As a consequence of leary hirsch theorem?

Thank you for the answer.you proved that the foliation gives us a fibre bundle, since the holonomy is trivial. Why this fibre bundle is globally trivia?

a good question about R2 plane. But just a question :why long exact homotopy sequence implies that there is no a fibration,We do not know what is the base space

@J.Martel, I agree with you. R^{3}-{0} is foliated by a one parameter familly of 2- spheres. Do you have any Idea on the main question:the foliation of R^3-{0} by torus? Thanks

Narutaka, Thanks for your comment. I will think to this modified version which you suggested.

I dont see how his counterexample gives an idea to proof commutativity. note that my question is the following; Let A be a C* algebra such that the spectrum of each matrix which entries are positive elements, has nontrivial intersection with non negative real number. Is A necessarilly commutative? A related question: what is a general formula for spectrum of an element of M_{n}(A), in term of the spectrum of entries?