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.

(symmetric, bilinear, char $\neq2$)
$$\langle v+w,v+w\rangle_1=\langle v+w,v+w\rangle_2$$
$$\langle v,v\rangle_1+\langle w,w\rangle_1+2\langle v,w\rangle_1=\langle v,v\rangle_2+\langle w,w\rangle_2+2\langle v,w\rangle_2$$
$$\langle v,w\rangle_1=\langle v,w\rangle_2$$
if the product is sesquilinear over $\mathbb{C}$ (linear in first coord, conjugate linear in the second) the above gives
$$
\langle v,w\rangle_1+\overline{\langle v,w\rangle}_1=\langle v,w\rangle_2+\overline{\langle v,w\rangle}_2
$$
ie
$$
{\rm Re}\langle v,w\rangle_1={\rm Re}\langle v,w\rangle_2.
$$
we can equate the imaginary parts by noting that
$${\rm Im}\langle v,w\rangle={\rm Re}\langle -iv,w\rangle$$
