So I'm working through this counterexample of a bilinear map. It is attached in the picture below.
So we have that $$S=\{ \varphi ( x,y) | x\in E, y\in F \}$$ the set of vectors that respect the bilinear form might not necessarily form a vector subspace. In the example below we have that long expression of $\varphi (x,y)$ in terms of its basis elements, similarly for $x$ and $y$ themselves. What I don't "easily" see is what $z$ is. I'm given a summation of $\varphi c_v$ what does this actually mean? The corresponding basis element when writing out the vector $z$ in its $\varphi$ form?
Now that I am reading and typing this out, am I supposed to also introduce the elements $\lambda, \gamma$ from the base and try to find out the bilinear form? I guess I don't feel like chugging along in the wrong direction only to find out what I was supposed to think about was right in my face.
So I guess my question is: Am I doing this right am I understanding the notation correctly? (If not then what am I doing wrong so I can check this counterexample correctly)
Thanks and all the best. I typed this from my phone so I might need to tweak the TeX and some formatting stuff.
See if this helps...
Yes, they are claiming that the image of a bilinear map is not in general a linear subspace, and they are trying to prove this claim with a counterexample. The point of the expression for $\phi(x, y)$ is that it defines a function $E \times F \to G$, because it tells you the value of $\phi$ on any $x$ and $y$.
So you don't have to "find out the bilinear form"; they give it to you. However, it is worth checking their claim that $\phi$ is bilinear. It should be easy to verify the required equations for any $\lambda$ and $\mu$. (Although their definition is a bit more complex than it needs to be: the Wikipedia definition of bilinear map, which is equivalent, makes it easier.)
The idea of the equation for $z$ is that that we are describing what the image $S$ of this bilinear map looks like: which elements of $G$ are in $S$? Any element of $G$ can be written uniquely as a linear combination of the $c_\nu$, and their claim is that the elements of $G$ which are in $S$ are precisely the ones whose components in that linear combination satisfy the $\phi^i$ equation. Unfortunately they make a very confusing choice of notation for those components: they use "$\phi^i$" even though the components are entirely different from the map $\phi$. The version of this passage in the other Stack Exchange question (see the comment from @blargoner) fixes this problem, so I suggest you look there.
Their claim that it's "easy to see" that the $\phi$ equation describes $S$ is, frankly, stretching it. The answers to that other Stack Exchange have some approaches on how to prove this.
Finally, they claim that this $S$ is not a linear subspace, which really is easy to check from their example.