Let two selfadjoint operators A and X in a Hilbert space with a common dense domain D(A)=D(X), A>0, satisfy in H the anti-commutator equation
{A,X}:=AX+XA=0.
It is true that X=0?
Dear Prof. Prykarpatski,
I think the answer is "no". For example. H=C^{2), A=Projection on x-axis and X=projection on y-axis. Then AX=XA=0. But I guess that you consider more restriction on, A, or X . Am I missing something?
Cite
Anatolij K. Prykarpatski
Lviv Polytechnic National University
Dear Ali,
you are right - a Hilbert space H should be 1) infinite dimensional, and 2) the operator A should be positive. To the regret, it is not the case in your example....
Regards!
Cite
Anatolij K. Prykarpatski
Lviv Polytechnic National University
Dear Peter,
under self-adjoint I mean ... self-adjoint, as the Hermitian operator is only symmetric and nothing more, not conserving the dense domain D(X)⊂H. Not always operators A and X possess the regularized determinants, as they can not belong to the trace-class ...Concerning the infinite dimensionality - it is well known that only compact operators are well approximated in Hilbert spaces by means of matrix operators, yet it is not for selfadjoint and not compact ones...
Sincerely, regards!
Cite
Anatolij K. Prykarpatski
Lviv Polytechnic National University
Dear Peter,
to my regret, - no, (Ax,y)=(x,Ay) means only that the operator A is symmetric (the same, it is Hermitian), yet not self-adjoint, - the self-adjointness in Physics means that A*=A and Dom(A)= Dom(A*) , that is the domain of the operator A , Dom(A) coincides with the domain Dom (A*)⊂H (see a good manual by M. Reed and B. Simon "Modern Mathematical Physics", vol. 1).
Next, you matrix A= (10;0-1) is not, to the regret, positive definite... All above fails then.
Regards!
]
Cite
Dinu Teodorescu
Valahia University of Targoviste
Dear Prof. Prykarpatski,
Clearly the Hilbert space H should be real, because the operator A is positive. From AX+XA=0, we obtain (AXy,y)=-(XAy,y) for all y in D(A)=D(X). Consequently (Xy,Ay)=-(Ay,Xy). Therefore (Xy,Ay)=0 for all y in D(A)=D(X). (1)
Now it is sufficient to prove that //Xy//=0 which is equivalent to (X2y,y)=0 for all y in D(A)=D(X). (2)
Maybe using (1) and A>0 it is posiible to prove that (2) is true.
I will think about.
Regards!
Cite
1 Recommendation
Anatolij K. Prykarpatski
Lviv Polytechnic National University
Dear Peter
you have missed the info - please read carefully my first post: there is written: the self-adjont operator A>0... (that is it is positive); remark: the Hilbert space, evidently, should NOT be real, it can be arbitrary...
Sincerely, regards!
Cite
Anatolij K. Prykarpatski
Lviv Polytechnic National University
Dear Dinu,
Remark: the Hilbert space, evidently, should NOT be real, it can be arbitrary... Anyway, I like you way of thinking!
Sincerely, regards!
Cite
Anatolij K. Prykarpatski
Lviv Polytechnic National University
Dear Peter,
As I could understand your post, you are joking in such a very amazing way! :-) :-) :-)
Anyway,
Sincerely regards!
P. S. Concerning the "infinitedimensionality", this property was added only owing to the fact that in the finite dimensional case the answer is trivial, and the problem has sense only in the case of an infinite dimensional Hilbert space H. :-) :-) :-)
Cite
Ali Taghavi
Qom University of Technology
Dear Prof. Prykarpatski.
Your very interesting question is a motivation to ask the following question in the context of C* algebras:
Question: Let A be a complex C* algebra and "a" is an invertible positive element. Define a linear map T on A with T(x)=ax+xa. Is T a positive operator?Is T completly positive?Is T an injective operator?
