<p> Definiteness Conditions in the Multiparameter Spectral Theory </p><p>AFGAN ASLANOV Department of Computer Engineering Fatih University Buyukcekmece, Istanbul TURKEY</p><p>Abstract: - Definiteness conditions for multiparameter eigenvalue problems are considered. Multiparameter eigenvalue problem n where A , B are self-adjoint, bounded (Aj-k =1a k B jk ) x j = 0 , j = 1 , 2 ..., n , j jk n operators on Hilbert space H j , xj� H j a=( a1 ,..., an ) C could be learned through the commuting system of self-adjoint operators acting on H= H1哪 H 2 ...� H if the tensor determinant D = det( B jk ) is strongly positive on the tensor product H (that is, if problem is right definite). It has been proved that if n dimH j < � j=1 ,..., n and det( Bjk x j, x j )�a j=1 ( x j , x j ) for some a >0 , j=1 , 2 ,..., n , then D >> 0 on H [3] (for infinite dimensional case see [4,6]). In general, the positivity of D in the case det( Bjk x j, x j ) > 0 , x j �0 is open. In the case n =2 , problem has been solved in [2]. In this paper we solved this problem for n = 3 . We solved some problems related with different definite conditions as well. </p><p>Key-Words: - positive operators, tensor determinants, multiparameter eigenvalue problems</p><p>1 Introduction he positivity of  in the case det(( Bjk x j, x j )) > 0 i The multi-parameter eigenvalue problem of finding s open. In this paper we solve this problem for n=3. =(1,…,n) such that n (A-a B ) x = 0 , j = 1 , 2 ..., n (1) 2 Non-negative tensor determinants jk =1 k jk j has been considered by many authors. All authors i (”weak” right definite case) mpose some kind of ”definiteness condition” on the Definition: A linear operator A: D( A ) H is array of operators(Bjk ). We form an operator  on th said to be (i) positive definite, denoted by A >0 , e tensor product H=H1…Hn by defining x for a if (Ax, x ) > 0 , (ii) strongly positive definite, decomposable tensor x=x …x as x=det(B x 1 n jk denoted by A >>0 , if $c > 0 such that j), j,k=1,2,…,n where the determinant is to be expanded formally using the tensor product.  is (Ax, x )� c ( x x ) , (iii) positive (or non-negative), then extended to all of H by linearity and denoted by A �0 if (Ax, x )� 0 for all continuity. We shall denote this operator simply by 0刮x D ( A ) . The next theorem seems new one =det(Bjk).Similarly, we define operators j, whic even in finite dimensional case. h obtained by replacing j-th column of  by the col Theorem 1. If det( Bjk x j, x j ) 0 for all umn (A1,A2,..,An). If  is strongly positive operator on H, then problem 0 刮xj H j , then D �0 (1) can be learned through the (commuting) system Proof. There are two possibilities: -1 of operators { j} on Hilbert Space a) det( Bjk x j, x j ) = 0 for all xj� H j H={ H , ( x , y ) = ( x ,D y )} . This leads to the stud D D b) det( B x, x ) 0 at least for one decomposed y of what may be appropriately called ”right definit jk j j e” problems. It has been proved that if tensor x= x1 � .. � xn det(( Bjk x j, x j ))�a 0 , for 0 xj H j , then  is Proof a). We use mathematical induction. If n =1 strongly positive on H [3,7]. In general, question of t this statement is trivial. Suppose that the statement is true for the (n- 1) x ( n - 1) determinants and we algebraic cofactor of C1k . Thus we proved that if must prove for