# the spectrum of matrix with positive entries

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?

• 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? – Nik Weaver Nov 28 '13 at 19:05
• 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 – Ali Taghavi Dec 2 '13 at 17:29

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.
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 $X$.
• Yes, and given Taka's counterexample where $A$ is the $2 \times 2$ matrices, I think it is likely to be true. – Nik Weaver Dec 3 '13 at 18:05
• Ali, you need to modify the conclusion of your question to that $A$ has a nonzero commutative quotient. It's plausible. – Narutaka OZAWA Dec 4 '13 at 0:29