Consider the bijective linear map: $\alpha : K^2 \otimes K^2 \to Mat(2 \times 2,K), \alpha(v \otimes w) = vw^t, v, w \in K^2$ , where $K$ is an arbritrary field.
First I want to show that every element of $K^2 \otimes K^2$ can be written as the sum of two elementary tensors.
I know that every element of $K^2 \otimes K^2$ can be written as $a_{11}e_1 \otimes e_2 + a_{12}e_1 \otimes e_2 + a_{21}e_2 \otimes e_1 + a_{22}e_2 \otimes e_2$ where $a_{ij}, (i,j) \in \{1,2\} \times \{1,2\}$ is uniquely determined because $e_i \otimes e_j$ is a basis. How can I use this to show that every element of $K^2 \otimes K^2$ can be displayed as $v \otimes w + x \otimes y, x,y,v,w \in K^2$ ?
I a second step I want to show that
$ det(\alpha(x)) = 0 \Leftrightarrow$ x is elementary tensor, so there exist $v,w$ with $ x= v\otimes w$
I don't really know how to start here. Can anybody help me? Thanks in advance!
For the first question, one could rewrite the general element $a_{11} e_1 \otimes e_1 + a_{12} e_1 \otimes e_2 + a_{21} e_2 \otimes e_1 + a_{22} e_2 \otimes e_2$ of $K^2 \otimes K^2$ e.g. in the form $e_1 \otimes (a_{11} e_1 + a_{12} e_2) + e_2 \otimes (a_{21} e_1 + a_{22} e_2)$.
For the second question, suppose $A \in \operatorname{Mat}(2\times 2, K)$ is a non-zero matrix with $\det A = 0$. Then, $A$ has rank $1$ and hence the linear map $K^2 \to K^2$ represented by $A$ (which we will denote also by $A$) has a one-dimensional image $L \subset K^2$. Let $v$ be a non-zero vector in $L$. Then, for any $x \in K^2$ we have $$A(x) = f(x)v$$ for a (uniquely defined) function $f : K^2 \to K$. Since $A$ is $K$-linear, it follows that $f$ is also $K$-linear. Thus, $f(x) = u_1x_1 + u_2x_2$ for some $u_1,u_2 \in K$ independent of $x = (x_1,x_2) \in K^2$. Putting $u = (u_1,u_2) \in K^2$, we deduce that $A = \alpha(v \otimes u)$. Conversely, the fact that $\det(\alpha(v\otimes w)) = 0$ for any $v,w \in K^2$ is a matter of direct calculation.