Let $\mathbb{K}$ be a field and let $V$ be a $\mathbb{K}$-vector space.
Let $\phi,\psi:V\rightarrow V$ be linear maps, such that $\phi\circ\psi=\psi\circ\phi$.
I have shown using induction that if $U\leq_{\phi}V$ (i.e. it $U$ is a subspace and $\phi$-invariant), then $U\leq_{\phi^k}V$ for all $k\in \mathbb{N}$.
Now I want to show that if $\phi$ is invertible and if $U\leq_{\phi}V$, then $U\leq_{\phi^z}V$ for all $z\in \mathbb{Z}$.
For that we have to take cases, if $z\in \mathbb{Z}_{>0}$ then we would get the previous case and if $z\in \mathbb{Z}_{<0}$ we use the inverse linear map of $\phi$ and then we get the previous result.
My idea is the following:
If $z=:n\in \mathbb{Z}_{> 0}$, so $z=n\in \mathbb{N}$, the from the previous result it follows that $U$ $\ \phi^n$-invariant.
Since $\phi$ isinvertible there is a linear map $\chi$ such that $\chi:=\phi^{-1}$.
If $z=:-n\in \mathbb{Z}_{< 0}$, with $n\in \mathbb{N}$, then $\phi^{-n}=\left (\phi^{-1}\right )^n=\chi^n$. Then if we show that $U$ is $\chi$-invariant then it follows from the previous result that $U$ is $\ \phi^{-n}$-invariant.
Is that correct?
Or do we show that in an other way?
The result does not hold in general. First consider the level of sets: Doubling and halfing are inverse to each other on $ℚ$, so take $φ \colon ℚ → ℚ,~x ↦ 2x$ and $χ\colon ℚ → ℚ,~x ↦ \frac x 2$. Then $φ(ℤ) ⊆ ℤ$ but $χ(ℤ) \not\subseteq ℤ$.
You can carry this over by taking vector spaces generated by these sets: Let $K$ be a field. For $V = \bigoplus_{x ∈ ℚ} K$ and $U = \bigoplus_{m ∈ ℤ} K$, we have $U ⊆ V$ and since $φ$ and $χ$ are given on a basis of $V$, we can extend both of them them to $V → V$. Now $φ(U) ⊆ U$, but $χ(U) \not\subseteq U$.
The result does however hold if $U$ is finite dimensional. For this observe that if $U$ is $φ$-invariant, $φ$ can be restricted to $φ \colon U → U$. Now $φ$ is injective because it’s the restriction of an invertible map. What can you conclude? How does this relate to $χ$?