Can we say $TT^{*}=T^{2}$ implies $T=T^{*}$?

Question

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

Answer 1

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 $H$.