Why can we identify the spaces $\operatorname{Hom}(\mathbb{R}^m,\mathbb{R}^n)$ and $\mathbb{R}^{m\ast} \otimes \mathbb{R}^n$? Why and how can we identify the space of linear maps $\operatorname{Hom}(\mathbb{R}^m,\mathbb{R}^n)$ with the tensor product $\mathbb{R}^{m\ast} \otimes \mathbb{R}^n$? 
 A: More generally, consider two finite dimensional vector spaces $V$ and $W$. There is a bilinear map
$$
\varphi\colon V^*\times W\to\operatorname{Hom}(V,W)
$$
defined by
$$
\varphi(\xi,w)\colon v\mapsto \xi(v)w
$$
It's easy to verify that this is well defined. Hence this defines a linear map $\hat\varphi\colon V^*\otimes W\to\operatorname{Hom}(V,W)$. Now it remains to show the map is surjective, because equality of dimensions will yield also injectivity.
If $\{v_1,v_2,\dots,v_m\}$ is a basis for $V$ and $\{w_1,\dots,w_n\}$ is a basis for $W$, a basis for $\operatorname{Hom}(V,W)$ is given by the linear maps $f_{hk}$ defined on the basis by
$$
f_{hk}(v_i)=
\begin{cases}
w_k & \text{if $i=h$} \\
0 & \text{if $i\ne h$}
\end{cases}
$$
Can you express $f_{hk}$ as $\varphi(\xi,w)$ for some $\xi\in V^*$ and some $w\in W$?
A: That's because the $\;\DeclareMathOperator{\Hom}{Hom}\Hom(\mathbf R^m, \boldsymbol{\cdot})\;$ functor commutes with direct sums:
$$\Hom(\mathbf R^m,\mathbf R^n)\simeq \bigoplus_{i=1}^n\Hom(\mathbf R^m,\mathbf R)\simeq((\mathbf R^m)^*)^n\simeq((\mathbf R^m)^*\oplus\mathbf R^n.$$
