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Let $a$ be a positive element in $A$, where $A$ is a $C^*$-algebra. Let $p\in A$ a projection and suppose $a\leq p$.

Is it true that $ap=pa$? If yes, shouldn't we have $ap=pa=a$, since $(1-p)a(1-p)=0$?

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  • $\begingroup$ What does $a \le p$ mean in a $C^\ast$ algebra? $\endgroup$ – Robert Lewis Feb 4 '14 at 23:31
  • $\begingroup$ $p-a$ positive i.e. there is $x\in A$ such that $p-a=xx^*$. $\endgroup$ – Alessandro Vignati Feb 4 '14 at 23:44
  • $\begingroup$ Thanks. Just wiki'ed up as well, but I like your definition better. $\endgroup$ – Robert Lewis Feb 4 '14 at 23:48
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Yes, like you said, $a$ lies in the corner $pAp$, for which $p$ is the unit.

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  • $\begingroup$ and that's hereditary? That would solve it.. $\endgroup$ – Alessandro Vignati Feb 5 '14 at 1:26
  • $\begingroup$ Corners are hereditary. $\endgroup$ – Michael Feb 5 '14 at 16:02

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