Uniqueness of bilinear transformation $E\times F\to G$ vector spaces 
Let $E,F,G$ be vector spaces, with $dim \ E = m, \ dim \ F = n$ and
   $\{u_1,\cdots,u_n\}\subset E$ and $\{v_1,\cdots,v_n\}\subset F$ bases
  for $E$ and $F$ respectively. Show that:
a) Given a mn-uple of vectors $w_{ij}\in G$, with $i\in
 \{1,\cdots,m\}$ and $j\in \{1,\cdots,n\}$, there is an unique bilinear
  application $B:E\times F\to G$ such that $B(u_i,v_j) = w_{ij}$, for
  all $i\in \{1,\cdots,m\}$ and $j\in \{1,\cdots,n\}$.

Is this question about the uniqueness of a matrix representation for a linear transformation? It kinda looks like, but the word 'bilinear' bugs me.
I think it has to do with matrices because of the $mn$ of the 'mn-uple'. A matrix could be represented as an mn-uple... But I don't remember matrices defining bilinear transformations.
How to show that such 'matrix' exist, if that's not the case of a common matrix?
 A: You're almost on the right track, but you have the wrong $mn$ in mind. Matrices of shape $m \times n$ correspond to $mn$-uples of coordinates, but this question is talking about $mn$-uples of vectors. Here is your question, simplified to the linear operator case:
Let $E, G$ be vector spaces with $\dim E = m$, and $(u_1, \ldots, u_m)$ an ordered basis for $E$. Show that, given an $m$-uple of vectors $w_i \in G$, that there is a unique linear map $A: E \to G$ such that $A(u_i) = w_i$ for all $i$.
You are probably quite familiar with this result, but let's recall how this is usually done. Firstly show that linear maps from $E$ to $G$ form a vector space, and using this structure show that a linear map is determined by where it sends the basis vectors. For example, since any $u \in E$ is uniquely expressed as $u = a_1 u_1 + \cdots + a_m u_m$, you can find that
$$ A(u) = a_1 A(u_1) + \cdots + a_m A(u_m)$$
and so the whole map $A$ is determined by the $A(u_i) = w_i$. There is a little bit to check here, but it mostly comes from what you know about linear independence in vector spaces. You can use the vector space structure on the space of linear maps to go the other way, and build such an $A$ from a given $m$-uple of $w_i$.
It is straightforward to do almost exactly the same thing for bilinear maps $E \times F \to G$. They form a vector space, and by applying bilinearity we can see that if $u \in E$ and $v \in F$ are expressed in the bases as $u = a_1 u_1 + \cdots + a_m u_m$, $v = b_1 v_1 + \cdots + b_n v_n$, then
$$B(u, v) = a_1 b_1 B(u_1, v_1) + a_2 b_1 B(u_2, v_1) + \cdots + a_m b_n B(u_m, v_n)$$
Applying the same arguments as in the linear operator case will get you the correct answer: a bilinear map is determined by where it sends pairs of basis vectors $(u_i, v_j)$, and so is determined by specifying an $mn$-uple of vectors in $G$.
A: No, it is not asking about linear functions; it is about the uniqueness of a bilinear or 2-linear transformation. So, I shall define the concept you are probably missing. If need more clarification, please ask.
Consider a finite sequence of vector spaces $\mathrm{V}_1, \ldots, \mathrm{V}_d$ and their Cartesian product $\mathrm{V} = \prod\limits_{k = 1}^d \mathrm{V}_k.$ Endow $\mathrm{V}$ with coordinatewise sum and scalar multiplication; that is, $$(v_1, \ldots, v_d) + \lambda (w_1, \ldots, w_d) = (v_1 + \lambda w_1, \ldots, v_d + \lambda w_d).$$
This defines on $\mathrm{V}$ the structure of a vector space. For sake of simplicty, assume $d = 2$ (as is your case). A linear transformation $\mathrm{V} \to \mathrm{W},$ being the latter another vector space, is a function $L$ that obeys the following rule (or an equivalent version): $$L((v_1, v_2) + \lambda (w_1, w_2)) = L(v_1, v_2) + \lambda L(w_1, w_2).$$ For example, when everything is $\Bbb R,$ one can consider $L(x, y) = x + y.$ If you want to talk about a bilinear transformation on $\mathrm{V},$ one would need to think as it being linear in each coordinate; more specifically, $B:\mathrm{V} \to \mathrm{W}$ is bilinear if for each $v_1 \in \mathrm{V}$ the partial function $v_2 \mapsto B(v_1, v_2)$ is linear, and for each $v_2 \in \mathrm{V}$ the partial function $v_1 \mapsto B(v_1, v_2)$ is also linear. When everything is $\Bbb R,$ the classical example is $B(x,y) = xy$ (when $x$ is fixed, it is a linear function of $y$ and viceversa). Notice that for a bilinear $B$ is turns out that $$B((v_1, v_2) + \lambda (w_1, w_2)) = B(v_1 + \lambda w_1, v_2 + \lambda w_2) = B(v_1 + \lambda w_1, v_2) + \lambda B(v_1 + \lambda w_1, w_2) = B(v_1, v_2) + \lambda B(w_1, v_2) + \lambda B(v_1, w_2) + \lambda^2 B(w_1, w_2),$$
which is very different from the expresion for $L.$ If you wanted to express $B$ as some kind of array of numbers, you would need three dimensions and a box of size $\dim \mathrm{V}_1 \times \dim \mathrm{V}_2 \times \dim \mathrm{W}$ (for $L$ you need a two dimensional array of dimensions $(\dim \mathrm{V}_1 + \dim \mathrm{V}_2) \times \dim \mathrm{W}$).
A: This question is about the existence and uniqueness of a bilinear extension for a function $A : S \subseteq U \times V \to W$ where $U$, $V$ and $W$ are vector spaces over some fixed field and $S$ is a subset of $U \times V$. A bilinear extension of such $A$ is a bilinear function $B : U \times V \to W$ such that, $B|_{S}=A$, i.e., $B(u,v)=A(u,v)$ for all $(u,v) \in S \subseteq U \times V$.
It's not hard to show that if a bilinear extension exists, it is unique. If $S$ is a product $\mathcal{B}_{U} \times \mathcal{B}_{V}$ of basis, the extension always exists, and therefore, is unique.
Let me sketch the proof for the finite dimensional case. So we have $\mathcal{B}_{U}=\{u_{1},\dots,u_{m}\}$ and $\mathcal{B}_{V}=\{v_{1},\dots,v_{n}\}$. Let $(w_{ij})_{i,j=1}^{m,n}$ be the $nm$-tuple of images of $A$ in the product of basis, that is, $w_{ij}=A(u_{i},v_{j})$.
Now define $B : U \times V \to W$ by
$$B(x,y):=\sum_{i=1}^{m} \sum_{j=1}^{n} x^{i} y^{j}w_{ij}$$
for each $(x,y) \in U \times V$, with $x=\sum_{i=1}^{m} x^{i}u_{i}$ and $y=\sum_{j=1}^{n} y^{j}v_{j}$. This function is well defined, since the expression of $x$ and $y$ in the basis are unique, and its immediate to check that $B(u_{i},v_{j})=A(u_{i},v_{j})$ for each $i=1,\dots,m$ and $j=1,\dots,n$, that is, $B|_{S}=A$.
