BOUNDS FOR DETERMINANTS WITH POSITIVE DIAGONALS

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1 BOUNDS FOR DETERMINANTS WITH POSITIVE DIAGONALS BY EMILIE V. HAYNSWORTH(i) 1. Itroductio. I this paper upper ad lower bouds are foud for the determiat of a real, X matrix A = (a<,-)> with positive diagoal elemets satisfyig (1) a ^ X) I a.71. i = 1,2,,, ad a lower boud is foud for determiats whose elemets satisfy (2) a ^ At ^ ««, i = 1, 2,,, j **«1 / \ At = I max a + max a# 1. 2 \ j*i j*i / G. B. Price [6], A. Ostrowski [4] ad [5], J. L. Breer [l] ad [2], ad H. Scheider [7] have give lower ad upper bouds for the absolute value of determiats satisfyig more geeral coditios tha (1). However, the followig theorem, proved i 2, is ot implied by ay of the above results : Theorem 1. If A = (o ) has elemets satisfyig (l),itis possible to defie Lt ad Rt such that a = Li + Ri,,,. Li ^ X) I o.71. (i = 1,, w). The, for ay choice of Lt ad R{ satisfyig (3), (4) E(l[i. R) det A II (Li + 2Ri)Lk f[ Rh Zc=0 \»=1 >=i+l / Ji=.0 j=l i=k+l a empty product is defied to be 1. Preseted to the Society, Jauary 28, 1960; received by the editors November 2, (') The author is presetly at Aubur Uiversity, Aubur, Alabama. 395

2 396 E. V. HAYNSWORTH [September Ostrowski prove Theorem [5] has show that if A satisfies (2) the det A 2;0. I 3 we 2. If A satisfies (2), the det A ^ II ( E a - At). «=i V j=i ' I provig this we prove a result which may be used to improve ay boud depedig oly o the odiagoal elemets i the row, whe the diagoal elemets are positive. 2. Proof of Theorem 1. To prove Theorem 1 we eed the followig boud give by Price [6]: If (1) holds, the (5) II (««- r.) ^ det A g II (<*«+ r<) r,= E>>< o,j. We proceed by iductio o. Let > represet det A, whe ^4 is of order, ad suppose the elemets of A satisfy (3). 1. For m = 2, >2 = Li + Ri ai2 #21 l-»2 I -*^2 Expadig Z?2 by the diagoal elemets, Therefore Li «i2 Li 0 i?x 0 jri ai2 A> = L,2 0 i?2 021 Ri #21 '-'2 Z-iL2 + Z.1^2 + R1R2 ^ Z>2 ^ R1R2 + LA + (Li + 2Ri)U, sice 0 g g (Ui + a12)jl2 < 2RiLt, by (3) ad (5). 2. Assume that for ay matrix of order w 1 with elemets satisfyig (3), (6) e ( l, r) ^ z? _i ^ e ( a*+m^u r). k=0 \ i=l «'-*+l / i-0 \ i-1 t=*+l / If Z> = det A, A = (0,7), i, j=l, (3), partitio D as follows:,, ad the elemets ay satisfy

3 19601 BOUNDS FOR DETERMINANTS WITH POSITIVE DIAGONALS 397 D = Ai a2 03 L + R Ai=(aij),i,j=l,, \; a2 is the colum vector with compoets ai, i=l,,«1;o3 is the row vector with compoets aj, j = 1,, 1; ad, as i (3), J-> "T* -K- == O, -1 L ^ X) I Oi I, J?^ 0. The we ca write > as the sum of two determiats, i.e., (7) D = A + Rdet Ax A = Ai a-i But the elemets of A satisfy (1), hece, by (3) ad (5), (8) A ^ fl (a -ri)^f[ (a - Rt) = jj L{,»=i t 1 t-i -l -1 ad A ^ L II (o» + u) ^ i II {U + 22?,). i l t 1 ff3 Also, by the iductive assumptio, sice A\ is of order 1, ad, by (3), L we have, usig (6), (7) ad (8), ad Ri^L t, \ai,-\ ^ a*, 1 / fc 1 \ / ft V D * l* + R z( ii &) = z( < *.), D g L (l,+2*0+* z ( ( <+2Ri)Lk *.) = il(l[(li+2ri)lk rx k 0 \ t=l»'-*:+1 /

4 398 E. V. HAYNSWORTH [September 3. Proof of Theorem 2. Suppose we have a lower boud for the determiat of A which holds whe the diagoal elemets are idividually bouded from below. That is, suppose (9) a ; a(a) 0, i = 1, 2, «, implies det A ^ m(a) ^ 0, ct(a) is a fuctio of the odiagoal elemets i the ith row, with Ci{A) =0 if atj = 0 for all j^i, ad m(a) is a fuctio of all the odiagoal elemets of A. Let Z) = diag(di, dit, d) with tf\^0. We shall show that (10) detu + D) ^ w(^) + II <* ' t=i It is this result which we will use to improve Ostrowski's lower boud for determiats satisfyig (2). Sice du^ci(d) =0 for all i l, 2,,, (10) is a direct cosequece of the followig theorem which we proved i [3]: // two matrices A ad B of the same order satisfy the same row hypothesis (i.e., (9)) which is, i tur, sufficiet to prove det.4^0 ad det B^0, the det (A +B) ^det A +det B. A determiat whose elemets satisfy (2) does ot ecessarily have a domiat diagoal uless A* = 0 (» = 1,,»). Ostrowski's result shows, however, that such determiats are ever egative, ad, as a corollary he proved that, if X is ay root of A, = mi f E a ~ Af\. Sice the determiat is the product of the roots, this would imply det A ^ ". This boud is, i geeral, less tha that give by Theorem 2. If A satisfies (2), let bij = O, bu = Af E a- j ** h The B?=At, ad B satisfies (2). Thus det 5^0. Let

5 19601 BOUNDS FOR DETERMINANTS WITH POSITIVE DIAGONALS 399 di = a ba = zl ai3- Af, i = 1, 2,,, 3-1 ad set D = dia.g(di, d2,, d). Sice (2) implies au^bu each i=l, 2,,. The (10) implies we have di?z0 for which proves Theorem 2. det A = det (B + D) ^ II o\ t=i Refereces 1. J. L. Breer, A boud for a determiat with domiat mai diagoal, Proc. Amer. Math. Soc. vol. 5 (1954) pp , Bouds for determiats. II, Proc. Amer. Math. Soc. vol. 8 (1957) pp E. Haysworth, Note o bouds for certai determiats, Duke Math. J. vol. 24 (1957) pp A. Ostrowski, Note o bouds for determiats with domiat pricipal diagoal, Proc. Amer. Math. Soc. vol. 3 (1952) pp , Note o bouds for some determiats, Duke Math. J. vol. 22 (1955) pp G. B. Price, Determiats with domiat pricipal diagoal, Proc. Amer. Math. Soc. vol. 2 (1951) pp H. Scheider, A iequality for latet roots applied to determiats with domiat pricipal diagoal, J. Lodo Math. Soc. vol. 28 (1953) pp Natioal Bureau of Stadards, Washigto, D. C.

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