the nxn determinants. Let det( Bjk x j, x j ) = 0 for all xj� H j then D =0 . det( B x, x ) = 0 , j , k = 1 ,... n jk j j . If there exists b) Again we use the mathematical induction. The n case n =1 is trivial. Let its true for the 0� (a1, a 2 ..., an ) R such that akB jk = 0 , (n- 1) x ( n - 1) determinants and we prove the j=1 , 2 ,..., n , then it is clear that D = 0 (One statement for the nxn determinants. There exists at column may be replaced by zero operators). So we least one v= v1哪 v 2 ... 奈 vn H such that assume that akB jk 0 for all det( Bjk v j, v j )� 0 j, k =1 , 2 ,..., n . Consider new n 0� (a1, a 2 ..., an ) R and at least for one operator Dv(e ) = det(B jk + e c jk ) , where j�{1, 2 ...,. n } cjk=( B jk v j , v j ) , j, k =1 , 2 ,..., n . Let’s prove that Let x1 be any fixed non-zero element from H1, ( Dv (e )x , x ) > 0 for all decomposable tensors and b1k=( B 1 k x 1 , x 1 ) . Let, for simplicity, b11 0 x� H Indeed, D (e ) can be represented in the (if all b = 0 for all x� H then immediately we v 1k 1 1 form have B1k = 0 for all k=1 , 2 ,..., n and D =0) . ec11 I... e c 1n I By induction we have Dx = 1 Dv(e ) = D + B21 ... B 2 n ...... b11.. b 1n b 11 0. 0 B11... B 1ne c 11 ... e c 1 n =�detB21 .. B 2n � , det B 21 B ' 22 . B ' 2 n 0 +ec... e c + ...+ e c ... e c ...... 21 2n 21 2 n ...... since D is (n- 1) x ( n - 1) determinant in fact, an x1 All determinants on the right hand side are non- (Dy , y ) = 0 y d x1 for all decomposed tensors fro negative on decomposed tensors and 0 by induction. But last one is greater than zero by the m H2 �.. � Hn where definition of v . Thus Bjk= B jk -( b1 k / b 11 ) B j 1 , k=2 ,..., n . Let’s consid (Dv (e )x , x )� det( B jk v j , v j )( x , x ) on decomposed er the determinant D = �det(Bjk ) where x, . tensors and therefore, Dv (e ) 0 [4,6] On the B= B . The cofactor of first entry in this deter j1 j 1 D =s -lim Dv (e ) . other hand we have e 0 This y minant is zero by the induction. Now we find 1 fr means that D �0 Theorem 1 is proved. ⅱ om H1, such that b1k=( B 1 k y 1 , y 1 ) 0 at least for Now we consider the relationship between different definite conditions. one of k=2 ,..., n (otherwise, we obtain immedia Theorem 2. If for any e=( e , e ,..., e ) , D = 0) det(B ) 1 2 n tely, . The replacement jk n e k = 1 there exists 0� (a1, a 2 ..., an ) R det(B ) changes just the cofactor of B in the jk 11 such that for all j=1,2,...,n first row. So we can apply the same procedure agai n (2) ejk =1 a kB jk �0 and 0 n for the determinant �det(B jk ) Let’s assume th Then D 0 or 0 in H. ⅱ ⅱ ⅱ . at b12 0 and replace Bjk= B jk - ( b jk / b j2 ) B j 2 Proof We’ll use the next mathematical induction: ⅱ Denote by N the number of non-strong positive op for k 2 and B= B . After this replacement t j2 j 2 erators in (2). If N = 0 , that is the number of oper he cofactor of the second entry in the first row will b ators with non-zero kernel is 0, then it is clear that e zero, but all other cofactors are the same. Continui H (m ) ng this process we make all cofactors of first row ze D 0 or 0 : Indeed if j any finite dimen ro (or operator itself in the first row is zero). This m (m ) sional subspace of H j and Pj is the orthoproject C (m ) (m ) ( m ) eans that D =C1k腄 1 k = 0 , where 1k is the li or on to H j , then P1 �..