TITLE: $TT^*=T^2$, show that $T$ is self-adjoint
QUESTION [8 upvotes]: Let $V$ be an inner product space, finitely generated over $\mathbb{C}$, $T\in \operatorname{End}(V)$ that satisfies $TT^*=T^2$, show that $T$ is self-adjoint. 

I know that $TT^*$ is positive so has positive square-root, thus the square root is positive definite. But I don't feel like this is a proof. Can anyone prove it without considering square root?

REPLY [13 votes]: If $Tx=0$, then $T^*x=0$ because
$$
          \|T^*x\|^2=(T^*x,T^*x)=(TT^*x,x)=(T^2x,x)=0.
$$
You are given that $T(T^*-T)=0$, which then forces $T^*(T^*-T)=0$ by the above result, and, hence, also $(T-T^*)(T^*-T)=0$. Therefore, for all $x$,
\begin{align}
    \|(T^*-T)x\|^2&=((T^*-T)x,(T^*-T)x) \\
         &=((T-T^*)(T^*-T)x,x) = 0.
\end{align}
Thus, $\|(T^*-T)x\|=0$ for all $x$. So $T^*x=Tx$ for all $x$, or $T^*=T$.