Topics on a generalization of Gershgorin's theorem

NORTlt- ~ Topics on a Generalization of Gershgorin's Theorem F. O. F a r i d Department of Mathematics and Statistics University of Calgary Calgary, Alberta, Canada T2N 1N4 Submitted by Rajendra Bhatia ABSTRACT We construct two classes of 3 × 3 and 4 × 4 real symmetric matrices, and establish sufficient conditions for the spectrum of a matrix A in each class to be disjoint from its kth order Gershgorin region. This provides a partial answer to a question raised by Newman and Thompson. The problem of providing sufficient conditions for the localization of the spectrum of a matrix in its kth order Gershgorin region is also discussed. © 1998 Elsevier Science Inc. 1. INTRODUCTION AND NOTATION T h e classical G e r s h g o r i n t h e o r e m provides an inclusion region for the eigenvalues o f any n x n matrix with complex entries: if A = (aq) is an n X n matrix with complex entries a n d o - ( A ) = {'~1. . . . . )t n} is the s p e c t r u m o f A, t h e n for every i ~ {1 . . . . . n}, hi ~ Cl(m) = 0 {z ~ ~ : Iz j=l ajjl ~ aj}, ~laul. The w h e r e ~" d e n o t e s the set o f c o m p l e x n u m b e r s and R i = E " j = I , j , sets {z ~ ~ ' : l z - ajjl <~ Rj}, j = 1. . . . . n, are called the G e r s h g o r i n discs of LINEAR ALGEBRA AND ITS APPLICATIONS 268:91-116 (1998) © 1998 Elsevier Science Inc. All rights reserved. 655 Avenue of the Americas, New York, NY 10010 0024-3795/98/$19.00 PII S0024-3795(97)00030-X 92 F. O. FARID A. A. Brauer [1] showed that the eigenvalues of A lie in the set i <j It is natural to ask whether it is possible to generalize this result in the following sense. For every k ~ {2 . . . . . n}, define ~k = {P = (Pl ..... for a l l i = Pk): Pi ~ {1 . . . . . 1 . . . . . k, p ~ < p j i f i n} (1.1) <j} and k Gk(A) = U z~':I--[lz-ap,p,] P~gk i= 1 i=1 0 = (el ..... . (1.2) We call Gk(A) the kth order Gershgorin region. Recently, M. Newman and R. C. Thompson [4] found a 3 x 3 matrix A with the property that none of its eigenvalues lies in G3(A). They also posed the problem: For what values of n and k, where 3 <~ k <~ n, does there exist an n x n matrix A with or(A) A Gk(A) = Q ? Here Q denotes the empty set. We provide a partial answer to this question by constructing for every n ~ {3, 4} a class 5~,, of n x n real symmetric matrices, where for each integer k, 3 <~k <~n, sufficient conditions for any A ~ d n to satisfy Gk(A) A o-(A) = O are established. The case n = k = 3 is studied in Section 2. Section 3 deals with the two cases n = 4, k = 3 and n = k = 4. We also study in Section 4 the related problem of finding sufficient conditions for the spectrum o-(A) of an n X n matrix A to satisfy o-(A) c Gk(A), where 3 ~< k ~< n. The sets of positive integers and real numbers are denoted by///, and ~ , respectively. For ever positive integer n, the set of all n x n matrices with complex entries is denoted by ~¢'~. The identity matrix in ~¢,, is denoted by I n. The permutation matrix obtained from I n by interchanging rows i and j of In, where i < j , is denoted by En(i,j). For every A = ( a i j ) ~ ¢ ~ n and every i ~ {1. . . . n} we denote ~ =j 1 , J ,,~lai,IJ by R i, or by R i ( A ) when it is needed to specify the matrix for which this sum is taken. I f A ~ l / , and 9:3 GERSHGORIN'S THEOREM /9/1 a k are distinct integers in {1 . . . . , n}, we denote by A({a L. . . . . ak}) the principal submatrix of A obtained by deleting row i and column i for all i ~ {1. . . . . n } - {a 1. . . . . c~k}. For every A ~Me,,, the transpose of A is denoted by A T and the set of eigenvalues of A (the spectrum of A) is denoted by tr(A). The cardinal number of a set X is denoted by card X. . . . . 2. , T H E 3 × 3 CASE DEFINITION 2.1. matrices of the form Let d 3 be the class of all 3 × 3 real symmetric A = all al2 0 / a12 0 0 a23 a23 I " a33 where all < 0, and al2, a23 , and aa:~ are all positive. The lbllowing remark shows how the smallest and largest eigenvalues of A ~ : 3 are related to those of A({1, 2}) and A({2, 3}). REMARK 2.1. Let A = (aq) ~ 3 " I f t 1 and tx 1 are the smallest eigenvalues of A, and A({1, 2}), respectively, then 11 ~< /z I < all (2.1) and t1(t, - a l 0 (2.2) Also, if Aa and v 3 are the largest eigenvalues of A and A((2, 3}), respectively, then 1:~ ~> v 3 > a33 (2.3) 2 1 3 ( t 3 - a33) >1 a23. (2.4) and 94 F . O . FARID To prove (2.1) and (2.3), we notice that since A is Hermitian, from T h e o r e m 4.3.15 of [3] We have hl~l and u 3 ~ h 3. (2.5) Also, ~1 and u 3 are given by /Zl= 1w 1+ a121] j u3 = 1 + 1 + and a-'-~--~ " (2.6) Since all < 0 < a33, from (2.5) and (2.6) we obtain hi ~</zl < all and h a /> I)3 > a33. Since all < 0, it follows from (2.1) that ~k,(/~l -- a11) ~ ~1(/~1 -- all) >/ /L/'l( ~'/~1 -- all)- (2.7) all)= But since /zl( ~ 1 a212, (2.2) follows. Similarly, (2.4) follows from (2.3), a33 > 0 and u3(u 3 - a33) = a~3. The following theorem establishes sufficient conditions for a matrix A ~,a¢3 to satisfy or(A) f3 G3(A) = •. THEOREM 2.1. Let A = ( a / j ) E d 3 , and let h 1, h 2, and h 3 be the eigenvalues of A, where h 1 <~ A 2 <~ A 3. Define the function f : (0, oo) ~ (0, 1) by f(z) 2z 2),/2 . (1 + 4z + 1 = (2.8) Suppose ¢ ( aiz[f[t a,2I]211' min lal,I, T[ aaz > 1 a12 ~] + a23 , + f ( i--~ll{]]a12 (2.9) (2.10) 95 GERSHGORIN’S THEOREM zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHG and +J (a231+ lullI + (2.11) %(a,,+~~~1. zyxwvutsrqponmlkjihgfedcbaZY a33 > a33 Then hj e G,(A) Proof. Since and aB > 0. Let (T(C) = {V2’ v3), are given by (2.6). for allj = 1,2,3, and h, > 0. A = (aij) E_w’~, we have a,i < 0 = a2a < a33, al2 > 0, B = A({l, 21) and C = A({2,3}). Let c+(B) = { pl, ~2) and where Z..Q< ~~ and vz < v3. The eigenvalues / .L~and v3 Since A is Hermitian, we infer from Theorem 4.3.15 of [31 that Al < p1 hl < v2 G A, G P2 6 (2.12) A3 and x = kt < A2 < v3 < (2.13) A,. a12/lallland zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA y = a23/a33.Then x, y > 0. It follows from (2.6) that and v3 - a33 = a23f(y ), where f is the function defined by (2.8). The theorem will be established in five steps. a,, - w1 = a,,f(x) 1. Zf zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA a,,f( zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA y) + lull1 + a33 > (a12/a23Xa12 + a23), then 4 e It follows from a,, < a22 < a33 and (2.3), that nj= ,lA3 - ajjI 2 n,3, Il v3 - ajjI. Thus step G3( A). if fiIV3 - aljl j=l > Since a12a23(a12 (v3 - a33)v3 = ai, and + ‘23) = (2.14) RlR2R3. Iv3 - allI = vg + lull1 = a,,_f(y) + a33 + lullI, from (2.141, A3 e G3( A) if &(a23f( y) + lullI + a33) ’ a12a23(a12 “23). (2.15) 96 F . O . FARID But since a2a > 0, using (2.15)we see that Aa ff G a ( A ) i f aeaf(y) + laiN[ + aa3 > (a12/aea)(a12 + ae3). Step 2. Ga(A). I f alef(x) + ]a11] + aaa > (a23/ale)(a12 + aea), then The p r o o f o f this step is similar to that of step 1. Step 3. If aaa >7 ale, then a l e f ( x ) - a 2 3 y a=, aj, from ace = 0 we have A2 = ( a l l ~< Ae < a a a . -- AI) q- (a33 -- A3). A1 Since t r A = (2.16) H e n c e from (2.1) and a n - #1 = al,2f(x), we obtain A2 > a~ef(x ) + (a3a - aa). (2.17) To show a 2 > / a l 2 f ( x ) - a23, we first prove IA3 - aa31 ~< a23. Suppose Ia a a331 > a,ga. H e n c e from (2.3), we have I13 - aa31 = a a - a33 > a23. Thus from the hypothesis aaa > a12, it follows that a 3 > a12 + a23. Also since a,ga > 0 and all < 0, from ,t a > a12 + a23 we have la3 - a l l l = aa + la~ll > a12. Since a12 , a23 > 0, this shows that if [aa - a33[ > a2a, then aa ~ GI(A). This contradicts Gershgorin's theorem. H e n c e we must have l a3 - a331 ~ a23. Thus from Aa > aaa [see (2.3)] and (2.17), we get a 2 >1 a12f(x) - ae3. It follows from a 2 ~< /z 2 [see (2.12)] and /z 2 = a u - / * 1 = a12f(x) that A2 a12f(x). But since f ( x ) < 1 and al.2 > 0, we obtain from A2 ~ al,gf(x) that A2 < a12. H e n c e from the hypothesis a33 >~ a12, we have a 2 < aaa. This completes the p r o o f of step 3. Step 4. If a23 <<,min{lau],(al2/2)[f(x)] e} and a33 > [1 + f(x)]a,e + a23, then A2 > 0 and Ae ~ Ga(A). It follows from aa3 > [1 +f(x)]ale + a23, ai.i+ 1 > 0 for i = 1,2 and 0 < f ( x ) < 1 that a33 > ate. H e n c e from step 3, we obtain a2 >1 a~2f( x) - a23 (2.18) A2 < aa3. (2.19) and Since f ( x ) ~ (0, 1) and a~2 > 0, from the hypothesis a23 ~< (a~J2)[f(x)] 2 it follows that a23 < a l 2 f ( x ) . (2.20) 9T GERSHCORIN'S THEOREM Thus from (2.18), we have a e > 0. It is clear that )t 2 ~ G3(A) if IAel IA~ - alll > a12a23 (2.21) IA2 - a3,31 > ale + a23. (2.22) and We now prove that (2.21) and (2.22) hold. It follows from (2.18), (2.2(/) and a u < 0 that [X/tlX_~ - am[ = X,z(X 2 + laul ) H e n c e from (2.20) and the hypothesis ]all] >~ a,z3, we have IXel Ix~ - aHI > a~2a2a if [av2f(x) - ae:3]a,2f(x) > a,2a2:3. (2.2.3) But since a~2 > 0, we may deduce from (2.23) that 1~211A2 - a~[ > a12a23 if a~2[f(x)] 2 > a2~[1 + f ( x ) ] . (2.24) It follows from the hypothesis az~ ~< (a~J2)[f(x)] 2, f ( x ) ~ (0, 1), and ale > 0 that ale[f(x)] e > a23[1 + f ( x ) ] . H e n c e from (2.24), we have [I,~21 IA,2 - alll > aleaz3. This proves (2.21). It follows from (2.19) and A,2 ~</x e [see (2.12)] that [A2 - a331 = a33 - a~, >/ a33 - /x e. H e n c e from /x 2 = all - /~1 = aver(x), we have - a3 l >/a 3 - a ef(x). (2.'25) Thus from the hypothesis a33 > ale[1 + f ( x ) ] + a23 and (2.25), we obtain IA2 - a a z [ > ale + a23. This proves (2.22), and the proof of step 4 is complete. Step 5. If the entries aq satisfy the conditions (2.9), (2.10), and (2.11), then A~ > O and aj q~ G3(A) for aUj = 1,2,3. F r o m the condition (2.11) and step 1, we have A3 ~ G3(A). Also it follows from the conditions (2.9), (2.10) and step 4 that )t 2 :> 0 and )t 2 ~ Ga(A). Since 0 < f ( x ) < 1, from 98 F . O . FARID (2.9) and ale ~> 0 we have ae3 < a l J 2 . Also, it follows from (2.11), a23 > 0, and f ( y ) ~ (0, 1) that laHI + a33 > (ave/a23)(a12 + a23) - ae3. Thus from 0 < a23 < a l J 2 we get a12 la111 + a33 > 2(a12 + ae3) 3> ~(ale + a23) 2 ae3 ( ). - a l e q- a23 )._ a12 Hence from a12 > 0 and f ( x ) ~ ( O , 1), we get a12f(x)+la111+a33 > (az3/a12)(ale + a23). Thus from step 2 we obtain A1 ¢~ G3(A), and this completes the proof of the theorem. • Similar conditions on the entries of a matrix A ~ ~'3 may be stated to ensure that o-(A) A G3(A) = • and the middle eigenvalue of A is negative. THSOREM 2.2. Let A =(aij) E5~"3, and let A1, A2, and A3 be the eigenvalues of A, where A1 ~< A2 ~< As. If aij satisfy the conditions a12 ~< rain a33, ~ [a111> l+f~ -- f az3 ] , (2.26) a23 -1- t/12 , and .