TITLE: If $T^2=TT^*$ then can i conclude that $T=T^*$?
QUESTION [5 upvotes]: let $B(H)$ be all bounded operator on Hilbert space H. If $T^2=TT^*$ then can i conclude that $T=T^*$? I think this is true if T is one to one. Can i construct an example that shows it is not true for any T?

REPLY [4 votes]: For a bounded linear operator $T$, we have that $H = \mathrm{ker}(T) \oplus \overline{\mathrm{Im}(T^*)}$.
The two operators $T$ and $T^*$ coincide on $\mathrm{Im}(T^*)$ thanks to the hypothesis: we can rewrite it as $T^* T^* = T T^*$, so if $y = T^*x$, then $$Ty = TT^*x = T^*T^*x = T^*y$$
so they also coincide on its closure (being continuous).
It remains to show that they coincide on $\mathrm{Ker}(T)$, in other words that $T^*$ is zero on $\mathrm{Ker}(T)$. But if $Tx = 0$, then:
$$\langle T^*x, T^*x \rangle = \langle x, TT^*x \rangle = \langle x, T^2x \rangle = 0$$
Therefore $T^*x = 0$. QED.