Let $A$ be a $C^∗$-algebra, $a \in A$ and let $p,q \in A$ be orthogonal projections (i.e. selfadjoint idempotents with $pq = 0$). Suppose that a is positive and $pap = 0$. Show that $paq = 0$
Could you check my solution of following problem ( i am stressed a little due to the operator $q$)?
My solution: so, there is the positive element $b$ that $bb=a$. Therefore $pap=pbbp=pb(bp)^*=0$, then using the $ C^*$ property we can conclude $pb=0$
The only thing I can see that is wrong when you claim $pbbp=pb(bp)^*$. What you should write is $pbbp=pb(pb)^*$. (It's not actually wrong, because you actually prove that $pb=0=0^*=bp$, but, you know, deduction.)
And as far as my proof-writing style is concerned (and to make this more than a two-line answer), you should write $$pap=pbbp=pb(pb)^*=0$$ as $$0=pap=pbbp=pb(bp)^*,$$ so that it's logically consistent.
Your argument is the standard way to deal with this, with the minor exception mentioned by Aweygan.
The probable reason the problem is formulated this way is, that if you write $a$ as a $2\times 2$ matrix with respect to $p$, you are showing that if $a_{11}=0$ then $a_{12}=a_{21}=0$.
