My attempt in this question: (Here, $B_{ij}=\pi\circ B|_E{_{j}}:E_j\to E_i$)
We want to exhibit an invariant space that which contracts and another one that expands. The idea is to use the "graph transform", where we establish a function $P$ in which $gr(P)$ is our invariant space that contracts or expands.
Let $P:E_1\to E_1$ linear. For all $v\in E_1$ we have that $$\pmatrix{B_{11}&0\\B_{21} & B_{22}} \pmatrix{v\\Pv}=\pmatrix{B_{11}v\\B_{21}v+B_{22}Pv}$$
Hence $B(gr(P))\subseteq gr(P)\iff PB_{11}v=B_{21}v+B_{22}Pv \iff PB_{11}=B_{21}+B_{22}P \iff P=(B_{21}+B_{22}P)B_{11}^{-1}$
Then, define a map $T:L(E_1,E_2)\to L(E_1,E_2)$ such that, $T(P)=(B_{21}+B_{22}P)B_{11}^{-1}$
The claim is that $T$ is a contraction: $$T(P)B_{11}=B_{21}+B_{22}P$$ $$T(Q)B_{11}=B_{21}+B_{22}Q$$ \
$\implies (T(P)-T(Q))B_{11}=B_{22}(P-Q) \implies T(P)-T(Q)=B_{22}(P-Q)B_{11}^{-1}$
$ \implies |T(P)-T(Q)|=|B_{22}||(P-Q)||B_{11}^{-1}|<|P-Q| $ (by hypothesis)
Hence, $T$ is in fact a contraction. By Banach fixed point theorem, exists an unique $P\in L(E_1,E_2)$ such that $T(P)=P$. We fixe this $P$ and then $gr(P)$ is invariant by $B$.
Now, we have to show that $gr(P)$ contracts or expands:
$|B(v,Pv)|=\max\{|B_{11}v|,|B_{21}v+B_{22}Pv|\}=\max\{|B_{11}||v|,|B_{21}+B_{22}P||v|\}$
We knows that $|B_{11}|<1$ if it is the max we are done and B contracts on $gr(P)$;
If $|B_{21}+B_{22}P|<1$ then both options contracts and we are done;
If $|B_{21}+B_{22}P|>1$ it is the max and expands.
Questions:
Can $|B_{21}+B_{22}P|=1?$
How can I get the other invariant space that contracts or expands?
All help coments will be welcome, I am just starting my studies in Dynamical systems