For what k is $\mathcal{M}_{m \times n}$ isomorphic to $\mathbb{R}^{k}$? For what  k  is  $\mathcal{M}_{m \times n}$  isomorphic to  $\mathbb{R}^{k}$ ?
I get a feel but am unable to prove it.
 A: Hint for different approaches: 


*

*Concatenate the columns of a matrix $m \times n$ one under the other. You will get a big vector, what is its size?
$$ \begin{pmatrix} \color{red}{ a_{1,1}} & \color{blue}{a_{1,2}} & \ldots & a_{1,n} \\ \color{red}{a_{2,1}} & \color{blue}{a_{2,2} }& \ldots & a_{2,n} \\ \color{red}{\vdots }& \color{blue}{\vdots }& \ddots  & \vdots \\ \color{red}{a_{m,1}} & \color{blue}{a_{m,2}} & \ldots & a_{m,n}  \end{pmatrix}\longrightarrow \begin{pmatrix}\color{red}{a_{1,1} }\\\color{red}{ a_{2,1}}\\ \color{red}{\vdots} \\ \color{red}{a_{m,1}} \\ \color{blue}{a_{1,2}} \\ \color{blue}{\vdots }\\ \color{blue}{a_{m,2}} \\ \vdots \\ a_{m,n} \end{pmatrix}$$

*Find a basis $B=\{M_1, M_2, \ldots, M_k\}$ of $\mathcal{M}_{m \times n}$. The canonical one is usually the most easy to find. Here are the two first ones:
$$M_1 = \begin{pmatrix} 1 & 0 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \\ \vdots & \vdots& \ddots  & \vdots \\ 0 & 0 & \ldots & 0 \end{pmatrix}, M_2 = \begin{pmatrix} 0 & 1 & \ldots & 0 \\ 0 & 0 & \ldots & 0 \\ \vdots & \vdots& \ddots  & \vdots \\ 0 & 0 & \ldots & 0 \end{pmatrix},\ldots $$
It is now clear that the dimension of $\mathcal{M}_{m \times n}$ equals the amount of entries in one matrix $M \in\mathcal{M}_{m \times n}$. Then you can use that a vector space of dimension $k< \infty$ is isomorphic to $\mathbb{R}^k$.

