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This is an exercise in Murphy's book:

Let $A$ be a unital $C^*$-algebra and $a,b$ are positive elements in $A$. Then $\sigma(ab)\subset\mathbb{R}^+$.

The problem would be trivial if the algebra is abelian. On the other hand I do not have a clue for the non-abelian case. I guess one needs to use the fact $\sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\}$ and then maybe use some algebraic manipulation.

Anyway, I wonder whether someone has a hint on this. I guess I am missing a trick.

Better though, maybe someone has some general insight on the many techniques concerning positive elements and approximate identities. For me they all seem very tricky and mysterious. For instance, how can they think of those strange functions when proving the existence of approximate identities and quasicentral approximate identities.

Thanks!

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"...how can they think of those strange functions...": You could probably turn that into a separate question, with added detail. –  Jonas Meyer Jul 9 '12 at 2:16
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I agree with @JonasMeyer. My own slightly cynical view (as someone from Banach algebras rather than C*-algebras) is that one eventually sees enough of these tricks that one starts to use them on one's own. Either that, or you find that someone like Arens had the clever idea first in the 1950s. –  user16299 Jul 9 '12 at 2:32
    
@YemonChoi Well, I guess trying to see enough tricks is the only way, so do you have some good books to recommend? I am reading Murphy's and I think it's a pretty good book, just at the right level for me. But it always helps me to read several books on the same topic. I checked Arveson's An Invitation to $C^*$ Algebras and felt it might be too deep for me now. I wonder whether there are good books between them. I have a solid background in analysis but only have average experience with algebra. Thanks! –  Hui Yu Jul 9 '12 at 3:03
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2 Answers

up vote 5 down vote accepted

A common trick is to write a positive element of a C*-algebra as a square of a positive element: $b=\sqrt b\sqrt b$. Thus $\sigma(ab)\cup\{0\}=\sigma(a\sqrt b\sqrt b)\cup\{0\}=\sigma(\sqrt b a\sqrt b)\cup \{0\}$. If you haven't already seen it, you can think about why if $a$ is positive, then $xax^*$ is positive for all $x$.

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Ha! Got it! Thanks! –  Hui Yu Jul 9 '12 at 2:57
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$$\sigma(ab)\subset \sigma(a)\sigma(b)\subset \mathbb R^+$$

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This is typically incorrect when $ab\neq ba$. For example, consider the $2$-by-$2$ matrices $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}$. As noted in the question, "The problem would be trivial if the algebra is abelian." –  Jonas Meyer Feb 5 '13 at 17:25
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