[ a12 a23 a12J| 7"--7| + la111 + a33 > - - ( a 1 2 + a23), lalll ] (2.28) a12 where f is the function defined by (2.8), then Aj ~ G3( A) for all j = 1, 2, 3 and A2 < O. Proof. Let x = a12/la111 and y = a23/a33. The theorem is established in five steps. The proofs of steps 1, 2, 3, and 5 are similar to those of steps 1, 2, 3, and 5 in Theorem 2.1. We will indicate the proof of step 4 only. GERSHGORIN'S THEOREM 99 Step 1. If a,2J(y) + laH[ + a:~3 > (a~Jaz~Xal2 + a2a), then A3 Step 2. If a12f(x) + laul + a:~3 > (a2Jar2)(al2 + a23), then A~ G~(A). G~(A). Step 3. If laHI >1 a23, then all < A2 <~ ale - ae3f( y). Step 4. If at2 <<.min{a33,(a~23/2)[f(y)] e} and lalll > [1 + f(y)]a23 + arz, then A,2 < 0 and A2 ~ G3( A). It follows from the hypothesis (2.27), 0 <f(y) we have < 1 and a~,i+ l > 0 for i = 1,2 that lalll > a2,3- Hence from step 3, "2 <~ al2 - a2~f( y ) . Since 0 < f ( y ) it follows that (2.2~) < 1 and a23 > 0, from the hypothesis a12 ~< ( a 2 3 / / 2 ) [ f ( y ) ] 2 al2 < a 2 3 f ( y ) . (2.30) Thus from (2.29) and (2.30), we get A,2 < 0. The rest of the proof is similar to that of step 4 in T h e o r e m 2.1. Step 5. If the entries aij of A satisfy the conditions (2.26), (2.27), and (2.28), then A,2 < 0 and Aj fli G3(A) .for allj = 1, 2, 3. • The classes of matrices that satisfy the hypotheses of Theorems 2.1 and 2.2 are nonempty. This is shown in the following example. EXAMPLE 2.1. The matrix A = (aij) with entries a u = = 4, a23 = a32 = 1, a33 = 20, and a22 = al3 = aal = 0 is in ~3 all the hypotheses of T h e o r e m 2.1. The matrix B = (bij) b u = - 2 0 , bi2 = bzl = b33 = 1, b23 = b32 = 4, and bz2 = bl3 in ~¢3 and satisfies all the hypotheses of T h e o r e m 2.2. 1, al2 = a21 and satisfies with entries = b31 = 0 is The following remark, which follows from Theorems 2.1 and 2.2, will be used in T h e o r e m 3.1. REMARK 2.2. Let A = (aij) E,~¢3, and let A2 be its middle eigenvalue. (i) If a23 <~ min{laul,(a12/2)[f(alJla111)] 2} and a33 >t a12, then A2 > 0 and Xz( A2 - a u ) > alzaz3 , where f is the function defined by (2.8). (ii) If a12 <~ min{a33,(azJ2)[f(a23/aa~)] z} and laul >i a23, then Ae < 0 and IAzl I)t 2 - az31 > alzaz3. 100 F . O . FARID It follows from the hypothesis a3a >~ a12 and step 3 of T h e o r e m 2.1 that A2 >1 a12f(a12/la111)- a23 [which is the inequality (2.18)]. In step 4 of T h e o r e m 2.1 we have shown that the inequalities A2 > 0 and Az(A 2 - a11) > al2a2z (2.21) follow from (2.18), all < 0, a12 > 07 and the hypothesis a23 ~< min{la111, (alJ2)[f(a12/laH[)]2}. This proves (i). Similarly, (ii) follows from steps 3 and 4 of T h e o r e m 2.2. 3. T H E 4 × 4 CASE T h e spectrum of a matrix A ~ , K 4 may be disjoint from either Ga(A) or G4(A) (or both). However, we have the inclusion relation o ' ( A ) N G4(A) c tr(A) ¢3 Gz(A). This follows from the following lemma, which we state without proof. LEMMA 3.1. Let A = (aij) ~J-¢'n, where n >~ 2, and let k ~ { 1 , . . . , n - 1}. Then Gk+I(A) c Gk(A). COROLLARY 3.1. Let A ~ t " n, where n >1 4. If A ~ ~r( A) and A f~ Gk( A) for some k ~ {3 . . . . . n - 1}, then A f~ Gj( A) for all j satisfying k<j<<,n. DEFINITION 3.1. matrices of the form Let ~¢4 be the class of all 4 × 4 real symmetric all 2 a12 0 0 0 a23 0 a23 a33 a34 0 a34 a44 where all < 0 = a22 < a33 < a44 and ai. i + 1 > 0 for all i = 1, 2, 3. It is clear that for every A ~ ~¢4, A({1, 2, 3}) and A({2, 3, 4}) - a3313 are both in ~¢3. Relations b e t w e e n tr(A), o-(A({1, 2, 3})), and o-(A({2, 3, 4})) are given in the following remark. REMARK 3.1. Let A @ 5~4. Suppose o r ( A ) = {/~1' /~2' a3' /~4 }' tr(A({1, 2, 3})) = {/xl,/z2,/.t3}, and o-(A({2, 3, 4})) = {u 2, u 3, u4}, where Xi ~< hi+ 1, tzj <<.Ixj+l and uk <~ uk+ 1 for all i = 1,2,3, j = 1,2, and k = 2, 3, respectively. Since A is Hermitian, from Theorem 4.3.15 of [3] we have 101 GERSHGORIN'S THEOREM the interlacing inequalities (:3.1) and A 1 ~ b'9 ~ A 2 ~ lP3 ~ A 3 ~ P4 ~ h4" (3:2) For A = (a 0 ~M/4, one may write [~3(A) = (~1,2,3(A) U (~I+2.4(A) [,_) C+1.3.4(A) U G2,3.4(a), where for any distinct integers i, j, and k in {1, 2, 3, 4}, The following theorem establishes sufficient conditions for a matrix A ~ o% to satis~ ~r(A) A G3(A) = Q. THEOREM 3.1. Let A = (ai/) ~.~4, and let h l, h2, h 3, and h 4 be the eigenvalues of A, where Aa ~< A2 ~< A~ ~< A4. Define the function f : (0, w) (0, 1) by f(z) 2= = (1 + 4z2) 1/~ + l " (3.3) Suppose { a23 m~x{a12, a34 } ~< rain a3.3, ~2 f a2_~3 a33 ] ' (3.4} (al~ + a23)(a23 + a:~) fa11[ ~ a:~:3 + ('/23 (3.5) 102 F. O. FARID la~l[ + a33 >/ max{ (a~2 +a~3)(a~3+a34 ) a l ~ ax2(a12 + a23)(a23 + a34 ) a23a34 (3.6) a44 >~ a3~ + a~3 + (a12 + a23)(a23 + a~4) a23 , (3.7) and a34(a12 + a23)(a23 + a34 ) maxf (a12 + az3)(a23 + a34) a44 >~ a34 ' a12a23 (3.8) f Then Aj f~ G3(A) for all j = 1,2,3,4. NOTE. If max{a12, a34} < a23 , then either the right hand side of the inequality (3.6) is L = (a12 + a23)(a23 + a34)/a12 or the right hand side of the inequality (3.8) is M = (a12 + a23)(a23 + a34)/a34. If a12 ~< a3a, then from 0 < a12 < az3 we see that the right hand side of (3.6) is L. Similarly, if a~ ~< a12 , then from 0 < a34 < a23 it follows that the right hand side of (3.8) is M. Proof. Let A = (aij) ~ ~¢4. Denote the principal submatrices A({1, 2, 3}) and A({2, 3, 4}) of A by B and C, respectively. Let /zl, /x2, and /z3 be the eigenvalues of B, where /z 1 ~</x,2 ~</xz, and let u 2, uz, and u4 be the eigenvalues of C, where u 2 <~ u 3 <~ u4. Since A is Hermitian, from Remark 3.1 we have the interlacing inequalities (3.1) and (3.2). The theorem will be established in nine steps. Step 1. I f [alll + a33 >1 (ax2 + a23)(a23 + a34)/a12, then A t f~ G1,2,3(A) U G1,2,4(A) U G2,3,4(A ). Since B ~ ¢ 3 , from Remark 2.1 it follows that/z 1 < all and Iq~_ll/z 1 - ajjl >~ a~2. Thus from A1 ~< /z 1 [see (3.1)], all < 0 = a22 < a33 , and al2 > 0, we have /~1 < all and 1-I~=I[A1 - aij I > a2 • ~J 12(lalll + a33). Hence from the hypothesxs Falll + a33 >1 (al,2 + a23)(a23 + a34)/a12, it follows that Iq}=llA 1 - ajjl > a12(a12 + a23)(az3 + a34). Hence /~1 ~ GI,z,3(A)- 103 GERSHGORIN'S THEOREM It follows f r o m A.1 ~ G 1 . z , 3 ( A ) a n d A~ < a l l < 0 < a33 < a44 that 3 Ia 1 - alll lal - a=l Ix1 - a441 > I - I la, - %1 j= 1 > a12((112 -}- a23)(a23 "[- a34 ) . (3.9) H e n c e f r o m a12 > 0 a n d a,2a > 0, w e get IA l - a1111A 1 - a2211A i - a44i > alea34(a12 + a23). This p r o v e s A1 ~ GI,2,4(A). N o w we p r o v e a 1 q~ G2,3,4(A). W e first s h o w 1-I9 zlA 1 - %jl > al2(a,~3 + a34). W e have e i t h e r IX1 - a111 ~< ale or IX 1 - alil > a12. If IA1 - alll ~< al,2, t h e n f r o m a 1 ff G1,2,a(A) a n d aii+ ~ > 0 for i = 2 , 3 w e have IAal IA1 - aaal > alz(a2a + aa4). In the case IX 1 - alll > ale, w e d e d u c e from Al < all < 0 < a33 < a44, the h y p o t h e s i s lalll + aaa >/(a12 + a2:3)(a23 + aa4)/al, 2, and aci+l > 0 for all i = 1 , 2 , 3 that 12t1 - a j j l > a23 +a34 > a 3 4 for j = 3 , 4 . H e n c e if Ial - alll > ale, t h e n f r o m the classical G e r s h g o r i n t h e o r e m we m u s t have Iall ~ a12 + a23. T h u s w e d e d u c e from )t 1 ~ G1,2,a(A) that I a l - alll I,tl - aaal > a12(a23 + a34). H e n c e from a 1 < a l l < 0 we get Iall Ia 1 - aaal > al2(a2a + a34). This c o m p l e t e s the p r o o f t h a t Fl~=~la~ - %1 > a12(a23 + a34). T h e n w e d e d u c e f r o m A I < all < 0 < aa3 < a44 , the h y p o t h e s i s l a n l + a33 > (al2 + a2aXa23 + a34)/a12, a n d ai,i+ 1 > 0 for all i = 1, 2, 3 that 4 _ (a23 + a34). H e n c e a 1 4~ - %1 > a34(a12 + a~3) FIj=21A~ Ge, 3,4(A) • Step 2. Ifaa4 <-~a23 and lalll A- a33 >j (al2 -+- a23Xa23 A- aa4)/a12 , then a 1 ~ Ga(A). F r o m step 1, w e have a 1 ~ G1,2,a(A ) tO GI, e,4(A ) tO G2, a,4(A). It follows f r o m ~l < a H < 0 = ae2 < aaa < a44 that I& - aHI lai - a3311& - a441 > IA1 - a l l l Ial - a2211& - a441. H e n c e f r o m (3.9), w e get Iai - aHI Ia~ - a331 Ixl - a441 > a l 2 ( a i z + a23)(a23 + a34). T h e n from ale,a34 > 0 a n d the a s s u m p t i o n a23 ~> a34, w e o b t a i n I & - a ~ l I& - aa311al - a441 > a12aa4(ae3 + a34). H e n c e a 1 ~ GI,a,4(A). Step 3. If a44 >~ (a12 + a2a)(a23 + aa4)/a34, then a 4 ~ G2,a,4(A) tO GI, a,4(A) to G1,2,a(A). T h e p r o o f is similar to step 1, so w e only m e n t i o n the m a i n steps. W e first use t h e fact that C - aaa 13 ~E5a~3 to c o n c l u d e f r o m R e m a r k 2.1 that v4 > a44 a n d ( u 4 - a33)(v 4 - a44) >1 a~4. T h u s f r o m A4 >~ u4 [see (3.2)], a34 > 0, a n d a44 > aaa > aze = 0, w e o b t a i n l--I~=z[a 4 - ajj[ > a44a~34. H e n c e from the h y p o t h e s i s a44 > (alz + a2a)(a2a + a34)/a34, it follows that a 4 ~ Gz, a,4(A). This t o g e t h e r with A4 > a44 > aa3 > a2~ = 0 > a n p r o v e s A4 ~ G~,G,4(A). T o p r o v e a 4 ~ G~,2,3(A), w e first s h o w that 3 I-Ij=elA 4 - ajjl > a34(a12 + a23). T h e n w e d e d u c e f r o m a l l < 0 < a44 < a 4, the h y p o t h e s i s a44 >i (a12 + a23)(ae3 + a 3 4 ) / a 3 4 , a n d ai, i+ 1 > 0 for all i = 1 , 2 , 3 that l--Iaa= 11A4 - %1 > a~e(a~2 + a,23)(az3 + aa4). 104 F . O . FARID Step 4. I f a12 ~< a23 and a44 >~ (ale + a23)(a23 + a34)/a34, then A 4 qE: Gz(A). F r o m step 3, it remains to show A4 ~ GI,z,4(A). It follows from A4 > a44 > a33 > 0 = a22 > a l l and A4 ~ C2,3,4(A) that IA4 - a l l l IAn - aeel IA4 - a441 > a34(a12 + az3)(aez + az4). N o w t h e result follows from ale, a34 > 0 and the assumption a23 >~ al2. Step 5. If ale, a23 , a33 , a34 , and all satisfy the conditions al2 ~< rain a3a, T f a33 ]1 ) ' a34 < aa3, (3.10) (3.11) and (ai + + a3,) a23 then A2 ~ G1,2,3(A) U GI,e,4(A) L/ G1,3,4(A). Since ai, i+ 1 > 0 for all i = 1,2,3, from (3.12) we have la111 > 2a23 + a12. (3.13) Since B ~ 5 ¢ 3, it follows from (3.10), (3.13), and R e m a r k 2.2 that the eigenvalue/x 2 of B satisfy/x 2 < 0 and I ix211 tL2 - a331 > a12a23. H e n c e from A2 ~< /xa [see (3.1)] and a33 > 0, we have A2 < 0 and IAel [Aa - a331 > a12a23. (3.14) Also, it follows from u 2 ~< A2 [see (3.2)] and A2 < 0 [see (3.14)] that the smallest eigenvalue u 2 of C satisfies u 2 < 0. Now we show ]u 2] ~< az3. Suppose Iv21 > a23- Then from u 2 < 0, a44 > a33 > 0, and (3.11), we have I v ~ " %1 = lye1 + % > a23 + a34 f o r j = 3,4. W h e n j = 4, we may deduce from az3 > 0 that l u 2 - a44[ > a34. This shows that if 1~21 > a23 then u 2 ¢~ GI(C), which contradicts the classical Gershgorin theorem. H e n c e we must have 1~21 < a23. Thus from u 2 ~< A2 [see (3.2)] and A2 < 0 [see (3.14)], it follows that 0 < IA=I < a23- (3.15) GERSHGORIN'S THEOREM 105 H e n c e f r o m (3.13), (3.15), a n d azo > 0, w e have (3.16) lau[ -IA,~I >/laall - a_o,3 > 0. But since a H < 0 a n d X2 < 0, w e o b t a i n from (3.15) a n d (3.16) that (3.17) IA2 - a~,l >/ lalll - a e 3 > 0. H e n c e from (3.14) a n d (3.17), w e get 3 FI j=l IA2 - aj, I > a, aea(lal,I- (3.1S) T h u s from (3.12), (3.18), a n d ai, i+ 1 > 0 for i = 1,2, we o b t a i n 3 El*2 j=l -- ajjl > a12(a12 + (19_3)(a2:] + a34 ) • (3.19) H e n c e A2 ~ Gl, e,3(A), Since