It is well known that for a bilinear form over an n-dimensional vector space, $n^2$ values (on all pairs of basis-vectors) determine it uniquely.


How many values do we need to specify in order to uniquely determine a norm?
The first trivial observation is that homogeneity implies it's enough to specify all the values on the unit sphere. 


Then, if we indeed assume $V$ is a finite-dimensional space, every two norms on $V$ are equivalent, and in particular are continuous w.r.t each other. Hence their set of mutual agreement is closed, so it is enough to specify a dense subset of the unit sphere. since a the unit sphere of a finite dimensional normed space is separable, we conclude it is enough to specify a countable number of values.


Is it possible to use only a finite number points? (the constant may depend on the dimension ofcourse).



How many values determine a norm?
Well, as you have certainly already remarked (reading your post, I assume this), bilinearity makes a big difference. For the "only norm" case, what you are looking for, if I understand correctly your question, is a **set of uniqueness** for the admissible norms on a given vector space $V$. Your demonstration establishes that values on a dense set $U$ (on the unit sphere) is sufficient (and then, we can reduce to countable). You **can go further** by choosing a set $U$ such that $U\cup (-U)$ is dense, then **you cannot go further** as, on $V$, a norm $p$ is the jauge function of the balanced convex
$$
C=\{x\in V|\, p(x)\leq 1\}
$$
then, if your set $U\cup (-U)$ is not dense, there is a point $M\in S\_V$ (the unit sphere) and a neighbourhood $W$ of $\{M,-M\}$ such that $U\cap W=\emptyset$. Now, deforming the sphere around $\{M,-M\}$ in a convex way (you can find a one parameter deformation $C\_t$ such), one obtains an infinite family of norms which coincide with $p$ on $U$ and differ from it. If the space $V$ is complex, just replace the "real balanced saturation" $U\mapsto U\cup (-U)$ by its analogue (the orbit of $U$ under the group of complex numbers of modulus one $\mathbb{U}=\{z\in\mathbb{C} | |z|=1\}$). 



> 
> **Proposition** In order $U\subset S\_V$ be a uniqueness set for the collection of all norms, it is necessary and sufficient that the orbit $G.U$ be dense in $S\_V$ ($G=\{−1,1\}$ for the real case and $G=\mathbb{U}$ for the complex case).
> 
> 
> 


[$G.U$ dense$\Longrightarrow$ $U$ is a uniqueness set] is clear. 


Now, if $G.U$ is not dense, it exists a point $w\in S\_V$ which does not belong to $\overline{G.U}=G.\overline{U}$ and then 
$$
max\_{v\in U}\,\{|\langle w|v\rangle|\}=M<1
$$
Now, the convex sets 
$$
C\_t=\{v\in V|\, ||v||\,\leq 1\ \mathit{and}\ |\langle w|v\rangle|\leq (1-t)M+t\}
$$ 
for $0<t\leq 1$ give the desired deformation (for each $C\_t$ contains $U$).



> 
> **A bit more** (Local implies global ?) For $p$ a norm, we will say that set $U\_p\subset S\_V$ is a *relative uniqueness set w.r.t. $p$* if, for any norm $q$, $p|\_{U\_p}=q|\_{U\_p}\Longrightarrow p=q$. We have just seen that, if $p$ is the euclidean (resp. hermitian) norm then each $U\_p$ is a uniqueness set for **all** norms. This is not the case for every norm (especially if it has many *linear zones*) for example take the two dimensional euclidean space $\mathbb{C}$ with the norm $|z|$ (real euclidean space). It can be seen easily that, with $p=l\_1$ (see below), the set 
>  $$
> U\_{l\_1}=\{e^{i\theta}\,|\, \theta\in T=[-\frac{\pi}{6},\frac{\pi}{6}]\cup [\frac{\pi}{3},\frac{2\pi}{3}]\}
> $$  
> 
>  is a relative uniqueness set w.r.t. the $l^1$ norm $||z||=|\Re(z)|+|\Im(z)|$ ; but $U\cup\, (-U)$ is not dense in the unit sphere(for example $e^{i\frac{\pi}{4}}$ is not in the closure of it).
> 
> 
> **Remark** If you want $U\_{l\_1}$ to be denumerable replace $T$ by $T\cap \mathbb{Q}$, with the same closure. 
> 
> 
> 


