If $p^{2}=p^{*}=p$ and $pa^{*}a=paa^{*}=0$, then $a^{*}p+pa=0$.

Question

Suppose that $p$ is a projection (i.e. $p^{2}=p^{*}=p$) in a C*-algebra $A$. Let $a\in A$ be an element such that $pa^{*}a=paa^{*}=0$. I want to prove that $$a^{*}p+pa=0.$$ I tried to express $a$ in terms of $p$, but I dont know how to split $a$ from the given identities $pa^{*}a=paa^{*}=0$. Also I think that it possible to prove the (stronger) statement that $a^{*}p=pa=0$. Any suggestions would be greatly appreciated.

Answer 1

Note that $(pa)(pa)^* = paa^*p = 0$. However, we have $$ \|(pa)(pa)^*\| = \|pa\|^2 = 0. $$ Thus, we have $pa = 0$, as desired.