2
$\begingroup$

It is well known that a matrix which all entries are positive real numbers, has a positive eigenvalue.(see algebraic topology, by Allen Hatcher). Now is the following generalization, true? Let A be a C* algebra and X is a matrix which entries are positive elements of A. Does sp(X) has nonempty intersection with positive real line?

$\endgroup$
  • 7
    $\begingroup$ Ali, I notice that you never accept answers to your questions. Why don't you show some courtesy to people who are going out of their way to help? $\endgroup$ – Nik Weaver Nov 28 '13 at 19:05
  • 1
    $\begingroup$ Nik, I was not realy aware of "accept" mark on answers.I did not have intension to not show courtesy to people who answer my question.I thank you for inform me of this "accept" mark. but just a question: could not you send me a personal message(email) for this subject? thanks again for your comment $\endgroup$ – Ali Taghavi Dec 2 '13 at 17:29
5
$\begingroup$

The answer is no. For example, the 4-by-4 matrix $\begin{pmatrix} p & e \\ f & q \end{pmatrix}$, with $p=\begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix}$, $q=1-p$, $e=\begin{pmatrix} 1/2 & 1/2\\ 1/2 & 1/2\end{pmatrix}$, and $f=1-e$ has the characteristic polynomial $x^4-2x^3+x^2+1/4=(x^2-x)^2+1/4\geq1/4$ and has no real eigenvalues.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

However, the answer is yes if the entries of $X$ commute. Then you can treat them as continuous functions on some LCH space, and evaluating at any point of that space gives you a scalar matrix with positive entries. Any eigenvalue of any of these matrices will belong to the spectrum of $X$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So it is natural to ask :Is the property under question(for all matrix of all size) is a sufficient condition for commutativity? $\endgroup$ – Ali Taghavi Dec 3 '13 at 6:38
  • $\begingroup$ Yes, and given Taka's counterexample where $A$ is the $2 \times 2$ matrices, I think it is likely to be true. $\endgroup$ – Nik Weaver Dec 3 '13 at 18:05
  • $\begingroup$ 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? $\endgroup$ – Ali Taghavi Dec 3 '13 at 18:19
  • 1
    $\begingroup$ Ali, you need to modify the conclusion of your question to that $A$ has a nonzero commutative quotient. It's plausible. $\endgroup$ – Narutaka OZAWA Dec 4 '13 at 0:29
  • $\begingroup$ Narutaka, Thanks for your comment. I will think to this modified version which you suggested. $\endgroup$ – Ali Taghavi Dec 4 '13 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.