MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that a $C^*$-algebra $A$ has real rank zero iff every self-adjoint element of $A$ can be approximated in norm by self-adjoint element with finite spectra. My question is:

If we have two self-adjoint elements $T, S$ in $A$ with the same spectrum(may be infinite), Now can we also find two self-adjoint elements in $A$ with the same finte spectrum approximated $T, S$ in norm( within the same $\epsilon>0$)?

Hope some help or suggestion, thanks!

share|cite|improve this question
Do you want any relation between $S$ and $T$ and between the two approximating elements? Are you referring to the joint spectrum of commuting elements? If not, then there is no difficulty approximating two elements instead of one, so I suspect I am not following you. – Terry Loring May 8 '12 at 15:45
Thanks, I can prove this. – Aviv May 9 '12 at 9:36
Do not edit your questions so that they remove meaningful information. I am reverting this to your previous version. – Yemon Choi May 20 '12 at 12:08
If you have since solved the problem, add this extra information as an update. Please do not delete the record of what you originally asked. – Yemon Choi May 20 '12 at 12:11

Yes, you can jiggle the two approximating operators so that they have the same finite spectrum. Use functional calculus.

share|cite|improve this answer
Thanks, Can you say any more, or some reference? – Aviv May 8 '12 at 16:14
You need to learn about functional calculus. If you didn't really care that the two approximating operators have the same spectrum (as I gather from your response to Terry's answer) then you don't need functional calculus, you need to brush up on epsilon-delta arguments. – Nik Weaver May 8 '12 at 18:09
Thanks, I can prove this. – Aviv May 9 '12 at 9:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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