Let $H$ be a Hilbert space. An operator $T\in L(H)$ is said to be normal if $TT^*=T^*T$, where $T^*$ is the adjoint operator. I have to prove that $T$ is normal if and only if $||Tx||=||T^*x||$, for any $x\in H$.
I was able to solve it in the case when $H$ is complex, using the property $$Tx\cdot x=0,\forall x\in H\Longrightarrow T=0.$$ But this is not true in $\mathbb{R}^2$, for example. The rotation of $\pi/2$ would be a counterexample.
My question is: is this still true when $H$ is a real Hilbert space? Maybe there is another simple way of proving it, but I have not been able to figure that out.
Thank you!
Suppose that $H$ is a real Hilbert space $TT^*=T^*T$, $\langle T(x),T(x)\rangle=\langle T^*T(x),x\rangle=\langle TT^*(x),x\rangle=\langle T^*(x),T^*(x)\rangle$.
On the other hand, suppose that $\|T(x)\|^2=\|T^*(x)\|^2$. You have $2\langle x,y\rangle =\|x+y\|^2-\|x\|^2-\|y\|^2$. This implies that for every $x,y\in H$,
$\|T(x)+T(y)\|^2-\|T(x)\|^2-\|T(y)\|^2=\|T^*(x)+T^*(y)\|-\|T^*(x)\|-\|T^*(y)\|$
$=2\langle T(x),T(y)\rangle=2\langle T^*(x),T^*(y)\rangle$. We deduce that $\langle T^*T(x),y\rangle=\langle TT^*(x),y\rangle$ for every $y\in H$ and henceforth $T^*T=TT^*$.