Az < 0 < aaa < a44, w e have IA2 - a441 > tA 2 - a331. T h u s from (3.19) a n d ale, ae3 > 0, w e o b t a i n Ia 2 - a111 IX2 - a2zl [A,2 - a441 > alea:34(a12 + a23). H e n c e A2 ~ G1,2,4(A). It follows from (3.15) a n d (3.19) that: ( 3. o) IA2 - a l l l l A 2 - a331 > a 23 Since A,~ < 0 < a33 < a44 , f r o m (3.11) w e get IA2 - a441 > a34. H e n c e f r o m (3.20) a n d ai, i+ 1 > 0 for all i = 1 , 2 , 3 , w e o b t a i n IA 2 - alll IA 2 - a:~3t IA2 - a441 > al2a34(a23 + a34). This p r o v e s A2 ¢ G1 3.4(A). Step 6. If a12, a23, aa3, a34, and all satisftt the conditions (3.10), (3.11), and (3.12), and a44 satisfies the condition a44 ~> a 4(a 2 + a23)(a, + a 4) (3.21) a12 a23 then A,2 ~ G 3 ( A ) . F r o m step 5, w e have A2 ~ G I , e , 3 ( A ) tO G1,2,4(A) to GL3,4(A), a n d (3.14) is satisfied. T h e n from (3.14) a n d a44 > 0, we get 106 F . O . FARID Vl4_zlAe - ajjl > alea23a44. Thus from a12 > 0, az3 > 0, and (3.21), we ob~tain VI~_elA2 - ajjl > a34(a12 + a23Xa23 + a34). H e n c e h e ~ G e 3 4(A)This compi-etes the p r o o f that A2 ~ G3(A). ' ' Step 7. If a23, a33, a34, a12, and a44 satisfy the conditions ale ~< az3, (3.23) and a44 >/ a33 + a23 + (ale + a23)(a23 + a34 ) a23 , (3.24) then A3 f~ Ge,3,4(A) U G1,3,4( A) U G1,2,4( A). The p r o o f is similar to step 5, so we mention only the main steps. F r o m (3,24) and a~,i+ 1 > 0 for all i = 1, 2, 3, we have a44 > a33 + 2a33 + a34. (3.25) W e use the fact C - a33 13 ~ ~¢3 to conclude from (3.22), (3.25), and Remark (2.2) that u 2 - a33 > 0 and ru 3 - a3311~31 > a23a34 > 0. Thus from A3 >/ u 3 [see (3.2)] and a ~ > 0, we have A3 > a33 > 0 and IA311)t 3 - a331 = )t3()t 3 - a33 ) > a23a34 > 0. (3.26) Next we prove ]/x 3 - a~31 < a23, and conclude from it, A3 ~</z 3 [see (3.1)] and A3 > a33 [see (3.26)] that 0 < )t~ - a3z ~< a23. (3.27) H e n c e from (3.24) and ai, i+ 1 > 0 for all i = 1, 2, 3, we get )t 3 < a44 and IAz - a441 >t (a12 + a23Xa23 + a34)/a23 > 0. Thus from I-I~=21A3 - ajjr > az3a34 [see (3.26)], we obtain 4 V I I A 3 - ajj] > a34(a12 + a23)(az3 + a34 ). j=2 (3.28) 107 GERSHGORIN'S THEOREM Hence h a ~ G2,a,4(A). It follows from a u < a22 = 0 < h 3, (3.28), and a23, aa4 > 0 that h a ~ GI,a,4(A). To prove h a ~ G1,2,4(A), we infer from a u <0<aaa < h a, and (3.23) that Ih 3 - a l l l > a l , z. Hence from (3.27), (3.28), and aci+l > 0 for all i = 1,2,3, we obtain Ih 3 - alll Ihal Ih:~ - a44t > avzaa4(a12 + a23). Step 8. If the entries a2a, aaa, aa4, a12, and a44 satisfy the conditions (3.22), (3.23), and (3.24), and a n satisfies lalll + a% > al,2(a12 + aza)(a23 + aa4)/a2aa34, then h a ~ Ga(A). Step 8 follows from step 7 in a similar way as step 6 follows from step 5. Step 9. If the entries a~j satisfy the conditions (3.4), (3.5), (3.6), (3.7), and (3.8), then hj ~ Ga(A) for allj = 1 , 2 , 3 , 4 . Since f(a23/aa3) ~ (0, 1) and a2a > 0, we have from (3.4) that max{a>> a34 } < a,23. Thus from (3.6) and step 2 we have h 1 ~ Ga(A), and from (3.8) and step 4 we have h 4 ~ G3(A). It follows from (3.4), (3.5), (3.8), and step 6 that h e ~ G3(A). Finally, we deduce from (3.4), (3.6), (3.7), and step 8 that h a ~ Ga(A). This completes the proof of the theorem. • The following theorem establishes sufficient conditions {br the eigenvalues of a matrix in a¢4 not to be in its fourth order Gershgorin region. THEOREM 3.2. Let A = (aij) E&¢4, and let I~1, 1~2' A3' and Z 4 be the eigenvalues of A, where A1 <~ t,2 <~ Z~ <. h 4. Define the function f : (0, 2) (o, 1) by f(-) 2z = (1 + 4z'2) 1/2 + 1" (3.29) Suppose { max{a12,aa4 } ~ min a3~ - 7 [ ]alll ) \a3a] (3.3o) , + a34 ) max{a23+ (al2 + ae3)(a~3 a23 (al2 + a23)(a23 + aa4 ) a12 -- a33 } , (3.31) F. O. FARID 108 and max[ (al2 + a23) (a23 + a34) , a33 + a2a + (ale + a23)(a23 + 334 ) a44 >~ 334 a23 (3.32) Then Aj ~ G4(A) forall j = 1,2,3,4. Proof. The theorem will be established in two steps. Step 1. A 1 ~ G4(A) and t 4 ~[= G4(A). Since 0 < f(a23/a33) < 1 and a23 > 0, from (3.30) we have max{a12, a34} < a23, Thus from 1a111 + aaa >/ (a12 + a2a)(a23 + a34)/alz [see (3.31)] and step 2 of Theorem 3.1 we obtain A 1 f~ G3(A). Hence from Corollary 3.1, we get A1 ff G4(A). Also, it follows from a~2 < a23, a44 >~ (a12 + a23Xa23 + a34)/334 [see (3.32)], and step 4 of Theorem 3.1 that A4 ff G3(A). Hence from Corollary 3.1, we get A4 ff G4(A). Step 2. 12 ~ G4(A) and A3 ~ G4(A). It follows from (3.30), ]al~[/> a23 + (a12 + a23Xa23 + a34)/a23 [see (3.31)] and step 5 of Theorem 3.1 that 12 f~ C1,2,3(A) and 12 < 0 [see (3.14)]. So if 12 were in G4(A), then from 12 ff G1,2,3(A) it follows that ]A2 - 3441 < a34. But then from A2 < 0 < a33 < a44 , we obtain a33 < a34, which contradicts (3.30). This proves A4 G4(A). It follows from (3.30), 344 ~ a33 -I- 323 q- (a12 q- a23)(a23 + a34)/a23 [see (3.32)], and step 7 of Theorem 3.1 that A3 ff Ge,3,4(A)and A3 > 333 > 0 [see (3.26)]. If 13 ~ G4(A), then from 13 ~ G2,3.4(A) we get [A3 - ajll < ape. Thus from a l l < 0 < a33 < 13, we obtain 333 < a12 , which contradicts (3.30). Hence we must have A3 ~ G4(A). This completes the proof of the theorem. • The following example shows that the classes of matrices satisfying the hypotheses of Theorems 3.1 and 3.2 are nonempty. EXAMPLE 3.1. Let A = (aij) ~ t " 4 be defined by all = - 2 4 , al2 --- 321 = 333 = a34 = a43 = 1, az3 --- a32 = 4, 344 = 25, and all other a~j = 0. Then A ~ d 4 and satisfies all the hypotheses of Theorems 3.1 and 3.2. The matrix B =(b~j) ~ ' 4 given by hi1 = - 2 0 , b12 =b21 =b33 = 1, b23 =b3z = 4 , b ~ = b4a = 0.2, b44 = 105, and all other b~j = 0 is in ~4 and satisfies all the hypotheses of Theorem 3.2, but it does not satisfy the condition (3.6) of Theorem 3.1. GERSHGORIN'S THEOREM 4. 