Can we say $TT^{*}=T^{2}$ implies $T=T^{*}$? Let $A$ be a $C^{*}$-algebra, Can we say  $TT^{*}=T^{2}$ implies $T^{*}=T$? for $T\in A$
I am looking for a counterexample!
Thanks

Via a faithful state, we can think of $A$ as represented in some $B(H)$. We have $$ H=\ker T\oplus \overline{\text{ran} T^*}. $$
For $x\in\ker T$, we have $Tx=0$ and then
$$
\|T^*x\|^2=\langle T^*x,T^*x\rangle=\langle TT^*x,x\rangle=\langle T^2x,x\rangle=0;
$$
so $T^*x=0$ and $T=T^*$ on $\ker T$. 
Taking adjoints on $TT^*=T^2$ we have $TT^*=T^*T^*$, that is $(T-T^*)T^*=0$. This shows that $T-T^*=0$ on $\text{ran}\,T^*$, and by continuity on its closure. That is, $T=T^*$ on $\overline{\text{ran}\,T^*}$.
So $T=T^*$ on all of $H$. 
