a question on linear relations

Question

Given that $R\subseteq \mathbb{C}^n\times \mathbb{C}^n$ is called linear relation $\Leftrightarrow R$ is a linear space.

Its inverse relation is $R^{-1}:=\{(y,x)\in \mathbb{C}^n\times \mathbb{C}^n:(x,y)\in \mathbb{C}^n\times \mathbb{C}^n $.

Multiplication with a relation $S\subseteq \mathbb{C}^n\times \mathbb{C}^n$ is $RS=\{(x,y)\in \mathbb{C}^n\times \mathbb{C}^n| \exists z\in \mathbb{C}^n:(x,z)\in S\wedge(z,y)\in R\}$

Now in context of matrix pencil $A+\lambda E\in \mathbb{C}^{n\times n}[\lambda]$, the matrices $A,E$ induces linear relations $$\mathcal{A}=\{(x,Ax):x\in\mathbb{C}^n\}$$

$$\mathcal{E}=\{(x,Ex):x\in\mathbb{C}^n\}$$

$$\mathcal{A}^{-1}=\{(Ax,x):x\in\mathbb{C}^n\}$$

$$\mathcal{E}^{-1}=\{(Ex,x):x\in\mathbb{C}^n\}$$

now my question is why

1. $\mathcal{A}^{-1}\mathcal{E}=\{(x,y):x\in\mathbb{C}^n\times \mathbb{C}^n: Ex=Ay\}$,$\mathcal{E}^{-1}\mathcal{A}=\{(x,y):x\in\mathbb{C}^n\times \mathbb{C}^n: Ex=Ay\}$ in particular why this $Ax=Ey,Ex=Ay$ is coming in the picture?

I understand $R$ is a relation on $\mathbb{C}^n$ and $x\sim y$ iff $y=\lambda x$, am I right? and furthermore $R$ is an equivalence relation as $x\sim x\forall x,x\sim y\Rightarrow y\sim x, x\sim y,y\sim z\Rightarrow x\sim z$, I can prove.

thanks for helping

Answer 1

1. By defintion, $RS$ is the set of all tuples $(x,y)$ where there's some $(x,z)\in S$ and some $(z,y) \in R$. If $S$ and $R$ would be linear maps, this would translate to $S$ maps $x$ to $z(=Sx)$, and in turn $R$ maps $z$ to $y$ so that $RS$ maps $x$ to $y$.

For your particular example, $E=\{(x,Ex), x\in \mathbb C^n\}$ and $A^{-1}=\{(Ay,y), y \in \mathbb C^n\}$. There's a 'connecting element' $z$ for $(x,y)$ iff $z=Ex=Ay$.