The latest question is actually your question. As we know it has an affirmative answer in finite dimensional case that is for A=Matrix algebra M_{n}(C). There is an alternative prove for matrix algebra, in terms of tensor product: The matrix representation of T is in the form $a\otimes Id+Id \otimes a$, this is a sum of two positive invertible element of M_{n}(C) \otimes M_{n}(C). So it is invertible, hence the operator T is invertible, in particular is injective. So in finite dimensional case, in your formulation of the problem, we do not need to extra assumption "X is selfadjoint".
BTW: The operators which I exampled in C^2(In my previous answer) are positive but not strictly positive. On the other hand this example can be modifies on infinit dimension since H is isomorphic to H+H(The first H play the role of x-axis, and the second plays the role of y-axis.
Best Regards
Dear Peter,
Please, explain what modification in the question are You speaking about??? I did Not find them! :-) :-) :-)
I can really not understanf You! :-) :-) :-) Or, You don't like the problem posed and you are continuing to joking such an original way? :-) :-) :-)
My remarks about the infinite dimensionality are Not necessary, they only served to help You to look at the problem a bit deepper! Because in the finite dimensional (matrix) case the question is TRIVIAL!
Concerning positivity of the operator A in H, that is A>0, this condition is formulated in the post and there is no sense to speak more about!
My cordial regards!
Dear Ali,
I like your extension of the post question - it really can be a root of a nice problem!
Sincerely,
Regards!
P. S. Concerning your finite dimensional projector operators, they by definition are always digenerate, so they can be in no way positive, being nondegenerate.:-)
Dear Peter,
Thanks for your kind remark. Then, if the problem, really interesting and important, is trivial for finite dimensional Hilbert spaces, it would be nice to analyze it from general point of view in the infinite dimensional case of H, and the algebraic approach suggested above by Prof. Ali, is looking very promising.
Thanks once more and regards!
Dear Peter,
What are you writing about??? I could not catch your so original way of thinking... We are discussing here interesting math problems, not some mistifications, appearing in mind... I am filling bad reading all that above - I have met for the first time during past hundred years such a strange and far from being professional behavior. Sorry, let's stop this out- of-mathematics activity!-:)
All the best
and sincerely regards!
P.S. Dear Colleagues,
There are also looking interesting enough articles attached!
Regards!
Dear Prof. Prykarpatski,
According to my previous answer, I think that we can not expect that the operator T(x)=ax+xa is a positive operator where a is a positive element. I realize that there is a counterexample in M_{2}(C). There are two positive matrices A and B such that AB+BA is not positive. Ex: A=\pmatrix(1&0 / /0&2) and B=\pmatrix(10.01& 10 \\ 10& 10.01). In fact the following general statement is true:
Statement: If A is a C* algebra such that ab+ba is a positive element for all positive elements a and b then A is a Commutative C* algebra.
This is a consequence of the following theorem(which I forget the paper and its Author but it is quoted in the book "K-theory and C* algebra by W. Olsen):
Theorem: Assume that A is a C* algebra such that 0
P.S. Dear Ali,
yet, matrix B is looking not positive.... Any way, thanks for your kind remarks.
A.
Dear Prof. Prykarpatski
May be I am missing some thing. But I think B is positive since it is symmetric and has positive determinant and positive trace.
What you wish to prove is that the map
W_A: X->AX + XA
has empty kernel. For convenience consider the case in which A has discrete spectrum. Then there is an orthonormal eigenbasis phi_n of A. X can then be described by its matrix elements X_{mn} wrt this basis. In this representation the map becomes
X_{mn}->(lambda_m + lambda_n) X_{mn}
which, due to the positivity of A, clearly has empty kernel.
The extension to continuous spectrum should be straightforward using the spectral theorem.
Dear Prof. Prykarpatski
The following MO conversation is, somehow, related to the subject of our conversation:
http://mathoverflow.net/questions/246827/for-positive-definite-a-b-why-does-abba-tend-to-be-positive-definite/246844#246844
Best Regards
Ali Taghavi
- Badges
- Science topic
- Similar topics
- Linguistics
- Composition Studies
- Publication