$\dim \mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$ I know we can identify the set $\mathscr L(\mathbb R^m;\mathbb R^n)$ of the linear transformations $f:\mathbb R^m\to \mathbb R^n$ with a matrix $M_{n\times m}$ in the canonical way. Then, we have $\dim \mathscr L(\mathbb R^m;\mathbb R^n)=mn$.
So, I'm trying to use a similar technique to prove the set $\mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$ of the bilinear transformations $\varphi:\mathbb R^m\times \mathbb R^n\to \mathbb R^p$ has dimension $mnp$. In this way, I would like to find a identification between  $\mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$ and some set I already know.
Any suggestions?
Thanks
 A: Since you said you are not familiar with tensor products, let me try to sidestep the notion altogether. There are some details and checks I have left to you. Please ask if you need any clarification.
Just like a normal linear map is determined completely by its value on a basis, a bilinear map $f:\mathbb{R}^n\times \mathbb{R}^m\rightarrow \mathbb{R}^p$ is determined by the bases of $\mathbb{R}^n$ and $\mathbb{R}^m$. To see this, note that if we know the value of $f$ on pairs of basis elements, then for $x=\Sigma_i a_i x_i $ and $y=\Sigma_j b_j y_j$ in $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, where $x_i$ and $y_i$ are bases for $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, we have
$$f(x,y)=f(\Sigma_i a_ix_i,\Sigma_j b_jy_j)=\sum_{i,j} a_ib_j f(x_i,y_j)$$
and it is not difficult to see here that this completely determines $f$ and that two bilinear maps agree if and only if they agree on pairs of basis elements. Again, each of the $mn$ values $f(x_i,y_j)$ is an element of a $p$-dimensional space $\mathbb{R}^p$, so the dimension of $\mathscr{L}(\mathbb{R}^n,\mathbb{R}^m;\mathbb{R}^p)$ must be $mnp$.
Edit:
Whenever you have a function $f$ into $\mathbb{R}^p$, you can decompose it into $p$ functions $f_k$ taking values in $\mathbb{R}$, so each $f(x_i,y_j)$ is given by a linear combination $f(x_i,y_j)=\Sigma_k f_k(x_i,y_i)$. $f_k(x_i,y_i)$ is just a real number, say $c_{ijk}$, so each $f(x_i,y_j)$ is given by a $p$-tuple of real numbers. There are $mn$ of these $p$-tuples, parameterized by the pairs of basis elements of $\mathbb{R}^n$ and $\mathbb{R}^m$.
In conclusion, each $f\in\mathscr{L}(\mathbb{R}^n,\mathbb{R}^m;\mathbb{R}^p)$ is determined by the $mnp$-tuple $c_{ijk}$ defined above.
A: An “abstract” way for getting at $mnp$ is finding an isomorphism
$$\def\cL{\mathscr{L}}
\cL(U,V;W)\to\cL(U;\cL(V;W))
$$
This is easily generalizable to an isomorphism
$$
\cL(U_1,U_2,\dots,U_n,U_{n+1};W)\to\cL(U_1,U_2,\dots,U_n;\cL(U_{n+1};W))
$$
Note that $\cL(U,V;W)$ is a vector space in a very natural way.
So, let $f\in\cL(U,V;W)$; then we define $\widehat{f}\colon U\to\cL(V;W)$ by
$$
\widehat{f}(u)\colon v\mapsto f(u;v)
$$
It's just computation showing that $\widehat{f}(u)\in\cL(V;W)$ for all $u\in U$ and that the map $f\to\widehat{f}$ is linear. It's bijective because we can find its inverse. Namely, if $g\in\cL(U;\cL(V;W))$, define
$$
\check{g}\colon (u,v)\mapsto g(u)(v)
$$
(that is, $g(u)$, which is a map from $V$ to $W$, evaluated at $v$). Again, verifications are just a bit boring, but easy.
Note that finite dimension is not necessary here (and hypotheses can be relaxed even more, to modules over not necessarily commutative rings).
In the case of finite dimension, the computation of $\dim\cL(U,V;W)$ is reduced to first computing $\dim\cL(V;W)$ (which you already know to be $np$) and then
$\dim\cL(U;\cL(V;W))$ which is, by the same reason $m(np)$.
