Problem-Show that $\Bbb{F}^{m×n}$ is isomorphic to $\Bbb{F}^{mn}$.
Attempt-Deffine a mapping $T:\Bbb{F}^{m×n}\rightarrow \Bbb{F}^{mn}$ by $T(E_{ij})=e_{n(i-1)+j}$ ,where $1≤i≤m$ and $1≤j≤n$.
I assume $E_{ij}$ ,$1≤i≤m$ and $1≤j≤n$ whose non-zero entry $(E_{ij})_{ij}=1$. It is easy to check that $E_{ij}$ form a basis for $\Bbb{F}^{m×n}$. From here I can check that $T$ is bijective and linear transformation. Am I on the right track?
Also I would like to know that how it will come to mind to choose $n(i-1)+j$? I just take it as hit and trial.
Any help or suggestions would be appreciable. Thanks in advance.
Yes, you are on the right track. Alternatively, it suffices to note that any bijective function from a basis of one vector space to another uniquely extends to an isomorphism between the two spaces.
As for how this particular ordering would come to mind, the assignment $f(i,j) = n(i-1) + j$ is the result of counting the tuples $\{(i,j): 1 \leq i \leq m,\ 1 \leq j \leq n\}$ in lexicographical order.