Prove $K^n \otimes K^m \cong K^{nm}$

Question

I defined a function $g:K^n\times K^m \rightarrow K^{nm}$ s.t $$ g(u,v) = uv= \begin{cases} (u_1v_1,\dots,u_nv_m) &\quad\text{if $n=m$},\\ (u_1v_1,\dots,u_mv_m, u_{m+1},\dots,u_n) &\quad\text{if $n>m$},\\ (u_1v_1,\dots,u_nv_n, v_{n+1},\dots,v_m) &\quad\text{if $m>n$} \\ \end{cases} $$ where $(u_1,\dots,u_n) \in K^n$ and $(v_1,\dots,v_m)\in K^m$. Now, there exists a unique function $h:K^n\otimes K^m \rightarrow K^{nm}$ s.t $h \circ f = g$.

Then, I have to prove that $h$ is an isomorphism.

$h$ is surjective:

Let $\lambda\eta \in K^{nm}$, note that $\lambda\eta$ is generated uniquely by $g(\lambda,\eta) \Rightarrow g$ is surjective, and because $g = h \circ f$, then $h$ is surjective.

$h$ is injective:

Let $\lambda, a \in K^n$ and $\eta,b\in K^m$, now suppose that $g(\lambda, \eta) = g(a,b) \Rightarrow \lambda\eta=ab \Rightarrow \lambda =a$ and $\eta = b$, because $\lambda, a \in K^n$ and $\eta,b\in K^m$.

$\therefore h$ is bijective, i.e., $h$ is an isomorphism. Is it right?

Answer 1

Think of the coordinates of $K^{mn}$ as an $m\times n$ grid, so there is a basis consisting of elements $w_{ij}$ where $1\leq i\leq m$ and $1\leq j\leq n$. Then given a basis $u_1,\ldots,u_m$ of $K^m$ and a basis $v_1,\ldots,v_n$ of $K^n$, define $$g(u_i,v_j)=w_{ij},$$ extending this to make it bilinear. Can you prove that this has the required properties? It would help to show that given any bilinear map $g':K^m\times K^n\to A$ into some $K$-module $A$, $g'$ factors through $g$.

Answer 2

You can also just note that $\{v_1,...,v_n\}$ is a basis for the n-dim vector space and $\{w_1,...,w_m\}$ is a basis for the m dimensional vector space. Now, $\{v_i \otimes w_j\}_{I,j}$ is a basis for the tensor product. Of course this vector space has dimension nm and since finite dimensional vector spaces of the same dimension are isomorphic he result follows.