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.
 A: 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.
A: Hint: Note that the associated norms satisfy $\|v\|_1 = \sqrt{\langle v,v\rangle_1} = \sqrt{\langle v,v\rangle_2} = \|v\|_2$ and then use the polarization identity to recover the scalar products and see that they are equal.
A: (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
$${\rm Im}\langle v,w\rangle={\rm Re}\langle -iv,w\rangle$$
