Trouble understanding proof that linear map $T$ is normal if and only if $\lVert Tv\rVert=\lVert T^*v\rVert$ for all $v$

Question

I'm having trouble understanding the last step in the proof given in Axler's LADR. He proceeds as follows:

Here (7.16) is the following:

I don't understand 2 things:

a)Do we even need (7.16) to establish the 2nd equivalence? By definition of normality and $T$ being normal (from the statement of the theorem), $T^*T-TT^*$ is the $0$ operator and hence the inner product is equal to $0$ (because it is a product of a vector with $0$). I don't see what the point of (7.16) here is. And even if it were, how exactly WAS it used? The property says that if the operator $S$ is self-adjoint (in our case $T^*T-TT^*$) and $\langle Sv,v\rangle$, the operator is $0$. Well the operator is already established to be $0$ by $T$'s normality and $\langle (T^*T-T^*T)v,v\rangle=0$ as a consequence of that, so what's the intent here?

b)It is completely unclear to me how the final equivalence is established (i.e. how we get from $\langle T^*Tv,v\rangle=\langle TT^*v,v\rangle$ to $\lVert Tv\rVert^2=\lVert T^*v\rVert^2$

Answer 1

Credits to @Cardioid_Ass_22 for the complete answer in the comments:

As the author says, both directions of the proof are being done at the same time i.e. alongside showing that $T$ is normal implies $||Tv||=||T^*v||\forall v$, the author is also proving the converse: that $||Tv||=||T^*v||\forall v$ implies that $T$ is normal. As for your second, question - in the expression $\langle T^*Tv,v\rangle$, denote $Tv$ by $a$, so you have $\langle T^*a,v\rangle$. Now recall that, by definition, we must have $\langle T^*a,v\rangle=\langle a,Tv\rangle$. Substitute back in $Tv$ for $a$.