1. Oct 12, 2008

### 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 $$x^*\neq x$$?

2. Oct 12, 2008

### morphism

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.

3. Oct 13, 2008

### jostpuur

I see.

4. Oct 13, 2008

### HallsofIvy

Staff Emeritus
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5. Oct 13, 2008

### 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:
Real Eigenvalues of Unsymmetric Matrices

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.

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Last edited: Oct 13, 2008
6. Oct 13, 2008

### jostpuur

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.

7. Oct 13, 2008

### HallsofIvy

Staff Emeritus
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!