. If m = tr A tr A s = m2, the ad m s )m + s/ ) det A m + s )m s/ ). Proof of Theorem 3.5. Let A = I +B so that A is positive semidefiite. A calculatio gives m = tr A = ad s 2 tr A2 = m2 tr B2 =. Thus, by Lemma 3.8, we have where s = tr B 2 deti + B) + s ) s/ ), 3). Note that tr B2 = i j B2 ij < 2 for 3, so s <. O the other had, tr B2 by Lemma 3.6, we have, so s. By Lemma 3.7, we kow fs) = + s ) s/ ) is decreasig with respect to s [, ). Thus, fs) f ) = 2 ). 4) Iequality 2) ow follows from 3) ad 4). Takig B ij = for all i j, equality i 2) holds. This proves the sharpess of 2). Remark 3.9. The lower boud of det A i 3) does ot give a useful lower boud for deti +B) i Theorem 3.5. Ideed, defie gs) = s ) + s/ ) tr B for s = 2. The i order that ) gs) 0, we must have s, but g = 0. Remark 3.0. I the proof of Theorem 3.5, we do ot require that the etries of B be positive. Thus Theorem 3.5 is also valid for diagoally balaced symmetric matrices I + B with etries of B egative. Refereces [] J.M. Borwei, G.P.H. Stya, H. Wolkowicz. Some iequalities ivolvig statistical expressios: Solutio to problem 8-0 of L. V. Foster, SIAM Review ) [2] J.W. Demmel, N.J. Higham, ad R. Schreiber. Block LU factorizatio, Research Istitute of Advaced Computer Sciece, 992. [3] B. Groe, C. Johso, E.M. De Sa, H. Wolkowicz. Improvig Hadamard s iequality, Liear Multiliear Algebra, 6 984) [4] C. Hillar ad A. Wibisoo. Maximum etropy distributios o graphs, arxiv e-prits:30.332,

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