$P=P^t$, proof with symmetric operators 


*Let $V$ be a finite dimensional space over $\mathbb{R}$, with a positive definitive scalar product. Let $P:V\to V$ be a linear map such that $PP=P$. Assume that $P^tP=PP^t$.  Prove that $P=P^t$ Linear Algebra, Serge Lang
SOLUTION. We have $\langle P^t(v),P^t(v)\rangle=\langle v,PP^t(v)\rangle=\langle v,P^tP(v)\rangle=\langle P(v),P(v)\rangle$
so we see that $\ker(P)=\ker(P^t)$ and
$\langle P(v)-P^t(v),P(v)-P^t(v)\rangle=2(\langle P(v),P(v)\rangle-\langle P(v),P^t(v)\rangle)$.
$\DeclareMathOperator{Im}{Im}V=\Im(P)\oplus\ker(P)$,so we can write $v=P(x)+w$, where $w\in\ker(P)=\ker(P^t)$. Then:
$\langle P(v),P(v)\rangle=\langle P(x),P(x)\rangle$
and
$\langle P(v),P^t(v)\rangle=\langle P(x),P^tP(x)\rangle=\langle P(x),P(x)\rangle$
The scalar product is positive definite, so $P(v)=P^t(v)$.Solutions manual to Lang´s Linear Algebra, by Rami Sharkarchi

Questions:
1) Given $\langle P^t(v),P^t(v)\rangle=\langle v,PP^t(v)\rangle=\langle v,P^tP(v)\rangle=\langle P(v),P(v)\rangle$, in which $P^tP=PP^t$ is an assumed property. Why is the theorem $P=P^t$ not proven with $\langle P^t(v),P^t(v)\rangle=\langle v,PP^t(v)\rangle=\langle v,P^tP(v)\rangle=\langle P(v),P(v)\rangle$? I guess $v$ is arbitrary and the property holds $P^tP=PP^t$  for all the $v\in V$
2) Why does the author write this $\langle P(v)-P^t(v),P(v)-P^t(v)\rangle=2(\langle P(v),P(v)\rangle-\langle P(v),P^t(v)\rangle)$? Is this later used in the proof? I could not see it.
3) Does this expression $\langle P(v),P(v)\rangle=\langle P(x),P(x)\rangle$ comes from the fact $P:V\to V$?
Thanks in advance
 A: Regarding 3):
Using the decomposition of $v$:
\begin{align}
v = P(x) + w \Rightarrow \\
\langle P(v), P(v) \rangle 
&= \langle P(P(x)+w), P(P(x)+w) \rangle \\
&= \langle P^2(x)+ P(w), P^2(x)+ P(w) \rangle \\
&= \langle P^2(x), P^2(x) \rangle \\
&= \langle P(x), P(x) \rangle
\end{align}
where the linearity of $P$, $P(w) = 0$ and $P^2 = P$ was used.
A: For 1) $\langle P^t(v),P^t(v)\rangle=\langle v,PP^t(v)\rangle=\langle v,P^tP(v)\rangle=\langle P(v),P(v)\rangle$ is not sufficient to deduce $P^t = P$ for example if we P was a unitary matrix ($P^tP = PP^t = I$) we would have $\langle P^t(v),P^t(v)\rangle=\langle v,PP^t(v)\rangle=\langle v,P^tP(v)\rangle=\langle P(v),P(v)\rangle$ but $P \neq P^t$.
For 2) $\langle P(v)-P^t(v),P(v)-P^t(v)\rangle=2(\langle P(v),P(v)\rangle-\langle P(v),P^t(v)\rangle)$ this formula combined with the two formulas $\langle P(v),P(v)\rangle=\langle P(x),P(x)\rangle$ and $\langle P(v),P^t(v)\rangle=\langle P(x),P(x)\rangle$ prove that $\langle P(v)-P^t(v),P(v)-P^t(v)\rangle = 0$ which implies that $P(v) = P^t(v)$. 
