Prove that $N$ is a normal operator if and only if $\Vert Nx \Vert= \Vert N^{*}x \Vert$, for any x.
For sufficiency, i can't solve. Here are the answers given by others:
$N N^{*}=N^{*}N$ iff $\langle N^{*}Nx,x \rangle= \langle NN^{*}x,x \rangle$ iff $\langle Nx,Nx \rangle =\langle N^{*}x,N^{*}x \rangle$ iff $\Vert Nx \Vert^{2}= \Vert N^{*}x \Vert^{2}$.
But I think '$N N^{*}=N^{*}N$ iff $\langle N^{*}Nx,x \rangle = \langle NN^{*}x,x \rangle$'is wrong, it should be "$N N^{*}=N^{*}N$ iff $\langle N^{*}Nx,y \rangle = \langle NN^{*}x,y \rangle$, for any $x,y$. So I don't have a good proof for sufficiency.