109 H I G H E R ORDER G E R S H G O R I N REGIONS In this section we study the problem of providing sufficient conditions tbr A ~ , , , where n >~ 3, to satisfy 6r(A) c G k ( A ) for some k ~ {3. . . . . n}. R. Brualdi [2] discussed the localization of the eigenvalues of a weakh irreducible matrix in its kth order Gershgorin region, where k ~ {2. . . . . n} (see Theorem 2.3 and Corollary 2.4 of [2]). Because of the importance of this result to this section, we state it in the following equivalent form (see Theorem 6.4.18 of [3]): is weakly irreducible, tvhere THEOREM 4.1 (Brualdi). I f A = (a~j) ~ , , n >1 2, t h e n f i ) r e v e r y A ~ o ' ( A ) t h e r e exists k ~ {2. . . . . n} (depending on A) .such that A ~ Gk ( A ). In Theorem 4.1 it is shown that if ,~ ~ o-(A) - {ajj :j = 1 . . . . . n} and x = [x 1. . . . . x,,] T is an eigenveetor of A corresponding to )t, then there is a cycle 3' : PqP~, . . . . P~P~k+, of length k >/2 in the directed graph F(A) of A with the following properties: Property, (1): xig 4 : 0 for all j = 1 . . . . . k. Property (2): For every j ~ {1. . . . . k}, r ~ {1 . . . . . n } - {ij} satisfying a~# 4: 0. Ix~j+,[ >/ [x~] for any Here x~ = x/k+l, since i I ik+ I. Also, from the definition of a cycle it is clear that aij/j+~ ¢ 0 for all j = 1 . . . . . k [for future reference we say that 7 is a cycle satisfying properties (1) and (2) with respect to the eigenvalue ,k and the eigenvector x]. It then follows from properties (1) and (2) that 2~ ~ {z ~:~: 1-I~_ llz -- aijij [ ~ I-I~=IRij}, which is a subset of Gk(A). For a given A, the largest possible integer k obtained from Brualdi's theorem that satisfies A E Gk(A) may equal 2, and this case is included in Brauer's theorem. An example where this situation occurs is a matrix A ~ ~:3 satisfying either the hypotheses of Theorem 2.1 or Theorem 2.2. In this case it is clear A is weakly irreducible, and any A ~ cr(A) is not a diagonal entry of A (see Remarks 2.1 and 2.2). The following proposition provides another example. = PROPOSITION 4.1. Suppose that A = (ar~) ~/{,,, n >1 3, satisfies at, 4= 0 f o r all r 4= s. Then A is weakly irreducible. In addition, assume that A is an eigenvalue o f A such that A ~ ajj fi)r all j = 1 . . . . . n and no more than two Gershgorin discs contain dr. Then f o r any eigenvector x = [ x 1. . . . . x,, ]7" of A corresponding to A and any cycle T: Pi Pi2, . . . . PikPi~+~ of length k in the directed graph F ( A ) of A that satisfies the preceding properties (1) and (2) with respect to )t and x, w e have k <<.2. 110 F . O . FARID Proof. Since all the off-diagonal entries of A are nonzeros, A is weakly irreducible. Now let A ~ tr( A ) - {ajj : j = 1 . . . . . n}, x = [ x 1. . . . . Xn] r be an ei~envector of A corresponding to A, and y : P/Pi . . . . ~ P/P/ k + l be a cycle 1 2 ¢ of length k in F ( A ) that satisfies the preceding properties I1) and (2) with respect to A and x (the existence of y is guaranteed by T h e o r e m 4.1). We prove the second part of the theorem by showing that if k >I 3, then A ~ N ~=l{z ~ ~ : l z - a i / I <<,R i ) . So suppose k >~ 3. The conclusion will be esta(31ished in four stepJsj. Step 1. I f r, s ~ {1 . . . . . k} a n d 0 < s - r < k - 1, then Ix, I >t Ixil. We have either r = 1 or r > 1. If r = 1, then it follows from the hypothesis aid # 0 for all j # i k, s 4: k, and property (2) that Ix/xl = [xi~+~l ~> Ix,/. If r > 1, then it follows from the hypothesis a/~ d 4~ 0 for all j # i~_ 1, s ~ k, and property (2) that Ix i I >~ Ix/I. Step 2. I f r, s ~ {1 . . . . . k} a n d s - r >~ 2, then Ix, I = Ix/I. From step 1, we have Ix/I >/Ixi,+xl and Ix/=+~l >/Ix/I. Hence Ix/I >1 Ix~I. (4.1) It also follows from the hypothesis ai~ ~ ~ 0 for all j # i s_ 1, s - 1 > r, and property (2) that Ix i I ~> Ixi I. Thus from (4.1), we get Ix, I = [x/]. Step 3. I f r ~ { 1 . . . . . rk - 1}, then ]X/rl = IX/r+,[. Since k >/3, there exists s E {1 . . . . . k} - {r, r + 1}. We have either s < r or s > r + 1. We prove the statement in the case s < r. The other case is proved similarly. It follows from step 1 that ]x,[>~ Ix/I and Ix/[/> Ix/+,l. But from step 2, we have Ix i ] = ]x i +l]. Hence We get [x i ] = Ix i ~l]. Step'4. A~ f3~_~{z~:lzZa~/l~'R,}. Sincea/o~S0forallj~ s i~, from property (2)-we have Ix/~l/> I~c~,lfor ~I1 j # i~. Thus max{lx,l:j = 1 . . . . . n} = max{lxi,], Ix,~l}. But from step 3 we have I x / ] = Ix,+ ]~for all j = 1 . . . . . k - 1. Then max(lxjl:j = 1 ..... n} = IxJ forJall r L ~ l . . . . . k. H e n c e from the classical Gershgorin theorem, we get A ~ {z E ~ ' : ] z - a%] <~ R/j} for all j = 1 . . . . . k. This proves the proposition. • We notice that the condition of weak irreducibility of the matrix in T h e o r e m 4.1 could be dropped and the same conclusion still obtained. (ars) E.~'n, w h e r e n >1 2. I f A ~ ( r ( A ) , then THEOREM 4.2. Let A there exist distinct integers i o . . . . . i k ~ {1 . . . . . n}, w h e r e k >>, 1, such that = { x~ k z~: k / I-IIz-a,,I.