Inner product on tangent space of projective space I just got stuck at a certain part of my book which concern projective spaces. Really need a help to understand this. 
First, their definition.

They define an inner product in $V$ in order to induce an inner product $\mathbb{P}(V)$.

Finally, the inner product is defined.

What I don't get is how this is independent of the choice of the representative? For instance, we can choose $\lambda v$ to be representative, for some $\lambda\in\mathbb{C}$ such that $|\lambda|=2$. Then 
$$\langle a,b \rangle_{\lambda v} = \frac{\langle a,b \rangle_\mathbb{R}}{\|\lambda v\|^2} = \frac{1}{4}\frac{\langle a,b \rangle_\mathbb{R}}{\| v\|^2} \neq \frac{\langle a,b \rangle_\mathbb{R}}{\|v\|^2}.$$
EDIT: Just to show all details, here is how they defined the charts and how they are working the tangent spaces.



 A: Given non-zero $v$ we want to  describe local coordinates of $P(V)$ in a neighborhood of the projective point $[v]={\Bbb C}v$. We do so   through the orthogonal complement:
$T_v=\{z: \langle a,v \rangle=0\}$. 
Except for projective points in $P(T_v)$ we will expres any $[x]$ in the form $[v+w]$ for a suitable  $w\in T_v$.  To get this map let $x\in V\setminus T_v$ and let us seek $\lambda$ and $w\in T_v$ for which
  $$ x = \lambda (v+w) $$
Taking scalar product with $v$ we get: $\langle x,v\rangle = \lambda \langle v,v\rangle$ from which we first isolate $\lambda$ and insert in the first to obtain $w$:
$$ \phi_v(x):= w = \frac{\langle v,v\rangle}{\langle x,v\rangle}x-v. $$
Now, for any non-zero $\mu$ we have $\phi_v(\mu x)=\phi_v(x)$ so this lifts to a well-defined map: $\hat{\phi}_v([x]) = \phi_v(x)$
and $\Psi_v^{-1}=\hat{\phi}_v:  P(V)\setminus P(T_v) \rightarrow T_v$ is the wanted coordinate map.
Exercise: Given $v,v'$ and corresponding $[x]\in A_v\cap A_{v'}$ find the change of coordinate map taking $w\in T_v \mapsto w'\in T_{v'}$
