Afgan aslanov
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- Bu səhifədəki naviqasiya:
- 1 Introduction
- 2 Non-negative tensor determinants (”weak” right definite case) Definition
- 3 Positive definite tensor determinants Theorem
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Definiteness Conditions in the Multiparameter Spectral Theory AFGAN ASLANOV Department of Computer Engineering Fatih University Buyukcekmece, Istanbul TURKEY
Abstract: - Definiteness conditions for multiparameter eigenvalue problems are considered. Multiparameter eigenvalue problem  where   are self-adjoint, bounded operators on Hilbert space    could be learned through the commuting system of self-adjoint operators acting on  if the tensor determinant  is strongly positive on the tensor product  (that is, if problem is right definite). It has been proved that if   and  for some   , then  on [3] (for infinite dimensional case see [4,6]). In general, the positivity of  in the case   is open. In the case  problem has been solved in [2]. In this paper we solved this problem for  . We solved some problems related with different definite conditions as well.
Key-Words: - positive operators, tensor determinants, multiparameter eigenvalue problems 1 Introduction The multi-parameter eigenvalue problem of finding =(1,…,n) such that   (1) has been considered by many authors. All authors impose some kind of ”definiteness condition” on the array of operators(Bjk ). We form an operator on the tensor product H=H1…Hn by defining x for a decomposable tensor x=x1…xn as x=det(Bjkxj), j,k=1,2,…,n where the determinant is to be expanded formally using the tensor product. is then extended to all of H by linearity and continuity. We shall denote this operator simply by =det(Bjk).Similarly, we define operators j, which obtained by replacing j-th column of by the column (A1,A2,..,An).
If is strongly positive operator on H, then problem (1) can be learned through the (commuting) system of operators {-1j} on Hilbert Space 2 Non-negative tensor determinants (”weak” right definite case) Definition: A linear operator  is said to be (i) positive definite, denoted by  if  , (ii) strongly positive definite, denoted by  if  such that  , (iii) positive (or non-negative), denoted by  if  for all  . The next theorem seems new one even in finite dimensional case.
Theorem 1. If  for all  then 
Proof. There are two possibilities: a) b) Proof a). We use mathematical induction. If  this statement is trivial. Suppose that the statement is true for the  determinants and we must prove for the  determinants. Let  . If there exists  such that   , , then it is clear that  (One column may be replaced by zero operators). So we assume that  for all and at least for one 
Let  be any fixed non-zero element from  and  Let, for simplicity,  (if all  for all  then immediately we have  for all  and  By induction we have 
 since  is determinant in fact, and  for all decomposed tensors  from  where   Let’s consider the determinant  where  The cofactor of first entry in this determinant is zero by the induction. Now we find  from such that  at least for one of  (otherwise, we obtain immediately,  . The replacement    changes just the cofactor of  in the first row So we can apply the same procedure again for the determinant  Let’s assume that  and replace  for  and  After this replacement the cofactor of the second entry in the first row will be zero, but all other cofactors are the same. Continuing this process we make all cofactors of first row zero (or operator itself in the first row is zero). This means that  , where  is the linear combination of  and  is the algebraic cofactor of  Thus we proved that if  for all then 
b) Again we use the mathematical induction. The case  
All determinants on the right hand side are non-negative on decomposed tensors and Now we consider the relationship between different definite conditions.
Proof  where  and  Then the number of non-strong inequalities in (2) for the determinant will be at most  and therefore  or  by induction. On the other hand determinant
 also will be 0 for >0 since the number of non-strong inequalities in (2) for this operator is less than or equal to  :  Thus we have that  or and the Theorem 2 is proved.
3 Positive definite tensor determinants Theorem 3. If 
then for any  and >0 at least for one j.
Proof. Let If there exist vectors Proof. For any given 
Let us show that 
if Now we consider the case Let’s consider the case b) and let 
 =0 and  Now we have     As we demonstrated at the beginning of the proof of this theorem if some linear combination of operators in some row is zero then positivity in decomposed tensors and in whole space are equivalent. That is  and Theorem is proved.
References:
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. This leads to the study of what may be appropriately called ”right definite” problems. It has been proved that if 
, for 0
, then is strongly positive on H [3,7]. In general, question of the positivity of in the case 
is open. In this paper we solve this problem for n=3. 
at least for one decomposed tensor 
such that 
. Consider new operator 
where 
for all decomposable tensors 
Indeed, 
can be represented in the form 
by induction. But last one is greater than zero by the definition of 
. Thus 
on decomposed tensors 
and therefore, 
[4,6]
This means that 
there exists 
(2)
or 
the number of non-strong positive operators in (2). If 
, that is the number of operators with non-zero kernel is 0, then it is clear that 
any finite dimensional subspace of 
and 
is the orthoprojector on to
then 
or 
on 
[3], and therefore 
and try to prove in the case 
. Let 
and for instance, let 
has non-zero kernel and consider determinant 
such that for all 
Then 
is strongly positive on 
and therefore, for 
such that 
on 
[5]. As 
, it is possible to find a convergent subsequence of a sequence 
for simplicity, let 
It’s clear that 
and 
. 
such that 
for all 
and, for instance, 
then first column in can be replaced by 
and so 
for all
.
, 
, there exists 
such that 
and 
at least for one j (for any fixed 
Consider the 3x4 matrix 
where 
In general, there are eight different possibilities for the operators 
but four of them could be received multiplying the columns of the matrix M by (-1). In the case 
for some j and k we can replace one of the columns in by the column 
and so we get in fact 2x2 determinantal 
and 
then 
=
tensors 
For any fixed
and <0 if 
is the 2x2 determinantal operator and 
always positive or always negative for all non-zero decomposed tensors 
. Thus 
or 
on 
and so 
for all 
and 
. There are at least four positive definite (>0) operators among 
and therefore there is at least one row in M with two positive definite operators. For example, let 
and 
Consider two cases a) 
and b) 
or 
In the case a) we take some 
and consider the next tensor determinant 
where 
[2]. There exists an operator 
, such that 
Let us show that 
for small enough 
Indeed, for all m =1,2,3,4 we have 
so that the conditions of the Theorem 2 holds for the operator 
This means that if 
then 
and considering that 
. There exists an operator 
such that 
and 
([1], (2.25)). Now if we consider the operator 
where again 
we have 
m=1,2,3,4. (We get 
using the fact that 
Theorem 2 implies that 
Thus 
and therefore, 
Then we have 
since 
Let 
is a projector onto the 
and consider operator 
