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 $TT^*=I$?
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 $T=T^*$.
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 $T=T^{\star}$.