< r= 0 r r I-In, r • 7"=0 (4.2) 111 GERSHGORIN'S THEOREM Proof. Let A b e an eigenvalue o f A, and x = [x l . . . . . x,,] r be an eigenvector of A c o r r e s p o n d i n g to A. I f A = a r t for some r ~ {1 . . . . . n}. t h e n for any distinct integers i 0 . . . . . i k ~ {1 . . . . . n} with i 0 = r, we haw~ A ~ {z ~ ~ : Fl~_0lz - ai,i] <~ 1-I~ oRi). N o w suppose A :~ at,_ for all r {1 . . . . . n}. F r o m Ax = Ax, we h~we tl IN - a~r[ Ix~l < ~ s=l. lar,I Ix,I, (4.:3) s~-r for all r = 1 . . . . . n. Since A # a ~ for all r = l. . . . . . n, it follows from (4.:3) that for every r ~ {1 . . . . . n} satisfying x~ v~ 0 the set A~ = {s ~ {1 . . . . . n} - { r } : a , , # 0, x~ # 0} (4.4) is nonempty. Thus A = {r ~ {1 . . . . . n} : x~ v~ 0} satisfies card A > 2. F o r every r ~ A, define a r e l a t i o n < r on A~: If" s t , s 2 ~ A , , then s~ % .s'~ if and only if IXsll ~< Ix~[. It is clear that -% is a p r e o r d e r on A~. Also, since A~ is n o n e m p t y and finite, by L e m m a 6.4.17 o f [3], A r has a maximal e l e m e n t for every r ~ A. Define, for every r ~ A, N~ = {s ~ A , : s is a maximal e l e m e n t o f A,}, and let f l = {A'r : r ~ A}. T h e n by the axiom o f choice, t h e r e exists a function g : g l --+ U , e AN~, w h e r e g(N~) ~ N, for every N ~ 11. Since N , c A e c A for every A'~ ~ f L the range o f the function g is a subset of A. D e f i n e the function f : A ~ A by f ( r ) = g(N~) for every r ~ A. L e t m = c a r d A and p ~ A. Let ~ = { p , f ( p ) , . . . . f r o - 1( p)}, w h e r e f~(p) = (f ..... f)(p) foreverv rEV/F. r times Since f ~ ( p ) ~ A for every r ~ , we have ~ c A . It follows from the definition o f f that for every q ~ A, f ( q ) 4= q. In particular, p 4: f ( p ) . Thus card 9 >~ 2. W e show the existence o f distinct integers f J ( p ) . . . . . fj+k(p) in such that fj+k+l(p) = i f ( p ) . W e have either ~ = A or 3 is a p r o p e r subset o f A. W e consider each case separately. Case 1: ~,@ = A. Since fro(p) ~ A t h e r e exists j ~ {0 . . . . . m - 2} such that Take k = m j 1. Since c a r d ~ = .... f j + k ( p ) are all distinct. Case 2: ~ is a proper subset of A. < m. H e n c e from the definition o f ~ r 1, r e ~ {0 . . . . . m - 1} such that f r , ( p ) (=9), from the definition of f J ( p ) = fro(p), w h e r e f 0 ( p ) = p. card A = m, the integers fJ(p), Since card A = m, we have card t h e r e exist two distinct integers =F:(P)" Define j = m i n { r ~ {0 . . . . . m - 3}: t h e r e exists s ~ { r + 2 . . . . . m - 1} with f ~ ( p ) =fr(p)} F. O. FARID 112 (m ~> 3, since 2 ~ c a r d ~ < m) and Jl = min{r ~ {j + 2 . . . . . m - 1} : f r ( p ) = i f ( p)}. We have that i f ( p ) . . . . . f f l - l ( p ) are all distinct. To prove this, assume that r and s are integers satisfying j ~< r < s < j l and f r ( p ) = i f ( p ) . I f j = r then we have j < s <j~ and f f ~ ( p ) = i f ( p ) = i f ( p ) , and this contradicts the definition of jl. If j < r, then by induction we have f t s - ( t 1)r+l(p) = f r + l ( p ) for all l = 0 . . . . . s - r - 1 and t ~ . But since j l ~ {ts (t- 1)r+l:l~{0 ..... s-r1}, t ~y//}, there exists l ~ { 0 . . . . . s t - 1} such that f r + l ( p ) . : f j ~ ( p ) = f j ( p ) , and this again contradicts the definition of jl, since r + l < s < j l - This proves that i f ( p ) . . . . . f j l - l ( p ) are all distinct. In this case we take k = Jl - j - 1. The arguments presented in the preceding existence of distinct integers i f ( p ) . . . . . f j + k ( p ) that f j + k + l ( p ) = i f ( p ) . Define i r = f j + r ( p ) f o r i 0. . . . . i k are distinct integers in {1. . . . . n} and definition of f it follows that for every r ~ cases 1 and 2 prove the in 9 , where k ~> 1, such all r = 0 , . . . , k + 1. Then ik+l = il. Also, from the {0. . . . . k}, airir+~ ¢= 0 and (4,1xi3)I,]>~we]xslobtainf°r all s ~ {1. . . . . n} -{i,,} satisfying air s --/=O. Hence from k k 1-~ I* -- airir[ IXG[ < 1-I a,~lxir+l[. r=O r=O (4.5) k k Now (4.2) follows from (4.5), since 17Ir= o lX GI = l--Ir=0lxir+ 1. This proves the theorem. • Although Theorem 4.2 shows that a matrix A ~ , , , does not have to be weakly irreducible so that each of its eigenvalues lies in a k th order Gershgorin region, the following theorem shows that a matrix A ~ r , must be weakly irreducible if none of its diagonal entries is an eigenvalue of A. THEOREM 4.3. Let A = (a 0 ~ ¢ , , , where n >~ 2. I f a , q~ tr( A) f o r all i = 1 . . . . . n, then A is weakly irreducible. Proof. Let P1 . . . . . Pn be the nodes in the directed graph F(A) of A. We consider the cases n = 2 and n >t 3 separately. GERSHGORIN'S THEOREM 113 C a s e 1: n = 2. As a u , a2,2 ~ o r ( A ) , the c h a r a c t e r i s t i c p o l y n o m i a l of A satisfies ( a - a n X a - a22) - a12a21 4= ( a - a H ) ( a -- a.22). T h u s ale =/= 0 a n d a21 =~ 0. H e n c e A is w e a k l y i r r e d u c i b l e . C a s e 2: n > 3. W e p r o v e the t h e o r e m in this case b y s h o w i n g that if A is not w e a k l y i r r e d u c i b l e , t h e n o - ( A ) N {a u . . . . . a,,,,} 4= Q . So s u p p o s e that A is not w e a k l y i r r e d u c i b l e . T h e n t h e r e exists a n o d e P~ in F ( A ) such that P< does not b e l o n g to a nontrivial cycle. L e t A 1 = E A E ~r = (a{})), w h e r e E = I,, if 0¢1 = 1, E,(I, al) otherwise. L e t p}l) . . . . . Pt} 1) b e the n o d e s in the d i r e c t e d g r a p h F ( A 1) of A p Since P~, d o e s not b e l o n g to