Normal Self-Invertible Operator is Self-Adjoint If $T\in B(H)$ for some Hilbert space $H$, is a normal operator and $T^2=I$, then $T=T^*$. It seemed simple when I first saw the claim, but I'm having trouble showing it. I know that it implies 
$I=(T^*)^2=(T^*T)^2=(TT^*)^2=(T^*T)(TT^*)=(TT^*)(T^*T)$, etc.  
Does the result come from looking at inner products, or is there a nice string of equalities to look at? Or is there a good way to show that $TT^*=I$?
 A: It suffices to show that $\|(T-T^*)x\|=0$, for all $x\in H$. We have
\begin{align}
\|(T-T^*)x\|^2&=
\langle (T-T^*)x,(T-T^*)x\rangle \\&
=\langle Tx,(T-T^*)x\rangle-\langle T^*x,(T-T^*)x\rangle\\&=
\langle x,T^*(T-T^*)x\rangle-\langle x,T(T-T^*)x\rangle \\&=\langle x,T^*Tx-x\rangle-
\langle x,x-TT^*x\rangle=2\|Tx\|^2-2\|x\|^2.
\end{align}
Hence $\|Tx\|\ge \|x\|$ for all $x$, and if  $\|Tx\|= \|x\|$, then $Tx=T^*x$.
On the other hand
\begin{align}
\|(I-TT^*)x\|^2&=
\langle (I-TT^*)x,(I-TT^*)x\rangle \\&
=\langle x,(I-TT^*)x\rangle-\langle TT^*x,(I-TT^*)x\rangle\\
&=\|x\|^2-2\langle x,TT^*x\rangle+\langle TT^*x,TT^*x\rangle\\
&=\|x\|^2-2\langle x,T^*Tx\rangle+\langle T^*Tx,TT^*x\rangle\\
&=\|x\|^2-2\langle Tx,Tx\rangle+\langle Tx,TTT^*x\rangle\\
&=\|x\|^2-2\langle Tx,Tx\rangle+\langle Tx,T^*x\rangle\\
&=\|x\|^2-2\langle Tx,Tx\rangle+\langle TTx,x\rangle\\
&=2\|x\|^2-2\langle Tx,Tx\rangle.
\end{align}
Hence $\|x\|\ge\|Tx\|$, for all $x$, and thus $\|Tx\|=\|x\|$, for all $x$, which implies
that $T=T^*$.
A: You have shown that $P=T^{\star}T$ satisfies $P^{2}=I$. Let $x \in X$ be given, and define $y=x-Px$. Notice that $Py=Px-P^{2}x=Px-x=-y$. Therefore,
$$
        0 \ge -\|y\|^{2}=(Py,y)=(T^{\star}Ty,y)=(Ty,Ty)=\|Ty\|^{2} \ge 0.
$$
So $y=0$, which implies $x=Px$. This is true for all $x$. Thus $P=I$, which means $T^{\star}T=I$. So $T^{2}=I$ becomes $T^{\star}T^{2}=T^{\star}$ or $T=T^{\star}$.
