I'd need clarification on one of the statements here below.
Given that V is the set of linear operators f: $\mathbb{R}^k \rightarrow \mathbb{R}^n $, and I'm trying to prove that V has the same dimension as $\mathbb{R} ^{kn}$ so that to prove it.
Now, why is $ m=kn$ ? And also why is V contained in $(\mathbb{R}^{n})^{(\mathbb{R}^k)} ?$
I'm really looking for the reasoning (the simpler the better) behind these statements.
To define a linear operator $f \in V$ you have to define the image of each of the $k$ basis vectors in $\mathbb{R}^k$. Each of these images is a point in $\mathbb{R}^n$ so it has $n$ components. So each $f \in V$ is defined by a set of $kn$ real numbers, and so $V$ is isomorphic to $\mathbb{R}^{kn}$. You can represent each $f \in V$ by a $k \times n$ matrix.
To make this informal argument more formal, you would have to prove that $V$ is a vector space and that the $kn$ linear functions $f_{ij}$ defined by $f_{ij}(\mathbf{e}_i) = \mathbf{e}_j$ form a basis of $V$.