Greub Multilinear Algebra bilinear mappings question In Greub, Chapter 1, page 1 it says:
Suppose $E$, $F$ and $G$ are vector spaces and consider a mapping $$\psi\colon E \times F \rightarrow G.$$ The set 
$$S = \{\psi (x,y) \in G \mid x \in E, y \in F \}$$
is not in general a subspace of $G$. 
For example, let $E=F$ be two-dimensional and $G$ be four-dimensional. Choose a basis $a_1$, $a_2$ in $E$ and a basis $c_1, \dots, c_4$ in $G$ and define the bilinear mapping $\psi$ by
$$\psi (x,y) = \zeta^1 \eta^1 c_1 + \zeta^1 \eta^2 c_2 + \zeta^2 \eta^1 c_3 + \zeta^2 \eta^2 c_4,$$ 
where $x = \zeta ^1 a_1 + \zeta^2 a_2$ and $y = \eta^1 a_1 + \eta^2 a_2$. 
Then it is easy to see that a vector 
$$z = \sum_v \lambda^v c_v$$ of $G$ is contained in $S$ if and only if $$\lambda^1 \lambda^4 - \lambda^2 \lambda^3 = 0.$$ 
My Question: Why? The reason for that last step is not obvious to me. Any clarification is much appreciated.
 A: Here's a more "direct" argument for the harder direction, in case it's helpful to you:
If $\lambda^1\lambda^4-\lambda^2\lambda^3=0$, we want $\zeta^1,\zeta^2,\eta^1,\eta^2$ with
$$\lambda^1=\zeta^1\eta^1\qquad\lambda^2=\zeta^1\eta^2\qquad\lambda^3=\zeta^2\eta^1\qquad\lambda^4=\zeta^2\eta^2$$
If $\lambda^1=0$, then $\lambda^2\lambda^3=\lambda^1\lambda^4=0$, so $\lambda^2=0$ or $\lambda^3=0$.

*

*If $\lambda^2=0$, we take $\zeta^1=0$, $\zeta^2=1$, $\eta^1=\lambda^3$, and $\eta^2=\lambda^4$.

*If $\lambda^3=0$, we take $\zeta^1=\lambda^2$, $\zeta^2=\lambda^4$, $\eta^1=0$, and $\eta^2=1$.

If $\lambda^1\ne 0$ and $\lambda^2=0$, then $\lambda^4=0$ and we take $\zeta^1=\lambda^1$, $\zeta^2=\lambda^3$, $\eta^1=1$, and $\eta^2=0$. If $\lambda^1\ne 0$ and $\lambda^2\ne 0$, we take $\zeta^1=1$, $\zeta^2=\lambda^3/\lambda^1=\lambda^4/\lambda^2$, $\eta^1=\lambda^1$, and $\eta^2=\lambda^2$.
A: $z=\psi(x,y)$ is equivalent to $\sum_v\lambda^vc_v=\zeta^1 \eta^1 c_1 + \zeta^1 \eta^2 c_2 + \zeta^2 \eta^1 c_3 + \zeta^2 \eta^2 c_4$, which in turn is equivalent to
$$
\begin{cases}
\lambda^1=\zeta^1 \eta^1,\\
\lambda^2=\zeta^1 \eta^2,\\
\lambda^3=\zeta^2 \eta^1,\\
\lambda^4=\zeta^2 \eta^2.
\end{cases}\tag{1}
$$
This clearly implies that $\lambda^1\lambda^4-\lambda^2\lambda^3=0$. Conversely, if $\lambda^1\lambda^4-\lambda^2\lambda^3=0$, then $A=\pmatrix{\lambda^1&\lambda^2\\ \lambda^3&\lambda^4}$ is singular. Hence its rank is at most $1$ and we may write $A=\underline{\zeta}\,\underline{\eta}^\top$ for some vectors $\underline{\zeta}$ and $\underline{\eta}$. Hence $(1)$ holds and $z=\psi(x,y)$ where $x=\zeta^1a_1,\zeta^2a_2$ and $\eta^1a_1,\eta^2a_2$.