腄 Pn > > 0 or (m ) ( m ) ( m ) near combination of B11, B 12,..., B 1n , and D1k is the << 0 on H1哪 H 2 ... H n [3], and there (m ) ( m ) (m ) ( m ) fore D =s -lim P1 � ..腄 Pn 0 or 0 . (a1,..., an )� ( a 1 ...,. a n ) It’s clear that n Now we assume that the statement of theorem is tru e a B 0 and |a |=1 . e if N= k and try to prove in the case N= k +1. jk =1 k jk k Let N= k +1 , and for instance, let If there exist vectors x j 0 such that n e a B x =0 , for all j=1 ,..., n , and, for e1 akB 1 k 0 has non-zero kernel and consider d jk =1 k jk j eterminant instance, a1 0, then first column in  can be b I... b I 11 1n j=1 ,.., n replaced by (1/a1 ) akB jk , and so D = detB ... B , where x1 21 2 n (Dx哪 x ...� x 哪 x x . ..�. x ) 0 This ...... 1 2n 1 2 n contradiction completes the proof of Theorem 3. n b=( B x , x ) and T Theorem 4. If det( Bjk x j, x j ) > 0 for all 1k 1 k 1 1 e1k =1 a k (B 1k x 1, x 1 ) > 0 . hen the number of non-strong inequalities in (2) for 0 刮xj H j , j,k=1,2,3, then D > 0 . the determinant D x will be at most k, and theref 1 Proof. For any given e= ( e1m, e 2 m , e 3 m ) , ore D x 0 or 0 by induction. On the other ha 3 1 e jm = 1, there exists (a1m, a 2 m , a 3 m ) R such nd determinant 3 that Cjm= e jm a km B jk �0 j =1 , 2 , 3, and B11+d b 11 I. . B 11 + d b 1n I k =1 B. . B C jm > 0, at least for one j (for any fixed D +d D = 21 2n al x1 m =1 , 2 , 3 , 4). Consider the 3x4 matrix B31. . B 3n M= ( C ), . . . . jk where e1m =1. In general, there are so will be 0 for >0 since the number of non-stron eight different possibilities for the operators g inequalities in (2) for this operator is less than or e 3 ejm a B jk , but four of them could be receive n k =1 km qual to k : e1k =1 ( akB 1 k+ d b 1 k I ) > 0 . Thus we ha d multiplying the columns of the matrix M by (-1).</p><p>D =s -lim( D +d D x ) 0 In the case C = 0 for some j and k we can replac ve that d 0 1 or 0 and the jm Theorem 2 is proved. e one of the columns in  by the column ^ (C1m, C 2 m , C 3 m ) and so we get in fact 2x2 determi 3 Positive definite tensor determinants nantal operator and this follows D > 0 [2]: If for in Theorem 3. If det( B x, x ) > 0, x �0 jk j j j stance, C13 = 0 and a33 0, then a33D = then for any e=( e1 , e 2 ,..., e n ) , e k = �1 $ n B22 C23 B 21 C 23 (a1,..., an ) R such that for all j=1 ,..., n =B11� � B 12 腄 B 11 - 腄 11 B 12 . 12 n B C B C and >0 at least for one j. 32 33 31 33 ejk =1 a kB jk �0 (m ) Proof. Let Pj be a projector onto some m- Let us show that a33D or -a33D is positive on dimensional subspace H (m ) of a space H and j j decomposed on H1哪( H 2 H 3 ) tensors x1 z, (m ) ( m ) (m ) ( m ) Bjk= P j B jk . Then D = det(Bjk ) is x1� H 1 z文 H2 H 3. For any fixed x1� H 1 (m ) ( m ) strongly positive on H1 �.. H n and a33(Dx 1� z � x 1 D z ) , b 11 ( - 11 z D z ) , b 12 ( = 12 z z ) therefore, for "e =( e1 , e 2 ,..., e n ) , e k = 1 there =((b11 D 11 - b 12 D 12 ) z , z ) > 0 , (m ) ( m ) ( m ) n if a >0 , and <0 if a >0 , since bD - b D exists (a1, a 2 ,..., an ) R such that 33 33 11 11 12 12 n (m ) is the 2x2 determinantal operator and (m ) ( m ) on H and ejk =1 a kB jk >> 0 j ((b11D 11 - b 12 D 12 ) z 1� z 2 z 1 z 2 ) always positive (m ) m  |ak |=1 [5]. As , it is possible to find or always negative for all non-zero decomposed a convergent