When is a vector in the image of a bilinear map? In this case I have a bilinear map defined as:
$\varphi:V\times V\rightarrow U$ with:  

$$\varphi(x,y)=x_1y_1c_1+x_1y_2c_2+x_2y_1c_3+x_2y_2c_4$$

Here $\{c_1,\ldots,c_4\}$ is a basis of $U$ and $V$ has dimension 2 with basis $\{b_1,b_2\}$ (so you can think of $x:=x_1b_1+x_2b_2$ and "" for $y$)
Given a vector, $z\in U$ I want to know under what conditions $z=\sum^4_{i=1}z_ic_i\in S$ where:
$S:=\{\varphi(x,y)\in U\ |\ x\in V\wedge y\in V\}$ (the image of the bilinear map)
I am looking for solving this in the general case
So please no specific-to-this-question tricks. If this question were different I'd like to have learned how to solve it (rather than learning a trick for this exact question). It annoys me I can't seem to "solve" the equations:
$$\begin{array}{ll}z_1=x_1y_1 &z_2=x_1y_2\\ z_3=x_2y_1 & z_4=x_2y_2\end{array}$$
into some level set $\{z\in Z|f(z_1,\ldots,z_4)=c\}$ form.
Suspected answer
I believe that $z\in S\iff z_1z_4-z_2z_3=0$ but "I pulled this out of a hat" if you will by thinking "it looks a bit like matrx multipication" and taking a guess. 
Context
I am trying to find a bilinear map whose image is not a vector subspace of $U$. This means that finding a basis of the image will not work (and indeed has not worked for me)
 A: This question is equivalent to asking what $2\times 2$ matrices have rank $1$. (Consider the matrix product $x\,y^{\textrm{t}}$ for two column vectors $x$ and $y$.)  In other words, what $2\times 2$ matrices are non-zero and have pairwise dependent columns (or rows)? This characterization also works well in higher dimensions. See segre embedding for pointers.
A: Notice for this particular case you can divide to cancel out variables and see that
$$\frac{z_{1}}{z_{2}} = \frac{z_{3}}{z_{4}} = \frac{y_{1}}{y_{2}}$$
and
$$\frac{z_{1}}{z_{3}} = \frac{z_{2}}{z_{4}} = \frac{x_{1}}{x_{2}}.$$
Each of these gives that $z_{1}z_{4}-z_{2}z_{3} = 0$.
You have essentially two cases: if $z_1=z_2=0$,
then take $x_1=0$, $x_{2}=1$, $y_{1}=z_{3}$, and $y_{2} = z_{4}$.
If $z_{1}=z_{4}=0$, then we either have that $x_{1}=y_{2}=0$ and so $z_{2}=0$, in which case take $x_{2}=1$ and $y_{1}=z_{3}$,
or else $y_{1}=x_{2}=0$, in which case $z_{3}=0$, and you can similarly find a solution.
In general, you will get a whole bunch of quadratic equations involving two variables, and the image will be the intersection of all the solution sets. This is explained in the link posted by WimC involving Segre varieties.
