Can a Non-Self-Adjoint Element Have a Real Spectrum?

[jostpuur]

Let X be a C^*-algebra. I know that if x\in X is self-adjoint, then its spectrum is real, \sigma(x)\subset\mathbb{R}. I haven't seen a claim about the converse, but it seems difficult to come up with a counter example for it. My question is, that is it possible, that some x\in X has a real spectrum, but still x^*\neq x?

 

[morphism]

My question is, that is it possible, that some x\in X has a real spectrum, but still x^*\neq x?

Yes it is. Take for instance the 2x2 matrix (so X=M_2(\mathbb{C}))

x = \begin{pmatrix}a & 1 \\ 0 & b\end{pmatrix},

where a and b are any real numbers. The spectrum of x is {a,b} but x is not selfadjoint.

 

[jostpuur]

I see.

(hmhmhmh... I didn't receive mail notification of your response...)

 

[HallsofIvy]

You will only get e-mail notification if you "subscribe" to a thread. To do that, clilck on "Thread Tools" at the top of the thread, then click on "Subscribe to this Thread".

 

[soarce]

I am facing the same problem (see https://www.physicsforums.com/showthread.php?t=257751)

On the internet I found a reference, however I don't have acces to it:
http://oai.dtic.mil/oai/oai?verb=getRecord&metadataPrefix=html&identifier=AD0736116"

In general you can not say anything about the eigenvalues of a real (unsymmetric) matrix. However, if you can write your matrix as a product of matrices then analyzing them you may say something about the eigenvalues of the big matrix.

I put here two articles, maybe you will find them usefull.

 

[jostpuur]

You will only get e-mail notification if you "subscribe" to a thread. To do that, clilck on "Thread Tools" at the top of the thread, then click on "Subscribe to this Thread".

But isn't the subscribing automatic, so that one has to unsubscribe a thread if one doesn't want notifications. I didn't do anything with thread tools, and I got the notification of your post now.

There is a non-zero probability for the possibility, that I casually destroyed the first notification without later remembering it. I cannot know it for sure, of course... I was merely mentioning the remark anyway.

 

[HallsofIvy]

When you initially join this forum you are offered the option of automatic "subscription" or not. I chose not because I don't want an e-mail everytime someone responds to one of the threads I responded to. I can't delete all those e-mails AND respond to questions!