subsequence of a sequence tensors z1奈 z 2 H 2 H 3 . Thus a33D > 0 or (m ) ( m ) ( m ) n {(a1, a 2 ,..., an )} �R for simplicity, let a33D < 0 on H= H1哪( H 2 H 3 ) and so D >0 . t t Now we consider the case C jm 0 for all KerC14 and consider operator P1D P 1 on the</p><p> j =1 , 2 , 3 and k =1 , 2 , 3 , 4 . There are at least four tensor product (PH1 1 ) 哪 H 2 H 3 : positive definite (>0) operators among (C ), and jk PB1 11 P 1 PB1 12 P 1 PB 1 13 P 1 therefore there is at least one row in M with two PtD P t = � B B B positive definite operators. For example, let 1 1 21 22 23 B31 B 32 B 33 C11 > 0 and C12 >0 . Consider two cases a) C13, C 14 > 0 and b) KerC13 or KerC14 {0} t t (Dx , x ) = 0轉 ( P1 P 1 x , x ) =0 and In the case a) we take some 0 刮x H and 1 1 0刮x ( PH ) 哪 H H . Now we have consider the next tensor determinant 1 1 2 3 C11= PC 1 11 P 1 >0 , C12= PC 1 12 P 1 >0 , b11 b12 b 13 D = �B B B C13= PC 1 13 P 1 �0 C14= PC 1 14 P 1 =0 . As we x1 21 22 23 where B B B demonstrated at the beginning of the proof of this 31 32 33 theorem if some linear combination of operators in b1k=( B 1 k x 1 , x 1 ) . As a 2x2 determinant on H2H3 some row is zero then positivity in decomposed tensors and in whole space are equivalent. That is D > 0 [2]. There exists an operator B > 0 , such x1 1 x = 0 and Theorem is proved. that C1k > B 1 , k =1 , 2 , 3 , 4 [1]. Consider operator B-d b B B - d b B B - d b B References: 11 11 1 12 12 1 13 13 1 [1] T. Ando, Extreme points of an intersection of D(d ) = � B21 B 22 B 23 operator intervals. Preceed of the Inter Math conference ’94 Kaohsiung, Taiwan, China, B31 B 32 B 33 World Scientific Publish Company, 1996 = D -d B 腄 . 1 x1 [2] A. A. Aslanov , H. A. Isaev, A theory of two- Let us show that D(d ) 0 for small enough d >0 . parameter spectral problems. Dokl. Akad. Nauk Indeed, for all m =1,2,3,4 we have SSSR 283 (1985) N 5, 1033-1035 [3] F. V. Atkinson, Multiparameter Eigenvalue C1m=e 1 m a km( B 1 k - d b 1 k B 1 ) > 0 , so that the Problems. Vol.1 Academic Press, New York conditions of the Theorem 2 holds for the operator 1972 D(d ) . This means that if (Dx , x ) = 0 , then [4] P. A. Binding, Another positivity results for determinantal operators. Proc. Roy. Soc. ((B腄 ) x, x ) = 0 and considering that D > 0 1 x1 x1 Edinburg, 86A (1980), 333-337. and B1 > 0 we have x=0. [5] P. A. Binding, Multiparameter definiteness conditions. Proc. Roy. Soc. Edinburg, Sect. A Let’s consider the case b) and let x KerC . 1 13 89 (1981), 319-332. There exists an operator B1 �0 [6] P. A. Binding and P. J. Browne, Positivity ran( B1/ 2 )= ran ( C 1 / 2 ) {0} such that C> B , results for determinantal operators, Proc. Roy. 1 14 11 1 Soc. Edinburg, 81A, (1978), 267-271 C12> B 1 , and C14> B 1 ([1], (2.25)). Now if we [7] B. D. Sleeman, Multiparameter spectral theory consider the operator D(d ) , where again in Hilbert Space. Pitman, London, 1978 b1k=( B 1 k x 1 , x 1 ) , k =1 , 2 , 3 , we have 3 m=1,2,3,4. C1m=e 1 mk =1 akm ( B 1 k - d b 1 k B 1 )� 0</p><p>(We get C13 �0 using the fact that 3 Theorem 2 implies k =1a k3b1k =( C 13 x 1 , x 1 ) = 0) . D(d ) 0 ((B腄 ) x, x ) = 0 , that � Thus 1 x1 and t t therefore, B1 x =0 . Then we have C14 x =0 , since 1/ 2 1 / 2 KerB1= KerC 14( = KerB 1 = KerC 14 ) t t � KerB1. KerC 14 Let P1 is a projector onto the</p>