Let $\{e_1,e_2,\cdots,e_n\}$ and $\{f_1,f_2, \cdots, f_m \}$ be the standard ordered bases for $\Bbb R^m$ and $\Bbb R^n$ respectively and let $\{E_{11},E_{12},\cdots,E_{mn} \}$ be the standard basis for $\Bbb R^{mn}$ appearing in some order. Consider the map $f : \Bbb R^m \times \Bbb R^n \longrightarrow \Bbb R^{mn}$ defined by $$\sum\limits_{i,j} a_{ij} (e_i,f_j) \longmapsto \sum\limits_{i,j} a_{ij} E_{ij},\ a_{ij} \in \Bbb R.$$
Is this map bilinear? Is $f$ onto? Our instructor suggests that this map is bilinear and onto. Suppose I take $(e_1+e_2,f_1) \in \Bbb R^n \times \Bbb R^m.$ Then how can I show that $f ((e_1+e_2,f_1) = E_{11} + E_{21}\ $? I can't see how it is the case. Can anybody please help me in this regard?
Thanks for your time.
EDIT $:$ Every element of $\Bbb R^n \times \Bbb R^m$ can be written as $\left (\sum\limits_{i=1}^{n} a_i e_i, \sum\limits_{j=1}^{m} b_j f_j \right ).$ So if $f$ acts bilinearly on such elements then we should have $$f \left ( \left (\sum\limits_{i=1}^{n} a_i e_i, \sum\limits_{j=1}^{m} b_j f_j \right ) \right ) = \sum\limits_{i,j} a_i b_j E_{ij}.$$
Now the question is whether $f$ is onto or not?
RE-EDIT $:$ I don't think $f$ thus defined is onto. For instance let us take $m=n=2.$ Let $\{e_1,e_2\}$ be the standard basis of $\Bbb R^2.$ Let $E_{11} = (1,0,0,0), E_{12} = (0,1,0,0), E_{21} = (0,0,1,0)$ and $E_{22} = (0,0,0,1).$ Now any element of $\Bbb R^2 \times \Bbb R^2$ is of the form $((x,y),(z,w)),$ $x,y,z,w \in \Bbb R.$ Then by the definition of $f$ it follows that $$f((x,y),(z,w)) = (xz,xw,yz,yw).$$ This is clearly not onto since there does not exist any element in $\Bbb R^2 \times \Bbb R^2$ which maps to $(1,2,3,4) \in \Bbb R^4$ under $f.$
Can anybody please check my argument above? Thanks.
If my argument above is fine then the bilinear map $\otimes : \Bbb R^2 \times \Bbb R^2 \longrightarrow \Bbb R^2 \otimes \Bbb R^2 \cong \Bbb R^4$ fails to become onto as well.
The equality $(e_1+e_2,f_1)=(e_1,0)+(e_2,f_1)$ should solve your problems. This is true due to the algebraic properties of $\mathbb{R}^n\times\mathbb{R}^m$.
There does seem to be something wrong with the map $f$ as it is written down in your post. There are no $E_{ij}$ basis vectors of $\mathbb{R}^{nm}$. You have taken the standard basis $\{E_1,\ldots,E_{nm}\}$. So a good map would be $$\sum_{i\leq n}a_i(e_i,0)+\sum_{j\leq m}b_j(0,f_j)\mapsto \sum_{i\leq n} a_i E_i+\sum_{j\leq m} b_jE_{j+n}. $$