If for every $v\in V$  $\langle v,v\rangle_{1} = \langle v,v \rangle_{2}$ then $\langle\cdot,\cdot \rangle_{1} = \langle\cdot,\cdot \rangle_{2}$ Let $V$ be a vector space with a finite Dimension above $\mathbb{C}$ or $\mathbb{R}$.
How does one prove that if $\langle\cdot,\cdot\rangle_{1}$ and $\langle \cdot, \cdot \rangle_{2}$ are two Inner products
and for every $v\in V$  $\langle v,v\rangle_{1}$ = $\langle v,v\rangle_{2}$ so $\langle\cdot,\cdot \rangle_{1} = \langle\cdot,\cdot \rangle_{2}$
The idea is clear to me, I just can't understand how to formalize it.
Thank you.

You can use the polarization identity.
$\langle \cdot, \cdot \rangle_1$ and $\langle \cdot, \cdot \rangle_2$ induces the norms $\| \cdot \|_1$ and $\| \cdot \|_2$ respectively, i.e.:
$$\begin{align}
\| v \|_1 = \sqrt{\langle v, v \rangle_1} \\
\| v \|_2 = \sqrt{\langle v, v \rangle_2} 
\end{align}$$
From this it is obvious that $\|v\|_1 = \|v\|_2$ for all $v \in V$, so we can write $\| \cdot \|_1 = \| \cdot \|_2 = \| \cdot \|$.
By the polarization identity we get (for complex spaces):
$$\begin{align}
\langle x, y \rangle_1 &=\frac{1}{4} \left(\|x + y \|^2 - \|x-y\|^2 +i\|x+iy\|^2 - i\|x-iy\|^2\right) \ \forall\ x,y \in V \ \\
\langle x, y \rangle_2 &=\frac{1}{4} \left(\|x + y \|^2 - \|x-y\|^2 +i\|x+iy\|^2 - i\|x-iy\|^2\right) \ \forall\ x,y \in V
\end{align}$$
since these expressions are equal, the inner products are equal.
