7.20 on page 213 of linear algebra done right,
An operator $T \in \mathscr{L}$ is normal if and only if $||Tv||=||T^*v||$ for all $v \in V$.
For one direction of the proof,
suppose $T$ is normal $\implies T^*T-T^*T=0\implies \langle(T^*T-T^*T)v,v\rangle=0\implies \langle T^*Tv,v\rangle=\langle TT^*v,v\rangle \implies||Tv||^2=||T^*v||^2$.
I can't understand the implication of $\langle T^*Tv,v\rangle=\langle TT^*v,v\rangle \implies||Tv||^2=||T^*v||^2$, I was think Im missing the piece that shows $\langle T^*Tv,v\rangle=\langle TT^*v,v\rangle\implies\langle Tv,Tv\rangle=\langle T^*v,T^*v\rangle$,
Here is what Ive tried, $\langle T^*Tv,v\rangle=\langle v,(T^*T)^*v\rangle=\langle TT^*v,(T^*T)^*v\rangle=\langle TT^*v,TT^*v\rangle=||TT^*v||^2=||T^*Tv||^2$
Below is the proof of this result from the fourth edition of Linear Algebra Done Right (to be published around December 2023, plus or minus a few months). Perhaps the extra details included here will be helpful to you.