a nontrivial cycle, it does not b e l o n g to a nontrivial cycle in F ( A I ) . W e c o n s i d e r the following two eases: (i) E i t h e r a~)) = 0 for all j = 2 . . . . . n or a}l ~ = 0 for all i = 2 . . . . . n. In this case w e have a < ~ , = a{~} ~ c r ( A 1) = o ' ( A ) . (ii) T h e r e exist r, s ~ {9,. . . . . n} such t h a t a~ti) # 0 a n d a ~ ~ # 0. In this case the sets O1 = {piO) : i # 1 a n d t h e r e exists a d i r e c t e d p a t h from p~l) to pi(i)} and 0 2 = {P}I) : i # 1 a n d t h e r e exists a d i r e c t e d p a t h from Pi(1~ to P}'~} are n o n e m p t y . Also, since p}l) does not b e l o n g to a O l c3 0 2 = Q. H e n c e c a r d O l + c a r d 0 2 ¢ n {p~l, . . . . . p,(])} w h e r e s 2 < ... < s k, a n d d e n o t e w h e r e r . . . . . < "-- < r , , . Since c a r d 0 1 + c a r d O ~ m ~< n - 1. F o r every, i e {2 . . . . . k} a n d j ~ {n - Ei = nontrivial cycle, w e have - 1. D e n o t e O t bv O o b y {E(1),, . . . . P},I;} K n - 1, w e have k + m . . . . . n} d e f i n e I,, if E,,( i, s i) otherwise ,s'i = i, I,, if E,,(rj, j) otherwise. and Fj = t)=j, 114 F . O . FARID L e t A 2 = F, • "" F, _ reek "'" E 2 A 1 E T "'" E~FT_m "'" F~T = ( a i(2) j ) , arid let el2) . . . . . Pn(z) be the nodes in the directed graph F(A 2) of A 2. Since O 1 ¢qO 2 = ~ , we have t~ij ~ ( 2 ) = 0 for 2 ~<i~<k and n - m ~ < j ~ < n . Also, since p~2) does not belong to a nontrivial cycle, we have a~.) = 0 for n-m~<j~<n and a~ ) = 0 f o r 2 ~<i ~<k. I f k + m = n 1, then A 2 has the block matrix form IAll o) Az ~ A21 A22 , where all the entries in the first column of A n below ",,t2) u are zeros. Hence a,,~ 1 = a~ ) ~ o'(A u ) c cr(A 2) = tr(A). If k + m < n - 1, then for every node p(1) in F ( A l) satisfying p(l)j ~ {p}l)} U ®1 tJ O 2 we have al}) = 0 and a}1) = 6 for all i ~ {1, s 2 . . . . . s k} and l ~ { r , _ . . . . . . rn}. Thus A 2 has the block matrix form A2 Au 0 0 / A21 Aa A22 A~2 O /' A33 where all the entries in the first column of A u below ~u"(2)are again zeros. Hence a .... = a~21) is a point of the spectrum of A. This completes the proof of the theorem. • The following theorem establishes sufficient conditions for k eigenvalues of A ~ r , n >/2, to be in its kth order Gershgorin region, where k {2 . . . . . n}. THEOREM 4.4. L e t A = ( a i j ) ~ , , ~ , w h e r e n >. 2. S u p p o s e there exist k (/> 2) distinct integers m 1. . . . . m k ~ {1. . . . . n} s u c h t h a t f o r every r {1. . . . . k} one has amr m +1 ~ 0 a n d amd = 0 f o r all j ~ {1. . . . . n} { m r , mr+ 1}, w h e r e m k + 1 Zr m p T h e n there exist k eigenvalues A1, . . . , A k o f A ( c o u n t i n g multiplicities) s u c h 1 . . . . . k} f o r all r = 1 . . . . . k. that 1~ r ~ Gk(A) and A r q~ {a,,,jmj: j = Proof. By taking E A E T, where E is a permutation matrix, we can assume without loss of generality that m r = r for all r = 1 . . . . . k. Thus ar, r + 1 5~: 0 and ars = 0 (4.6) 115 GERSHGORIN'S THEOREM for all r = 1 . . . . . k - 1 a n d s E {1 . . . . . n} - {r, r + 1}. Also w e have akl 4:0 and ak,~ = 0 (4.7) for all s ~ {1 . . . . . n} - {1, k}. I f k < n, t h e n A m a y b e w r i t t e n in the block matrix f o r m All A = 1 A21 0 ) A22 ' w h e r e Al l E.~¢k. In this case w e have o - ( A ) = o ' ( A u ) U cr(A22). If k t h e n A({1 . . . . . k}) = A. This p r o v e s o ' ( A ( { 1 . . . . . k})) c ~r(A). ~r(A({1 . . . . . k})) = {A 1. . . . . Ak}. W e show that A t ~ G k ( A ) and {an . . . . . akk} for all r = 1 . . . . . k. L e t r ~ {1 . . . . . k} a n d x = [x 1. . . . . b e an e i g e n v e c t o r o f A({1 . . . . . k}) c o r r e s p o n d i n g to Ar. T h e n A({1 . . . . . k})x = ArX, (4.6), a n d (4.7), w e get (A r -- aii)x i = ai,i+lXi+ 1 = n, Let Ar x~] 'r from (4.8) for all i = 1 . . . . . k - 1, a n d ( l~r -- akk)xk = aklXx" (4.9) T o p r o v e 1~r ~1~ {ajj : j = 1 . . . . . k}, s u p p o s e Ar = a r r for s o m e ri {1 . . . . . k}. T h e n f r o m (4.6) a n d (4.8) w e have Xr, + I = 0 if r / ~ {1 . . . . . k - 1}, a n d f r o m (4.7) a n d (4.9) w e have x 1 = 0 if r i = k. H e n c e b y induction, it follows f r o m (4.6), (4.7), (4.8), a n d (4.9) that x i = 0 for all i = 1 . . . . . k, w h i c h c o n t r a d i c t s t h a t x is an e i g e n v e c t o r o f A({1 . . . . . k}). This p r o v e s that Ar f~ {ajj : j = 1 . . . . . k}. Since t h e r e is at least o n e n o n z e r o c o m p o n e n t of x, it follows, b y k i n d u c t i o n , f r o m (4.8), (4.9), a n d /~r ~ {all . . . . . akk } t h a t I-Ij=llXjl ~ O. H e n c e from (4.8) a n d (4.9), w e get k H i=1 k-I l~r -- aii] = lakll 1-I lai,,+al. i=1 T h u s f r o m (4.6) a n d (4.7), w e o b t a i n Ar ~ G k ( A ) , a n d this c o m p l e t e s the p r o o f o f the t h e o r e m . • 116 F . O . FARID I would like to thank the referees for their careful reading of the manuscript and their useful suggestions. REFERENCES 1 A. Brauer, Limits for the characteristic roots of matrices II, Duke Math. J. 14:21-26 (1947). 2 R.A. Brualdi, Matrices, eigenvalues, and directed graphs, Linear and Multilinear Algebra 11:143-165 (1982). 3 R.A. Horn and C. R. Johnson, Matrix Analysis, Cambridge U.P., 1985. 4 Morris Newman and Robert C. Thompson, A counterexample connected with Gershgorin's theorem, Linear Algebra Appl. 201:165-169 (1994). Received 28 September 1995;final manuscript accepted 23 January 1997